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Full text of "Organic Chemistry"

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Chapter 



i 



Structure and 
Properties 



4.1 Organic chemistry 

Organic chemistry is the chemistry of the compounds of carbon. 

The misleading name "organic" is a relic of the days when chemical com- 
pounds were divided into two classes, inorganic and organic, depending upon 
where they had come from. Inorganic compounds were those obtained from 
minerals; organic compounds were those obtained from vegetable or animal 
sources, that is, from material produced by living organisms. Indeed, until about 
1850 many chemists believed that organic compounds must have their origin in 
living organisms, and .consequently could never be synthesized from inorganic 
material. 

These compounds from organic sources had this in common : they all contained 
the element carbon. Even after it had become clear that these compounds did not 
have to come from living sources but could be made in the laboratory, it was 
convenient to keep the name organic to describe them and compounds like them. 
The division between inorganic and organic compounds has been retained to 
this day. 

Today, although many compounds of carbon are still most conveniently 
isolated from plant and animal sources, most of them are synthesized. They are 
sometimes synthesized from inorganic substances like carbonates or cyanides, but 
more often from other organic compounds. There are two large reservoirs of 
organic material from which simple organic compounds can be obtained '.petroleum 
and coal. (Both of these are "organic" in the old sense, being products of the 
decay of plants and animals.) These simple compounds are used as building 
blocks from which larger and more complicated compounds can be made. 

We recognize petroleum and coal as the fossil fuels, laid down over millenia and non- 
replaceable, that are being consumed at an alarming rate to meet our constantly increas- 
ing demands for power. There is, fortunately, an alternative source of power nuclear 
energy but where are we to find an alternative reservoir of organic raw material? 



2 STRUCTURE AND PROPERTIES CHAP. 1 

What is so special about the compounds of carbon that they should be sepa- 
rated from compounds of all the other hundred-odd elements of the Periodic 
Table? In part, at least, the answer seems to be this: there are so very many 
compounds of carbon, and their molecules can be so large and complex. 

The number of compounds that contain carbon is many times greater than the 
number of compounds that do not contain carbon. These organic compounds 
have been divided into families, which generally have no counterparts among the 
inorganic compounds. 

Organic molecules containing thousands of atoms are known, and the 
arrangement of atoms in even relatively small molecules can be very complicated. 
One of the major problems in organic chemistry is to find out how the atoms are 
arranged in molecules, that is, to determine the structures of compounds. 

There are many ways in which these complicated molecules can break apart, 
or rearrange themselves, to form new molecules; there are many ways in which 
atoms can be added to these molecules, or new atoms substituted for old ones. 
Much of organic chemistry is devoted to finding out what these reactions are, how 
they take place, and how they can be used to synthesize compounds we want. 

What is so special about carbon that it should form so many compounds? 
The answer to this question came to August Kekule in 1854 during a London bus 
ride. 

"One fine summer evening, I was returning by the last omnibus, 'outside' as 
usual, through the deserted streets of the metropolis, which are at other times 
so full of life. I fell into a reverie and lo! the atoms were gambolling before my 
eyes. ... I saw how, frequently, two smaller atoms united to form a pair, how a 
larger one embraced two smaller ones; how still larger ones kept hold of three or 
even four of the smaller; whilst the whole kept whirling in a giddy dance. I saw 
how the larger-ones formed a chain. ... I spent part of the night putting on paper 
at least sketches of these dream forms." August Kekule", 1890. 

Carbon atoms can attach themselves to one another to an extent not possible 
for atoms of any other element. Carbon atoms can form chains thousands of 
atoms long, or rings of all sizes; the chains and rings can have branches and cross- 
links. To the carbon atoms of these chains and rings there are attached other 
atoms, chiefly hydrogen, but also fluorine, chlorine, bromine, iodine, oxygen, 
nitrogen, sulfur, phosphorus, and many others. (Look, for example, at cellulose 
on page 1126, chlorophyll on page 1004, and oxytocin on page 1 143.) 

Each different arrangement of atoms corresponds to a different compound, 
and each compound has its own characteristic set of chemical and physical prop- 
erties. It is not surprising that close to a million compounds of carbon are known 
today and that thousands of new ones are being made each year. It is not surpris- 
ing that the study of their chemistry is a special field. 

Organic chemistry is a field of immense importance to technology: it is the 
chemistry of dyes and drugs, paper and ink, paints and plastics, gasoline and 
rubber tires; it is the chemistry of the food we eat and the clothing we wear. 

Organic chemistry is fundamental to biology and medicine. Aside from water, 
living organisms are made up chiefly of organic compounds; the molecules of 
"molecular biology" are organic molecules. Ultimately, biological processes are a 
matter of organic chemistry. 



SEC. 1.3 THE CHEMICAL BOND BEFORE 1926 



1.2 The structural theory 

"Organic chemistry nowadays almost drives me mad. To me it appears like 
a primeval tropical forest full of the most remarkable things, a dreadful endless 
jungle into which one does not dare enter for there seems to be no way out.'* 
Friedrich Wohler, 1835. 

How can we even begin to study a subject of such enormous complexity? Is 
organic chemistry today as Wohler saw it a century and a half ago? The jungle is 
still there largely unexploredand in it are more remarkable things than Wohler 
ever dreamed of. But, so long as we do not wander too far too fast, we can enter 
without fear of losing our way, for we have a chart: the structural theory. 

The structural theory is the basis upon which millions of facts about hundreds 
of thousands of individual compounds have been brought together and arranged 
in a systematic way. It is the basis upon which these facts can best be accounted 
for and understood. 

The structural theory is the framework of ideas about how atoms are put 
together to make molecules. The structural theory has to do with the order in 
which atoms are attached to each other, and with the electrons that hold them 
together. It has to do with the shapes and sizes of the molecules that these atoms 
form, and with the way that electrons are distributed over them. 

A molecule is often represented by a picture or a model sometimes by several 
pictures or several models. The atomic nuclei are represented by letters or wooden 
balls, and the electrons that join them by lines or dots or wooden pegs. These 
crude pictures and models are useful to us only if we understand what they are 
intended to mean. Interpreted in terms of the structural theory, they tell us a 
good deal about the compound whose molecules they represent : how to go about 
making it; what physical properties to expect of it melting point, boiling point, 
specific gravity, the kind of solvents the compound will dissolve in, even whether it 
will be colored or not; what kind of chemical behavior to expect the kind of 
reagents the compound will react 'with and the kind of products that will be 
formed, whether it will react rapidly or slowly. We would know all this about a 
compound that we had never encountered before, simply on the basis of its 
structural formula and what we understand its structural formula to mean. 



1.3 The chemical bond before 1926 

Any consideration of the structure of molecules must begin with a discussion 
of chemical bonds, the forces that hold atoms together in a molecule. 

We shall discuss chemical bonds first in terms of the theory as it had developed 
prior to 1926, and then in terms of the theory pf today. The introduction of 
quantum mechanics in 1926 caused a tremendous change in ideas about how 
molecules are formed. For convenience, the older, simpler language and pictorial 
representations are often still used, although the words and pictures are given a 
modern interpretation. 

In 1916 two kinds of chemical bond were described: the ionic bond by Walther 
Kossel (in Germany) and the covalent bond by G. N. Lewis (of the University of 



STRUCTURE AND PROPERTIES 



CHAl 



California). Both Kossel and Lewis based their ideas on the following concept 
the atom. 

A positively charged nucleus is surrounded by electrons arranged in con- 
centric shells or energy levels. There is a maximum number of electrons that can be 
accommodated in each shell: two in the first shell, eight in the second shell, eight or 
eighteen in the third shell, and so on. The greatest stability is reached when the 
outer shell is full, as in the noble gases. Both ionic and covalent bonds arise from 
the tendency of atoms to attain this stable configuration of electrons. 

The ionic bond results from transfer of electrons, as, for example, in the 
formation of lithium fluoride. A lithium atom has two electrons in its inner shell 




Li 



e- 




F+ e- 



and one electron in its outer or valence shell ; the loss of one electron would leave 
lithium with a full outer shell of two electrons. A fluorine atom has two electrons 
in its inner shell and seven electrons in its valence shell; the gain of one electron 
would give fluorine a full outer shell of eight. Lithium fluoride is formed by the 
transfer of one electron from lithium to fluorine; lithium now bears a positive 
charge and fluorine bears a negative charge. The electrostatic attraction between 
the oppositely charged ions is called an ionic bond. Such ionic bonds are typical 
of the salts formed by combination of the metallic elements (electropositive 
elements) on the far left side of the Periodic Table with the non-metallic elements 
(electronegative elements) on the far right side. 

The covalent bond results from sharing of electrons, as, for example, in the 
formation of the hydrogen molecule. Each hydrogen atom has a single electron; 
by sharing a pair of electrons, both hydrogens can complete their shells of two. 
Two fluorine atoms, each with seven electrons in the valence shell, can complete 
their octets by sharing a pair of electrons. In a similar way we can visualize the 
formation of HF, H 2 O, NH 3 , CH 4 , and CF 4 . Here, too, the bonding force is 
electrostatic attraction : this lime between each electron and both nuclei. 



H- 



H 
F: 



H- + -F: 
2H- + -0: 
3H- + -N: 



H:H 
:F:F: 

H:F: 

H 
H:O: 

H 

H:N: 
H 



SEC. 1.5 ATOMIC OBITALS 

H 

^ 4H- + -C- > H:C:H 

H 



4:F- + -C- > :F:C:F:' 



The covalent bond is typical of the compounds of carbon; it is the bond of chief 
importance in the study of organic chemistry. 

Problem 1.1 Which of the following would you expect to be ionic, and which 
non-ionic? Give a simple electronic structure for each, showing only valence shell 
electrons. 

(a) KBr* (c) NF 3 V . (e) CaSCV (g) PH 3 

(b) H 2 S/ (d) CHC1/ (0 NH 4 Ch (h) CH 3 OH/ 

Problem 1.2 Give a likely simple electronic structure for each of the following, 
assuming them to be completely covalent. Assume that every atom (except hydrogen, 
of course) has a complete octet, and that two atoms may share more than one pair of 
electrons. 

(a) H 2 2 (c) HON0 2 (e) HCN (g) H,CO 3 

(b)N 2 (d)N0 3 - (f)C0 2 (h)C 2 H 6 

1.4 Quantum mechanics 

In 1926 there emerged the theory known as quantum mechanics, developed, in 
the form most useful to chemists, by Erwin Schrodinger (of the University of 
Zurich). He worked out mathematical expressions to describe the motion of an 
electron in terms of its energy. These mathematical expressions are called wave 
equations, since they are based upon the cjncept that electrons show properties 
not only of particles but also of waves. 

A wave equation has a series of solutions, called wave functions, each corre- 
sponding to a different energy level for the electron. For all but the simplest of 
systems, doing the mathematics is so time-consuming that at present and super- 
high-speed computers will some day change this only approximate solutions can 
be obtained. Even so, quantum mechanics gives answers agreeing so well with the 
facts that it is accepted today as the most fruitful approach to an understanding 
of atomic and molecular structure. 

"Wave mechanics has shown us what is going on, a d at the deepest possible 
level ... it has taken the concepts of the experimental chemist the imaginative 
perception that came to those who had lived in their laboratories and allowed their 
minds to dwell creatively upon the facts that they had found and it has shown ho 
they all fit together; how, if you wish, they all have one single rationale; and how 
this hidden relationship to each other can be brought out.*' C. A. Coulson, 
London, 1951. 



1.5 Atomic orbitals 

A wave equation cannot tell us exactly where an electron is at any particular 
moment, or how fast it is moving; it does not permit us to plot a precise orbit 



6 STRUCTURE AND PROPERTIES CHAP. 1 

about the nucleus. Instead, it tells us the probability of finding the electron at any 
particular place. 

The region in space where an electron is likely to be found is called an orbital. 
There are different kinds of orbitals, which have different sizes and different shapes, 
and which are disposed about the nucleus in specific ways. The particular kind 
of orbital that an electron occupies depends upon the energy of the electron. It is 
the shapes of these orbitals and their disposition with respect to each other that 
we are particularly interested in, since these determine or, more precisely, can 
conveniently be thought of as determining the arrangement in space of the atoms 
of a molecule, and even help determine its chemical .behavior. 

It is convenient to picture an electron as being smeared out to form a cloud. 
We might think of this cloud as a sort of blurred photograph of the rapidly mov- 
ing electron. The shape of the cloud is the shape of the orbital. The cloud is not 
uniform, but is densest in those regions where the probability of finding the electron 
is highest, that is, in those regions where the average negative charge, or electron 
density, is greatest. 

Let us see what the shapes of some of the atomic orbitals are. The orbital 
at the lowest energy level is called the 15 orbital. It is a sphere with its center at 
the nucleus of the atom, as represented in Fig. 1.1. An orbital has no definite 





(a) (h) 

Figure 1.1. Atomic orbitals: s orbital. Nucleus at center. 

boundary since there is a probability, although a very small one, of finding the 
electron essentially separated from the atom or even on some other atom! 
However, the probability decreases very rapidly beyond a certain distance from the 
nucleus, so that the distribution of charge is fairly well represented by the electron 
cloud in Fig. 1.10. For simplicity, we may even represent an orbital as in Fig. 
1.1, where the solid line encloses the region where the electron spends most (say 
95%) of its time. 

At the next higher energy level there is the 2s orbital. This, too, is a sphere 
with its center at the atomic nucleus. It is naturally larger than the \s orbital: 
the higher energy (lower stability) is due to the greater average distance between 
electron and nucleus, with the resulting decrease in electrostatic attraction. 
(Consider the work that must be done the energy put into the system -to move 
an electron away from the oppositely charged nucleus.) 

Next there are three orbitals of equal energy called 2p orbitals, shown in 



Preface 



In preparing this third edition, we have done just about all the things that one 
can do to revise a textbook: added new material, deleted old, corrected mistakes, 
rewritten, and reorganized. To increase the book's value as a teaching aid, we have 
made two major, related changes: a change in organization, and a change in con- 
tent. Our aim was to bring the book up-to-date not just in the chemistry of each 
topic, but in the selection of topics, so that it would reflect, to the extent that a 
beginning textbook can, the directions that organic chemistry is taking today. 

We have divided the book into three parts, and thus have explicitly recognized 
what has always been our practice and that of most other teachers: to assign for 
study the first twenty-odd chapters of our book and then, of the last ten or tjvelve 
chapters, to pick three or four or five. 

In the twenty-four chapters of Part 1, the student is introduced to the funda- 
mentals of organic chemistry. As before, these chapters are tightly woven together; 
although certain sections or even chapters can be omitted, or their sequence 
altered, the organization is necessarily a fairly rigid one. 

In Parts II and HI the student reinforces his understanding of the fundamentals 
by applying them to more complicated systems. It is not really so important just 
which of these later chapters are chosen although each of us may consider one 
or another of these topics absolutely essential as that some of them are studied. 
What is learned at the beginning of the course can evaporate pretty rapidly if the 
lid is not screwed down at the end. 

The new edition is about as long as the previous one, and contains the same 
number of chapters. But, through deletion and transfer of material, about 100 
pages have been cut from the early part of the book. Using these pages, and re- 
arranging some old material, we have written seven quite new chapters: Carban- 
ions I, Carbanions II, Macromolecules, Rearrangements and Neighboring Group 
Effects, Molecular Orbitals and Orbital Symmetry, Fats, and Biochemical Pro- 
cesses and Molecular Biology. 

vti 



viii PREFACE 

The teacher will most probably assign all or most of Part I. From Part II he 
may then select topics to take the student more deeply into " straight" organic 
chemistry: use of carbanion* m organic synthesis; conjugate addition; polynuclear 
and heterocyclic compounds; rearrangements and neighboring group effects; the 
application of the orbital symmetry concept to concerted reactions of various 
kinds. In Part HI, he may have the student learn something of the organic chemis- 
try of biomolecules; fats, carbohydrates, proteins, and nucleic acids. And, as be- 
fore, the brighter or more ambitious student may dip into chemistry not studied 
by the rest of the class. 

There are other changes in organization. Glycols are introduced. with alcohols, 
epoxides with ethers* and dicarboxylic acids with monocarboxylic acids. The aldol 
and Claisen condensations and the Wittig and Reformatsky reactions appear in 
Chapter 21, just after aldehydes and ketones and esters. These changes do not, we 
have found, place an undue burden on the student, but they do stimulate him and 
keep him busy in the middle part of the course: after the onslaught of new ideas 
in the beginning and before the complexities of the later topics. The student is 
ready for poly functional compounds at this time. He certainly finds epoxides more 
exciting than ordinary ethers, and can do more with phthalic and succinic an- 
hydrides than with acetic anhydride. With carbanion chemistry in Chapter 21, he 
has opened to him the really important routes to carbon-carbon bond formation. 

As before, we use problems as the best way to help the student to learn what 
he has been exposed to, and to let him broaden his acquaintance with organic 
chemistry beyond the bounds of the text. We have extended the practice intro- 
duced in our first edition of inserting problems within the chapter as checkpoints 
on the student's progress; nearly half of the more than 1300 problems are thus 
used to provide a kind of programmed instruction. 

Spectroscopic analysis, chiefly nmr and infrared, is again introduced in Chap- 
ter 13, and in subsequent chapters the spectral characteristics of each class of com- 
pound is outlined. Emphasis is on the study of spectra themselves there are 97 
nmr and infrared spectra to be analyzed in problems or of spectral data. 

There is much that is entirely new: new reactions, like thallation, solvomer- 
curation, Corey-House hydrocarbon synthesis, orgapoborane synthesis of acids 
and ketones; the use of enamines, 2-oxazoliites, tetrajhydropyranyl esters; and, of 
course, etectrocyclic reactions, cycloaddition, and slgmatropic shifts. In one of 
the new chapters, rearrangements and neighboring group effects are discussed as 
related, often indistinguishable, kinds of intramolecular nucteophilic attack. In 
another, the concept of orbital symmetry is applied to concerted reactions; the 
treatment is based on the roles played by the highest occupied and lowest unoccu- 
pied molecular orbitals, and the student learns the enormous power of the simple 
Wood ward- Hoffmann rules by applying them to dozens of examples: in the text, 
and in problems of graded difficulty. 

In teaching today, one must recognize that many organic chemists will end up 
working in biological fields, and even those calling themselves biologists must 
know much organic chemistry. It seems clear, then, that we must do an even better 
job of teaching the fundamentals of organic chemistry to all students, whatever 
their ultimate goal. At the same time, the student should be made aware of the 
rote of organic chemistry in biology, and it is with this in mind that we approach 
the study of biomolecules. Emphasis is on their structure and their chemistry in 



PREFACE 

the test tube the foundation for any further study. In addition, we have tried to 
give the student some idea of the ways in which their properties as organic mole- 
cules underlie their functions in biological systems, and how all this ultimately 
goes back to our fundamental ideas of structure. Biology, on the molecular level, 
15 organic chemistry, and we try to let the student see this. 

No changes have been made just for the sake of change. In our rewriting and 
in the selection of new topics, we have stuck to the principle we have always held: 
these are beginning students, and they need all the help they can get. Discussion of 
neighboring group effects or the Woodward-Hoffmann rules is pitched' at the same 
level as the chlorination of methane in Chapter 2. New material is introduced at 
the rate at which we have found students can absorb it. Once presented, a principle 
is used and re-used. In a beginning course, we cannot hope to cover more than a 
tiny fraction of this enormous field; but what we can hope for is to make a good 
job of what we do teach. 

ROBERT THORNTON MORRISON 
ROBERT NEILSON BOYD 



Acknowledgments 



Our thanks to Sadtler Research Laboratories for the infrared 
spectra labeled "Sadtler" and to the Infrared Data Committee of 
Japan for those labeled "IRDC," and to the following people for 
permission to reproduce material: Professor George A. Olah, Figure 
5.7; Irving Geis and Harper and Row,. Publishers, Figure 37.1; the 
editors of The Journal of the American Chemical Society, Figures 
5.7, 13.17, and 13.18; and Walt Disney Productions, Figure 9.13. 



SEC. 1.6 



ELECTRONIC CONFIGURATION 



Fig. 1.2. Each 2p orbital is dumbbell-shaped. It consists of two lobes with the 
atomic nucleus lying between them. The axis of each 2p orbital is perpendicular 
to the axes of the other two. They are differentiated by the names 2p x , 2p v , and 
2p a , where the x, y, and z refer to the corresponding axes. 




(a) 



.1. 





V 




Figure 1.2. Atomic orbitals: p orbitals. Axes mutually perpendicular. 
(a) Cross-section showing the two lobes of a single orbital, (b) Approximate 
shape as pairs of distorted ellipsoids, (c) Representation as pairs of not- 
quite-touching spheres. 



1.6 Electronic configuration. Pauli exclusion principle 

There are a number of "rules" that determine the way in which the electrons 
of an atom may be distributed, that is, that determine the electronic configuration 
of an atom. 

The most fundamental of these rules is the Pauli exclusion principle: only two 
electrons can occupy any atomic orbital, and to do so these two must have opposite 
spins. These electrons of opposite spins are said to be paired. Electrons of like 



8 



* STRUCTURE AND PROPERTIES 



CHAP. 1 



spin tend to get as far from each other as possible. This tendency is the most im- 
portant of all the factors that determine the shapes and properties of molecules. 

The exclusion principle, advanced in 1925 by Wolfgang Pauli, Jr. (of the Institute for 
Theoretical Physics, Hamburg, Germany), has been called the cornerstone of chemistry. 

The first ten elements of the Periodic Table have the electronic configurations 
shown in Table 1.1. We see that an orbital becomes occupied only if the orbitals 



25 

o 



Table 1.1 'ELECTRONIC CONFIGURATIONS 

15 

H O 

ID 

He 

Li 

Be 

B 





C 

Ne 



o 








o 


o 


o 


o 


o 





o 


o 





o 


o 


o 





o 


o 








o 



of lower energy are filled (e.g., 2s after Is, 2p after 2s). We see that an orbital is 
not occupied by a pair of electrons until other orbitals of equal energy are each 
occupied by one electron (e.g., the 2p orbitals). The Is electrons make up the first 
shell of two, and the 2s and 2p electrons make up the second shell of eight. For 
elements beyond the first ten, there is a third shell containing a 3s orbital, 3p 
orbitals, and so on. 

Problem 1.3 (a) Show the electronic configurations for the next eight elements 
in the Periodic Table (from sodium through argon), (b) What relationship is there 
between electronic configuration and periodic family? (c) Between electronic con- 
figuration and chemical properties of the elements? 



1.7 Molecular orbitals 

In molecules, as in isolated atoms, electrons occupy orbitals, and in accordance 
with much the same "rules." These molecular orbitals are considered to be cen- 
tered about many nuclei, perhaps covering the entire molecule; the distribution 
of nuclei and electrons is simply the one thkt results in the most stable molecule. 

To make the enormously complicated mathematics more workable, two 
simplifying assumptions are commonly made: (a) that each pair of electrons is 



SEC. 1.8 THE COVALENT BOND 9 

essentially localized near just two nuclei, and (b) that the shapes of these localized 
molecular orbitals and their disposition with respect to each other are related in a 
simple way to the shapes and disposition of atomic orbitals in the component 
atoms. 

The idea of localized molecular orbitals or what we might call bond orbitals 
is evidently not a bad one, since mathematically this method of approximation is 
successful with most (although not all) molecules. Furthermore, this idea closely 
parallels the chemist's classical concept of a bond as a force acting between two 
atoms and pretty much independent of the rest of the molecule; it can hardly be 
accidental that this concept has worked amazingly well for a hundred years. 
Significantly, the exceptional molecules for which classical formulas do not work 
are just those for which the localized molecular orbital approach does not work, 
either. (Even these cases, we shall find, can be handled by a rather simple adapta- 
tion of classical formulas, an adaptation which again parallels a method of mathe- 
matical approximation.) 

The second assumption, of a relationship between atomic and molecular 
orbitals, is a highly reasonable one, as discussed in the following section. It has 
proven so useful that, when necessary, atomic orbitals of certain kinds have been 
invented just so that the assumption can be retained. 

1.8 The covalent bond 

Now let us consider the formation of a molecule. For convenience we shall 
picture this as happening by the coming together of the individual atoms, although 
most molecules are not actually made this way. We make physical models of 
molecules out of wooden or plastic balls that represent the various atoms; the 
location of holes or snap fasteners tells us how to put them together. In the same 
way, we shall make mental models of molecules out of mental atoms; the location 
of atomic orbitals some of them imaginary will tell us how to put these together. 

For a covalent bond to form, two atoms must be located so that an orbital of 
one overlaps an orbital of the other; each orbital must contain a single electron. 
When this happens, the two atomic orbitals merge to form a single bond orbital 
which is occupied by both electrons. The two electrons that occupy a bond 
orbital must have opposite spins, that is, must be paired. Each electron has 
available to it the entire bond orbital, and thus may be considered to "belong to" 
both atomic nuclei. 

This arrangement of electrons and nuclei contains less energy that is, is more 
stable than the arrangement in the isolated atoms; as a result, formation of a 
bond is accompanied by evolution of energy. The amount of energy (per mole) 
that is given off when a bond is formed (or the amount that must be put in to break 
the bond) is called the bond dissociation energy. For a given pair of atoms, the 
greater the overlap of atomic orbitals, the stronger the bond. 

What gives the covalent bond its strength? It is the increase in electrostatic 
attraction. In the isolated atoms, each electron is attracted by and attracts 
one positive nucleus; in the molecule, each electron is attracted by two positive 
nuclei. 

It is the concept of "overlap" that provides the mental bridge between atomic 
orbitals and bond orbitals. Overlap of atomic orbitals means that the bond 



10 STRUCTURE AND PROPERTIES CHAP. 1 

orbital occupies much of the same region in space that was occupied by both atomic 
orbitals. Consequently, an electron from one atom can, to a considerable extent, 
remain in its original, favorable location with respect to "its" nucleus, and at the 
same time occupy a similarly favorable location with respect to the second nucleus; 
the same holds, of course, for the other electron. 

The principle of maximum overlap, first stated in 1931 by Linus Pauling (at the 
California Institute of Technology), has been ranked only slightly below the exclusion 
principle in importance to the understanding of molecular structure. 

As our first example, let us consider the formation of the hydrogen molecule, 
H 2 , from two hydrogen atoms. Each hydrogen atom has one electron, which occu- 
pies the \s orbital. As we have seen, this Is orbital is a sphere with its center at 
the atomic nucleus. For a bond to form, the two nuclei must be brought closely 
enough together for overlap of the atomic orbitals to occur (Fig. 1.3). For hy- 
drogen, the system is most stable when the distance between the nuclei is 0.74 A; 





(a) 





(c) (if) 

Figure 1.3. Bond formation: H 2 molecule, (a) Separate s orbitals. 
(b) Overlap of s orbitals. (c) and (d) The a bond orbital. 



this distance is called the bond length. At this distance the stabilizing effect of 
overlap is exactly balanced by repulsion between the similarly charged nuclei. The 
resulting hydrogen molecule contains 104 kcal/mole less energy than the hydrogen 
atoms from which it was made. We say that the hydrogen-hydrogen bond has a 
length of 0.74 A and a strength of 104 kcal. 

This bond orbital has roughly the shape we would expect from the merging 
of two s orbitals. As shown in Fig. 1.3, it is sausage-shaped, with its long axis 
lying along the line joining the nuclei. Ft is cylindrically symmetrical about this 
long axis; that is, a slice of the sausage is circular. Bond orbitals having this shape 
are called a orbitals (sigtna orbitals) and the bonds are called a bonds. We may 
visualize the hydrogen molecule as two nuclei embedded in a single sausage-shaped 
electron cloud. This cloud is densest in the region between the two nuclei, where 
the negative charge is attracted most strongly by the two positive charges. 

The size of the hydrogen molecule as measured, say, by the volume inside the 
95% probability surface is considerably smaller than that of a single hydrogen 
atom. Although surprising at first, this shrinking of the electron cloud is actually 
what would be expected. It is the powerful attraction of the electrons by two 



SEC. 1.9 HYBRID ORBIT ALS: sp 11 

nuclei that gives the molecule greater stability than the isolated hydrogen atoms; 
this must mean that the electrons are held tighter, closer, than in the atoms. 

Next, let us consider the formation of the fluorine molecule, F 2 , from two 
fluorine atoms. As we can see from our table of electronic configurations (Table 
1.1), a fluorine atom has two electrons in the Is orbital, two electrons in the 2$ 
orbital, and two electrons in each of two 2p orbitals. In the third 2p orbital there 
is a single electron which is unpaired and available for bond formation. Overlap 
of this/? orbital with a similar p orbital of another fluorine atom permits electrons 
to pair and the bond to form (Fig. 1.4). The electronic charge is concentrated 
between the two nuclei, so that the back lobe of each of the overlapping orbitals 







(a) 

OODO oo-o 

(/>) (r) 

Figure 1.4. Bond formation: F 2 molecule, (a) Separate p orbitals. 
(b) Overlap of/? orbitals. (c) The a bond orbital. 

shrinks to a comparatively small size. Although formed by overlap of atomic 
orbitals of a different kind, the fluorine-fluorine bond has the same general shape 
as the hydrogen-hydrogen bond, being cylindrically symmetrical about a line 
joining the nuclei; it, too, is given the designation of a bond. The fluorine-fluorine 
bond has a length of 1.42 A -and a strength of about 38 kcal. 

As the examples show, a covalent bond results from the overlap of two atomic 
orbitals to form a bond orbital occupied by a pair of electrons. Each kind of 
covalent bond has a characteristic length and strength. 

1.9 Hybrid orbitals: sp 

Let us next consider beryllium chloride, BeCl 2 . 
Beryllium (Table 1. 1) has no unpaired electrons. 

\s 2s 2p 



Be O O O 

How are we to account for its combining with two chlorine atoms ? Bond forma- 
tion is an energy-releasing (stabilizing) process, and the tendency is to form 
bonds and as many as possible even if this results in bond orbitals that bear 
little resemblance to the atomic orbitals we have talked about. If our method of 
mental molecule-building is to be applied here, it must be modified. We must 
invent an imaginary kind of beryllium atom, one that is about to become bonded 
to two chlorine atoms. 



12 



STRUCTURE AND PROPERTIES 



CHAP. 1 



To arrive at this divalent beryllium atom, let us do a little electronic book- 
keeping. First, we "promote" one of the 2s electrons to an empty/? orbital: 



Is 



Be 



25. 

O 



O O O 



One electron promoted: 
two unpaired electrons 



This provides two unpaired electrons, which are needed for bonding to two 
chlorine atoms. We might now expect beryllium to form one bond of one kind, 
using the p orbital, and one bond of another kind, using the s orbital. Again, this 
is contrary to fact: the two bonds in beryllium chloride are known to be equivalent* 
Next, then, we hybridize the* orbitals. Various combinations of one s and 



Be 



Be 



Is 



Is 



O O O 

1 I 

sp 2p 

(iTo cT^D 



sp Hybridization 



one p orbitals are taken mathematically, and the mixed (hybrid) orbitals with the 
greatest degree of directional character are found (Fig. 1 .5). The more an atomic 
orbital is concentrated in the direction of the bond, the greater the overlap and the 
stronger the bond it can form. Three highly significant results emerge from the 




/% * 

V&. :.::$ 

^f^f' 

( 





"180 



Figure 1.5. Atomic orbitals: hybrid sp orbitals. (a) Cross-section and 
approximate shape of a single orbital. Strongly directed along one axis. 
(b) Representation as a sphere, with small back lobe omitted, (c) Two or- 
bitals, with axes lying along a straight line. 



HYBRID ORBITALS: v 3 15 

^ j^i-vid orbitals are called sp 2 orbitals, since they are considered to arise 
rowir-^rttiw ig of one s orbital and two p orbitals. They He in a plane, which 
includes the atomic nucleus, and are directed to the corners of an equilateral 
triangle; the angle between any two orbitals is thus 120. Again we see the geo- 
metry that permits the orbitals to be as far apart as possible: here, a trigonal (three- 
cornered) arrangement, 

When we arrange the atoms for maximum overlap of each of the sp 2 orbitals 
of boron with ap orbital of fluorine, we obtain the structure shown in Fig. 1.8: 
a flat molecule, with the boron atom at the center of a triangle and the three 
fluorine atorus at the corners. Every bond angle is 120. 



Figure 1.8. BF 3 molecule. 



Experiment has shown that boron fluoride has exactly this flat, symmetrical 
'Structure calculated by quantum mechanics. 

1.11 Hybrid orbitals: s/> 3 

Now, let us turn to one of the simplest of organic molecules, methane, CH 4 . 

Carbon (Table 1.1) has an unpaired electron in each of the two p orbitals, 

and on this basis might be expected to form a compound CH 2 . (It docs, but 

1 2s 2p 



c O O O 

CH 2 is a highly reactive molecule whose properties center about the need to pro- 
vide carbon v ith two more bonds.) Again, we see the tendency to form as many 
bonds as posMble: in this case, to combine with four hydrogen atoms. 

To providi; four unpaired electrons, we promote one of the 2s electrons to 
the empty p orbital : 

\s 2s ' 2p 

One electron promoted: 

O O f?\ f?\ O f our un P a * re d electrons 

V* vlx vlx \+U vlx ^LJ 

Once more the most strongly directed orbitals are hybrid orbitals: this time, sp* 
orbilals, from the mixing of one s orbital and three p orbitals. Each one has the 

h 2s 2p 



O O O O 

I I sp 3 Hybridization 

lj Jp3 

O O O O 



16 STRUCTURE AND PROPERTIES 

shape shown in Fig. 1.9; as with sp and sp* orbitals, we shall 
back lobe and represent the front lobe as a sphere. 







(c) 



Figure 1 9. Atomic orbitals: hybrid sp* orbitals. (a) Cross- 
approximate shape of a single orbital. Strongly directed along 
(b) Representation as a sphere, with small back lobe omitted, 
orbitals, with axes directed toward corners of tetrahedron. 



Now, how are sp 3 orbitals arranged in space? The answer 
us: in the way that lets them get as far away from each other ; 
are directed to the corners of a regular tetrahedron. The angle - 
orbitals is the tetrahcdral angle 109.5 (Fig. 1.9). Just as mutual 
orbitals gives two linear bonds or three trigonal bonds, so it 
hedral bonds. 

Overlap of each of the sp* orbitals of carbon with an s orb>u. 1&f i, 
results in methane: carbon at the center of a regular tetrahedro 
hydrogens at the corners (Fig, 1.10). 

Experimentally, methane has been found to have the higU;> 
tetrahedral structure we have assembled. Each carbon-hydrcy, , 
exactly the same length, 1.10 A; the angle between any pair of boi <c 
hedral angle 109.5. It takes 104 kcal/mole to break one of the bor <* 

Thus, in these last three sections, we have seen that there are i 
covalent bonds not only characteristic bond lengths and bond di$J 
gies but also characteristic bond angles. These bond angles can 
related to the arrangement of atomic orbitals including hybrid! w>,wy 
volved in bond formation; t^v r**[ f ,, ;^v go back to the Pauli wu/i 
ciple and the tendrr^v tor uflD&iieu ^leciroiu to p.ot A% fr, r from'db&fttt! 





UNSHVRED PAIRS OF ELECTRONS 
H 



17 




() ( (c) 

Figure 1.10. Bond formation : CH 4 molecule, (a) Tetrahedral sp* orbitals. 
(b) Predicted shape: H nuclei located for maximum overlap, (c) Shape and 
size. 

Unlike the ionic bond, which is equally strong in all directions, the covalent 
.id is a directed bond. We can begin to see why the chemistry of the covalent 
ond is so mucr^concerned with moleculaj size and shape. 

1.12 Unshared pairs of electrons 

Two familiar compounds, ammonia (NH 3 ) and water (H 2 O), show how 
unshared pairs of electrons can affect molecular structure. 

In ammonia, nitrogen resembles the carbon of methane. Nitrogen is sp 3 - 
hybridized, but (Table 1.1) has only three unpaired electrons; they occupy three 



N 



O O O 

I I sp 3 Hybridization 



N 



000 



of the sp* orbitals. Overlap of each of these orbitals with the s orbital of a hydrogen 
atom results in ammonia (Fig. 1.11). The fourth sp* orbital of nitrogen contains 
a pair of electrons. 

If there is to be maximum overlap and hence maximum bond strength, the 
hydrogen nuclei must be located at three corners of a tetrahedron; the fourth 
corner is occupied by an unshared pair of electrons. Considering only atomic 
nuclei, we would expect ammonia to be shaped like a pyramid with nitrogen at the 
apex and hydrogen at the corners of a triangular base. Each bond angle should be 
the tetrahedral angle 109.5. 

Experimentally, ammonia is found to have the pyramidal shape calculated by 
quantum mechanics. The bond angles are 107, slightly smaller than the predicted 
value; it has been suggested that the unshared pair of electrons occupies more 
space than any of the hydrogen atoms, and hence tends to compress the bond 



18 STRUCTURE AND PROPERTIES CHAP. 1 






W W (c) 



Figure 1.11. Bond formation : NH 3 molecule, (a) Tetrahedral sp 3 orbitals. 
(b) Predicted shape, showing unshared pair: H nuclei located for maximum 
overlap, (c) Shape and size. 

angles slightly. The nitrogen-hydrogen bond length is 1.01 A; it takes 103 kcal/mole 
to break one of the bonds of ammonia. 

The sp 3 orbital occupied by the unshared pair of electrons is a region of high 
electron density. This region is a source of electrons for electron-seeking atoms 
and molecules, and thus gives ammonia its basic properties (Sec. 1.22). 

There are two other conceivable electronic configurations for ammonia, but neither 
fits the facts. 

(a) Since nitrogen is bonded to three other atoms, we might have pictured it as using 
jp 2 orbitals, as boron does in boron trifluoride. But ammonia is not a flat molecule, and j 
so we must reject this possibility. It is the unshared pair of electrons on nitrogen that 
makes the difference between NH 3 and BF 3 these electrons need to stay away from those 
in the carbon-hydrogen bonds, and the tetrahedral shape makes this possible. 

(b) We might have pictured nitrogen as simply using the p orbitals for overlap, 
since they would provide the necessary three unpaired electrons. But this would give 
bond angles of 90 remember, the p orbitals are at right angles to each other in con- 
trast to the observed angles of 107. More importantly, the unshared pair would be 
buried in an s orbital, and there is evidence from dipole moments (Sec. 1.16) that this 
is not so. Evidently the stability gained by using the highly directed sp 3 orbitals for bond 
formation more than makes up for raising the unshared pair from an s orbital to the 
higher-energy sp 3 orbital. 

One further fact about ammonia: spectroscopy reveals that the molecule 
undergoes inversion, that is, turns inside-out (Fig. 1.12). There is an energy barrier 



H 



il 



Figure 1.12. Inversion of ammonia. 



of only 6 kcal/mole between one pyramidal arrangement and the other, equivalent 
one. This energy is provided by molecular collisions, and even at room tempera- 
ture the fraction of collisions hard enough to do the job is so large that a rapid 
transformation between pyramidal arrangements occurs. 



SEC. 1.12 UNSHARED PAIRS OF ELECTRONS 19 

Compare ammonia with methane, which does not undergo inversion. The 
unshared pair plays the role of a carbon-hydrogen bond in determining the most 
stable shape of the molecule, tetrahedral. But, unlike a carbon-hydrogen bond, 
the unshared pair cannot maintain a particular tetrahedral arrangement; the pair 
points now in one direction, and the next instant in the opposite direction. 

Finally, let us consider water, H 2 O. The situation is similar to that for am- 
monia, except that oxygen has only two unpaired electrons, and hence it bonds 



I 

is 5/>3 



O 

I sp* Hybridization 



o O G 

with only two hydrogen atoms, which occupy two corners of a tetrahedron. The 
other two corners of the tetrahedron are occupied by unshared pairs of electrons 
(Fig. 1.13). 






Figure 1.13. Bond formation : H 2 O molecule, (a) Tetrahedral 5p 3 orbitals. 
(b) Predicted shape, showing unshared pairs: H nuclei located for maxi- 
mum overlap, (c) Shape and size. 

As actually measured, the H O Hjmglejs J05, smaller than thej:alculated 
tetrahedral angle, and even smaller than the angle in ammonia. Here there are 
two bulky unshared pairs of electrons compressing the bond angles. The oxygen- 
hydrogen bond length is 0.96 A ; it takes 1 1 8 kcal/mole to break one of the bonds of 
water. " =='-'"' 

Because of the unshared pairs of electrons on oxygen, water is basic, although 
less strongly so than ammonia (Sec. 1.22). 

Problem 1.4 Predict the shape of each of the following molecules, and tell 
how you arrived at your prediction : (a) the ammonium ion, NH 4 + ; (b) the hydro- 
nium ion, H 3 O+ ; (c) methyl alcohol, CH 3 OH; (d) methylamine, CHjNH 2 . 



20 STRUCTURE AND PROPERTIES CHAP. 1 

1.13 Intramolecular forces 

We must remember that the particular method of mentally building mole- 
cules that we are learning to use is artificial: it is a purely intellectual process in- 
volving imaginary overlap of imaginary orbitals. There are other, equally artificial 
ways that use different mental or physical models. Our method is the one. that so 
far has seemed to work out best for the organic chemist. Our kit of mental atomic 
models will contain just three "kinds" of carbon: tetrahedral C$/? 3 -hybridized), 
trigonal (sp 2 -hybridized), and digonal (,s/?-hybridized). By use of this kit, we shall 
find, one can do an amazingly good job of building hundreds of thousands of 
organic molecules. 

But, however we arrive at it, we see the actual structure of a molecule to be 
the net result of a combination of repulsive and attractive forces,, which are related 
to charge and electron spin. 

(a) Repulsive forces. Electrons tend to stay as far apart as possible because 
they have the same charge and also, if they are unpaired, because they have the 
same spin (Pauli exclusion principle). The like-charged atomic nuclei, too, repel 
each other. 

(b) Attractive forces. Electrons are attracted by atomic nuclei as are the 
nuclei by the electrons because of their opposite charge, and hence tend to occupy 
the region between two nuclei. Opposite spin permits (although, in itself, probably 
does not actually encourage) two electrons to occupy the same region. 

In methane, for example, the four hydrogen nuclei are as widely separated 
as they can be. The distribution of the eight bonding electrons is such that each 
one occupies the desirable region near two nuclei the bond orbital and yet, 
except for its partner, is as far as possible from the other electrons. We can picture 
each electron accepting perhaps reluctantly because of their similar charges 
one orbital-mate of opposite spin, but staying as far as possible from all other elec- 
trons and even, as it wanders within the loose confines of its orbital, doing its 
best to avoid the vicinity of its restless partner. 

1.14 Bond dissociation energy. Homolysis and heterolysis 

We have seen that energy is liberated when atoms combine to form a mole- 
cule. For a molecule to break into atoms, an equivalent amount of energy must be 
consumed. The amount of energy consumed or liberated when a bond is broken or 
formed is known as the bond dissociation energy, D. It is characteristic of the par- 
ticular bond: Table 1 .2 lists bond dissociation energies that have been measured 
for a number of bonds. As can be seen, they vary widely, from weak bonds like 
II (36 kcal/mole) to very strong bonds like H F (136 kcal/mole). Although the 
accepted values may change as experimental methods improve, certain trends are 
clear. 

We must not confuse bond dissociation energy (D) with another measure of bond 
strength called bond energy (E). If one begins with methane, for example, and breaks, 
successively, four carbon-hydrogen bonds, one finds four different bond dissociation 
energies: 

CH 4 > CH 3 + H- D(CH 3 -H) = 104 kcal/mole 
CH 3 > CH 2 + H- D(CH 2 -H) - 106 



SEC. 1.14 



BOND DISSOCIATION ENERGY 



21 



CH 2 
CH 



CH + H 
C+ H- 



D(CH-H) = 106 
T>(C-H) = 81 



The carbon-hydrogen bond energy in methane, E(C H), on the other hand, is a single 
average value : 

CH 4 > C + 4H- A// = 397 kcal/mole, E(C-H) = 397/4 = 99 kcal/hiole 

We shall generally find bond dissociation energies more useful for our purposes. 

Table 1.2 BOND DISSOCIATION ENERGIES, KCAL/MOLE 
A:B > A- -f -B A// = Bond Dissociation Energy or D(A-B) 



H H 


104 


CH 3 ~H 


104 


H F 


136 F F 38 


CH 3 F 


108 


H -Cl 


103 Cl -Cl 58 


CH 3 Cl 


84 


H Br 


88 Br Br 46 


CH 3 Br 


70 


H I 


71 I I 36 


CH 3 I 


56 


CH 3 H 104 


CH 3 CH 3 88 


CH,-- Cl 84 


CH 3 Br 70 


C 2 H 5 H 98 


C 2 H 5 -CH 3 85 


C 2 H 5 - -Cl 81 


C 2 H 5 Br 69 


w-C 3 H 7 -H 98 


if-C 3 H 7 -CH 3 85 


w-C,H 7 ~Cl 82 


W .C 3 H 7 Br 69 


z-C 3 H 7 -H 95 


i-CjHr- CH 3 84 


/-C 3 H7 Cl 81 


/-C 3 H 7 Br 68 


f-C 4 H 9 - H 91 


/-C 4 Hy CH 3 80 


f-C 4 H Cl 79 


r-C 4 H 9 Br 63 


H 2 C--CH-H 104 


H 2 C CH CH 3 92 


H 2 C^-CH Cl 84 




H 2 C-CHCH 2 ~H 88 


H 2 C CHCH 2 -CH 3 72 H 2 C 


-CHCH r -Cl 60 


H 2 O-CHCH 2 Br 47 


C 6 H 5 H 112 


CftHs CH 3 93 


C 6 H S Cl 86 


C 6 H 5 Br 72 


C 6 H 5 CH 2 -H 85 


C 6 H 5 CH 2 CH 3 70 


C 6 H 5 CH 2 Cl 68 


C 6 H 5 CH 2 Br 51 



So far, we have spoken of breaking a molecule into two atoms or into an 
atom and a group of atoms. Thus, of the two electrons making up the covalcnt 
bond, one goes to each fragment; such bond-breaking is called homolysis. We shall 
also encounter reactions involving bond-breaking of a different kind: heterolysis, 
in which both bonding electrons go to the same fragment. 



A.B 



A- + B 



A:B > A + :B 



Homolysis: 

one electron to 
each fragment 

Heterolysis: 

both electrons to 
one fragment 



(These words are taken from the Greek: homo and hetero, the same and 
different; and lysis, a loosing. To a chemist lysis means "cleavage" as in, for 
example, hydro-lysis, "cleavage by water.") 

Simple heterolysis of a neutral molecule yields, of course, a positive ion and 
a negative ion. Separation of these oppositely charged particles lakes a great deal 
of energy: 100 kcal/mole or so marc than separation of neutral particles. In the 
gas phase, therefore, bond dissociation generally takes place by the easier route, 
tyomolysis. In an ionizing solvent (Sec. 1.21), on the other hand, heterolysi.js the 
preferred kind of cleavaee 



22 STRUCTURE AND PROPERTIES CHAP. I 

1.15 Polarity of bonds 

Besides the properties already described^ certain covalent bonds have another 
property: polarity. Two atoms joined by a covalent bond share electrons; their 
nuclei are held by the same electron cloud. But in most cases the two nuclei do 
not share ihe electrons equally; the electron cloud is denser about one atom than 
the other. One end of the bond is thus relatively negative and the other end is 
relatively positive; that is, there is a negative pole and a positive pole. Such a bond 
is said to be a polar bond, or to possess polaritv. 

We can indicate polarity by using the symbols 8+ and S_, which indicate 
partial -t- and - charges. For example: 

8. S_ 

S + 8~ O N 

H-F &<,- ^ a, a f / \ \a + 

H H H H H 

8 + 
Polar bonds 

We can expect a covalent bond to be polar if it joins atoms that differ in their 
tendency to attract electrons, that is, atoms that diJTer in electroneganvity. Further- 
more, the greater the difference in electroncgativity, the more polar the bond will be.. 

The most electronegative elements are those located in the upper right-hand 
corner of the Periodic Table. Of the elements we are likely to encounter in organic 
chemistry, fluorine has the highest electronegativiiy, then oxygen, then nitrogen and 
chlorine, then bromine, and finally carbon. Hydrogen does not differ very much 
from carbon in electronegativity; it is not certain whether it is more or less electro- 
negative. 

Electronegativity F > O > Cl, N > Br > C, H 

Bond polarities are intimately concerned with both physical and chemical 
properties. The polarity of bonds can lead to polarity of molecules, and thus pro- 
foundly affect melting point, boiling point, and solubility. The polarity of a bond 
determines the kind of reaction that can take place at t]ial bond, and even affects 
reactivity at nearby bonds. 

1.16 Polarity of molecules 

(A molecule is polar if the center of negative charge does not coincide with the 
center of positive charge. Such a molecule constitutes a dipole: two equal and 
opposite charges separated in space. A dipole is often symbolized by -f->, where 
the arrow points from positive to negative?) The molecule possesses a dipole 
moment, /z, which is equal to the magnitude of the charge, e, multiplied by the 
distance, </, between the centers of charge: 



in in in 

Debyc c.s.u. Angstroms 
units, D 

In a way that cannot be gone into here, it is possible to measure the dipole 
moments of molecules; some of the values obtained are listed in Table 1.3. We 



SEC. 1.16 POLARITY O* MOLECULES 23 

shall be interested in the values of dipQle moments as indications of the relative 
polarities of different molecules. 



Table 1.3 DIPOIF MOMENIS, D 

H 2 HF 1.75, CH 4 

2 H 2 1.84 CH 3 C1 1.86 

N 2 NH 3 1.46 CC1 4 

C1 2 NF 3 0.24 CO 2 

Br 2 BF 3 

It is the fact that some molecules arc polar which has given rise to the specula- 
tion that some bonds are polar. We have taken up bond polarity first simply be- 
cause it is convenient to consider that the polarity of a molecule is a composite 
of the polarities of the individual bonds. 

Molecules like H 2 , O 2 , N 2 , C1 2 , and Br 2 have zero dipole moments, that is, 
are non-polar. The two identical atoms of each of these molecules have, of course, 
the same electronegativity and share electron^ equally; e is ?CTO and hence /z is 
zero, too. 

A molecule like hydrogen fluoride has the large dipole moment of 1.75 D. 
Although hydrogen fluoride is a small molecule, the very high electronegative 
fluorine pulls the electrons strongly; although d is small, e is large, and hence /x is 
large, too. 

' Methane and carbon tctrachloride, CC1 4 , have zero dipole moments. We 
certainly would expect the individual bonds of carbon tetrachloride at least to be 
nolar; because of the very symmetrical tetrahedral arrangement, however, they 
exactly cancel each other out (Fig. 1.14). In methyl chloride, r HhCl, the polarity 



H F -^ 

H ^%, 






/i = 1.75D 



Hydrogen Methane Carbon Methyl chloride 

fluoride tetrachloride 

Figure 1.14. Dipole moments of some molecules. Polantv of bonds and 
of molecules. 



of the carbon -chlorine bond is not canceled, however, and methyl chloride has a 
dipole moment of 1.86 r>. Thus the polarity of a molecule depends not only upon 
the polarity of its individual bonds but also upon the way the bonds are directed, 
that is, upon the shape of the molecule. 

Ammonia has a dipole moment of 1.46 n. This could be accounted for as a 
net dipole moment (a vector sum) resulting from the three individual bond moments. 



24 



STRUCTURE AND PROPERTIES 



CHAP. 1 



and would be in the direction shown in the diagram. In a similar way, we could 
account for water's dipole moment of 1.84 D. 





Dipole moments 

expected from 

bond moments alone 



Water 



Now, what kind of dipole moment would we expect for nitrogen trifluoride, 
NFa, which, like ammonia, is pyramidal? Fluorine is the most electronegative 
element of all and should certainly pull electrons strongly from nitrogen; the N F 
bonds should be highly polar, and their vector sum should be large far larger 
than for ammonia with its modestly polar N H bonds. 



Large dipole moment 

expected from 
bond moments alone 



Nitrogen trifluoride 

What are the facts? Nitrogen trifluoride has a dipole moment of only 0.24 D. It is 
not larger than the moment for ammonia, but rather is much smaller. 

How are we to account for this? We have forgotten the unshared pair of 
electrons. In NF 3 (as in NH 3 ) this pair occupies an sp* orbital and must contribute 
a dipole moment in the direction opposite to that of the net moment of the N F 
bonds (Fig. 1.15); these opposing moments are evidently of about the same size, 





X-1.46D 



Ammonia 





M-0.24D 



Nitrogen 
trifluoride 



Figure 1.15. Dipole moments of some molecules. Contribution from un- 
shared pairs. In NF 3 , moment due to unshared pair opposes vector sum 
of bond moments. 



SEC. 1.17 STRUCTURE AND PHYSICAL PROPERTIES 25 

and the result is a small moment, in which direction we cannot say. In ammonia 
the observed moment is probably due chiefly to the unshared pair, augmented 
by the sum of the bond moments. In a similar way, unshared pairs of electrons 
must contribute to the dipole moments of water and, indeed, of any molecules in 
which they appear. 

Dipole moments can give valuable information about the structure of mole- 
cules. For example, any structure for carbon tetrachloride that would result in a 
polar molecule can be ruled out on the basis of dipole moment alone. The evi- 
dence of dipole moment thus supports the tetrahedral structure for carbon 
tetrachloride. (However, it does not prove this structure, since there are other 
conceivable structures that would also result in a non-polar molecule.) 



Problem 1.5 Which of the following conceivable structures of CCU would also 
have a zero dipole moment ? (a) Carbon at the center of a square with a chlorine at 
each corner, (b) Carbon at the apex of a pyramid with a chlorine at each corner of 
a square base. 

Problem 1.6 Suggest a shape for the CO 2 molecule that would account for its 
zero dipole moment. 

Problem 1.7 In Sec. 1.12 we rejected two conceivable electronic configurations 
for ammonia, (a) If nitrogen were s/? 2 -hybridized, what dipole moment would you 
expect for ammonia? What is the dipole moment of ammonia? (b) If nitrogen used 
p orbitals for bonding, how would you expect the dipole moments of ammonia and 
nitrogen trifluoride to compare? How do they compare? 

The dipole moments of most compounds have never been measured. For 
these substances we must predict polarity from structure. From our knowledge of 
electronegativity, we can estimate the polarity of bonds; from our knowledge of 
bond angles, we can then estimate the polarity of molecules, taking into account 
any unshared pairs of electrons. 



1.17 Structure and physical properties 

We have just discussed one physical property of compounds: dipole moment. 
Other physical properties like melting point, boiling point, or solubility in a 
particular solvent are also of concern to us. The physical properties of a new 
compound give valuable clues about its structure. Conversely, the structure of a 
compound often tells us what physical properties to expect of it. 

In attempting to synthesize a new compound, for example, we must plan a 
series of reactions to convert a compound that we have into the compound that 
we want. In addition, we must work out a method of separating our product from 
all the other compounds making up the reaction mixture: unconsumed reactants, 
solvent, catalyst, by-products. Usually the isolation and purification of a product 
take much more time and effort than the actual making of it. The feasibility of 
isolating the product by distillation depends upon its boiling point and the boiling 
points of the contaminants; isolation by recrystallization depends upon its solu- 
bility in various solvents and the solubility of the contaminants. Success in the 
laboratory often depends upon making a good prediction of physical properties 
from structure. 



26 



STRUCTURE AND PROPERTIES 



CHAP. 1 



We have seen that there are two extreme kinds' of chemical bonds: ionic 
bonds, formed by the transfer of electrons, and covalent bonds, formed by the 
sharing of electrons. The physical properties of a compound depend largely upon 
which kind of bonds hold its atoms together in the molecule. 



1.18 Melting point 

In a crystalline solid the particles acting as structural units -ions or mole- 
cules-are arranged in some very regular, symmetrical \vay, there is a geometric 
pattern repeated over and over \vithin a crystal. 

Melting is the change from ihe highly ordered arrangement of particles in the 
cr>stalline lattice to the more random arrangement that characterizes a liquid (see 
Figs. 1.16 and- 1.1 7). Melting occurs when a temperature is reached at which the 
thermal energv of the particles is great enough to overcome the intracryslalline 
forces that hold them in position. 

An Ionic compound forms crvstals in which the structural units are ions. 
Solid sodium chloride, for example, is made up of positive sodium ions and nega- 
tive chloride ions alternating in a verv regular way. Surrounding each positive 





Figure 1.16. Melting of an ionic crystal. Units are ions. 



ion and equidistant from it are six negative ions: one on each side of it, one above 
and one below, one in front and one in back. Each negative ion is surrounded in a 
similar wav by si\ positive ions. There is nothing that we can properly call a 
molecule of sodium chloride. A particular sodium ion does not "belong" to any 
one chloride ion; it is equally attracted to six chloride ions. The crystal is an 
extremely strong, rigid structure, since the electrostatic forces holding each ion in 
position are powerful. These powerful interionic forces are overcome only at a 
ver> high temperature; sodium chloride has a melting point of 801 . 

Crystals of other ionic compounds resemble crystals of sodium chloride in 
having an ionic lattice, although the exact geometric arrangement may be different. 
As a result, these other ionic compounds, too, have high melting points. Many 
molecules contain both ionic and covalent bonds. Potassium nitrate, KNOj, for 
example, is made up of K 4 ions and NO 3 ions; the oxygen and nitrogen atoms of 
the NCV ion are hc ld to cacn olner ty covalent bonds. The physical properties of 
compounds like these are largely determined by the ionic bonds; potassium nitrate 
has very much the same sort of physical properties as sodium chloride. 

A non-ionic compound, one whose atoms are held to each other entirely by 
covalent bonds, forms crystals in which the structural units are molecules. It is the 



SEC. 1.19 



1NTERMOLECULAR FORCES 



27 



forces holding these molecules to each other that must be overcome for melting to 
occur. In general, these intermodular forces are very weak compared with the 





Figure 1.17. Melting of a non-ionic crystal. Units are molecules. 

forces holding ions to each other. To melt sodium chloride we must supply 
enough energy to break ionic bonds between Na + and Cl~. To melt methane, 
CH 4 , we do not need to supply enough energy to break covalent bonds between 
carbon and hvdrogen; we need only supply enough energy to break CH 4 molecules 
av\ay from each other. In contrast to sodium chloride, methane melts at 183. 



1.19 Intel-molecular forces 

What kind of forces hold neutral molecules to each other? Like interionic 
forces, these forces seem to be electrostatic in nature, involving attraction of 
positive charge for negative charge. There are two kinds of intermolccular forces: 
dipole-dipole interactions and van der Waals forces. 

Dipole- dipolc interaction is the attraction of the positive end of one polar 
molecule for the negative end of another polar molecule. In hydrogen chloride, 
for example, the relatively positive hydrogen of one molecule is attracted to the 
relatively negative chlorine of another: 




As a result of dipolc- dipole interaction, polar molecules are generally held to each 
other more strongly than are non-polar molecules of comparable molecular weight; 
this difference in strength of intermolecular forces is reflected in the physical 
properties of the compounds concerned. 

An especially strong kind of dipole dipole attraction is hydrogen bonding, 
in which a hydrogen atom screes as a bridge between two electronegative atoms, 
holding one by a covalent bond and the other by purely electrostatic forces. When 
hydrogen is attached to a highly electronegative atom, the electron cloud is 
greatly distorted toward the electronegative atom, exposing the hydrogen nucleus. 
The strong positive charge of the thinly shielded hydrogen nucleus is strongly 
attracted by the negative charge of the' electronegative atom of a second mole- 
cule. This attraction has a strength of about 5 kcal/mole, and is thus much weaker 
than the covalent bond about 50-100 kcal/mole that holds it to the first electro- 
negative atom. It is, however, much stronger than other dipole-dipole attractions. 



28 STRUCTURE AND PROPERTIES CHAP. 1 

Hydrogen bonding is generally indicated in formulas by a broken line: 

H H H H 

H F-H F H O H O H-rN- H N H-N-H-O 

II II I 

H H H H H 

For hydrogen bonding to be important, both electronegative atoms must come 
from the group: F, O, N. Only hydrogen bonded to one of these three elements is 
positive enough, and only these three elements are negative enough, for the 
necessary attraction to exist. These three elements owe their special effectiveness 
to the concentrated negative charge on their small atoms. 

Hydrogen bonding, we shall find, not only exerts profound effects on the boil- 
ing point and solubility properties of compounds, but also plays a key role in 
determining the shapes of large molecules like proteins and nucleic acids, shapes 
that in a very direct way determine, in turn, their biological properties: the size 
of the "pockets" in the hemoglobin molecule, just big enough to hold heme 
groups with their oxygen-carrying iron atoms (p. 1152); the helical shape of 
a-keratin and collagen molecules that makes wool and hair strong, and tendons 
and skin tough (p. 1 158). It is hydrogen bonding that makes the double helix of 
DNA double and thus permits the self-duplication of molecules that is the basis of 
heredity (p. 1179). 

There must be forces between the molecules of a non-polar compound, since 
even such compounds can solidify. Such attractions are called van der Waals 
forces. The existence of these forces is accounted for by quantum mechanics. 
We can roughly visualize them arising in the following way. The average distri- 
bution of charge about, say, a methane molecule is symmetrical, so that there is no 
net dipole moment. However, the electrons move about, so that at any instant of 
time the distribution will probably be distorted, and a small dipole will exist. This 
momentary dipole will affect the electron distribution in a second methane molecule 
nearby. The negative end of the dipole tends to repel electrons, and the positive 
end tends to attract electrons; the dipole thus induces an oppositely oriented dipole 
in the neighboring molecule: 



Although the momentary dipoles and induced dipoles are constantly changing, the 
net result is attraction between the two molecules. These van der Waals forces have 
a very short range; they act only between the portions of different molecules that 
are in close contact, that is, between the surfaces of molecules. As we shall see, 
the relationship between the strength of van der Waals forces and the surface areas 
of molecules (Sec. 3.12) will help us to understand the effect of molecular size and 
shape on physical properties. We must not underestimate the power of these 
weakest intermolecular forces; acting between non-polar chains of phospholipids, 
for example, they are the mortar in the walls of living cells. 

With respect to other atoms to which it is not bonded whether in another molecule 
or in another part of the same molecule every atom has an effective "size," called its 



SEC. 1.20 



BOILING POINT 



29 



van der Waals radius. As two non-bonded atoms are brought together the attraction 
between them steadily increases, and reaches a maximum when they are just " touching** 
that is to say, when the distance between the nuclei is equal to the sum of the van der Waals 
radii. Now, if the atoms are forced still closer together, van der Waals attraction is very 
rapidly replaced by van der Waals repulsion. Thus, non-bonded atoms welcome each 
other's touch, but strongly resist crowding. 

We shall find both attractive and repulsive van der Waals forces important to our 
understanding of molecular structure. 

1.20 Boiling point 

Although the particles in a liquid are arranged less regularly and are freer to 
move about than in a crystal, each particle is attracted by a number of other 
particles. Boiling involves the breaking away from the liquid of individual mole- 
cules or pairs of oppositely charged ions (see Figs. 1.18 and 1.19). This occurs 





Figure 1.18. Boiling of an ionic liquid. Units are ions and ion pairs. 

when a temperature is reached at which the thermal energy of the particles is 
great enough to overcome the cohesive forces that hold them in the liquid. 

In the liquid state the unit of an ionic compound is again the ion. Each ion is 
still held strongly by a number of oppositely charged ions. Again there is nothing 
we could properly call a molecule. A great deal of energy is required for a pair of 
oppositely charged ions to break away from the liquid; boiling occurs only at a 
very high temperature. The boiling point of sodium chloride, for example, is 
1413. In the gaseous state we have an ion pair, which can be considered a sodium 
chloride molecule. 

In the liquid state the unit of a non-ionic compound is again the molecule. 
The weak intermolecular forces here dipole-dipole interactions and van der Waals 
forces are more readily overcome than the strong interionic forces of ionic 
compounds, and boiling occurs at a very much lower temperature. Non-polar 
methane boils at - 161.5, and even polar hydrogen chloride boils at only -85. 





Figure 1.19. Boiling of a non-ionic liquid. Units are molecules. 



30 STRUCTURE AND PROPERTIES CHAP. 1 

Liquids whose molecules are held together by hydrogen bonds are called 
associated liquids. Breaking these hydrogen bonds takes considerable energy, and 
so an associated liquid has a boiling point that is abnormally high for a compound 
of its molecular weight and dipole moment. Hydrogen fluoride, for example, 
boils 100 degrees higher than the heavier, non-associated hydrogen chloride; 
water boils 160 degrees higher than hydrogen sulfide. 

The bigger the molecules, the stronger the van der Waals forces. Other things 
being equal polarity, hydrogen bonding boiling point rises with increasing 
molecular size. Boiling points of organic compounds range upward from that of 
tiny, non-polar methane, but we seldom encounter boiling points much above 
350; at higher temperatures, covalent bonds within the molecules start to break, 
and decomposition competes with boiling. It is to lower the boiling point and thus 
minimize decomposition that distillation of organic compounds is often carried 
out under reduced pressure. 

Problem 1.8 Which of the following organic compounds would you predict to 
be associated liquids? Draw structures to show the hydrogen bonding you would 
expect. (a)CH 3 OH; (b) CH,OCH 3 ; (c) CH 3 F; (d) CH 3 Cl; (e) CH 3 NH,; 
(f)(CH 3 ) 2 NH; (g)(CH 3 ) 3 N. * 



1.21 Solubility 

When a solid or liquid dissolves, the structural units ions or molecules 
become separated from each other, and the spaces in between become occupied by 
solvent molecules. In dissolution, as in melting and boiling, energy must be 
supplied to overcome the interionic or intermolecular forces. Where does the 
necessary energy come from? The energy required to break the bonds between 
solute particles is supplied by the formation of bonds between the solute particles 
and the solvent molecules: the old attractive forces are replaced by new ones. 

A great deal of energy is necessary to overcome the powerful electrostatic 
forces holding together an ionic lattice. Only water or other highly polar solvents 
are able to dissolve ionic compounds appreciably. What kind of bonds are formed 
between ions and a polar solvent? By definition, a polar molecule has a positive 
end and a negative end. Consequently, there is electrostatic attraction between a 
positive ion and the negative end of the solvent molecule, and between a negative 
ion and the positive end of the solvent molecule. These attractions are called ion- 
dipole bonds. Each ion-dipole bond is relatively weak, but in the aggregate they 
supply enough energy to overcome the interionic forces in the crystal. In solution 
each ion is surrounded by a cluster of solvent molecules, and is said to be solvated; 
if the solvent happens to be water, the ion is said to be hydrated. In solution, as in 
the solid and liquid states, the unit of a substance like sodium chloride is the ion, 
although in this case it is a solvated ion (see Fig. 1.20). 

To dissolve ionic compounds a solvent must also have a high dielectric con- 
stant, that is, have high insulating properties to lower the attraction between 
oppositely charged ions once they are solvated. 

Water owes its superiority as a solvent for ionic substances not only to its 
polarity and its high dielectric constant, but to another factor as well: it contains 
the OH group and thus can form hydrogen bonds. Water solvates both cations 





SEC. 1.21 SOLUBILITY 31 



Figure 1.20. lon-dipole inter- 
actions: solvated cation and anion. 



and anions: cations, at its negative pole (its unshared electrons, essentially); anions, 
through hydrogen bonding. 

The solubility characteristics of non-ionic compounds are determined chiefly 
by their polarity. Non-polar or weakly polar compounds dissolve in non-polar 
or weakly polar solvents; highly polar compounds dissolve in highly polar solvents. 
"Like dissolves like" is an extremely useful rule of thumb. Methane dissolves in 
carbon tetrachloride because the forces holding methane molecules to each other 
and carbon tetrachloride molecules to each other are replaced by very similar 
forces holding methane molecules to carbon tetrachloride molecules. 

Neither methane nor carbon tetrachloride is readily soluble in water. The 
highly polar water molecules are held to each other by very strong dipole-dipole 
interactions hydrogen bonds; there could be only very weak attractive forces 
between water molecules on the one hand and the non-polar methane or carbon 
tetrachloride molecules on the other. 

In contrast, the highly polar organic compound methanol, CH 3 OH, is quite 
soluble in water. Hydrogen bonds between water and methanol molecules can 
readily replace the very similar hydrogen bonds between different methanol mole- 
cules and different water molecules. 

As we shall see, much of organic chemistry is concerned with reactions be- 
tween non-ionic compounds (generally organic) and ionic compounds (inorganic 
and organic), and it is necessary to select a solvent in which both the reagents will 
dissolve. Water is a poor solvent for most organic compounds, but this difficulty 
can be overcome by addition of a second solvent like methanol. 

Solvents like water or methanol are called protic solvents: solvents containing 
hydrogen that is attached to oxygen or nitrogen, and hence is appreciably acidic 
(Sec. 1.22). Through hydrogen bonding such solvents tend to solvate anions par- 
ticularly strongly; and anions, as it turns out, are usually the important half of 
an ionic reagent. Thus, although protic solvents dissolve 'the reagent and bring it 
into contact with the organic molecule, at the same time they stabilize the anions 
and lower their reactivity drastically; their basicity is weakened and, with it, the 
related property, nucleophilic power (Sec. 14.5). 

Recent years have seen the development and widespread use ofaprotic solvents: 
polar solvents of moderately high dielectric constants, which do not contain 
acidic hydrogen. For example: 



,-, 
CH 3 -S-CH 3 H -~ N x ( ) 



* CHJ 

Dimethyl sulfoxide N,N-Dimethylformamide Sulfolane 

DMSO DMF 



32 STRUCTURE AND PROPERTIES 



CHAP. 



These solvents dissolve both organic and inorganic reagents but, in dissolving ibnic 
compounds, solvate cations most strongly, and* leave the anions relatively un- 
encumbered and highly reactive; anions are more basic and more nucleophilic. 

Since about 1958, reports of dramatic solvent effects on a wide variety of 
reactions have appeared, first about dimethylformamide (DMF) and more recently 
about dimethyl sulfoxide (DMSO): reactions that, in most solvents, proceed slowly 
at high temperatures to give low yields may be found, in an aprotic solvent, to 
proceed rapidly often at room temperature to give high yields. A change of 
solvent may cause a million-fold change in reaction rate. A solvent is not simply a 
place a kind of gymnasium where solute molecules may gambol about and 
occasionally collide; the solvent is intimately involved in any reaction that takes 
place in it, and we are just beginning to find out how much it is involved, and in 
what way. 

Individual molecules may have both polar and non-polar parts and, if the molecules 
are big enough, these- parts display their individual solubility properties. The polar parts 
dissolve in water; the non-polar parts dissolve in a non-polar solvent or, if there is 
none about, cluster together in effect, dissolve in each other. Such dual solubility 
behavior gives soaps and detergents their cleansing power, and controls the alignment of 
molecules in cell membranes; a globular protein molecule an enzyme, say coils up to 
expose its polar parts to the surrounding water and to hide its non-polar parts, and in 
doing this takes on the particular shape needed for its characteristic biological properties. 

1.22 Acids and bases 

Turning from physical to chemical properties, let us review briefly one familiar 
topic that is fundamental to the understanding of organic chemistry: acidity and 
basicity. 

The terms acid and base have been defined in a number of ways, each defini- 
tion corresponding to a particular way of looking at the properties of acidity and 
basicity. We shall find it useful to look at acids and bases from two of these view- 
points; the one we select will depend upon the problem at hand. 

According to the Lowry-Brensted definition, an acid is a substance that gives up 
a proton, and a base is a substance that accepts a proton. When sulfuric acid dis- 
solves in water, the acid H 2 SO 4 gives up a proton (hydrogen nucleus) to the base 
H 2 O to form a new acid H 3 O + and a new base HSO 4 ~. When hydrogen chloride 
reacts with ammonia, the acid HC1 gives up a proton to the base NH 3 to form the 
new acid NH 4 + and the new base Cl". 

H 2 SO 4 + H 2 O ^= H 3 O+ + HS<V 

Stronger Stronger Weaker Weaker 

acid base acid base 

HC1 + NH 3 ^= NH 4 + + Cl- 

Stronger Stronger Weaker Weaker 

acid base acid base 

According to the Lowry-Br0nsted definition, the strength of an acid depends 
upon its tendency to give up a proton, and the strength of a base depends upon its 
tendency to accept a proton. Sulfuric acid and hydrogen chloride are strong acids 
since they tend to give up a proton very readily; conversely, bisulfate ion, HSO 4 ~, 
and chloride ion must necessarily be weak bases since they have little tendency to 



SEC. 1.22 ACIDS AND BASES 33 

hokLgn to protons. In each of the reactions just described, the equilibrium favors 
the formation of the weaker acid and the weaker base. 

If aqueous H 2 SO 4 is mixed with aqueous NaOH, the acid H 3 O* (hydronium 
ion) gives up a proton to the base OH" to form the new acid H 2 O and the new 
base H 2 O. When aqueous NH 4 C1 is mixed with aqueous NaOH, the acid NH 4 + 



H 3 O + H 

Stronger 
acid 


- OH- ^= H 2 - 

Stronger Weaker 
base acid 


f- H 2 O 

Weaker 
base 


NH 4 + H 

Stronger 
acid 


- OH- ^= H 2 O - 

Stronger Weaker 
base acid 


i- NH 3 

Weaker 
base 



(ammonium ion) gives up a proton to the base OH" to form the new acid H 2 O 
and the new base NH 3 . In each case the strong base, hydroxide ion, has accepted 
a proton to form the weak acid H 2 O. If we arrange these acids in the order shown, 
we must necessarily arrange the corresponding (conjugate) bases in the opposite 
order. 

H *\O 

Acid strength JJg^ 4 > H 3 O + > NH 4 + > H 2 O 

H^IO ~ 

Base strength "j_ 4 < H 2 O < NH 3 < OH~ 

Like water, many organic compounds that contain oxygen can act as bases and 
accept protons; ethyl alcohol and ethyl ether, for example, form the oxonium ions 
I and II. For convenience, we shall often refer to a structure like I as a protonated 
alcohol and a structure like II as a protonated ether. 


C 2 H 5 OH + H 2 SO 4 ^= C 2 H 5 OH + HSO 4 

Ethyl alcohol H 

I 

An oxonium ion 
Protonated ethyl alcohol 

e 

(C 2 H 5 ) 2 6: + HC1 ^= (C 2 H 5 ) 2 6:H + CT 
Ethyl ether H 

An oxonium ion 
Protonated ethyl ether 

According to the Lewis definition, a base is a substance that can furnish an 
electron pair to form a covalent bond, and an acid is a substance that can take up an 
electron pair to form a covalent bond. Thus an acid is an electron-pair acceptor 
and a base is an electron-pair donor. This is the most fundamental of the acid-base 
concepts, and the most general; it includes all the other concepts. 

A proton is an acid because it is deficient in electrons, and needs an electron 
pair to complete its valence shell. Hydroxide ion, ammonia, and water are bases 
because they contain electron pairs available for sharing. In boron trifluoride, 
BF 3 , boron has only six electrons in its outer shell and hence tends to accept an- 
other pair to complete its octet. Boron trifluoride is an acid and combines with 



34 STRUCTURE AND PROPERTIES CHAP. 1 

F F 

1 ^ el e 

F-B + :NH 3 ^= F-B:NH 3 

I i 

Acid Base 

F F 

I v 0! $ 

F-B + :0(C 2 H 5 ) 2 ^= F~B:0(C 2 H 5 ) 2 

F F 

Acid Base 

such bases as ammonia or ethyl ether. Aluminum chloride, AlClj , is an acid, and 
for the same reason. In stannic chloride, SnCl4, tin has a complete octet, but can 
accept additional pairs of electrons (e.g., in SnCl 6 ~ ") and hence it is an acid, too. 

We write a formal negative charge on boron in these formulas because it has one more 
electron half-interest in the pair shared with nitrogen or oxygen than is balanced by 
the nuclear charge; correspondingly, nitrogen or oxygen is shown with a formal positive 
charge. 

We shall find the Lewis concept of acidity and basicity fundamental to our 
understanding of organic chemistry. To make it clear that we are talking about 
this kind of acid or base, we shall often use the expression Lewis acid (or base), 
or sometimes acid (or base) in the Lewis sense. 

Chemical properties, like physical properties, depend upon molecular struc- 
ture. Just what features in a molecule's structure tell us what to expect about its 
acidity or basicity? We can try to answer this question in a general way now, 
although we shall return to it many times later. 

To be acidic in the Lowry-Bronsted sense, a molecule must, of course, contain 
hydrogen. The degree of acidity is determined largely by the kind of atom that 
holds the hydrogen and, in particular, by that atom's ability to accommodate the 
electron pair left behind by the departing hydrogen ion. This ability to accom- 
modate the electron pair seems to depend upon several factors, including (a) the 
atom's electronegativity, and (b) its size. Thus, within a given row of the Periodic 
Table, acidity increases as electronegativity increases: 

H-CHj < H -NH 2 < H-OH < H -F 
Acidity 

H~SH < H-C1 

And within a given family, acidity increases as the size increases: 

H -F < H Cl < H Br < H I 
Acidity 

H-OH < H SH < H-SeH 

Among organic compounds, we can expect appreciable Lowry-Br0nsted acidity 
from those containing Q--H, N -H, and S H groups. 

To be acidic in the Lewis sense, a molecule must be electron-deficient; in 
particular, we would look for an atom bearing only a sextet of electrons. 

Problem 1.9 Predict the relative acidity of: (a) methyl alcohol (CH 3 OH) and 
mcthylaminc (CH 3 NH 2 ); (b) methyl alcohol (CH 3 OH) and methanethiol (CH 3 SH); 



SEC. 1.23 ELECTRONIC AND STERIC EFFECTS 35 

Problem 1.10 Which is the stronger acid v of each pair: (a) H 3 O + or H 2 O; 
(b) NH 4 + or NH 3 ; (c) H 2 S or HS~; (d) H 2 O or OHT ? (e) What relationship is 
there between charge and acidity ? 

To be basic in either the Lowry-Br0nsted or the Lewis sense, a molecule must 
have an electron pair available for sharing. The availability of these unshared 
electrons is determined largely by the atom that holds them: its electronegativity, 
its size, its charge. The operation of these factors here is necessarily opposite to 
what we observed for acidity; the better an atom accommodates the electron pair, 
the less available the pair is for sharing. 

Problem 1.11 Arrange the members of each group in order of basicity: 
(a) F-, OH~, NH,~, CH 3 ~ ; (b) HF, H 2 O, NH 3 ; (c) CT , SH~ ; (d) F , Cl~, Br , I~ ; 
(e)OH~,SH-,SeH-. 

Problem 1.12 Predict the relative basicity of methyl fluoride (CH,F), methyl 
alcohol (CH 3 OH), and methylamine (CH 3 NH 2 ). 

Problem 1.13 Arrange the members of each group in order of basicity: 
(a) H 3 O + , H : O, OH~; (b) NH 3 , NH 2 ; (c) H 2 S, HS~, S". (d) What relationship 
is there between charge and basicity ? 



1.23 Electronic and steric effects 

Like acidity and basicity, other chemical properties, too, depend upon molec- 
ular structure. Indeed, most of this book will be concerned with finding out what 
this relationship is. 

A particular compound is found to undergo a particular reaction. Not sur- 
prisingly, other compounds of similar structure are also found to undergo the 
same reaction but faster or slower, or with the equilibrium lying farther to the 
right or left. We shall, first of all, try to see how it is that a particular kind of struc- 
ture predisposes a compound to a particular reaction. Then and much of our 
time will be spent with this we shall try to see how it is that variations in molec- 
ular structure give rise to variations in reactivity: to differences in rate of reaction 
or in position of equilibrium. 

To do all this is a complicated business, and to help us we shall mentally 
analyze the molecule: we shall consider that a molecule consists of a reaction 
center to which are attached various substituents. The nature of the reaction 
center determines what reaction occurs. The nature of the substituents determines 
the reactivity. 

A substituent affects reactivity in two general ways: (a) by its electronic effect, 
that is, by its effect on the availability of electrons at the reaction center; and 
(b) by its steric effect, that is, by its effect on crowding in the molecule. Since hydro- 
gen is the element most commonly attached to carbon, it is used as the standard of 
reference. We consider a substituent G, which may be an atom or a group of 
atoms, to be attached to a carbon in place of a hydrogen, and we ask the ques- 
tion: How does G C compare with H O~? 

Let us look first at electronic effects. At some stage of most reactions, a posi- 
tive or negative charge develops in the reacting molecule. Reactivity usually de- 
pends upon how easily the molecule can accommodate that charge. Accommoda- 



36 STRUCTURE AND PROPERTIES CHAP. 1 

tion of charge depends, in turn, on the electronic effects of the substituents. 
Compared with hydrogen, a substituent may tend either to withdraw electrons 
(G--G~) or to release electrons (G-*C~ ). An electron-withdrawing substituent 
will help to accept the surplus of electrons that constitutes a negative charge; an 
electron-releasing substituent will help offset the deficit of electrons that consti- 
tutes a positive charge. 

Just how does a substituent exert its electronic effect? Despite the vast amount 
of work that has been done and is still being done on this problem, there is 
no general agreement, except that at least two factors must be at work. We shall 
consider electron withdrawal and release to result from the operation of two fac- 
tors: the inductive effect and the resonance effect. 

The inductive effect depends upon the "intrinsic*' tendency of a substituent 
to withdraw electrons by definition, its electronegativity acting either through 
the molecular chain or through space. The effect weakens steadily with increasing 
distance from the substituent. Most elements likely to be substituted for hydrogen 
in an organic molecule are more electronegative than hydrogen, so that most 
substituents exert electron-withdrawing inductive effects: for example, F, Cl, 
_Br, -I, -OH, -NH 2 , -NO 2 . 

The resonance effect involves delocalization of electrons typically, those 
called TT (pi) electrons. It depends upon the overlap of certain orbitals, and there- 
fore can operate only when the substituent is located in certain special ways rela- 
tive to the charge center. By its very nature, as we shall see (Sec. 6.25), the reso- 
nance effect is a stabilizing effect, and so it amounts to electron withdrawal from a 
negatively charged center, and electron release to a positively charged center. 

A substituent can influence reactivity not only by its electronic effect (induc- 
tive and/or resonance), but also, in some cases, by its steric effect: an effect due to 
crowding at some stage of the reaction, and dependent therefore on the size of the 
substituent. 

Problem 1.14 Predict the relative basicity of NH 3 and NF 3 . 



1.24 Isomerism 

Before we start our systematic study of the different kinds of organic com- 
pounds, let us look at one further concept which illustrates especially well the 
fundamental importance of molecular structure: the concept of isomerism. 

The compound ethyl alcohol is a liquid boiling at 78. Analysis (by the 
methods described later, Sec. 2.26) shows that it contains carbon, hydrogen, and 
oxygen in the proportions 2C:6H:1O. Measurement of its mass spectrum shows 
that it has a molecular weight of 46. The molecular formula of ethyl alcohol 
must therefore be C 2 H 6 O. Ethyl alcohol is a quite reactive compound. For exam- 
ple, if a piece of sodium metal is dropped into a test tube containing ethyl alcohol, 
there is a vigorous bubbling and the sodium metal is consumed; hydrogen gas is 
evolved and there is left behind a compound of formula C 2 H 5 ONa. Ethyl alcohol 
reacts with hydriodic acid to form water and a compound of formula C 2 H 5 I. 

The compound methyl ether is a gas with a boiling point of 24. It is clearly 
a different substance from ethyl alcohol, differing not only in its physical properties 
but also in its chemical properties. It does not react at all with sodium metal. Like 



PROBLEMS 37 

ethyl alcohol, it reacts with hydriodic acid, but it yields a compound of formula 
CH 3 I. Analysis of methyl ether shows that it contains carbon, hydrogen, and oxy- 
gen in the same proportions as ethyl alcohol, 2C:6H: 1O. It has the same molec- 
ular weight as ethyl alcohol, 46. We conclude that it has the same molecular for- 
mula, C 2 H 6 O. 

Here we have two substances, ethyl alcohol and methyl ether, which have the 
same molecular formula, C 2 H 6 O, and yet quite clearly are different compounds. 
How can we account for the existence of these two compounds? The answer is: 
they differ in molecular structure. Ethyl alcohol has the structure represented by I, 
and methyl ether the structure represented by IT. As we shall see, the differences in 
physical and chemical properties of these two compounds can readily be accounted 
for on the basis of the difference in structure. 

H H H H 

H-C-C-0-H H-C-0-C-H 

iU A A 

I II 

Ethyl alcohol Methyl ether 

Different compounds that have the same molecular formula are called isomers 
(Gr.: isos, equal; meros, part). They contain the same numbers of the same kinds 
of atoms, but the atoms are attached to one another in different ways. Isomers 
are different compounds because they have different molecular structures. 

This difference in molecular structure gives rise to a difference in properties; 
it is the difference in properties which tells us that we are dealing with different 
compounds. In some cases, the difference in structure and hence the difference 
in properties is so marked that the isomers are assigned to different chemical 
families, as, for example, ethyl alcohol and methyl ether. In other cases the dif- 
ference in structure is sp subtle that it can be described only in terms of three- 
dimensional models. Other kinds of isomerism fall between these two extremes. 



PROBLEMS 

1. Which of the following would you expect to be ionic, and which non-ionic? Give 
a simple electronic structure (Sec. 1.3) for each, showing only valence shell electrons. 

(a) MgCl 2 (c) IC1 (e) KC1O 4 (g) BaSO 4 

(b) CH 2 C1 2 (d) NaOCl (f) SiCl 4 (h) CH 3 NH 2 

2. Give a likely simple electronic structure (Sec. 1.3) for each of the following, assum- 
ing them to be completely covalent. Assume that every atom (except hydrogen, of course) 
has a complete octet, and that two atoms may share more than one pair of electrons. 

(a) N 2 H 4 (d) COC1 2 (g) CO 3 " (j) CH 2 O 

(b) H 2 SO 4 (e) HONO (h) C 2 H 4 (k) CH 2 O 2 

(c) HS0 4 - (f) N0 2 ~ (i) C 2 H 2 (1) C 3 H 8 

3. What shape would you expect each of the following to have? 

(a) (CH 3 ) 3 B (e) the amide ion, NH 2 - 

(b) the methyl anion, CH 3 :~ (f) methyl ether 

(c) the methyl cation, CH 3 + (g) the fluoborate ion, BF 4 ~ 

(d) H 2 S (h) (CH 3 ) 3 N 



38 STRUCTURE AND PROPERTIES CHAP. 1 

4. In many complex ions, e.g., Co(NH 3 ) 6 4 + +, the bonds to the central atom can 
be pictured as utilizing six equivalent sp*d 2 (or d 2 sp 3 ) hybrid orbitals. On the basis of 
maximum separation of orbitals, what geometry would you expect these complexes to 
have? 

5. Indicate the direction of the dipole moment, // any, that you would expect for 
each of the following: 

(a) HBr (d) CH 2 C1 2 (g) methyl ether 

(b) IC1 (e) CHC1 3 (h) (CH 3 ) 3 N 

(c) I 2 (f) CH 3 OH (i) CF 2 C1 2 

6. (a) Although HC1 (1.27 A) is a longer molecule than HF (0.92 A), it has a smaller 
dipole moment (1.03 D compared to 1.75 D), How do you account for this fact? (b) The 
dipole moment of CH 3 F is 1.847 D, and of CD 3 F, 1.858 D. (D is 2 H, deuterium.) Com- 
pared with the C H bond, what is the direction of the C D dipole? 

7. What do the differences in properties between lithium acetylacetonate (m.p. 
very high, insoluble in chloroform) and beryllium acetylacetonate (m.p. 108, b.p. 270, 
soluble in chloroform) suggest about their structures? 

8. //-Butyl alcohol (b.p. 118') has a much higher boiling point than its isomer ethyl 
ether (b.p. 35 J ), yet both compounds show the same solubility (8 g per 100 g) in water. 

H H H H HH HH 

! I ! I I I II 

H--C CC-C O-H H- C-C-O C C-H 

I 1 I I II II 

HHHH HH HH 

n-Butyl alcohol Ethyl ether 

How do you account for these facts ? 

9. Rewrite the following equations to show the Lowry-Br0nsted acids and bases 
actually involved. Label each as stronger or weaker, as in Sec. 1.22. 

(a) HCl(aq) -f NaHCO 3 (aq) ~^ H 2 CO 3 4 NaCl 

(b) NaOH(aq) + NaHCO,(aq) "Z? Na 2 CO 3 -f H 2 O 

(c) NH 3 (aq) -f HNO,(aq) "T" NH 4 NO 3 (aq) 

(d) NaCN(aq) !!_ HCN(aq) + NaOH(aq) 

(e) NaH -f H 2 O > H 2 + NaOH 

(f) CaC 2 4- H 2 > Ca(OH) 2 + C 2 H 2 

Calcium carbide Acetylene 

10. What is the Lowry-Bronsted acid in (a) HC1 dissolved in water; (b) HC1 (union- 
ized) dissolved in benzene? (c) Which solution is the more strongly acidic? 

11. Account for the fact that nearly every organic compound containing oxygen 
dissolves in cold concentrated sulfuric acid to yield a solution from which the compound 
can be recovered by dilution with water. 

12. For each of the following molecular formulas, draw structures like those in Sec. 
1.24 (a line for each shared pair of electrons) for all the isomers you can think of. Assume 
that every atom (except hydrogen) has a complete octet, and that two atoms may share 
more than one pair of electrons. 

(a) C 2 H 7 N (c) C 4 H, (e) C 3 H 8 O 

(b) C 3 H 8 (d) C 3 H 7 C1 (f) C 2 H 4 

13. In ordinary distillation, a liquid is placed in a flask and heated, at ordinary or 
reduced pressure, until distillation is complete. In the modification called flash distilla- 
tion, the liquid is dripped into a heated flask at the same rate that it distills out, so that 
there is little liquid in the flask at any time. What advantage might flash distillation 
have, and under what conditions might you use it? 



PROBLEMS 39 



About Working Problems 

Working problems is a necessary part of your work for two reasons: it will 
guide your study in the right direction, and, after you have studied a particular 
chapter, it will show whether or not you have reached your destination. 

You should work all the problems that you can; you should get help with 
the ones you cannot work yourself. The first problems in each set are easy, but 
provide the drill in drawing formulas, naming compounds, and using reactions 
that even the best student needs. The later problems in each set are the kind 
encountered by practicing chemists, and test your ability to use what you have 
learned. 

You can check your answers to many of the problems in the answer section 
in the back of the book, and by use of the index. 



Chapter 



Methane 

Energy of Activation. 
Transition State 



2.1 Hydrocarbons 

Certain organic compounds contain only two elements, hydrogen and carbon, 
and hence are known as hydrocarbons. On the basis of structure, hydrocarbons 
are divided into two main classes, aliphatic and aromatic. Aliphatic hydrocarbons 
are further divided into families: alkanes, alkenes, alkynes, and their cyclic ana- 
logs (cycloalkanes, etc.). We shall take up these families in the order given. 

Hydrocarbons 



Aliphatic 



Aromatic 



Alkanes Alkenes Alkynes 



Cyclic 
aliphatic 



The simplest member of the alkane family and, indeed, one of the simplest of 
all organic compounds is methane, CH 4 . We shall sUidy this single compound at 
some length, since most of what we learn about it can be carried over with minor 
modifications to any alkane. 



2.2 Structure of methane 

As we discussed in the previous chapter (Sec. 1.11), each of the four hydrogen 
atoms is bonded to the carbon atom by a covalent bond, that is, by the sharing of a 
pair of electrons. When carbon is bonded to four other atoms, its bonding orbitals 
(sp* orbitals, formed by the mixing of one s and three/? orbitals) are directed to the 
corners of a tetrahedron (Fig. 2. la). This tetrahedral arrangement is the one that 
permits the orbitals to be as far apart as possible. For each of these orbitals to 



SEC. 2.3 



PHYSICAL PROPERTIES 



41 



overlap most effectively the spherical s orbital of a hydrogen atom, and thus to form 
the strongest bond, each hydrogen nucleus must be located at a corner of this 
tetrahedron (Fig.- 2. 1 b). 






(c) 



Figure 2.1. Methane molecule, (a) Tetrahedral sp* orbitals. (b) Predicted 
shape: H nuclei located for maximum overlap, (c) Shape and size. 



The tetrahedral structure of methane has been verified by electron diffraction 
(Fig. 2.1c), which shows beyond question the arrangement of atoms in such simple 
molecules. Later on, we shall examine some of the evidence that led chemists to 
accept this tetrahedral structure long before quantum mechanics or electron 
diffraction was known. 

We shall ordinarily write methane with a dash to represent each pair of elec- 
trons shared by carbon and hydrogen (I). To focus our attention on individual 
electrons, we may sometimes indicate a pair of electrons by a pair of dots (II). 
Finally, when we wish to consider the actual shape of the molecule, we shall use a 
simple three-dimensional picture (HI). 



H-C H 
H 



H 

H:C:H 
H 

H 




2 3 Physical properties 

As we discussed in the previous chapter (Sec. 1.18), the unit of such a non-ionic 
compound, whether solid, liquid, or gas, is the molecule. Because the methane 
molecule is highly symmetrical, the polarities of the individual carbon-hydrogen 
bonds cancel out; as a result, the molecule itself is non-polar. 

Attraction between such non-polar molecules is limited to van der Waals 
forces; for such small molecules, these attractive forces must be tiny compared 
with the enormous forces between, say, sodium and chloride ions. It is not sur- 
prising, then, that these attractive forces are easily overcome by thermal energy, 
so that melting and boiling occur at very low temperatures: m.p -183, b.p. 
- 101.5. (Compare these values with the corresponding ones for sodium chloride: 
m.p. 801. b.p. 1413.) As a consequence, methane is a gas at ordinary tempera- 
tures. 



42 METHANE . CHAP. 2 

Methane is colorless and, when liquefied, is less dense than water (sp.gr. 0.4). 
In agreement with the rule of thumb that "litfe dissolves like," it is only slightly 
soluble in water, but very soluble in organic liquids such as gasoline, ether, and 
alcohol. In its physical properties methane sets the pattern for the other members 
of the alkane family. 



2.4 Source 

Methane is an end product of the anaerobic ("without air") decay of plants, 
that is, of the breakdown of certain very complicated molecules. *As such, it is the 
major constituent (up to 97 ) of natural gas. It is the dangerousy/m/ow/? of the 
coal mine, and can be seen as marsh gas bubbling to the surface of swamps. 

If methane is wanted in very pure form, it can be separated from the other 
constituents of natural gas (mostly other alkanes) by fractional distillation. Most 
of it, of course, is consumed as fuel without purification. 

According to one theory, the origins of life go back to a primitive earth 
surrounded by an atmosphere of methane, water, ammonia, and hydrogen. 
Energy radiation from the sun, lightning discharges- -broke these simple mole- 
cules into reactive fragments (free radicals, Sec. 2.12); these combined to form 
larger molecules which eventually yielded the enormously complicated organic 
compounds that make up living organisms. (Recent detection of organic mole- 
cules in space has even led to the speculation that "organic seeds for life could 
have existed in interstellar clouds/') 

Evidence that this could have happened was found in 1953 by the Nobel Prize 
winner Harold C. Urey and his student Stanley Miller at the University of Chicago. 
They showed that an electric discharge converts a mixture of methane, water, 
animonia, and hydrogen into a large number of organic compounds, including 
amino acids, the building blocks from which proteins, the "stuff of life*' (Chap. 36), 
are made. (It is perhaps appropriate that we begin this study of organic chemistry 
with methane and its conversion .into free radicals.) 

The methane generated in the final decay of a once-living organism may well 
be the very substance from which in the final analysis the organism was derived. 
**. . . ear tlf to earth, ashes to ashes, dust to dust. . . ." 



2.5 Reactions 

In its chemical properties as in its physical properties, methane sets the pat- 
tern for the alkane family (Sec. 3.18). Typically, it reacts only with highly reactive 
substances or under very vigorous conditions, which, as we shall see, amounts to 
the same thing. At this point we shall take up only its oxidation: by oxygen, by 
halogens, and even by water. 

2.6 Oxidation. Heat of combustion 

Combustion to carbon dioxide and water is characteristic of organic com- 
pounds; under special conditions it is used to determine their content of carbon 
and hydrogen (Sec. 2.26). 

Combustion of methane is the principal reaction taking place during the 



SEC. 2.7 CHLORINATION: A SUBSTITUTION REACTION 43 



REACTIONS OF METHANE 

1. Oxidation 

CH 4 + 2O 2 -5^> CO 2 + 2H 2 O + heat (213 kcal/mole) Combustion 

2HC-CH + 2CO + 10H 2 Discussed in Sec. 8.5. 
Acetylene 

CO + 3H 2 

2. Halogenation 

HX HX HX HX 

CH 4 -^-> CH 3 X ^> CH 2 X 2 ^ CHX 3 -^-> CX 4 light required 

Reactivity of X 2 F 2 > C1 2 > Br 2 (> I 2 ) 

Unreactive 




burning of natural gas. It is hardly necessary to emphasize its importance in the 
areas where natural gas is available; the important product is not carbon dioxide or 
water but heat. 

Burning of hydrocarbons takes place only at high temperatures, as provided, 
for example, by a flame or a spark. Once started, however, the reaction gives off 
heat which is often sufficient to maintain the high temperature and to permit 
burning to continue. The quantity of heal evolved when one mole of a hydrocarbon 
is burned to carbon dioxide and water is called (he heat of combustion; for methane 
its value is 213 kcal. 

Through controlled partial oxidation of methane and the high-temperature 
catalytic reaction with water, methane is an increasingly important source of 
products other than heat: of hydrogen, used in the manufacture of ammonia; 
of mixtures of carbon monoxide and hydrogen, used in the manufacture of 
methanolznd other alcohols; and of acetylene (Sec. 8.5), itself the starting point of 
large-scale production of many organic compounds. 

Oxidation by halogens is of particular interest to us partly because we know 
more about it than the other reactions of methane and, in one way or another, is 
the topic of discussion throughout the remainder of this chapter. 



2.7 Chlorination: a substitution reaction 

Under the influence of ultraviolet light or at a temperature of 250-400 a 
mixture of the two gases, methane and chlorine, reacts vigorously to yield hydrogen 
chloride and a compound of formula CH 3 C1. We say that methane has undergone 
chlorination, and we call the product, CH 3 C1, chloromethane or methyl chloride 
(CH 3 = methyl). 

Chlorination is a typical example of a broad class of organic reactions known 
as substitution. A chlorine atom has been substituted for a hydrogen atom of 



44 METHANE CHAP. 2 

methane, and the hydrogen atom thus replaced is found combined with a second 
atom of chlorine. 

H H 

Ii8htorheat > H--C-C1 + H-C1 



Ji Chlorine A Hydrogen 

w M chloride 

Methane Methyl chloride 

(Chloromethane) 

The methyl chloride can itself undergo further substitution to form more 
hydrogen chloride and CH 2 C1 2 , dichloromethane or methylene chloride (CH 2 = 
methylene). 

H H 

1 H C Cl + H Cl 



I I 

H H 



Methylene chloride 
(Dichloromethane) 



HC1 


HC1 


HC1 






-f 


+ 


Heat or light 


CH 2 C1 2 ~ 


~> CHC1 3 - 


^> CC1 4 


required 


Methylene 


Chloroform 


Carbon 




chloride 




tetrachloride 





In a similar way, chlorination may continue to yield CHCIa , trichloromethane 
or chloroform, and CC1 4 , tetrachloromethane or carbon tetrachloride. These last 
two compounds are already familiar to us, chloroform as an anesthetic, and carbon 
tetrachloride as a non-flammable cleaning agent and the fluid in certain fire 
extinguishers. 

HC1 

+ 

CH 4 ^L> CH 3 C1 

Methane Methyl 

chloride 



2.8 Control of chlorination 

Chlorination of methane may yield any one of four organic products, depend- 
ing upon the stage to which the reaction is carried. Can we control this reaction 
so that methyl chloride is the principal organic product? That is, can we limit 
the reaction to the first stage, wowochlorination ? 

We might at first expect naively, as it turns out to accomplish this by pro- 
viding only one mole of chlorine for each mole of methane. But let us see what 
happens if we do so. At the beginning of the reaction there is only methane for 
the chlorine to react with, and consequently only the first stage of chlorination 
takes place. This reaction, however, yields methyl chloride, so that as the reaction 
proceeds methane disappears and methyl chloride takes its place. 

As the proportion of methyl chloride grows, it competes with the methane for 
the available chlorine. By the time the concentration of methyl chloride exceeds 
that of methane, chlorine is more likely to attack methyl chloride than methane, and 
the second stage of chlorination becomes more important than the first. A large 
amount of methylene chloride is formed, which in a similar way is chlorinated to 
chloroform and this, in turn, is chlorinated to carbon tetrachloride. When we 
finally work up the reaction product, we find that it is a mixture of all four 
chlorinated methanes together with some unreacted methane. 



SEC. 2.10 RELATIVE REACTIVITY 45 

The reaction may, however, be limited almost entirely to monochlorination if 
we use a large excess of methane. In this case, even at the very end of the reaction 
unreacted methane greatly exceeds methyl chloride. Chlorine is more likely to 
attack methane than methyl chloride, and thus the first stage of chlorination is the 
principal reaction. 

Because of the great difference in their boiling points, it is easy to separate the 
excess methane (b.p. -161.5) from the methyl chloride (b.p. -24) so that the 
methane can be mixed with more chlorine and put through the process again. 
While there is a low conversion of methane into methyl chloride in each cycle, the 
yield of methyl chloride based on the chlorine consumed is quite high. 

The use of a large excess of one reactant is a common device of the organic 
chemist when he wishes to limit reaction to only one of a number of reactive sites 
in the molecule of that reactant. 



2.9 Reaction with other halogens: halogenation 

Methane reacts with bromine, again at high temperatures or under the 
influence of ultraviolet light, to yield the corresponding bromomethanes: methyl 
bromide, methylene bromide, bromoform, and carbon tetrabromide. 

HBr 

+ 

C H 4 -"^ CH 3 Br 

Methane Methyl 

bromide 

Bromination takes place somewhat less readily than chlorination. 

Methane does not react with iodine at all. With fluorine it reacts so vigor- 
ously that, even in the dark and at room temperature, the reaction must be care- 
fully controlled : the reactants, diluted with an inert gas, are mixed at low pressure. 

We can, therefore, arrange the halogens in order of reactivity. 

Reactivity of halogens F 2 > C1 2 > Br 2 O I 2 ) 

This same order of reactivity holds for the reaction of the halogens with other 
alkanes and, indeed, with most other organic compounds. The spread of reac- 
tivities is so great that only chlorination and bromination proceed at such rates as 
to be generally useful. 

2.10 Relative reactivity 

Throughout our study of organic chemistry, we shall constantly be interested 
in relative reactivities. We shall compare the reactivities of various reagents toward 
the same organic compound, the reactivities of different organic compounds to- 
ward the same reagent, and even the reactivities of different sites in an organic 
molecule toward the same reagent. 

It should be understood that when we compare reactivities we compare rates 
of reaction. When we say that chlorine is more reactive than bromine toward 
methane, we mean that under the same conditions (same concentration, same 
temperature, etc.) chlorine reacts with methane faster than does bromine. From 
another point of view, we mean that the bromine reaction must be carried out under 



HBr 
CH 2 Br 2 - 


HBr 
^V CHBr 3 - 


HBr 

f 
-^> CBr 4 


Heat or light 
required 


Methylene 
bromide 


Bromoform 


Carbon 
tetrabromide 





46 METHANE CHAP. 2 

more vigorous conditions (higher concentration or higher temperature) if it is to 
take place as fast as the chlorine reaction. When we say that methane and iodine 
do not react at all, we mean that the reaction is too slow to be significant. 

We shall want to know not only what these relative reactivities are, but also, 
whenever possible, how to account for them. To see what factors cause one 
reaction to be faster than another, we shall take up in more detail this matter of the 
different reactivities of the halogens toward methane. Before we can do this, 
however, we must understand a little more about the reaction itself. 

2.11 Reaction mechanisms 

It is important for us to know not only what happens in a chemical reaction 
but also how it happens, that is, to know not only the facts but also the theory. 

For example, we know that methane and chlorine under the influence of heat 
or light form methyl chloride and hydrogen chloride. Just how is a molecule of 
methane converted into a molecule of methyl chloride? Does this transformation 
involve more than one step, and. if so, what are these steps? Just what is the func- 
tion of heat or light ? 

The answer to questions like these, that is, the detailed, step-by-step description 
of a chemical reaction, is called a mechanism. It is only a hypothesis; it is advanced 
to account for the facts. As more facts are discovered, the mechanism must also 
account for them, or else be modified so that it does account for them; it may even 
be necessary to discard a mechanism and to propose a new one. 

It would be difficult to say that a mechanism had ever been proved. If, how- 
ever, a mechanism accounts satisfactorily for a wide variety of facts; if we make 
predictions based upon this mechanism and find these predictions borne out; if the 
mechanism is consistent with mechanisms for other, related reactions; then the 
mechanism is said to be well established, and it becomes part of the theory of 
organic chemistry. 

Why are we interested in the mechanisms of reactions? As an important part 
of the theory of organic chemistry, they help make up the framework on which 
we hang the facts we learn. An understanding of mechanisms will help us to see a 
pattern in the complicated and confusing multitude of organic reactions. We shall 
find that many apparently unrelated reactions proceed by the same or similar 
mechanisms, so that most of what we have already learned about one reaction may 
be applied directly to many new ones. 

By knowing how a reaction takes place, we can make changes in the experi- 
mental conditions not by trial and error, but logically that will improve the 
yield of the product we want, or that will even alter the course of the reaction 
completely and give us an entirely different product. As our understanding of 
reactions grows* so does our power to control them. 

2.12 Mechanism of chlorination. Free radicals 

It will be worthwhile to examine the mechanism of chlorination of methane 
in some detail. The same mechanism holds for bromination as well as chlorina- 
tion, and for other alkanes as well as methane; it even holds for many compqunds 
which, while not alkanes, contain alkane-like portions in their molecules. Closely 



SEC. 2.12 MECHANISM OF CHLORINATION 47 

related mechanisms are involved in oxidation (combustion) and other reactions of 
alkanes. More important, this mechanism illustrates certain general principles 
that can be carried over to a wide range of chemical reactions. Finally, by studying 
the evidence that supports the mechanism, we can learn something of how a chemist 
finds out what goes on during a chemical reaction. 

Among the facts that must be accounted for are these: (a) Methane and 
chlorine do not react in thr dark at room temperature, (b) Reaction takes place 
readily, however, in the dark at temperatures over 250, or (c) under the influence 
of ultraviolet light at room temperature, (d) When the reaction is induced by light, 
many (several thousand) molecules of methyl chloride are obtained for each 
photon of light that is absorbed by the system, (e) The presence of a small amount 
of oxygen slows down the reaction for a period of time, after which the reaction 
proceeds normally; the length of this period depends upon how much oxygen is 
present. 

The mechanism that accounts for these facts most satisfactorily, and hence is 
generally accepted, is shown in the following equation : 

(1) C1 2 heatorlight > 2C1- 



(2) Cl- + CH 4 > HC1 + CH 3 - 

(3) CH 3 - + C1 2 > CH 3 C1 + Ci- 

then (2), (3), (2), (3), etc. 

The first step is the breaking of a chlorine molecule into two chlorine atoms. 
Like the breaking of any bond, this requires energy, the bond dissociation energy, 
and in Table 1.2 (p. 21) we find that in this case the value is 58 kcal/mole. The 
energy is supplied as either heat or light. 

energy + :CI:C1: > :C1- + -Cl: 

The chlorine molecule undergoes homolysis (Sec. 1.14): that is, cleavage of 
the chlorine-chlorine bond takes place in a symmetrical way, so that each atom 
retains one electron of the pair that formed the covalent bond. This odd electron 
is not paired as are all the other electrons of the chlorine atom; that is, it does not 
have a partner of opposite spin (Sec. 1.6). An atom or group of atoms possessing an 
odd (unpaired) electron is called a free radical. In writing the symbol for a free 
radical, we generally include a dot to represent the odd electron just as we include 
a plus or minus sign in the symbol of an ion. 

Once formed, what is a chlorine atom most likely to do? Like most free 
radicals, it is extremely reactive because of its tendency to gain an additional electron 
and thus have a complete octet; from another point of view, energy was supplied 
to each chlorine atom during the cleavage of the chlorine molecule, and this 
energy-rich particle tends strongly to lose energy by the formation of a new chemical 
bond. 

To form a new chemical bond, that is, to react, the chlorine atom must collide 
with some other molecule or atom. What is it most likely to collide with? Ob- 
viously, it is most likely to collide with the particles that are present in the" highest 
concentration: chlorine molecules and methane molecules. Collision with another 



48 METHANE CHAP. 2 

chlorine atom is quite unlikely simply because there are very few of these reactive, 
short-lived particles around at any time. Of the likely collisions, that with a 
chlorine molecule causes no net change; reaction may occur, but it can result 
only in the exchange of one chlorine atom for another: 

: Cl- + : Cl . Cl : > : Cl : Cl : + : Cl- Collision probable but not productive 

Collision of a chlorine atom with a methane molecule is both probable and 
productive. The chlorine atom abstracts a hydrogen atom, with one electron, to 
form a molecule of hydrogen chloride : 

H H 

H : C : H + Cl : > H : Cl : + H : C - Collision probable and productive 

H H 

Methane Methyl radical 

Now the methyl group is left with an odd, unpaired electron; the carbon atom has 
only seven electrons in its valence shell. One free radical, the chlorine atom, has 
been consumed, and a new one, the methyl radical, CH 3 -, has been formed in its 
place. This is step (2) in the mechanism. 

Now, what is this methyl radical most likely to do? Like the chlorine atom, 
it is extremely reactive, and for the same reason: the tendency to complete its octet, 
to lose energy by forming a new bond. Again, collisions with chlorine molecules 
or methane molecules are the probable ones, not collisions with the relatively 
scarce chlorine atoms or methyl radicals. But collision with a methane molecule 
could at most result only in the exchange of one methyl radical for another: 

H H H H 

H : C : H + C : H > H : C J -f H : C : H Collision probable but not productive 

H H H H 

The collision of a methyl radical with a chlorine molecule is, then, the impor- 
tant one. The methyl radical abstracts a chlorine atom, with one of the bonding 
electrons, to form a molecule of methyl chloride: 

H H 

H : C - -f : Cl : Cl : > H : C : Cl : + : Cl Collision probable and productive 

H H 

Methyl Methyl chloride 

radical 

The other product is a chlorine atom. This is step (3) in the mechanism. 

Here again the consumption of one reactive particle has been accompanied by 
the formation of another. The new chlorine atom attacks methane to form a 
methyl radical, which attacks a chlorine molecule to form a chlorine atom, and so 
the sequence is repeated over and over. Each step produces not only a new reactive 
particle but also a molecule of product: methyl chloride or hydrogen chloride. 

This process cannot, however, go on forever. As we saw earlier, union of two 

r^^rt \\*, a A folotlyglv crarrf* nartirh*c is not Itkelv ! but CVerV SO often it dOCS happen 



SEC. 2.14 INHIBITORS 49 

and when it does, this particular sequence of reactions stops. Reactive particles 
are consumed but not generated. 

:C1- + -Cl: > :C1:C1: 

CH 3 - + -CH 3 > CH 3 :CH 3 

CH 3 - + -Cl: > CH 3 :CJ: 

It is clear, then, how the mechanism accounts for facts (a), (b), (c), and (d) 
on page 47 : either light or heat is required to cleave the chlorine molecule and form 
the initial chlorine atoms; once formed, each atom may eventually bring about the 
formation of many molecules of methyl chloride. 

2.13 Chain reactions 

The chlorination of methane is an example of a chain reaction, a reaction that 
involves a series of steps, each of which generates a reactive substance that brings 
about the next step. While chain reactions may vary widely in their details, they all 
have certain fundamental characteristics in common. 

(1) C1 2 heatorli8ht > 2C1 Chain-initiating step 

(2) Cl- + CH 4 HC1 + CH 3 - 1 

> Chain-propagating steps 

(3) CH r + C1 2 > CH 3 C1 + Cl- J 

then (2), (3), (2), (3), etc., until finally: 



(4) Cl- + -Cl > C1 2 

or 

(5) CH 3 - + -CH 3 > CH 3 CH 3 }> Chain-terminating steps 

or 

(6) CH 3 + -CI > CH 3 C1 

First in the chain of reactions is a chain-initiating step, in which energy is 
absorbed and a reactive particle generated; in the present reaction it is the cleavage 
of chlorine into atoms (step 1). 

There are one or more chain-propagating steps, each of which consumes a 
reactive particle and generates another; here they are the reaction of chlorine atoms 
with methane (step 2), and of methyl radicals with chlorine (step 3). 

Finally, there are chain-terminating steps, in which reactive particles are con- 
sumed but not generated; in the chlorination of methane these would involve the 
union of two of the reactive particles, or the capture of one of them by the walls 
of the reaction vessel. 

Under one set of conditions, about 10,000 molecules of methyl chloride are 
formed for every quantum (photon) of light absorbed. Each photon cleaves one 
chlorine molecule to form two chlorine atoms, each of which starts a chain. On 
the average, each chain consists of 5000 repetitions of the chain-propagating cycle 
before it is finally stopped. 

2.14 Inhibitors 

Finally. h^ w ^^ the mechanism of chlorination account for fact (e), that a 



50 METHANE CHAP. 2 

small amount of oxygen slows down the reaction for a period of time, which de- 
pends upon the amount of oxygen, after which the reaction proceeds normally? 
Oxygen is believed to react with a methyl radical to form a new free radical: 

CH 3 - + 2 > CH 3 -O-0- 

The CH 3 OO- radical is much less reactive than the CH 3 - radical, and can do little 
to continue the chain. By combining with a methyl radical, one oxygen molecule 
breaks a chain, and thus prevents the formation of thousands of molecules of 
methyl chloride; this, of course, slows down the reaction tremendously. After all 
the oxygen molecules present have combined with methyl radicals* the reaction is 
free to proceed at its normal rate. 

A substance that slows down or stops a reaction even though present in small 
amount is called an inhibitor. The period of time during which inhibition lasts, and 
after which the reaction proceeds normally, is called the inhibition period. Inhibition 
by a relatively small amount of an added material is quite characteristic of chain 
reactions of any type, and is often one of the clues that first leads us to suspect that 
we are dealing with a chain reaction. It is hard to see how else a few molecules 
could prevent the reaction of so many. (We shall frequently encounter the use of 
oxygen to inhibit free-radical reactions.) 

2.15 Heat of reaction 

In our consideration of the chlorination of methane, we have so far been con- 
cerned chiefly with the particles involved molecules and atoms- and the changes 
that they undergo. As with any reaction, however, it is important to consider also 
the energy changes involved, since these changes determine to a large extent how 
fast the reaction will go, and, in fact, whether it will take place at all. 

By using the values of bond dissociation energies given in Table 1.2 (p. 21), 
we can calculate the energy changes that take place in a great number of reactions. 
In the conversion of methane into methyl chloride, two bonds are broken, CH 3 -H 
and CI Cl, consuming 104 f 58, or a total of 162 kcal/mole. At the same time 
two new bonds are formed, CH 3 - Cl and H Cl, liberating 84 f 103, or a total 
of 187 kcal/mole. The result is the liberation of 25 kcal of heat for every mole of 

CH 3 -H + Cl -CI > CH 3 -C! + H-C1 

104 58 84 103. 

162 187 A// = -25 kcal 

methane that is converted into methyl chloride; this is, then, an exothermic reac- 
tion. (This calculation, we note, does not depend on our knowing the mechanism 
of the reaction.) 

When heat is liberated, the heat content (enthalpy), //, of the molecules them- 
selves must decrease; the change in heat content, A//, is therefore given a negative* 
sign. (In the case of an endothermic reaction, where heat is absorbed, the increase 
in heat content of the molecules is indicated by a positive A//.) 

Problem 2.1 Calculate A// for the corresponding reaction of methane with: 
(a) bromine, (b) iodine, (c) fluorine. 



SEC. 2.16 ENERGY OF ACTIVATION 51 

The value of -25 kcal that we have just calculated is the netkH for the overall 
reaction. A more useful picture of the reaction is given by the A//'s of the indi- 
vidual steps. These are calculated below: 

(1) C1-C1 * 2C\> Atf=+58kcal 

(58) 

(2) Cl- + CH 3 H > CH 3 - + H Cl A# = +1 

(104) (103) 

(3) CH 3 - + C1-C1 > CH 3 -C1 + Cl- A// = -26 

(58) (84) 

It is clear why this reaction, even though exothermic, occurs only at a high tempera- 
ture (in the absence of light). The chain-initiating step, without which reaction 
cannot occur, is highly endothermic, and takes place (at a significant rate) only at a 
high temperature. Once the chlorine atoms are formed, the two exothermic chain- 
propagating steps occur readily many times before the chain is broken. The 
difficult cleavage of chlorine is the barrier that must be surmounted before the 
subsequent easy steps can be taken. 

Problem 2.2 Calculate A/7 for the corresponding steps in the reaction of meth- 
ane with: (a) bromine, (b) iodine, (c) fluorine. 

We have assumed so far that exothermic reactions proceed readily, that is, are 
reasonably fast at ordinary temperatures, whereas endothermic reactions proceed 
with difficulty, that is, are slow except at very high temperatures. This assumed 
relationship between A// and rate of reaction is a useful rule of thumb when other 
information is not available; it is not, however, a necessary relationship, and there 
are many exceptions to the rule. We shall go on, then, to a discussion of another 
energy quantity, the energy of activation* which is related in a more exact way to 
rate of reaction. 

2.16 Energy of activation 

To see what actually happens during a chemical reaction, let us look more 
closely at a specific example, the attack of chlorine atoms on methane: 

Cl- -I- CH 3 H > H Cl 4- CH 3 - \H = + 1 kcal E^ 4 kcal 
(104) (103) 

This reaction is comparatively simple: it occurs in the gas phase, and is thus not 
complicated by the presence of a solvent; it involves the interaction of a single atom 
and the simplest of organic molecules. Yet from it we can learn certain principles 
that apply to any reaction. 

Just what must happen if this reaction is to occur? First of all, a chlorine 
atom and a methane molecule must collide. Since chemical forces are of extremely 
short range, a hydrogen-chlorine bond can form only when the atoms are in close 
contact. 

Next, to be effective, the collision must provide a certain minimum amount of 
energy. Formation of the H Cl bond liberates 103 kcal/mole; breaking the 



52 METHANE CHAP. 2 

CHj H bond requires 104 kcal/mole. We might have expected that only 1 kcal/ 
mole additional energy would be needed for reaction to occur; however, this is 
not so. Bond-breaking and bond-making evidently are not perfectly synchronized, 
and the energy liberated by the one process is not completely available for the 
other. Experiment has shown that if reaction is to occur, an additional 4 kcal/mole 
of energy must be supplied. 

The minimum amount of energy that must be provided by a collision for reaction 
to occur is called the energy of activation, " acl . Its source is the kinetic energy of 
the moving particles. Most collisions provide less than this minimum quantity 
and are fruitless, the original particles simply bouncing apart. Only solid collisions 
between particles one or both of which are moving unusually fast are energetic 
enough to bring about reaction. In the present example, at 275, only about one 
collision in 40 is sufficiently energetic. 

Finally, in addition to being sufficiently energetic, the collisions must occur 
when the particles are properly oriented. At the instant of collision, the methane 
molecule must be turned in such a way as to present a hydrogen atom to the full 
force of the impact. In the present example, only about one collision in eight is 
properly oriented. 

In genera], then, a chemical reaction requires collisions of sufficient energy 
(a c t) an d of proper orientation. There is an energy of activation for nearly every 
reaction where bonds are broken, even for exothermic reactions, in which bond- 
making liberates more energy than is consumed by bond-breaking. 

The attack of bromine atoms on methane is more highly endothermic, with 
a A// of +16 kcal. 

'* Br- + CH 3 H > H Br + CH 3 - A/f = +16 kcaJ E act = 18 kcal 
*< (104) (88) 

Rr liking the CH 3 H bond, as before, requires 104 kcal/mole, of which only 88 
kcal is provided by formation of the H Br bond. It is evident that, even if this 
88 kcal were completely available for bond-breaking, at least an additional 16 
kcal/mole would have to be supplied by the collision. In other words, the act 
of an endothermic reaction must be at least as large as the A//. As is generally 
true, the &ct of the present reaction (18 kcal) is actually somewhat larger than 
the A//. 

2.17 Progress of reaction: energy changes 

These energy relationships can be seen more clearly in diagrams like Figs. 2.2 
and 2.3. Progress of reaction is represented by horizontal movement from reac- 
tants on the left to products on the right. Potential energy (that is, all energy except 
kinetic) at any stage of reaction is indicated by the height of the curve. 

Let us follow the course of reaction in Fig. 2.2. We start in a potential energy 
valley with a methane molecule and a chlorine atom. These particles are moving, 
and hence possess kinetic energy in addition to the potential energy shown. The 
exact amount of kinetic energy varies with the particular pair of particles, since 
some move faster than others. They collide, and kinetic energy is converted into 
potential energy. With this increase in potential energy, reaction begins, and we 
move up the energy hill. If enough kinetic energy is converted, we reach the top 
of the hill and start down the far side. 



SEC. 2.17 



PROGRESS OF REACTION: UN ERG Y CHANGES 



53 




A//- 4-1 kcai CHh- + HCI 



CHi-f-CI- 



CHi+Cl- 



+HC1 



Progress of reaction > 

Figure 2.2. Potential energy changes during progress of reaction: the 
methane-chlorine atom reaction. 

During the descent, potential energy is converted back into kinetic energy, 
until we reach the level of the products. The products contain a little more poten- 
tial energy than did the reactants, and we find ourselves in a slightly higher valley 
than the one we left. With this net increase in potential energy there must be a 
corresponding decrease in kinetic energy. The new particles break apart, and sin< 
they are moving more slowly than the particles from which they were for* 



CHr + HBr 



act =I8kcal 




+ Br 



CH 3 - + HBr 



Progress of reaction 



Figure 2.3. Potential energy changes during progress of reaction : the 
methane-bromine atom reaction. 



54 METHANE CHAP. 2 

we observe a drop in temperature. Heat will be taken up from the surroundings. 

In the bromine reaction, shown in Fig. 2.3, we climb a much higher hill and 
end up in a much higher valley. The increase in potential energy and the corre- 
sponding decrease in kinetic energy is much larger than in the chlorine reaction; 
more heat will be taken up from the surroundings. 

An exothermic reaction follows much the same course. (Take, for example, 
the reverse of the bromine reaction: that is, read from right to left in Fig. 2.3.) 
In this case, however, the products contain less potential energy than did the reac- 
tants so that we end up in a lower valley than the one we left. Since this time the 
new particles contain more kinetic energy than the particles from which they 
were formed, and hence move faster, we observe a rise in temperature. Heat will 
be given offio the surroundings. 

In any reaction there are many collisions that provide too little energy for us 
to reach the top of the hill. These collisions are fruitless, and we slide back to our 
original valley. Many collisions provide sufficient energy, but take place when the 
molecules are^improperly oriented. We then climb an energy hill, but we are off 
the road; we may climb very high without finding the pass that leads over into the 
next valley. 

The difference in level between the two valleys, is, of course, the A//; the dif- 
ference in level between the reactant valley and the top of the hill is the act . 
We are concerned only with these differences, and not with the absolute height at 
any stage of the reaction. We are not even concerned with the relative levels of the 
reactant valleys in the chlorine and bromine reactions. We need only to know that 
in the chlorine reaction we climb a hill 4 kcal high and end up in a valley 1 kcal 
higher than our starting point; and that in the bromine reaction we climb a hill 18 
kcal hign and end up in a valley 16 kcal higher than our starting point. 

As we shall see, it is the height of the hill, the act that determines the rate of 
reaction, and not the difference in level of the two valleys, A//. In going to a lower 
valley, the hill might be very high, but could be very low. or even non-existent. 
In climbing to a higher valley, however, the hill can be no lower than the valley to 
which we are going; that is to say, in an endothermic reaction the E ACl must be at least 
as large as the A//. 

An energy diagram of the sort shown in Figs. 2.2 and 2.3 is particularly useful 
because it tells us not only about the reaction we are considering, but also about the 
reverse reaction. Let us move from right to left in Fig. 2.2, for example. We see 
that the reaction 

CH 3 - + H-C1 > CH 3 H + Cl- A# = -1, aot = 3 

(103) (104) 

has an energy of activation of 3 kcal, since in this case we climb the hill from the 
higher valley. 'fUfcis, of course, an exothermic reaction with a A// of 1 kcal. 
In the same way we can see from Fig. 2.3 that the reaction 

CH 3 . + H Br > CH 3 ^~H + Br A// =-16, ;<* = 2 

(88) (104) 

has an energy of activation of 2 kcal, and is exothermic with a A// of - 16 kcal. 
(We notice that, even though exothermic, these last two reactions have energies of 
activation.) 



SEC. 2.18 RATE OF REACTION 55 

Reactions like the cleavage of chlorine into atoms fall into a special category: 

Cl Cl > Cl- + -Cl A#= +58, * = 58 

(58) 

C. 

a bond is broken but no bonds are formed. The reverse of this reaction, the union" 
of chlorine atoms, involves no bond-breaking and hence would be expected to 



Cl- + -Cl 



Cl-Cl 

(58) 



= -58, E ft ct = 



take place very easily, in fact, with no energy of activation at all. This is con- 
sidered to be generally true for reactions involving the union of two free radicals. 
If there is no hill to climb in going from chlorine atoms to a chlorine molecule, 
but simply a slope to descend, the cleavage of a chlorine molecule must involve 
simply the ascent of a slope as shown in Fig. 2.4. The act for the cleavage of a 
chlorine molecule, then, must equal the A//, that is, 58 kcal. This equality of ^. t 
and A/7 is believed to hold generally' for reactions in which molecules dissociate 
into radicals. 




Progress of reaction > 

Figure 2.4. Potential energy changes during progress of reaction: simple 
dissociation. 

2.18 Rate of reaction 

A chemical reaction is the result of collisions of sufficient energy and proper 
orientation. The rate of reaction, therefore, must be the rate at which these effec- 
tive collisions occur, the number of effective collisions, let us say, that occur 
during each second within each cc of reaction space. We can then express the 
rate as the product of three factors. (The number expressing the probability that a 



METHANE 



CHAP. 2 



collision will have the proper orientation is commonly called the probability factor.) 
Anything that affects any one of these factors affects the rate of reaction. 



number of 


total number 


fraction of 
collisions 


fraction of 
collisions 


effective _ 
collisions 
per cc pet sec 


of collisions, 
per cc per sec 


x that have x 
sufficient 
energy 


that have 
proper 
orientation 








probability 


_ 


collision 


x energy x 


factor 


rate 


frequency 


factor 


(orientation 








factor) 



The collision frequency depends upon (a) how closely the particles are crowded 
together, that is, concentration or pressure; (b) how large they are; and (c) how 
fast they are moving, which in turn depends upon their weight and the temperature. 

We can change the concentration and temperature, and thus change the rate. 
We are familiar with the fact that an increase in concentration causes an increase 
in rate; it does so, of course, by increasing the collision frequency. A rise in 
temperature increases the collision frequency; as we shall see, it also increases the 
energy factor, and this latter effect is so great that the effect of temperature en 
collision frequency is by comparison unimportant. 

The size and weight of the particles are characteristic of each reaction and 
cannot be changed. Although they vary widely from reaction .to reaction, this 
variation does not affect the collision frequency greatly. A heavier weight makes 
the particle move more slowly at a given temperature, and hence tends to decrease 
the collision frequency. A heavier particle is, however, generally a larger particle, 
and the larger size tends to increase the collision frequency. These two factors thus 
tend to cancel out. 

The probability factor depends upon the geometry of the particles and the kind 
of reaction that is taking place. For closely related reactions it does not vary 
widely. 

Kinetic energy of the moving molecules is not the only source of the energy needed 
for reaction ; energ> can also be provided, for example, from vibrations among the various 
atoms within the molecule. Thus the probability factor has to do not only with what 
atoms in the molecule suffer the collision, but also with the alignment of the other atoms 
in the molecule at the time of collision. 

By far the most important factor determining rate is the energy factor: the 
fraction of collisions that are sufficiently energetic. This factor depends upon the 
temperature, which we can control, and upon the energy of activation, which is 
characteristic of each reaction. 

At a given* temperature the molecules of a particular compound have an average 
velocity and hence an average kinetic energy that is characteristic of this system; 
in fact, the temperature is a measure of this average kinetic energy. But the indi- 
vidual n, )lecules do not all travel with the same velocity, some moving faster 
than the average and some slower. The distribution of velocities is shown in 
Fig. 2,5 by the familiar bell-shaped curve that describes the distribution among 
individuals of so many qualities, for example, height, intelligence, income, or even 
Jife expectancy. The number of molecules with a particular velocity is greatest 



SEC. 2.18 



RATE OF REACTION 



57 




Collisions with energy > i 
Collisions with energy > 2 




Average 
Energy *> 

Figure 2.5. Distribution of kinetic en- 
ergy among molecules. 



Energy > 

Figure 2.6. Distribution of kinetic en- 
ergy among collisions. 



for a velocity near the average and decreases as the velocity becomes larger or 
smaller than the average. 

The distribution of collision energies, as we might expect, is described by a 
similar curve, Fig. 2.6. Let us indicate collisions of a particular energy, act , by a 
vertical line. The number of collisions with energy equal to or greater than ftct 
is indicated by the shaded area under the curve to the right of the vertical line. The 
fraction of the total number of collisions that have this minimum energy, ' act , 
is then the fraction of the total area that is shaded. It is evident that the greater 
the value 0/ act , the smaller the fraction of collisions that possess that energy. 

The exact relationship between energy of activation and fraction of collisions 
with that energy is: 

e~ E *ct /RT = fraction of collisions with energy greater than 
where 

e = 2.7 1 8 (base of natural logarithms) 
R = 1.986 (gas constant) 
T = absolute temperature. 

Using P for the probability factor and Z for the collision frequency, we arrive at 
the rate equation : 

rate = 



This exponential /relationship is important to us in that it indicates that a 
small difference in act has a large effect on the fraction of sufficiently energetic 
collisions, and hence on the rate of reaction. For example, at 275, out of every 
million collisions, 10,000 provide sufficient energy if act = 5 kcal, 100 provide 
sufficient energy if act = 10 kcal, and only one provides sufficient energy if 
act = 15 kcal. This means that (all other things being equal) a reaction with 
acl = 5 kcal will go 100 times as fast as one with ftct = 10 kcal, and 10,000 
times as fast as one with aot = 15 kcal. 



58 



METHANE 



CHAP. 2 



We have so far considered a system held at a given temperature. A rise in 
temperature, of course, increases the average kinetic energy and average velocities, 
and hence shifts the entire curve to the right, as shown in Fig. 2.7. For a given 
energy of activation, then, a rise in temperature increases the fraction of sufficiently 
energetic collisions, and hence increases the rate, as we already know. 

The exponential relationship again leads to a large change in rate, this time for 
a small change in temperature. For example, a rise from 250 to 300, which is 
only a 10 increase in absolute temperature, increases the rate by 50",', if act = 
5 kcal, doubles the rate if act = 10 kcal, and trebles the rate if act = 15 kcal. 
As this example shows, the greater the act , the greater the effect of a given change in 
temperature; this follows from the e~ E *" IRT relationship. Indeed, it is from the 
relationship between rate and temperature that the act of a reaction is determined: 
the rate is measured at different temperatures, and from the results act is calculated. 

We have examined the factors that determine rate of reaction. What we have 
learned may be used in many ways. To speed up a particular reaction, for 
example, we know that we might raise the temperature, or increase the concentra- 
tion of reactants, or even (in ways that we shall take up later) lower the acL . 

Of immediate interest, however, is the matter of relative reactivities. Let us 
see, therefore, how our knowledge of reaction rates can help us to account for the 
fact that one reaction proceeds faster than another, even though conditions for the 
two reactions are identical. 



15 
3 



Collisions at T\ with energy E Act 
Collisions at T-i with energy & et 




act 

Energy > 
Figure 2.7. Change in collision energies with change in temperature. 

2.19 Relative rates of reaction 

We have seen that the rate of a reaction can be expressed as a product of three 
factors : 

rate = collision frequency x energy factor x probability factor 

Two reactions could proceed at different rates because of differences in any or all 
these factors. To account for a difference in rate, we must first see in which of 
these factors the difference lies. 



SEC. 2.20 RELATIVE REACTIVITIES OF HALOGENS TOWARD METHANE 59 

As an example, let us compare the reactivities of chlorine and bromine atoms 
toward methane; that is, let us compare the rates, under the same conditions, of 
the two reactions: 

Cl- + CH 3 -H > H-C1 + CH r A// = +1, act = 4 
Br . 4- CH 3 H > H-Br + CH 3 - A# = +16, ac t = 18 

Since temperature and concentration must be the same for the two reactions 
if we are to compare them under the same conditions, any difference in collision 
frequency would have to arise from differences in particle weight or size. A bro* 
mine atom is heavier than a chlorine atom, and it is also larger; as we have seen, 
the effects of these two properties tend to cancel out. In actuality, the collision 
frequencies differ by only a few per cent. It is generally true that for the same 
temperature and concentration, two closely related reactions differ but little in 
collision frequency. A difference in collision frequency therefore cannot be the 
cause of a large difference in reactivity. 

The nature of the probability factor is very poorly understood. Since our .two 
reactions are quite similar, however, we might expect them to have similar prob- 
ability factors. Experiment has shown this to be true: whether chlorine or bromine 
atoms are involved, about one in every eight collisions with methane has the 
proper orientation for reaction. In general, where closely related reactions are con- 
cerned, we may assume that a difference in probability factor is not likely to be 
the cause of a large difference in reactivity. 

We are left with a consideration of the energy factor. At a given temperature, 
the fraction of collisions that possess the amount of energy required for reaction 
depends upon how large that amount is, that is, depends upon the act . In our 
example act is 4 kcal for the chlorine reaction, 18 kcal for the bromine reaction. 
As we have seen, a difference of this size in the act causes an enormous difference 
in the energy factor, and hence in the rate. At 275, of every 10 million collisions, 
250,000 are sufficiently energetic when chlorine atoms are involved, and only one 
when bromine atoms are involved. Because of the difference in act alone, then, 
chlorine atoms are 250,000 times as reactive as bromine atoms toward methane. 

As we encounter, again and again, differences in reactivity, we shall in general 
attribute them to differences in " act ; in many cases we shall be able to account 
for these differences in act on the basis of differences in molecular structure. // must 
be understood that we are justified in doing this only when the reactions being com- 
pared are so closely related that differences in collision frequency and in probability 
factor are comparatively insignificant. 

2.20 Relative reactivities of halogens toward methane 

With this background, let us return to the reaction between methane and the 
various halogens, and see if we can account for the order of reactivity given before, 
p2 > C\2 > Br 2 > 12 , and in particular for the fact that iodine does not react at 
all. 

From the table of bond dissociation energies (Table -1 .2, p. 21) we can calculate 
for each of the four halogens the A A/ for each of the three steps of halogenation. 
Since act has been measured for only a. few of these reactions, let us see what 
tentative conclusions we can reach using only A//. 



60 METHANE CHAP. . 

X- F Cl Br I 

(1) X 2 > 2X- Atf=+38 +58 +46 +36 

(2) X- + CH 4 > HX + CH r -32 +1 +16 +33 

(3) CH 3 - + X 2 > CH 3 X + X- -70 -26 -24 -20 

Since step (1) involves simply dissociation of molecules into atoms, we may 
quite confidently assume (Sec. 2.17 and Fig. 2.4) that A// in this case is equal to 
-fact- Chlorine has the largest att , and should dissociate most slowly; iodine has 
the smallest act , and should dissociate most rapidly. Yet this does not agree 
with the observed order of reactivity. Thus, except possibly for fluorine, dissocia- 
tion of the halogen into atoms cannot be the step that determines the observed 
reactivities. 

Step (3), attack of methyl radicals on halogen, is exothermic for all four 
halogens, and for chlorine, bromine, and iodine it has very nearly the same A/f. 
For these reactions, act could be very small, and does indeed seem to be so; prob- 
ably only a fraction of a kcal. Even iodine has been found to react readily with 
methyl radicals generated in another way, e.g., by the heating of tetramethyllead. 
In fact, iodine is sometimes employed as a free-radical "trap" or "scavenger" in 
the study of reaction mechanisms. The third step, then, cannot be the cause of 
the observed relative reactivities. 

This leaves step (2), abstraction of hydrogen from methane by a halogen 
atom. Here we see a wide spread of A/Ts, from the highly exothermic reaction 
with the fluorine atom to the highly endothermic reaction with the iodine atom. 
The endothermic bromine atom reaction must have an act of at least 16 kcal; as 
we have seen, it is actually 18 kcal. The slightly endothermic chlorine atom reaction 
could have a very small act ; it is actually 4 kcal. At a given temperature, then, the 
fraction of collisions of sufficient energy is much larger for methane and chlorine 
atoms than for methane and bromine atoms. To be specific, at 275 the fraction 
is about 1 in 40 for chlorine and 1 in 10 million for bromine. 

A bromine atom, on the average, collides with many methane molecules 
before it succeeds in abstracting hydrogen ; a chlorine atom collides with relatively 
few. During its longer search for the proper methane molecule, a bromine atom 
is more likely to encounter another scarce particle a second halogen atom or a 
methyl radical or be captured by the vessel wall ; the chains should therefore be 
much shorter than in chlorination. Experiment has shown this to be so: where 
the average chain length is several thousand for chlorination, it is less than 100 for 
bromination. Even though bromine atoms are formed more rapidly than chlorine 
atoms at a given temperature because of the lower act of step (1), overall bromina- 
tion is slower than chlorination because of the shorter chain length. 

For the endothermic reaction of an iodine atom with methane, E Mt can be no 
less than 33 kcal, and is probably somewhat larger. Even for this minimum value 
of 33 kcal, an iodine atom must collide with an enormous number of methane 
molecules (10 12 or a million million at 275) before reaction is likely to occur. 
Virtually no iodine atoms last this long, but instead recombine to form iodine 
molecules; the reaction therefore proceeds at a negligible rate. Iodine atoms are 
easy to form; it is their inability to abstract hydrogen from methane that prevents 
iodination from occurring. 



SEC. 2.20 



RELATIVE REACTIVITIES OF HALOGENS TOWARD METHANE 



61 



We cannot predict the act for the highly exothermic attack of fluorine atoms 
on methane, but we would certainly not expect it to be any larger than for the 
attack of chlorine atoms on methane. It appears actually to be smaller (about 1 
kcal), thus permitting even longer chains. Because of the surprising weakness of 
the fluorine-fluorine bond, fluorine atoms should be formed faster than chlorine 
atoms; thus there should be not only longer chains in fluorination but also more 
chains. The overall reaction is extremely exothermic, with a A// of 102 kcal, 
and the difficulty of removing this heat is one cause of the difficulty of control of 
fluorination. 

Of the two chain-propagating steps, then, step (2) is more difficult than step 
(3) (see Fig. 2.8). Once formed, methyl radicals react easily with any of the halo- 
gens; it is how fast methyl radicals are formed that limits the rate of overall 
reaction. Fluorination is fast because fluorine atoms rapidly abstract hydrogen 
atoms from methane; act is only 1 kcal. lodination does not take place because 
iodine atoms .find it virtually impossible to abstract hydrogen from methane; /? act 
is more than 33 kcal. 

Values of E ACt for step (2), we notice, parallel the values of A//. Since the same 
bond, CH 3 -H, is being broken in every case, the differences in A// reflect differ- 
ences in bond dissociation energy among the various hydrogen-halogen bonds. 
Ultimately, it appears, the reactivity of a halogen toward methane depends upon 
the strength of the bond which that halogen forms with hydrogen. 

One further point requires clarification. We have said that an " act of 33 kcal 
is too great for the reaction between iodine atoms and methane to proceed at a 
significant rate; yet the initial step in each of these halogenations requires an even 



^Difficult step 



1 




CH 3 C1 + Cl 



Progress of reaction 



Figure 2.8. Potential energy changes during progress of reaction: 
chlorination of methane. Formation of radical is difficult step. 



62 METHANE CHAP. 2 

greater act . The difference is this: since halogenation is a chain reaction, dissocia- 
tion of each molecule of halogen gives rise ultimately to many molecules of methyl 
halide; hence, even though dissociation is very slow, the overall reaction can be 
fast. The attack of iodine atoms on methane, however, is a chain-carrying step and 
if it is slow the entire reaction must be slow; under these circumstances chain- 
terminating steps (e.g., union of two iodine atoms) become so important that 
effectively there is no chain. 

2.21 Structure of the methyl radical, sp 2 Hybridization 

We have spent a good part of this chapter discussing the formation and 
reactions of the methyl free radical CH 3 - . Just what is this molecule like? What 
is its shape? How are the electrons distributed and, in particular, where is the odd 
electron ? 

These are important questions, for the answers apply not only to this simple 
radical but to any free radical, however complicated, that we shall encounter. 
The shape, naturally, underlies the three-dimensional chemistry the stereo- 
chemistry of free radicals. The location of the odd electron is intimately involved 
with the stabilization of free radicals by substituent groups. 

As we did when we "made" methane (Sec. 1.11), let us start with the elec- 
tronic configuration of carbon, 

Is 2s 2p 



c O O O 

and, to provide more than two unpaired electrons for bonding, promote a 2s 
electron to the empty 2p orbital: 

Is 2s 2p , 

, * One electron promoted: 



O O O O 



four unpaired electrons 



Like boron in boron trifluoride (Sec. 1.10), carbon here is bonded to three 
other atoms. Hybridization of the 2s orbital and two of the p orbitals provides the 

is 2s 2p 



O O O O 

| | sp 2 Hybridization 

I 

\S Sp 2 In 



c O O O O 

necessary orbitals: three strongly directed sp 2 orbitals which, as we saw before, 
lie in a plane that includes the carbon nucleus, and are directed to the corners of 
an equilateral triangle. 



SBC. 2JB TRANSITION STATE 63 

If we arrange the carbon and three hydrogens of a methyl radical to permit 
maximum overlap of orbitals, we obtain the structure shown in Fig. 2.9a. It is 





Figure 2.9. Methyl radical, (0) Only a bonds shown, (b) Odd electron in 
p orbital above and below plane of a bonds. 

flat, with the carbon atom at the center of a triangle and the three hydrogen atoms 
at the corners. Every bond angle is 120. 

Now where is the odd electron? In forming the sp 2 orbitals, the carbon atom 
has used only two of its three p orbitals. The remaining p orbital consists of two 
equal lobes, one lying above and the other lying below the plane of the three sp 2 
orbitals (Fig. 2.9b); it is occupied by the odd electron. 

This is not the only conceivable electronic configuration for the methyl radi- 
cal: an alternative treatment would lead to a pyramidal molecule like that of am- 
monia, except that the fourth sp 3 orbital contains the odd electron instead of an 
electron pair (Sec. 1.12). Quantum mechanical calculations do not offer a clear- 
cut decision between the two configurations. Spectroscopic studies indicate that 
the methyl radical is actually flat, or nearly so. Carbon is trigonal, or not far from 
it; the odd electron occupies a/? orbital, or at least an orbital with much/? charac- 
ter. 

Compare the shapes of three molecules in which the central atom is bonded to three 
other atoms: (a) boron trifluoride, with no unshared electrons, trigonal; (b) ammonia, 
with an unshared pair, tetrahedral; and (c) the methyl radical, with a single unshared 
electron, trigonal or intermediate between trigonal and tetrahedral. 

There is stereochemical evidence (for example, Sec. 7.10) that most other free 
radicals are either flat or, if pyramidal, undergo rapid inversion like that of the 
ammonia molecule (Sec. 1.12). 

Problem 2.3 Besides free radicals, we shall encounter two other kinds of reac- 
tive particles, carbonium ions (positive charge on carbon) and carbanions (nega- 
tive charge on carbon). Suggest an electronic configuration, and from this predict 
the shape, of the methyl cation, CH 3 + ; of the methyl anion, CH 3 :~. 



2.22 Transition state 

Clearly, the concept of act is to be our key to the understanding of chemical 
reactivity. To make it useful, we need a further concept: transition state, 

A chemical reaction is presumably a continuous process involving a gradual 
transition from reactants to products. It has been found extremely helpful, how- 
ever, to consider the arrangement of atoms at an intermediate stage of reaction as 



64 



METHANE 



CHAP. 2 



though it were an actual molecule. This intermediate structure is called the tran- 
sition state; its energy content corresponds to the top of the energy hill (Fig. 2.10). 

Transition state 



a 

1 




Products 



Progress of reaction > 

Figure 2.10. Potential energy changes during progress of reaction : transi- 
tion state at top of energy hump. 



The reaction sequence is now: 

reactants > transition state 



products 



Just as A// is the difference in energy content between reactants and products, so 
act is the difference in energy content between reactants and transition state. 

The transition state concept is useful for this reason: we can analyze the struc- 
ture of the transition state very much as though it were a molecule, and attempt 
to estimate its stability. Any factor that stabilizes the transition state relative to 
the reactants tends to lower the energy of activation; that is to say, any factor that 
lowers the top of the energy hill more than it lowers the reactant valley reduces the 
net height we must climb during reaction. Transition state stability will be the 
basis whether explicit or implicit of almost every discussion of reactivity in 
this book. 

But the transition state is only a fleeting arrangement of atoms which, by its 
very nature lying at the top of an energy hill cannot be isolated and examined. 
How can we possibly know anything about its structure? Well, let us take as an 
example the transition state for the abstraction of hydrogen from methane by a 
halogen atom, and see where a little thinking will lead us. 

To start with, we can certainly say this: the carbon-hydrogen bond is stretched 
but not entirely broken, and the hydrogen-halogen bond has started to form but 
is not yet complete. This condition could be represented as 



H f H 

| 18- 8- 

H-C-H + -X > H-C--H-X 

A I A . 

Reactants Transition state 



H 
H-C- + H-X 






Products 
where the dashed lines indicate partly broken or partly formed bonds. 




SEC. 2.23 REACTIVITY AND DEVELOPMENT OF THE TRANSITION STATE 65 

Now, what can we say about the shape of the methyl group in this transition 
state? In the reactant, where methyl holds the hydrogen, carbon is tetrahedral 
(spMiybridized); in the product, where methyl has lost the hydrogen, carbon is 
trigonal (,sp 2 -hybridized). In the transition state, where the carbon-hydrogen 
bond is partly broken, hybridization of carbon is somewhere between sp 3 and sp 2 . 
The methyl group is partly but not completely flattened; bond angles are greater 
than 109.5 but less than 120. 



+ H X 



Reactant Transition state Product 

Tetrahedral Becoming trigonal Trigonal 

Finally, where is the odd electron? It is on chlorine in the reactants, on the 
methyl group in the products, and divided between the two in the transition state. 
(Each atom's share is represented by the symbol 8-.) The methyl group partly 
carries the odd electron it will have in the product, and to this extent has taken on 
some of the character of the free radical it will become. 

Thus, in a straightforward way, we have drawn a picture of the transition state 
that shows the bond-making and bond-breaking, the spatial arrangement of the 
atoms, and the distribution of the electrons. 

(This particular transition state is intermediate between reactants and products 
not only in the time sequence but also in structure. Not all transition states are 
intermediate in structure: as shown on page 462, reactant and product in S N 2 
reactions are tetrahedral, whereas the transition state contains pentavalent carbon.) 

In Sec. 2.18, we looked at the matter of reaction rates from the standpoint of the 
collision theory. An alternative, more generally useful approach is the transition state 
(or thermodynamic) theory of reaction rates. An equilibrium is considered to exist be- 
tween the reactants and the transition state, and this is handled in the same way as 
true equilibria of reversible reactions (Sec. 18.11). Energy of activation (**) and prob- 
ability factor are replaced by, respectively, heat {enthalpy) of activation (A//t) and entropy 
of activation (A5J), which together make up free energy of activation (AGJ). 



AGJ - AtfJ - 7ASJ 

The smaller (the less positive) the A// J and the larger (the more positive) the AS:):, the 
smaller AG{ will be, and the faster the reaction. 

Entropy corresponds, roughly, to the randomness of a system; equilibrium tends to 
favor the side in which fewer restrictions are placed on the atoms and molecules. Entropy 
of activation, then, is a measure of the relative randomness of reactants and transition 
state; the fewer the restrictions that are placed on the arrangement of atoms in the 
transition staterelative to the reactants the faster the reaction will go. We can see, 
in a general way, how probability factor and entropy of activation measure much the 
same thing. A low probability factor means that a rather special orientation of atoms is 
required on collision. In the other language, an unfavorable (low) entropy of activation 
means that rather severe restrictions are placed on the positions of atoms in the transition 
state. 

2.23 Reactivity and development of the transition state 

For the abstraction of hydrogen from methane by a halogen atom, we have 



66 



METHANE 



CHAP. 2 



just seen that the transition state differs from the reactants and this difference is, 
of course, what we are looking for chiefly in being like the products. This is 
generally true for reactions in which free radicals (or, for that matter, carbonium 
ions or carbanions) are formed. 

But just how much does this particular transition state resemble the products? 
How far have bond-breaking and bond-making gone? How flat has the methyl 
group become, and to what extent does it carry the odd electron? 

Surprisingly, we can answer even questions like these, at least in a relative 
way. In a set of similar reactions, the higher the E &ct , the later the transition state 
is reached in the reaction process. Of the theoretical considerations underlying 
this postulate, we shall mention only this: the difference in electronic distribution 
that we call a difference in structure corresponds to a difference in energy; the 
greater the difference in structure, the greater the difference in energy. If ,& 
is high, the transition state differs greatly from the reactants in energy and, pre- 
sumably, also in electronic structure; if E^ is low, the transition state differs 
little from the reactants in energy and, presumably, also in electronic structure 
(see Fig. 2.11). 

Practically, this postulate has been found extremely useful in the interpretation 
of experimental results; among other things, as we shall see, it enables us to account 
for the relationship between reactivity and selectivity (Sec. 3.28). 



A B C 

Transition state 
reached late 



A B -C 

Transition state 
reached early 




Difficult reaction 



A B C 

Reactants 



Easy reaction 



A B C 

Products 



Progress of reaction > 

Figure 2.11. Potential energy changes during progress of reaction: reac- 
tivity and development of the transition state. Difficult reaction: transition 
state reached late, resembles products. Easy reaction: transition state 
reached early, resembles reactants. 



SEC. 2.25 QUALITATIVE ELEMENTAL ANALYSIS 67 

Abstraction of hydrogen by the highly reactive chlorine atom has a low " ac - . 
According to the postulate, then, the transition state is reached before the reaction 
has proceeded very far, and when the carbon-hydrogen bond is only slightly 
stretched. Atoms and electrons are still distributed much as they were in the 
reactants; carbon is still nearly tetrahedral. The methyl group has developed 
little free-radical character. 

Abstraction of hydrogen by the less reactive bromine atom, in contrast, has 
a very high act . The transition state is reached only after reaction is well along 
toward completion and when the carbon-hydrogen bond is more nearly broken. 
The geometry and electron distribution has begun to approach that of the products, 
and carbon may well be almost trigonal. The methyl group has developed much 
free-radical character. 

Thus, in the attack by a reagent of high reactivity, the transition state tends to 
resemble the reactant; in the attack by a reagent of low reactivity, the transition state 
tends to resemble the products. 



2.24 Molecular formula: its fundamental importance 

,In this chapter we have been concerned with the structure of methane: the 
way In which atoms are put together to form a molecule of methane. But first 
we had to know what kinds of atoms these are and how many of them make up the 
molecule; we had to know that methane is CH 4 . Before we can assign a structural 
formula to a compound, we must first know its molecular formula^ 

Much of the chapter has been spent in discussing the substitution of chlorine 
for the hydrogen of methane. But first we had to know that there is substitution, 
that each step of the reaction yields a product that contains one less hydrogen 
atom and one more chlorine atom than the reactant; we had to know that CH 4 
is converted successively into CH 3 C1, CH 2 C1 2 , CHC1 3 , and CC1 4 . Before we can 
discuss the reactions of an organic compound, we must first know the molecular 
formulas of the products. 

TLet us review a little of what we know about the assigning of a molecular 
formula to a compound. We must carry out: 

(a) a qualitative elemental analysis, to find out what kinds of atoms are 
present in the molecule; 

(b) a quantitative elemental analysis, to find out the relative numbers of the 
different kinds of atoms, that is, to determine the empirical formula', 

(c) a molecular weight determination, which (combined with the empirical 
formula) shows the actual numbers of the different kinds of atoms, that is, gives us 
the molecular formula^ 

Most of this should be familiar to the student from previous courses in 
chemistry. What we shall concentrate on here will be the application of these 
principles to organic analysis. 

2.25 Qualitative elemental analysis 

The presence of carbon or hydrogen in a compound is detected by combustion: 
heating with copper oxide, which converts carbon into carbon dioxide and 
hydrogen into water. (Problem: How could each of these products be identified?) 



68 METHANE CHAP. 2 

(C,H) + CuO -^U Cu + CO 2 + H 2 O 

Covalently bonded halogen, nitrogen, and sulfur must be converted into 
inorganic ions, which can then be detected in already familiar ways* This con- 
version is accomplished in either of two ways: (a) through sodium fusion, treatment 
with hot molten sodium metal; 

(C,H,X,N,S) + Na -^-> Na+X~ -I- Na+CN- + Na+S--Na+ 
or (b> through Schoniger oxidation by oxygen gas. 

(C,H,X,N,S) + O 2 ^^> Na+X- + Na+NO 2 - + Na+SOj- -Na+ 

(A simpler method of detecting halogen in some organic compounds is dis- 
cussed in Sec. 14.24.) 

By these methods, we could show, for example, that.methane contains carbon 
and hydrogen, or that methyl chloride contains carbon, hydrogen, and chlorine. 

Further tests would show the absence of any other element in these compounds, 
except possibly oxygen, for which there is no simple chemical test; presence or 
absence of oxygen would be shown by a quantitative analysis. 

Problem 2.4 (at How would you detect halide ion as a product of sodium fu- 
sion or oxidation? (b) If sulfur and/or nitrogen is also present in an organic mole- 
cule, this test cannot be carried out on a sodium fusion mixture until it has been 
acidified and boiled. Why is this so? 

Problem 2.5 Only carbon and hydrogen were detected by a qualitative elemental 
analysis of the compound ethyl alcohol; quantitative analysis gave 52.1% carbon 
and 13.1% hydrogen, (a) Why would it be assumed that ethyl alcohol contains oxy- 
gen? (b) What percentage of oxygen would be assumed? 



2.26 Quantitative elemental analysis: carbon, hydrogen, and halogen 

Knowing what elements make up a compound, we must next determine the 
proportions in which they are present. To do this, we carry out very much the 
same analysis as before, only this time on a quantitative basis. To find out the 
relative amounts of carbon and hydrogen in methane, for example, we would 
completely oxidize a measured amount of methane and weigh the carbon dioxide 
and water formed. 

In a quantitative combustion, a weighed sample of the organic compound is 
passed through a combustion train: a tube packed with copper oxide heated to 
600-800, followed by a tube containing a drying agent (usually Dehydrite, 
magnesium perchlorate) and a tube containing a strong base (usually Ascarite, 
sodium hydroxide on asbestos). The water formed is absorbed by the drying 
agent, and the carbon dioxide is absorbed by the base; the increase in weight of 
each tube gives the weight of product formed. 

For example, we might find that a sample of methane weighing 9.67 mg 
produced 26.53 mg of CO 2 and 21.56 mg of H 2 O. Now, only the fraction 
C/CO 2 = 12.01/44.01 of the carbon dioxide is carbon, and only the fraction 
2H/H 2 O = 2.016/18.02 of the water is hydrogen. Therefore 

wt. C 26.53 x 12.01/44.01 wt. H - 21.56 x 2.016/18.02 



SEC. 2.27 EMPIRICAL FORM I LA 69 

wt. C (in sample) = 7.24 mg wt. H (in sample) = 2.41 mg 
and the percentage composition is 

% C = 7.24/9.67 x 100 % H = 2.41/9.67 x 100 

% C (in sample) - 74.9 % H (in sample) 24.9 

Since the total of carbon and hydrogen is 100 /, within the limits of error of the 
analysis, oxygen (or any other element) must be absent. 

In quantitative, as in qualitative, analysis, covalently bonded halogen must 
be converted into halide ion. The organic compound is heated either (a) in a 
bomb with sodium peroxide or (b) in a sealed tube with nitric acid (Can'us method)^ 
The halide ion thus formed is converted into silver haltde, which can be weighed. 

Problem 2.6 When 7.36 mg of methyl chloride was heated in a bomb with 
sodium peroxide, the chloride ion liberated yielded 20.68 mg of silver chloride, 
(a) What percentage of chlorine is indicated by this analysis? (b) What percentage of 
chlorine would be expected from a compound of formula CH 3 C1 ? (c) What weight of 
silver chloride would you expect from 7.36 mg of methylene chloride? (d) Of chloro- 
form? (e) Of carbon tetrachloride? 

(We shall lake up other quantitative analytical methods when we need them: 
nitrogen and sulfur analysis, Sec. 10.12; methoxyl determination, Sec. 17.16; 
neutralization equivalent, Sec. 18.21; saponification equivalent, Sec. 20.24.) 

t s 

2.27 . Empirical formula 

Knowing the percentage composition of a compound, we can now calculate 
the empirical formula: the simplest formula that shows the relative numbers of the 
different kinds of atoms in a molecule. For example, in 100 g (taken for convenience) 
of methane there are 74.9 g of carbon and 24.9 g of hydrogen, according to our 
quantitative analysis. Dividing each quantity by*the proper atomic weight gives 
the number of gram-aioms of each element. 

74 9 
C: ' = 6.24 gram-atoms 

249 
H : ' = 24.7 gram-atoms 

l.UUo 

Since a gram-atom of one element contains the same number of atoms as a gram- 
atom of any other element, we now know the relative number of carbon and 
hydrogen atoms in methane: C 6 24^47. Conversion to smallest whole numbers 
gives the empirical formula CH 4 for methane. 

C: 6.24/6.24 = 1 

H: 24.7/6.24 = 3.96, approximately 4 

Problem 2.7 Calculate the percentage composition and then the empirical 
formula for each of the following compounds: (a) Combustion of a 3.02-mg sample 
of a compound gave 8.86 mg of carbon dioxide and 5.43 mg of water, (b) Combus- 
tion of an 8.23-mg sample of a compound gave 9.62 mg of carbon dioxide and 3.94 
mg of water. Analysis of a 5.32-mg sample of the same compound by the Carius 
method gave 13.49 mg of silver chloride. 



70 METHANE CHAP. 2 



2.28-7 Molecular weight. Molecular formula 

At this stage we know what kinds of atoms make up the molecule we are 
studying, and in what ratio they are present. This knowledge is summarized in 
the empirical formula. 

But this is not enough. On the basis of just the empirical formula, a molecule 
of methane, for example, might contain one carbon and four hydrogens, or two 
carbons and eight hydrogens, or any multiple of CH 4 . We still have to find the 
molecular formula: the formula that shows the actual number of each kind of atom 
in a molecule. 

To find the molecular formula, we must determine the molecular weight: 
today, almost certainly by mass spectrometry, which gives an exact value (Sec. 
13.2). Ethane, for example, has an empirical formula of CH 3 . A molecular weight 
of 30 is found, indicating that, of the possible molecular formulas, C 2 H 6 must 
be the correct one. 

Problem 2.8 Quantitative elemental analysis shows that the empirical formula 
of a compound is CH. The molecular weight is found to be 78. What is the molecular 
formula? 

Problem 2.9 Combustion of a 5.17-mg sample of a compound gives 10.32 mg 
of carbon dioxide and 4.23 mg of water. The molecular weight is 88. What is the 
molecular formula of the compound? 



PROBLEMS 

1. Calculate the percentage composition of A, B, and C from the following analytical 
data: 

wt. sample wt. CO 2 wt. H 2 O wt. AgCl 

A 4.37 mg 15.02 mg 2.48 mg 

B 5.95 mg 13.97mg 2.39 mg 7.55 mg 

C 4.02 mg 9.14 mg 3.71 mg 

2. What is the percentage composition of: 

(a) C 3 H 7 C1 (c) C 4 H 8 2 (e) CH 4 ON 2 

(b) C 2 H 6 (d) C 6 H 8 2 N 2 S (f) C 6 H 8 NC1 

3. What is the empirical formula of an organic compound whose percentage com- 
position is: 

(a) 85.6% C, 14.4% H (d) 29.8% C, 6.3% H, 44.0% Cl 

(b) 92.2% C, 7.8% H (e) 48.7% C, 13.6% H, 37.8% N 

(c) 40.0% C, 6.7% H (f) 25.2% C, 2.8% H, 49.6% Cl 
(Note: remember that oxygen often is not determined directly.) 

4. A qualitative analysis of papaverine, one of the alkaloids in opium, showed 
carbon, hydrogen, and nitrogen. A quantitative analysis gave 70.8% carbon, 6.2%. 
hydrogen, and 4.1% nitrogen. Calculate the empirical formula of papaverine. 

5. Methyl orange, an acid-base indicator, is the sodium salt of an acid that contains 
carbon, hydrogen, nitrogen, sulfur, and oxygen. Quantitative analysis gave 51.4% 
carbon, 4.3% hydrogen, 12.8% nitrogen, 9.8% sulfur, and 7.0% sodium. What is the 
empirical formula of methyl orange? 

6. Combustion of 6.51 mg of a compound gave 20.47 mg of carbon dioxide and 



PROBLEMS 71 

8.36 mg of water. The molecular weight was found to be 84. Calculate: (a) percentage 
composition; (b) empirical formula; and (c) molecular formula of the compound. 

7. A liquid of molecular weight 60 was found to contain 40.0% carbon and 6.7% 
hydrogen. What is the molecular formula of the compound? 

8. A gas of the same empirical formula as the compound in Problem 7 has a molec- 
ular weight of 30. What is its molecular formula? 

9. Indigo, an important dyestuflf, gave an analysis of 73.3% carbon, 3.8% hydrogen, 
and 10.7% nitrogen. Molecular weight determination gave a value of 262. What is the 
molecular formula of indigo ? 

10. The hormone insulin contains 3.4% sulfur, (a) What is the minimum molecular 
weight of insulin? (b) The actual molecular weight is 5734; how many sulfur atoms are 
probably present per molecule? 

11. Calculate A// for: 

(a) (d) H 2 + X 2 > 2HX, where X = F, Cl, Br, I 

(e) C 2 H 6 + Br 2 > C 2 H,Br + HBr 

(f) C 6 H 5 CH 3 -f Br 2 > C\H 5 CH 2 Br + HBr 

(g) H 2 C -CHCH 3 + Br 2 > H 2 O= CHCH 2 Br + HBr 

(h) Reactions (e), (f), and (g) proceed by the same free radical mechanism as halo- 
genation of methane. Calculate A// for each step in these three reactions. 

12. A conceivable mechanism for the chlorination of methane involves the following 
steps : 

(1) C1 2 v 2C1- 

(2) Cl- + CH 4 > CH 3 C1 + H. 

(3) H- + C1 2 > HC1 + Cl- 

then (2), (3), (2), (3), etc. 

(a) Calculate Af/ for each of these steps, (b) Why does this mechanism seem less likely 
than the accepted one given in Sec. 2.12? (Additional, conclusive evidence against this 
alternative mechanism will be presented in Sec. 7,10.) 

13. (a) Free methyl radicals react with methane as follows: 

(/) CH 3 - + CH 4 > CH 4 + CH r 

On the basis of the bond strengths involved, show why the above reaction takes place 
rather than the following: 

(//) CH r + CH 4 > CH 3 -CH 3 + H- 

(b) Reaction (/) has an E ac t of 13 kcal. In Sec. 2.12 it was listed as probable (but un- 
productive) on grounds of collision probability. In actuality, how probable is reaction 
(/) in, say, a 50:50 mixture of CH 4 and C1 2 ? (Hint: See Sees. 2.20 and 2.18.) 

14. Bromination of methane is slowed down by addition of fairly large amounts of 
HBr. (a) Suggest a possible explanation for this. (Hint: See Sec. 2.17.) (b) Account for 
the fact that HC1 does not have a similar effect upon chlorination. (c) Any reaction tends 
to slow down as reactants are used up and their concentrations decrease. How do you 
account for the fact that bromination of methane slows down to an unusually great 
extent, more than, say, chlorination of methane? 

15. A mixture of H 2 and C1 2 does not react in the dark at room temperature. At 
high temperatures or under the influence of light (of a wavelength absorbed by chlorine) 
a violent reaction occurs and HC1 is formed. The photochemical reaction yields as 
many as a million molecules of HC1 for each photon absorbed. The presence of a small 
amount of oxygen slows down the reaction markedly, (a) Outline a possible mechanism 
to account for these facts, (b) Account for the fact that a mixture of H 2 and I 2 does not 



72 METHANE CHAP. 2 

behave in the same way. (Hydrogen iodide is actually formed, but by an entirely different 
mechanism.) 

16. A stream of tetramethyllead vapor, (CH 3 ) 4 Pb, was passed through a quartz 
tube which was heated at one spot; a mirror of metallic lead was deposited at the hot 
point, and the gas escaping from the tube was found to be chiefly ethane. The tube was 
next heated upstream of the lead mirror while more tetramethyllead was passed through; 
a new mirror appeared at the hot point, the old mirror disappeared, and the gas escaping 
from the tube was now found to be chiefly tetramethyllead. Experiments like this, done 
by Fritz Paneth at the University of Berlin, were considered the first good evidence for 
the existence of short-lived free radicals like methyl, (a) Show how these experimental 
results can be accounted for in terms of intermediate free radicals, (b) The farther up- 
stream the tube was heated, the more slowly the old mirror disappeared. Account for this. 

17. When a small amount (0.02%) of tetraethyllead, (C 2 H 5 ) 4 Pb, is added to a mix- 
ture of methane and chlorine, chlorination takes place at only 140 instead of the usual 
minimum of 250. In light of Problem 16, show how this fact strengthens the mechanism 
of Sec. 2.12. 



Chapter 



Alkanes 

Free- Radical Substitution 



3.1 Classification by structure: the family 

The basis of organic chemistry, we have said, is the structural theory. We 
separate all organic compounds into a number of families on the basis of structure. 
Having done this, ,we find that we have at the same time classified the compounds 
as to their physical and chemical properties. A particular set of properties is thus 
characteristic of a particular kind of structure. 

Within a family there are variations in properties. All members of the family 
may, for example, react with a particular reagent, but some may react more readily 
than others. Within a single cdmpound there may be variations in properties, 
one part of a molecule being more reactive than another part. These variations 
in properties correspond to variations in structure. 

As we take up each family of organic compounds, we shall first see what 
structure and properties are characteristic of the family. Next we shall see how 
structure and properties vary within the family. We shall not simply memorize 
these facts, but. whenever possible, shall try to understand properties in terms of 
structure, and to understand variations in properties in terms of variations in 
structure. 

Having studied methane in s.ome detail, let us now look at the more compli- 
cated members of the alkane family. These hydrocarbons have been assigned 
to the same family as methane on the basis of their structure, and on the whole 
their properties follow the pattern laid down by methane. However, certain new 
points will arise simply because of the greater size and complexity of these com- 
pounds. 



3.2 Structure of ethane 

Next in size after methane is ethane, C 2 H 6 . If we connect the atoms of this, 
molecule by covalent bonds, following the rule of one bond (one pair of electron?* 

73 



74 



ALKANES 



CHAP. 3 



for each hydrogen and four bonds (four pairs of electrons) for each carbon, we 
arrive at the structure 

HH ^ *f 

H:C:C:H H C C H 

HH ^ h 

Ethane 
Each carbon is bonded to three hydrogens and to the other carbon. 

Since each carbon atom is bonded to four other atoms, its bonding orbitals 
(sp* orbitals) are directed toward the corners of a tetrahedron. As in the case of 
methane, the carbon-hydrogen bonds result from overlap of these*^ 3 orbitals 
with the s orbitals of the hydrogens. The carbon-carbon bond arises from over- 
lap of two sp* orbitals. 

The carbon-hydrogen and carbon-carbon bonds have the same general 
electron distribution, being cylindrically symmetrical about a line joining the 
atomic nuclei (see Fig. 3.1); because of this similarity in shape, the bonds are given 
the same name, a bonds (sigma bonds). 





Figure 3.1. Ethane molecule. Carbon- 
carbon single bond: a bond. 



Figure 3.2. Ethane molecule : shape and 
size. 



In ethane, then, the bond angles and carbon-hydrogen bond lengths should 
be very much the same as in methane, that is, about 109.5 and about 1.10 A, 
respectively. Electron diffraction and spectroscopic studies have verified this 
structure in all respects, giving (Fig. 3.2) the following measurements for the 
molecule: bond angles, 109.5; C H length, 1.10 A; C C length, 1.53 A. Simi- 
lar studies have shown that, with only slight variations, these values are quite 
characteristic of carbon-hydrogen and carbon-carbon bonds and of carbon bond 
angles in alkanes. 



3.3 Free rotation about the carbon-carbon single bond. Conformations. 
Torsional strain 

This particular set of bond angles and bond lengths still does not limit us to a 
single arrangement of atoms for the ethane molecule, since the relationship 
between the hydrogens of one carbon and the hydrogens of the other carbon is 
not specified. We could have an arrangement like I in which the hydrogens exactly 
oppose each other, an arrangement like II in which the hydrogens are perfectly 
staggered, or an infinity of intermediate arrangements. Which of these is the 
actual structure of ethane ? The answer is : all of them. 

We have seen that the a bond joining the carbon atoms is cylindrically sym- 
netrical about a line joining the two carbon nuclei; overlap and hence bond 



SEC. 3.3 FREE ROTATION ABOUT THE CARBON-CARBON SINGLE BOND 



75 




I 
Eclipsed conformation 



H ii 

Staggered conformation 



Ethane 



strength should be the same for all these possible arrangements. If the various 
arrangements do not differ in energy, then the molecule is not restricted to any 
one of them, but can change freely from one to another. Since the change from 
one to another involves rotation about the carbon-carbon bond, we describe this 
freedom to change by saying that there is free rotation about the carbon-carbon 
single bond. 

Different arrangements of atoms that can be converted into one another by 
rotation about single bonds are called conformations. I is called the eclipsed, con- 
formation; II is called the staggeted conformation. (The infinity of intermediate 
conformations are called skew conformations.) 

The highly useful representations of the kind 




after M. S. Newman, of The Ohio State University, who 



are called 

first propos"ecTtheir use. 

The picture is not yet complete. Certain physical properties show that rota- 
tion is not quite free: there is an energy barrier of about 3 kcal/mole. The potential 
energy of the molecule is at a minimum for the staggered conformation, increases 
with rotation, and reaches a maximum at the eclipsed conformation (Fig. 3.3). 
Most ethane molecules, naturally, exist in the most stable, staggered conforma- 
tion; or, put differently, any molecule spends most of its time in the most stable 
conformation. 

How free are ethane molecules to rotate from one staggered arrangement to 
another? The 3-kcal barrier is not a very high one; even at room temperature 
the fraction of collisions with sufficient energy is large enough that a rapid inter- 
conversion between staggered arrangements occurs. For most practical purposes, 
we may still consider that the carbon-carbon single bond permits free rotation. 

The nature of the rotational barrier in ethane is not understood or what is not 
exactly the same thing -is not readily explained. It is too high to be due merely to van 
der Waals forces (Sec. 1.19): although thrown closer together in the eclipsed conforma- 
tion than in the staggered conformation, the hydrogens on opposite carbons are not big 
enough for this to cause appreciable crowding. The barrier is considered to arise in some 
way from interaction among the electron clouds of the carbon-hydrogen bonds. Quantum 



76 



ALKANES 



CHAP. 3 



Eclipsed 
H H 



H 







Rotation > 

Figure 3.3. Potential energy changes during rotation about carbon-car- 
bon single bond of ethane. 



mechanical calculations show that the barrier should exist, and so perhaps "lack of 
understanding" amounts to difficulty in paraphrasing the mathematics in physical terms. 
Like the bond orbitals in methane, the two sets of orbitals in ethane tend to be as far 
apart as possible to be staggered, 

The energy required to rotate the ethane molecule about the carbon-carbon 
bond is called torsional energy. We speak of the relative instability of the eclipsed 
conformation or any of the intermediate skew conformations as being due to 
torsional strain. 

As the hydrogens of ethane are replaced by other atoms or groups of atoms, 
other factors affecting the relative stability of conformations appear: van der 
Waals forces, dipole-dipte interactions, hydrogen bonding. But the tendency 
for the bond orbitals on adjacent carbons to be staggered remains, and any rotation 
away from the staggered conformation is accompanied by torsional strain. 



3.4 Propane and the butanes 

The next member of the alkane family is propane, C 3 H 8 . Again following the 
rule of one bond per hydrogen and four bonds per carbon, we arrive at structure I. 

Here, rotation can occur about two carbon-carbon bonds, and again is 
essentially free. Although the methyl group is considerably larger than hydrogen, 
the rotational barrier (3.3 kcal/mole) is only a little higher than for ethane. Evi- 
dently there is still not significant crowding in the eclipsed conformation, and the 
rotational barrier is due chiefly to the same factor as the barrier in ethane: tor- 
sional strain. 



SEC. 3.4 



PROPANE AND THE BUTANES 



77 





H H H 

H- C-C-C-H 

I ! I 
H H H 



I 
Propane 

When we consider butane, C 4 H} , we find that there are two possible structures, 
II and III. II has a four-carbon chain and III has a three-carbon chain with a 



H H H H 

III. 
H--C C-C-C H 

! ! 1 I 
H H H H 

II 
//-Butane 



H H H 

i I I 

H-C C C-H 

A i k 

H C-H 

A 

III 
Isobutane 

one-carbon branch. There can be no doubt that these represent different struc- 
tures, since no amount of moving, twisting, or rotating about carbon-carbon 
bonds will cause these ->nuctures to coincide. We can see that in the straight-chain 
structure (II) each carbon possesses at least two hydrogens, whereas in the branched- 
chain structure (III) one carbon possesses only a single hydrogen; or we may 
notice that in the branched-chain structure (III) one carbon is bonded to three 
other carbons, whereas in the straight-chain structure (II) no carbon is bonded to 
more than two other carbons. 

In agreement with this prediction, we find that two compounds of the same 
formula, C 4 H 10 , have been isolated. There can be no doubt that these two sub- 
stances are different compounds, since they show definite differences in their 
physical and chemical properties (see Table 3.1); for example one boils at and 
the other at \2\ By definition, they are isomers (Sec. 1.24). 

Table 3.1 PHYSICAL CONSTANTS OF THE ISOMERIC BUTANES 

n-Butane Isobutane 



b.p. 





-12 


m.p. 


-138 


-159 


sp.gr. at -20 


0.622 


0.604 


solub. in 100 ml alcohol 


1813 ml 


1320ml 



Two compounds of formula C 4 Hi are known and we have drawn two struc- 
tures to represent them. The next question is: which structure represents which 



78 ALKANES CHAP. 3 

compound ? For the answer we turn to the evidence of isomer number. Like methane, 
the butanes can be chlorinated; the chlorination can be allowed to proceed until 
there are two chlorine atoms per molecule. From the butane of b.p. 0\ six isomeric 
products of formula C 4 H 8 CN are obtained, from the butane of b.p. - 12, only 
three. We find that we can draw just six dichlorobutanes containing a straight chain 
of carbon atoms, and just three -containing a branched chain. Therefore, the bu- 
tane of b.p. (T must have the straight chain, and the butane of b.p. 12 must 
have the branched chain. To distinguish between these two isomers, the straight- 
chain structure is called //-butane (spoken "normal butane") and the branched- 
chain structure is called isobutane. 

Problem 3.1 Draw the structures of all possible dichloro derivatives of: 
(a) //-butane; (b) isobutane. 

Problem 3.2 Could we assign structures to the isomeric butanes on the <,-..' .iis 
of the number of isomeric monoMoro derivatives? 



3.5 Conformations of /r-butane. Van der Waals repulsion 

Let us look more closely at the /z-butane molecule and the conformations in 
which it exists. Focusing our attention on the middle C- C bond, we sec a molecule 



CH 






H^ ^p ^H U ^\ H H 

CH 3 H 

1 II 

Anti conformation Gauche conformations 

/?- Butane 

similar to ethane, but with a methyl group replacing one hydrogen on each carbon. 
As with ethane, staggered conformations have lower torsional energies and hence 
are more stable than eclipsed conformations. But, due to the presence of the 
methyl groups, two new points are encountered here: first, there are several 
different staggered conformations; and second, a factor besides torsional strain 
comes into play to affect conformational stabilities. 

There is the anti conformation, I, in which the methyl groups are as far apart 
as they can be (dihedral angle 180). There are two gauche conformations, II and 
III, in which the methyl groups are only 60 apart. (Conformations II and III are 
mirror images of each other, and are of the same stability; nevertheless, they are 
different. Make models and convince yourself that this is so.) 

The anti conformation, it has been found, is more stable (by 0.8 kcal/mole) 
than the gauche (Fig. 3.4). Both are free of torsional strain. But in a gauche con- 
formation, the methyl groups are crowded together, that is, are thrown together 
closer than the sum of their van der Waals radii; under these conditions, van der 
Waals forces are repulsive (Sec. 1.19) and raise the energy of the conformation. 
We say that there is van der Waals repulsion (or steric repulsion) between the 



SEC. 3.6 



HIGHER ALKANES. THE HOMOLOGOUS SERIES 



79 




Rotation > 

Figure 3.4. Potential energy changes during rotation about C 2 3 bond 
of /7-butane. 

methyl groups, and that the molecule is less stable because of van der Waals 
strain (or steric strain). 

Van der Waals strain can affect not only the relative stabilities of various 
staggered conformations, but also the heights of the barriers between them. The 
energy maximum reached when two methyl groups swing past each other rather 
than past hydrogensis the highest rotational barrier of all, and has been estimated 
at 4.4-6.1 kcal/mole. Even so, it is low enough that at ordinary temperatures, at 
least the energy of molecular collisions causes rapid rotation; a given molecule 
exists now in a gauche conformation, and the next instant in the ami conforma- 
tion. 

We shall return to the relationships among conformations like these of 
/i-butane in Sec. 4.20. 

Problem 3.3 Both calculations and experimental evidence indicate that the 
dihedral angle between the methyl groups in the gauche conformation of /t-butane 
is actually somewhat larger than 60. How would you account for this? 

Problem 3.4 Considering only rotation about the bond shown, draw a potential 
energy tu. rotation curve like Fig. 3.4 for: (a) (CH 3 ) 2 CH-CH(CH 3 ) 2 ; (b) (CH 3 ) 2 CH- 
CH 2 CH 3 ; (c) (CH 3 ) 3 C C(CH 3 ) 3 . (d) Compare the heights of the various energy 
barriers with each other and with those in Fig. 3.4. 



3.6 Higher alkanes. The homologous series 

If we examine the molecular formulas of the alkanes we have so far considered, 
we see that butane contains one caflki and two hydrogens more than propane, 



80 ALKANES CHAP. 3 

which in turn contains one carbon and two hydrogens more than ethane, and so on. 
A series of compounds in which each member differs from the next member by a 
constant amount is called a homologous series, and the members of the series are 
called homologs. The family of alkanes forms such a homologous series, the 
constant difference between successive members being CH 2 . We also notice that 
in each of these alkanes the number of hydrogen atoms equals two more than 
twice the number of carbon atoms, so that we may write as a general formula for 
members of this series, C n H 2n + 2 . As we shall see later, other homologous series 
have their own characteristic general formulas. 

In agreement with this general formula, we find that the next alkane, pentane, 
has the formula C 5 H 12 , followed by hexane, C 6 H 14 , heptane, C 7 H J6 , and so on. 
We would expect that, as the number of atoms increases, so does the number of 
possible arrangements of those atoms. As we go up the series of alkanes, we find 
that this is true: the number of isomers of successive homologs increases at a 
surprising rate. There are 3 isomeric pentanes, 5 hexanes, 9 heptanes, and 75 
decancs (Ci ); for the twenty-carbon eicosane, there are 366,319 possible isomeric 
structures! The carbon skeletons of the isomeric pentanes and hexanes are shown 
below, ] 

C 
C-C-C-C-C c-C-C-C C-C-C Pentanes 

tt-Pentane Isopentane Neopentane 

b.p. 36 b.p. 28 b.p. 9.5 

i j 

C C 

b.p. 69 b.p. 60 b.p. 63' 7/c/.tj 

C 

c-c-~c~c c c-c-c 

! I I 

C C C 

b.p. 50 b p. 58 C 

It is important to practice drawing the possible isomeric structures that corre- 
spond to a single molecular formula. In doing this, a set of molecular models is 
especially helpful since it will show that many structures which appear to be dif- 
ferent when drawn on paper are actually identical. 

Problem 3.5 Draw the structures of: (a) the nine isomeric heptanes (C 7 Hi$); 
(b) the eight chloropentanes (C 5 H,,C1); <c) the nine dibromobutanes (C 4 H 8 Br 2 ). 



3.7 Nomenclature 

We have seen that the names methane, ethane, propane, butane, and pentane are 
used for alkanes containing respectively one, two, three, four, and five carbon 
atoms. Table 3.2 gives the names of madjjplarger alkanes. Except for the first 



SEC. 3.8 



ALKYL GROUPS 



81 



Table 3.2 NAMES OF ALKANES 



CH 4 

C 2 H 6 

C 3 H, 



nonane 
CioH 22 decane 
CnH24 undecane 
C] 2 H 2 o dodecane 
CI^HSO let ra decane 
hexadecane 
octadecane 



methane 
ethane 
propane 
butane 
CsHi2 pentane 
Coll 14 hcxanc 
C?Hi6 heptane 
CsH| 8 octane 



four members of the family, the name is simply derived from the Greek (or Latin) 
prefix for the particular number of carbons in the alkane; thus pentane for five, 
hexane for six, heptane for seven, octane for eight, and so on. 

The student should certainly memorize the names of at least the first ten 
alkanes. Having done this, he has at the same time essentially learned the names 
of the first ten alkenes, alkyncs, alcohols, etc., since the names of many families 
of compounds are closely related. Compare, for example, the names propane, 
propene, and propvne for the three-carbon alkane, alkene, and alkyne. 

But nearly every alkane can have a number of isomeric structures, and there 
must be an unambiguous name for each of these isomers. The butanes and pen- 
tanes are distinguished by the use of prefixes: jj-butane and isobutane ; j-rjentane, 
isopentanc, and neopentane^But there are 5 hexanes. 9Tieptanes, and 75 decanes; 
it would be difficult to devise, and even more difficult to remember, a different 
prefix for each of these isomers. It is obvious that some systematic method of 
naming is needed. 

As organic chemistry has developed, several different methods have been 
devised to name the members of nearly every class of organic compounds; each 
method was devised when the previously used system had been found inadequate 
for the growing number of increasingly complex organic compounds. Unfortu- 
nately for the student, perhaps, several systems have survived and are in current 
use. Even if we are content ourselves to use only one system, we still have to under- 
stand the names used by other chemists; hence it is necessary for us to learn more 
than one system of nomenclature. But before we can do this, we must first learn 
the names of certain organic groups. 



3.8 Alkyl groups 

In our study of inorganic chemistry, we found it useful to have names for 
certain groups of atoms that compose only part of a molecule and yet appear 
many times as a unit. For example, NH 4 * is called ammonium', NO 3 ~, nitrate; 
SO 3 ~ ~, sulfite\ and so on. 

In a similar way names are given to certain groups that constantly appear as 
structural units of organic molecules. We have seen that chloromethane, CH 3 C1, 
is also known as methyl chloride. The CH* group is called methyl wherever it 
appears, CH 3 Br being methyl bromide, CH 3 I, methyi iodide, and CH 3 OH, methyl 
alcohol. In an analogous way, the C 2 H 5 group is ethyl; C 3 H 7 , propyl; C 4 H 9 , 
butyl; and so on. 

These groups are named simply by dropping -ane from the name of the 
corresponding alkane and replacing it by -yl. They are known collectively as 



82 ALKANES CHAP. 3 

alkyl groups. The general formula for an alkyl group is C n H 2n +i , since it contains 
one less hydrogen than the parent alkane, C n H 2n + 2 . 

Among the alkyl groups we again encounter the problem of isomerism. 
There is only one methyl chloride or ethyl chloride, and correspondingly only one 
methyl group or ethyl group. We can see, however, that there are two propyl 
chlorides. I and II, and hence that there must be two propyl groups. These groups 

H H H - H H H 

H C-C CCI H-C-C C- H 

ill lit 

K H H H Cl H 

I U 

w-Propyl chloride Jsopropyl chloride 

both contain the propane chain, but differ in the point of attachment of the 
chlorine; they are called it-propyl and isopropyl: We can distinguish the two 

CH 3 CH 2 CH 2 CH 3 CHCH 3 

w-Propyl 

Isoprop> 1 

chlorides by the names n-propyl chloride and isopropyl chloride; we distinguish the 
two propyl bromides, iodides, alcohols, and so on in the same way. 

We find that there are four butyl groups, two derived from the straight-chain 
/f-butane, and two derived from the branched-chain isobutane. These are given 
the designations it- (normal), sec- (secondary), iso-, and tert- (tertiary), as shown 
below. Again the difference between w-butyl and sec-butyl and between isobutyl 
and /erf-butyl lies in the point of attachment of the alkyl group to the rest of the 
molecule. 

CH 3 CH 2 CH 2 CH 2 CH 3 CH 2 CHCH 3 

w-Butyi I 

.sec-Butyl 

<j 

CHCH 2 CH 3 C 

CH/ ^H 3 

Isobutyl ten-Butyl 

Beyond butyl the number of isomeric groups derived from each alkane 
becomes so great that it is impracticable to designate them all by various prefixes. 
Even though limited, this system is so useful for the small groups just described 
that it is widely used; a student must therefore memorize these names and learn to 
recognize these groups at a glance in whatever way they happen to be represented. 

However large the group concerned, one of its many possible arrangements 
can still be designated by this simple system. The prefix n- is used to designate 
any alkyl group in which all carbons form a single continuous chain and in which 
the point of attachment is the very end carbon. For example: 

CH 3 CH 2 CH 2 CH 2 CH 2 C1 CH 3 (CH 2 ) 4 CH 2 C1 

/i-Pentyl chloride /i-Hexyl chloride 

The prefix iso- is used to designate any alkyl group (of six carbons or less) that 



SEC. 3.10 IUPAC NAMES OF ALKANES 83 

has a single one-carbon branch on the next-to-Iast carbon of a chain and has the 
point of attachment at the opposite end of the chain. For example: 

CH 3 CH 3x 

CHCH 2 CH 2 C1 CH(CH 2 ) 2 CH 2 C1 

CH 3 X CH 3 X 

Isopentyl chloride Isohcxyl chloride 

If the branching occurs at any other position, or if the point of attachment is at any 
other position, this name does not apply. 

Now that we have learned the names of certain alkyl groups, let us return to 
the original problem : the naming of alkanes. 



3.9 Common names of alkanes 

As we have seen, the prefixes //-, iso-, and neo- are adequate to differentiate 
the various butanes and pentanes, but beyond this point an impracticable number 
of prefixes would be required. However, the prefix //- has been retained for any 
alkane, no matter how large, in which all carbons form a continuous chain with no 
branching: 

CH 3 CH 2 CH 2 CH 2 CH 3 CH 3 (CH 2 ) 4 CH 3 

H-Pentane n-Hexane 

An isoalkane is a compound of six carbons or less in which all carbons except one 
form a continuous chain and that one carbon is attached to the next-to-end 
carbon : 

CH 3 CH 3 

CHCH 2 CH 3 CH(CH 2 ) 2 CH 3 

CH/ CH/ 

Isopentanc Isohcxanc 

In naming any other of the higher alkanes, we make use of the IUPAC system, 
outlined in the following section. 

(It is sometimes convenient to name alkanes as derivatives of methane; see, 
for example, I on p. 129.) 

3. 10 IUPAC names of alkanes 

To devise a system of nomenclature that could be used for even the most 
complicated compounds, various committees and commissions representing the 
chemists of the world have met periodically since 1892. In its present modification, 
the system so devised is known as the IUPAC system (International Union of Pure 
and Applied Chemistry). Since this system follows much the same pattern for all 
families of organic compounds, we shall consider it in some detail as applied to the 
alkanes. 



84 ALKANES CHAP. 3 

Essentially the rules of the IUPAC system are: 

1. Select as the parent structure the longest continuous chain, and then 
consider the compound to have been derived from this structure by the replacement 
of hydrogen by various alkyl groups. Isobutane (I) can be considered to arise 



CH 3 CHCH, CH 3 CH 2 CH 2 CHCH 3 

CH 3 CH 3 CH 3 

1 I! Ill 

Methylpropane 2-Methylpentane 3-Methylpentane 

(Isobutane) 

from propane by the replacement of a hydrogen atom by a methyl group, and thus 
may be named methylpropane. 

2. Where necessary, as in the isomeric methylpentanes (II and III), indicate 
by a number the carbon to which the alkyl group is attached. 

3. In numbering the parent carbon chain, start at whichever end results in the 
use of the lowest numbers; thus II is called 2*methylpentane rather than 4-methyl- 
pentane. 

4. If the same alkyl group occurs more than once as a side chain, indicate this 
by the prefix <//-. /r/-, tetra-, etc., to show how many of these alkyl groups there are, 
and indicate by various numbers the positions of each group, as in 2,2^4-trimethyl- 
pentane (IV). 

r 

CH 3 CH 2 

CH 3 CHCH 2 CCH 3 CH 3 CH 2 CH 2 CH CH C~ CH 2 CH 3 

CH 3 CH 3 CH CH 3 CH 2 

CH 3 X >H 3 ^H 3 

IV V 

2.2,4-Trimethylpentane 4-Methyl-3,3-diethyI-5-isopropyloctane 

5. If there are several different alkyl groups attached to the parent chain, 
name them in order of increasing size or in alphabetical order; as in 4-methyl- 
SJ-diethyl-S-isopropyhctane ( V) . 

There are additional rules and conventions used in naming very complicated 
alkanes, but the five fundamental rules mentioned here will suffice for the com- 
pounds we are likely to encounter. 

Problem 3.6 Give the IUPAC names for: (a) the isomeric hexanes shown on 
page 80; (b) the nine isomeric heptanes (see Problem 3.5, p. 80). 

Problem 3.7 The IUPAC names for w-propyl and isopropyl chlorides arc 
1-chhropropane and 2-chloropropane. On this basis name: (a) the eight isomeric 
chloropentanes; (b) the nine isomeric dibromobutanes (see Problem 3.5, p. 80). 



3.11 Classes of carbon atoms and hydrogen atoms 

It has been found extremely useful to classify each caibon atom of an alkane 
with respect to the number of other carbon atoms to which it is attached. 



SEC. 3.12 PHYSICAL PROPERTIES 85 

A primary (7) carbon atom is attached to only one other carbon atom; a secondary 
(2) is attached to two others; and a tertiary (3) to three others. For example: 

jo 2 o 2 o jo jo 30 jo 

H H H H H H H 

H-CC-C-CH 

III! 
H H H H 



A I A 

H C H 




H 

1 

Each hydrogen atom is similarly classified, being given the same designation 
of primary, secondary, or tertiary as the carbon atom to which it is attached. 

We shall make constant use of these designations in our consideration of the 
relative reactivities of various parts of an alkane molecule. 

3.12 Physical properties 

The physical properties of the alkanes follow the pattern laid down by 
methane, and are consistent with the alkane structure. An alkane molecule is 
held together entirely by covalent bonds. These bonds either join two atoms of 
the same kind and hence are non-polar, or join two atoms that differ very little in 
electronegativity and hence are only slightly polar. Furthermore, these bonds 
are directed in a very symmetrical way, so that the slight bond polarities tend to 
cancel out. As a result an alkane molecule is either non-polar or very weakly 
polar. 

As we have seen ^Sec. 1.19), the forces holding non-polar molecules together 
(van der Waals forces) arc weak and of very short range; they act only between 
the portions of different molecules that are in close contact, that is, between the 
surfaces of molecules. Within a family, therefore, we would expect that the larger 
the molecule and hence the larger its surface area the stronger the intermolecular 
forces. 

Table 3.3 lists certain physical constants for a number of the w-alkanes. As 
we can see, the boiling points and melting points rise as the number of carbons 
increases. The processes of boiling and melting require overcoming the inter- 
molecular forces of a liquid and a solid; the boiling points and melting points rise 
because these intermolecular forces increase as the molecules get larger. 

Except for the very small alkanes, the boiling point rises 20 to 30 degrees for 
each carbon that is added to the chain ; we shall find that this increment of 20-30 
per carbon hofds not only for the alkanes but also for each of the homologous 
series that we shall study. 

The increase in melting point is not quite so regular, since the intermolecular 
forces in a crystal depend not only upon the size of the molecules but also upon how 
well they fit into a crystal lattice. 

The first four i-alkanes are gases, but, as a result of the rise in boiling point 
and melting point with increasing chain length, the next 13 (C$Cn) are liquids, 
and those- containing 18 carbons or more are solids. 



86 



ALKANES 

Table 3.3 ALKANES 



CHAP. 3 



Name 


Formula 


M.p., 
"C 


B.p., 
C 


Density 
(at 20) 


Methane 


CH 4 


-183 


-162 




Ethane 


CH 3 CH 3 


-172 


- 88.5 




Propane 


CH 3 CH 2 CH 3 


-187 


- 42 




//-Butane 


CH 3 (CH 2 ) 2 CH 3 


-138 







w-Pentane 


CH 3 (CH 2 ) 3 CH 3 


-130 


36 


0.626 


w-Hexane 


CH 3 (CH 2 ) 4 CH 3 


- 95 


69 , 


.659 


/f-Heptane 


CH 3 (CH 2 ) 5 CH 3 


- 90.5 


9f 


.684 


//-Octane 


CH 3 (CH 2 ) 6 CH 3 


- 57 


1*26 


.703 


w-Nonane 


CH 3 (CH 2 ) 7 CH 3 


- 54 


151 


.718 


w-Decane 


CH 3 (CH 2 ) 8 CH 3 


- 30 


174 


.730 


w-Undecane 


CH 3 (CH 2 )oCH 3 


- 26 


196 


.740 


w-Dodecane 


CH 3 (CH 2 ), CH 3 


- 10 


216 


.749 


/f-Tridecane 


CH 3 (CH 2 )i,CH 3 


- 6 


234 


.757 


w-Tetradecane 


CH 3 (CH 2 ), 2 CH 3 


5.5 


252 


.764 


w-Pentadecane 


CH 3 (CH 2 ), 3 CH 3 


10 


266 


.769 


w-Hexadecane 


CH 3 (CH 2 ) 14 CH 3 


18 


280 


.775 


-Heptadecane 


CH 3 (CH 2 ), 5 CH 3 


22 


292 




w-Qctadecane 


CH 3 (CH 2 ) 16 CH 3 


28 


308 




rt-Nonadecane 


CH 3 (CH 2 ) 17 CH 3 


32 


320 




w-Eicosane 


CH 3 (CH 2 ) 18 CH 3 


36 






Isobutane 


(CH 3 ) 2 CHCH 3 


-159 


- 12 




Isopentane 


(CH 3 ) 2 CHCH 2 CH 3 


-160 


28 


.620 


Neopentane 


(CH 3 ) 4 C 


_ 17 


9.5 




Isohexane 


(CH 3 ) 2 CH(CH 2 ) 2 CH 3 


-154 


60 


.654 


3-Methylpentane 


CH 3 CH 2 CH(CH 3 )CH 2 CH 3 


-118 


63 


.676 


2,2-Dimethylbutane 


(CH 3 ) 3 CCH 2 CH 3 


- 98 


50 


.649 


2,3-DimethyIbutane 


(CH 3 ) 2 CHCH(CH 3 ) 2 


-129 


58 


.668 



Problem 3.8 Using the data of Table 3.3, make a graph of: (a) b.p. vs. carbon 
number forjji^-alkanes; (b) m.p. vs. carbon number; (c) density vs. carbon num- 
ber. 

There are somewhat smaller differences among the boiling points of alkanes 
that have the same carbon number but different structures. On pages 77 and 80 
the boiling points of the isomeric butanes, pentanes, and hexanes are given. We 
see that in every case a branched-chain isomer has a lower boiling point than a 
straight-chain isomer, and further, that the more numerous the branches, the lower 
the boiling point. Thus w-butane has a boiling point of and isobutane 12. 
w-Pentane has a boiling point of 36, isopentane with a single branch 28, and 
neopentane with two branches 9.5. This effect of branching on boiling point is 
observed within all families of organic compounds. That branching should lower 
the boiling point is reasonable: with branching the shape of the molecule tends to 
approach that of a sphere; and as this happens the surface area decreases, with the 
result that the intermolecular forces become weaker and are overcome at a lower 
temperature. 

In agreement with the rule of thumb, "like dissolves like," the alkanes are 
soluble in non-polar solvents such as benzene, ether, and chloroform, and are 
insoluble in water and other highly polar solvents. Considered themselves as sol- 



SEC. 3.13 INDUSTRIAL SOURCE 87 

vents, the liquid alkanes dissolve compounds of low polarity and do not dissolve 
compounds of high polarity. 

The density increases with size of the alkanes, but tends to level off at about 
0.8; thus all alkanes are less dense than water. It is not surprising that nearly all 
organic compounds are less dense than water since, like the alkanes, they consist 
chiefly of carbon and hydrogen. In general, to be denser than water a compound 
must contain a heavy atom like bromine or iodine, or several atoms like chlorine. 

3.13 Industrial source 

The principal source of alkanes is petroleum, together with^th^accc^p^r^jng 
natural gas. Decay and millions of years of geologicaTstresses have transformed 
the complicated organic compounds that once jnade up living plants or animals 
into a mixture of alkanes ranging in size from one carbon to 30 or 40 carbons. 
Formed along with the alkanes, and particularly abundant in California petroleum, 
are cycloalkanes (Chap. 9), known to the petroleum industry as naphthenes. 

The other fossil fuel, coal, is a potential second source of alkanes: processes are 
being developed to convert coal, through hydrogenation, into gasoline and fuel oil, and 
into synthetic gas to offset anticipated shortages of natural gas. 

Natural gas contains, of course, only the more volatile alkanes, that is, those 
of low molecular weight; it consists chiefly of methane and progressively smaller 
amounts of ethane, propane, and higher alkanes. For example, a sample taken 
from a pipeline supplied by a large number of Pennsylvania wells contained 
methane, ethane, and propane in the ratio of 12:2: 1, with higher alkanes making 
up only 3% of the total. The propane-butane fraction is separated from the 
more volatile components by liquefaction, compressed into cylinders, and sold as 
bottled gas in areas not served by a gas utility. 

Petroleum is separated by distillation into the various fractions listed in 
Table 3.4; because of the relationship between boiling point and molecular weight, 
this amounts to a rough separation according to carbon number. Each fraction 
is still a very complicated mixture, however, since it contains alkanes of a range 
of carbon numbers, and since each carbon number is represented by numerous 
isomers. The use that each fraction is put to depends chiefly upon its volatility 
or viscosity, and it matters very little whether it is a complicated mixture or a 

Table 3.4 PETROLEUM CONSTITUENTS 

Distillation 
Fraction Temperature, C Carbon Number 





\ 




Gas 


Below 20 


Ci-C 4 


Petroleum ether 


20-60 


C 5 -C 6 


Ligroin (light naphtha) 


60-100 


Ce-C? 


Natural gasoline 


40-205 


Cs-Cio, and cycloalkanes 


Kerosene 


175-325 


Ci2-Ci8, and aromatics 


Gas oil 


Above 275 


12 and higher 


Lubricating oil 


Non-volatile liquids 


Probably long chains attached 






to cyclic structures 



Asphalt or petroleum coke Non-volatile solids Polycyclic structures 



88 ALKANES CHAP. 3 

single pure compound. (In gasoline, as we shall see in Sec. 3.30, the structures of 
the components are of key importance.) 

The chief use of all but the non- volatile fractions is as fuel. The gas fraction, 
like natural gas, is used chiefly for heating. Gasoline is used in those internal 
combustion engines that require a fairly volatile fuel, kerosene is used in tractor 
and jet engines, and gas oil is used in Diesel engines. Kerosene and gas oil are 
also used for heating purposes, the latter being the familiar "furnace oil." 

The lubricating oil fraction, especially that from Pennsylvania crude oil 
(paraffin-base petroleum), often contains large amounts of long-chain alkanes 
(C2o-C 34 ) that have fairly high melting points. If these remained in the oil, they 
might crystallize to waxy solids in an oil line in cold weather. To prevent this, the 
oil is chilled and the wax is removed by filtration. After purification this is sold 
as solid paraffin wax (m.p. 50-55) or used in petrolatum jelly (Vaseline). Asphalt 
is used in roofing and road building. The coke that is obtained from paraflin-base 
crude oil consists of complex hydrocarbons having a high carbon-to-hydrogen 
ratio; it is used as a fuel or in the manufacture of carbon electrodes for the electro- 
chemical industries. Petroleum ether and ligroin are useful solvents for many 
organic materials of low polarity. 

In addition to being used directly as just described, certain petroleum frac- 
tions are converted into other kinds of chemical compounds. Catalytic isomeriza- 
tion changes straight-chain alkanes into branched-chain ones. The cracking pro- 
cess (Sec. 3.31) converts higher alkanes into smaller alkanes and alkenes, and thus 
increases the gasoline yield; it can even be used for the production of "natural" 
gas. In addition, the alkenes thus formed are perhaps the most important raw 
materials for the large-scale synthesis of aliphatic compounds. The process of 
catalytic reforming (Sec. 12.4)converts alkanes and cycloalkanesinto aromatic hydro- 
carbons and thus helps provide the raw material for the large-scale synthesis of 
another broad class of compounds. 



3.14 Industrial source vs. laboratory preparation 

We shall generally divide the methods of obtaining a particular kind of 
organic compound into two categories: industrial source and laboratory preparation. 
We may contrast the two in the following way, although it must be realized that 
there are many exceptions to these generalizations. 

An industrial source must provide large amounts of the desired material at 
the lowest possible cost. A laboratory preparation may be required to produce 
only a few hundred grams or even a few grams; cost is usually of less importance 
than the time of the investigator. 

For many industrial purposes a mixture may be just as suitable as a pure 
compound; even when a single compound is required, it may be economically 
feasible to separate it from a mixture, particularly when the other components 
may also be marketed. In the laboratory a chemist nearly always wants a single 
pure compound. Separation of a single compound from a mixture of related 
substances is very time-consuming and frequently does not yield material of the 
required purity. Furthermore, the raw material for a particular preparation may 
well be the hard-won product of a previous preparation or even series of prepara- 



SEC. 3.15 PREPARATION 89 

tions, and hence he wishes to convert it as completely as possible into his desired 
compound. On an industrial scale, if a compound cannot be isolated from natur- 
ally occurring material, it may be synthesized along with a number of related 
compounds by some inexpensive reaction. In the laboratory, whenever possible, 
a reaction is selected that forms a single compound in high yield. 

In industry it is frequently worth while to work out a procedure and design 
apparatus that may be used in the synthesis of only one member of a chemical 
family. In the laboratory a chemist is seldom interested in preparing the same 
compound over and over again, and hence he makes use of methods that are 
applicable to many or all members of a particular family. 

In our study of organic chemistry, we shall concentrate our attention on 
versatile laboratory preparations rather than on limited industrial methods. In 
learning these we may, for the sake of simplicity, use as examples the preparation 
of compounds that may actually never be made by the method shown. We may 
discuss the synthesis of ethane by the hydrogenation of ethylene, even though we 
can buy all the ethane we need from the petroleum industry. However, if we know 
how to convert ethylene into ethane, then, when the need arises, we also know how 
to convert 2-methyl-l-hexene into 2-methylhexane, or cholesterol into cholestanol, 
or, for that matter, cottonseed oil into oleomargarine. 



3.15 Preparation 

Each of the smaller alkanes, from methane through -pentane and isopentane, 
can be obtained in pure form by fractional distillation of petroleum and natural 
gas; neopentane does not occur naturally. Above the pentanes the number of 
isorners of each homolog becomes so large and the boiling point differences become 
so small that it is no longer feasible to isolate individual, pure compounds; these 
alkanes must be synthesized by one of the methods outlined below. 

In some of these equations, the symbol R is used to represent any alkyl group. 
This convenient device helps to summarize reactions that are typical of an entire 
family, and emphasizes the essential similarity of the various members. 

In writing these generalized equations, however, we must not lose sight of one 
important point. An equation involving RC1, to take a specific example, has 
meaning only in terms of a reaction that we can carry out in the laboratory using 
a real compound, lii e methyl chloride or tert-butyl chloride. Although- typical of 
alkyl halides, a reaction may differ widely in rate or yield depending upon the par- 
ticular alkyl group actually concerned. We may use quite different experimental 
conditions for methyl chloride than for ten- butyl chloride; in an extreme case, a 
reaction that goes well for methyl chloride might go so slowly or give so many 
side products as to be completely useless for tert-butyl chloride. 



PREPARATION OF ALKANES 
1. Hydrogenation of alkenes. Discussed in Sec. 6.3. 

r w H * + Pt< Pd * or Ni ^ <~ w 

*- n r2n > *~n 2n + 2 

Alkene Alkane 



90 



ALKANES 



CHAP. 3 



2. Reduction of alkyl halides 

(a) Hydrolysis of Grignard reagent. Discussed in Sec. 3.16. 



RX + Mg -> RMgX 

Grignard 
reagent 



RH 



Example: 



CH 3 CH 2 CHCH 3 ~^-> CH 3 CH 2 CHCH 3 --* CH 3 CH 2 CHCH 3 

H 



Br 
jvButyI bromide 



MgBr 

Ac'c-Butylmagncsium 
bromide 



w-Butane 



(b) Reduction by metal and acid. Discussed in Sec. 3.15. 

RX-fZn + H* - > RH + Zn + +-fX- 
Example: 

>H *> CH 3 CH 2 CHCH 3 



CH 3 CH 2 CHCH 3 
Br 



H 

w-Butane 



sec-Butyl bromide 
3. Coupling of alkyl halides with organometallic compounds. Discussed in Sec. 3.17. 



RX 

Mav he 

1 ', r, 3 ' 



RLi 

Alkvllithium 



CuX 



Examples: 



CH 3 CH 2 C1 

Ethyl 
chloride 



CH 3 CH 2 Li 

Fthyllithium 



R 2 CuLi , 

Lithium 
dialkylcopper 

R'X I 

Should he 1 " 



> (CH 3 CH 2 ) 2 CuLi | 

Lithium 
dicthylcoppcr 

CH 3 (CH 2 ) 5 CH 2 Br - 
//-Hcptyl bromide 



R R 



CH 3 (CH 2 ) 7 CH 3 

/i-Nonane 



CH 3 CCH, -> ~^-^ (/-C 4 H 9 ) 2 CuLi _ 



Cl 
/pr/-Buty] chloride 



CH 3 CH 2 C'H 2 CH 2 CH 2 Br * 

//-Pcntyl bromide 



CH,C'CH 2 CH 2 C H,CH 2 

t'H 3 

2,2-Dimcthylheptane 



By far the most important of these methods is the hydrogenation of alkencs. 
When shaken under a slight pressure of hydrogen gas in the presence of a small 
amount of catalyst, alkenes are converted smoothly and quantitatively into alkanes 
of the same carbon skeleton. The method is limited only by the availability of the 
proper alkene. This is not a very serious limitation; as we shall see (Sec. 5.11), 
alkenes are readily prepared, chiefly from alcohols, which in turn can be readily 
synthesized (Sec, 15.7) in a wide variety of sizes and shapes. 



SEC. 3.16 THE GRIGNARD REAGENT 91 

Reduction of an alkyl halide, either via the Grignard reagent or directly with 
metal and acid, involves simply the replacement of a halogen atom by a hydrogen 
atom; the carbon skeleton remains intact. This method has about the same ap- 
plicability as the previous method, since, like alkenes, alkyl halides are generally 
prepared from alcohols. Where either method could be used, the hydrogenation 
of alkenes would probably be preferred because of its simplicity and higher yield. 

The coupling of alkyl halides with organometallic compounds is the only 
one of these methods in which carbon-carbon bonds are formed and a new, bigger 
carbon skeleton is generated. 

3.16 The Grignard reagent: an organometallic compound 

When a solution of an alkyl halide in dry ethyl ether, (C 2 H 5 ) 2 O, is allowed to 
stand over turnings of metallic magnesium, a vigorous reaction takes place: the 
solution turns cloudy, begins to boil, and the magnesium metal gradually dis- 
appears. The resulting solution is known as a Grignard reagent, after Victor Grig- 
nard (of the University of Lyons) who received the Nobel prize in 1912 for its 
discovery. It is one of the most useful and versatile reagents known to the organic 
chemist. 

CH 3 I + Mg eth - r > CH 3 MgI 
Methyl Methylmagnesium iodide 

iodide 

CH 3 CH 2 Br + Mg ether > CH 3 CH 2 MgBr 
Ethyl bromide Ethylmagnesium bromide 

The Grignard reagent has the general formula RMgX, and the general name 
alkylmagnesium halide. The carbon-magnesium bond is covalent but highly polar, 
with carbon pulling electrons from electropositive magnesium; the magnesium- 
halogen bond is essentially ionic. 

R:Mg+:X:- 

Since magnesium becomes bonded to the same carbon that previously held 
halogen, the alkyl group remains intact during the preparation of the reagent. 
Thus 7i-propyl chloride yields /i-propylmagnesium chloride, and isopropyl chloride 
yields isopropylmagnesium chloride. 

CH 3 CH 2 CH 2 C1 + Mg ether > CH 3 CH 2 CH 2 MgCl 
n-Propyl chloride /i-Propylmagnesium chloride 

CH 3 CHC1CH 3 + Mg ether > CH 3 CHMgClCH 3 
Isopropyl chloride Isopropylmagnesium chloride 

The Grignard reagent is the best-known member of a broad class of substan- 
ces, called organometallic compounds, in which carbon is bonded to a metal: 
lithium, potassium, sodium, zinc, mercury, lead, thallium almost any metal 
known. Each kind of organometallic compound has, of course, its own set of 
properties, and 'its particular uses depend on these. But, whatever the metal, it is 
less electronegative than carbon, and the carbon-metal bond like the one in the 

8. 8+ 
R M 



92 ALKANES CHAP. 3 

Grignard reagent is highly polar. Although the organic group is not a full- 
fledged carbanion an anion in which carbon carries negative charge it never- 
theless has considerable carbanion- character. As we shall see, organometallic 
compounds owe their enormous usefulness chiefly to one common quality: they 
can serve as a source from which carbon is readily transferred with its electrons. 

The Grignard reagent is highly reactive. It reacts with numerous inorganic 
compounds including water, carbon dioxide, and oxygen, and with most kinds of 
organic compounds ; in many of these cases the reaction provides the best way to 
make a particular class of organic compound. 

The reaction with water to form an alkane is typical of the behavior of the 
Grignard reagent and many of the more reactive organometallic compounds 
toward acids. In view of the marked carbanion character of the alkyl group, we 
may consider the Grignard reagent to be the magnesium salt, RMgX, of the 
extremely weak acid, R H. The reaction 

RMgX + HOH > R-H + Mg(OH)X 

Stronger Weaker 

acid acid 

is simply the displacement of the weaker acid, R H, from its salt by the stronger 
acid, HOH. 

An alkane is such a weak acid that it is displaced from the Grignard reagent 
by compounds that we might ordinarily consider to be very weak acids themselves, 
or possibly not acids at all. Any compound containing hydrogen attached to 
oxygen or nitrogen is tremendously more acidic than an alkane, and therefore can 
decompose the Grignard reagent: for example, ammonia or methyl alcohol. 

RMgX + NH 3 > R-H + Mg(NH 2 )X 

Stronger Weaker 

acid acid 

RMgX + CH 3 OH > R-H + Mg(OCH 3 )X 

Stronger Weaker 

acid acid 

For the preparation of an alkane, one acid is as good as another, so we naturally 
choose water as the most available and convenient. 

Problem 3.9 (a) Which alkane would you expect to get by the action of water 
on w-propylmagnesium chloride? (b) On isopropylmagnesium chloride? (c) Answer 
(a) and (b) for the action of deuterium oxide ("heavy water," D 2 O). 

Problem 3.10 On conversion into the Grignard reagent followed by treatment 
with water, how many alkyl bromides would yield: (a) //-pentane; (b) 2-methylbutane; 
(c) 2,3-dimethylbutane; (d) neopentane? Draw the structures in each case. 



3.17 Coupling of alkyl halides with organometallic compounds 

To make an alkane of higher carbon number than the starting material re- 
quires formation of carbon-carbon bonds, most directly by the coupling together 
of two alkyl groups. The most versatile method of doing this is through a synthesis 
developed during the late 1960s by E. J. Corey and Herbert House, working inde- 
pendently at Harvard University and Massachusetts Institute of Technology. 



: 3az COUPLING OF ALKYL HALIDES 03 

Coupling takes place in the reaction between a lithium dialkylcopper, R 2 CuLi, 
and an alkyl halide, R'X. (R' stands for an alkyl group that may be the same as, 
or different from, R.) 

R 2 CuLi + R'X -- > R- R' + RCu + LiX 

Lithium Alkyl Alkane 

dialkylcopper halide 

An alkyllithium, RLi, is prepared from an alkyl halide, RX, in much the same 
way as a Grignard reagent. To it is added cuprous halide, CuX, and then, finally, 
the second alkyl halide, R'X. Ultimately, the alkane is synthesized from the two 
alkyl halides, RX and R'X. 



Alkyl Lithium 

lithium dialkylcopper > R R' 

R'X 

For good yields, R'X should be a primary halide; the alkyl group R in the 
organometallic may be primary, secondary, or tertiary. For example: 



>CH 3 (CH 2 ) 7 CH 3 

w-Nonane 



CH 3 Br 1!~> CH 3 Li ~^-> (CH 3 ) 2 CuLi , 

Methyl Methyllithium Lithium 

bromide dialkylcopper 

CH 3 (Ch 2 ) 6 CH 2 I J 
. . w-Octyl iodide 

CH 3 CH 2 CHCH 3 ~^-> -^-> (CH 3 CH 2 CH ) 2 CnLi 

Cl ' CH 3 |-* CH 3 CH 2 CH(CH 2 ) 4 CH 3 

sec-Buiyl 



chloride CH 3 CH 2 CH 2 CH 2 CH 2 Br 



CH 



/i-Pentyl bromide 3-Methyloctane 

The choice of organometallic reagent is crucial. Grignard reagents or organo- 
lithium compounds, for example, couple with only a few unusually reactive organic 
halides. Organosodium compounds couple, but are so reactive that they couple, as 
they are being formed, with their parent alkyl halide; the reaction of sodium with alkyl 
halides (Wurtz reaction) is thus limited to the synthesis of symmetrical alkanes, R R. 

Organocopper compounds were long known to be particularly good at the forma- 
tion of carbon-carbon bonds, but are unstable. Here, they are generated in situ from the 
organolithium, and then combine with more of it to form these relatively stable organo- 
metallics. They exist as complex aggregates but are believed to correspond roughly to 
R 2 Cu~Li+. The anion here is an example of an ate complex, the negative counterpart 
of an onium complex (ammo/jfffm, oxonium). 

Although the mechanism is not understood, evidence strongly suggests this 
much: the alkyl group R is transferred from copper, taking a pair of electrons 
with it, and attaches itself to the alkyl group R' by pushing out halide ion (nucleo- 
philic aliphatic substitution. Sec. 14.9). 

Problem 3.11 (a) Outline two conceivable syntheses of 2-methylpentane from 
three-carbon compounds, (b) Which of the two \\ould you actually use? Why? 



94 ALKANES CHAP. 3 



3.18 Reactions 

The alkanes are sometimes referred to by the old-fashioned name of paraffins. 
This name (Latin: parum qffinis, not enough affinity) was given to describe what 
appeared to be the low reactivity of these hydrocarbons. 

But reactivity depends upon the choice of reagent. If alkanes are inert toward 
hydrochloric and sulfuric acids, they react readily with acids like HF-SbF 5 and 
FSOjH-SbFs ("magic acid") to yield a variety of products. If alkanes are inert 
toward oxidizing agents like potassium permanganate or sodium dichromate, 
most of this chapter is devoted to their oxidation by halogens. Certain yeasts feed 
happily on alkanes to produce proteins certainly a chemical reaction. As Pro- 
fessor M. S. Kharasch (p. 189) used to put it, consider the "inertness" of a room 
containing natural gas, air, and a lighted match. 

Still, on a comparative basis, reactivity is limited. "Magic acid" is, after all, 
one of the strongest acids known; halogenation requires heat or light; combustion 
needs a flame or spark to get it started. 

Much of the chemistry of alkanes involves free-radical chain reactions, which 
take place under vigorous conditions and usually yield mixtures of products. A 
reactive particle typically an atom or free radical is needed to begin the attack 
on an alkane molecule. It is the generation of this reactive particle that requires 
the vigorous conditions: the dissociation of a halogen molecule into atoms, for 
example, or even (as in pyrolysis) dissociation of the alkane molecule itself. 

In its attack, the reactive particle abstracts hydrogen from the alkane; the 
alkane itself is thus converted into a reactive particle which continues the reaction 
sequence, that is, carries on the chain. But an alkane molecule contains many 
hydrogen atoms and the particular product eventually obtained depends upon 
which of these hydrogen atoms is abstracted. Although an attacking particle may 
show a certain selectivity, it can abstract a hydrogen from any part of the molecule, 
and thus bring about the formation of many isomeric products. 



REACTIONS OF ALKANES 
.1. Halogenation. Discussed in Sees. 3.19-3.22. 

-^L-H + X, *""''', -i-X + HX 

Usually a 
mixture 

Reactivity X 2 : C1 2 > Br 2 

H: 3 > 2 > 1 > CHj-H 

Example: 

CH 3 CH 3 CH 3 

CH 3 -CH~CH 3 230 ^y > CH 3 -CH-CH 2 C1 and CH 3 -C-CH 3 

Isobutane Isobutyl chloride j 

ter/-Butyl chloride 



SEC. 3.19 HALOGENATION 95 

2. Combustion. Discussed in Sec. 3.30. 

CH 2n +2 + excess O 2 Hamc > CO 2 + (/i + 1)H 2 O 

A/f = heat of combustion 
Example: 

w-C 5 H u + 8 O 2 flame > 5CO 2 + 6H 2 O A# = -845 kcal 

3. Pyrolysis (cracking). Discussed in Sec. 3.31. 

400-600; with or 

alkane without catalysts ^ H 2 + smaller alkanes + alkenes 



3.19 Halogenation 

As we might expect, halogenation of the higher alkanes is essentially the same 
as the halogenation of methane. It can be complicated, however, by the formation 
of mixtures of isomers. 

Under the influence of ultraviolet light, or at 250-400, chlorine or bromine 
converts alkanes into chloroalkanes (alkyl chlorides) or bromoalkanes (alkyl 
bromides); an equivalent amount of hydrogen chloride or hydrogen bromide is 
formed at the same time. When diluted with an inert gas, and in an apparatus 
designed to carry away the heat produced, fluorine has recently been found to give 
analogous results. As with methane, iodi nation does not take place at all. 

Depending upon which nydrogen atom is replaced, any of a number of iso- 
meric products can be formed from a single alkane. Ethane can yield only one halo- 
ethane; propane, /z-butane, and isobutane can yield two isomers each; -pentane 
can yield three isomers, and isopentane, four isomers. Experiment has shown 
that on halogenation an alkane yields a mixture of all possible isomeric products, 
indicating that all hydrogen atoms are susceptible to replacement. For example, 
for chlorination : 

CH 3 CH 3 -- . "250*' CH 3 CH 2 Cl 
Ethane ' b.p. 13 

Chloroethane 
Ethyl chloride 

CH 3 CH 2 CH 3 l{ ^\ y > CH 3 CH 2 CH 2 -C1 and CH 3 CHCH 3 
Propane ' b.p. 47 L 

1-Chloropropane ^ 

W .Propylchloride 2 ^hloropropane 

Isopropyl chloride 

55% 

CH 3 CH 2 CH 2 CH 3 ligh ^2 S o> CH 3 CH 2 CH 2 CH 2 Cl and CH 3 CH 2 CHCH 3 
n-Butane ' b.p. 78.5 L 

1-Chlorobutane e 

-Butylchloride **! 

870 wc-Bulyl chloride 



96 



ALKANES 



CHAP. 3 



CH 3 

CH 3 CHCH 3 

Isobutane 



CH 3 

CH 3 CHCH 2 

b.p. 69 

l-Chloro-2- 
methylpropane 

Isobutyl chloride 

, 

' 



CI and 



CH 3 

CH 3 CCH 3 

I 

C 

*P* 5I 

2-Chloro-2- 

methylpropane 

/?/7-Butyl chloride 
36% 



Bromination gives the corresponding bromides but in different proportions: 

CH 3 CH 3 .JTur ' CH,CH 2 Br 
Ethane 



CH 3 CH 2 CH 3 j 27 ,> CH 3 CH 2 CH 2 Br and CH 3 CHCH 3 
Propane ' 3/ 1 

97% 

CH 3 CH 2 CH 2 CH 3 Il8h f ,370 > CH 3 CH 2 CH 2 CH 2 Br and CH 3 CH 2 CHCH 3 
/r-Butane ' 2% 1 

DT 

98% 



CH 3 



CH 3 



CH, 



CH 3 CHCH 3 hgh * r2 127 o > CH 3 CHCH 2 Br and CH 3 CCH 3 

Isobutane ' trace I 

ur 

over 99% 



Problem 3.12 Draw the structures of: (a) the three monochioro derivatives 
of ff-pentune; (h) the four monochioro derivatives of isopentane. 

Although both chlorination and bromination yield mixtures of isomers, the 
results given above show that the relative amounts of the various isomers differ 
markedly depending upon the halogen used. Chlorination gives mixtures in which 
no isomer greatly predominates; in bromination, by contrast, one isomer may 
predominate to such an extent as to be almost the only product, making up 97- 
99% of the total mixture. In bromination, there is a high degree of selectivity as 
to which hydrogen atoms are to be replaced. (As we shall see in Sec. 3.28, this 
characteristic of bromination is due to the relatively low reactivity of bromine 
atoms, ancl is an example of a general relationship between reactivity and selec- 
tivity.) 

Chlorination of an alkane is not usually suitable for the laboratory prepara- 
tion of an alkyl chloride; any one product is necessarily formed in low yield, and 
is- difficult to separate from its isomers, whose boiling points are seldom far from 
its own. Bromination, on the other hand, often gives a nearly pure alkyl bromide 
in high yield. As we shall see, it is possible to predict just which isomer will pre- 
dominate; if this product is the one desired, direct bromination could be a feasible 
synthetic route. 

On an industrial scale, chlorination of alkanes is important. For many 



SEC. 3.20 



MECHANISM OF HALOGENATION 



97 



purposes, for example, use as a solvent, a mixture of isomers is just as suitable as, 
and much cheaper than, a pure compound. It may be even worthwhile^ when 
necessary, to separate a mixture of isomers if each isomer can then be marketed. 

Problem 3.13 How do you account for the fact that not only bromination but 
also chlorination is a feasible laboratory route to a neopentyl halidc, 



3.20 Mechanism of halogenation 

Halogenation of alkanes proceeds by the same mechanism as halogenaiion 
of methane: 

Chain-initiating step 



Chain-propagating steps 



then (2), (3), (2), (3), etc., until finally a chain is terminated (Sec. 2.13 N , 

A halogen atom abstracts hydrogen from the alkane (RH) to form an alkyl radical 
(R-). The radical in turn abstracts a halogen atom from a halogen molecule to 
yield the alkyl halide (RX). 

Which alkyl halide is obtained depends upon which alkyl radical is formed. 



v*/ 

(2) 
(V\ 


X- + RH 

R . .J- V~ 


ultraviolet 
light 


HX 

RV 





CH 4 - 

Methane 

CH 3 CH 3 - 

Ethane 



abstraction 



CH 3 CH 2 CH 3 
Propane 



CH 3 * 

Methyl 
radical 

CH 3 CH 2 - 

Ethyl 
radical 



CH 3 CH 2 CH 2 - 

//-Propyl 
radical 



CH 3 X 

Methyl 
halide 



CH 3 CH 2 X 

Ethyl 
halide 



abstraction 



CH 3 CHCH 3 

Isopropyl 
radical 



CH 3 CH 2 CH 2 X 

/?-Prop>l 
halidc 



CH 3 CHCH 3 

X 

Isopropyl 
haltde 



This in turn depends upon the alkane and which hydrogen atom is abstracted from 
it. For example, w-propyl halide is obtained from a w-propyl radical, formed 
from propane by abstraction of a primary hydrogen; isopropyl halide is obtained 
from an isopropyl radical, formed by abstraction of a secondary hydrogen. 

How fast an alkyl halide is formed depends upon how fast the alkyl radical is 
formed. Here also, as was the case with methane (Sec. 2.20), of the two chain- 
propagating steps, step (2) is more difficult than step (3), and hence controls the 
rate of overall reaction. Formation of the alkyl radical is difficult, but once formed 
the radical is readily converted into the alkyl halide (see Fig. 3.5). 



98 



ALKANES 



CHAP. 3 



Difficult step 




RCI +C1- 



Progress of reaction > 

Figure 3.5. Potential energy changes during progress of reaction: 
chlorination of an alkane. Formation of radical is rate-controlling step. 



3.21 Orientation of halogenation 

With this background let us turn to the problem of orientation; that is, let us 
examine the factors that determine where in a molecule reaction is most likely to 
occur. It is a problem that we shall encounter again and again, whenever we study 
a compound that offers more than one reactive site to attack by a reagent. It is an 
important problem, because orientation determines what product we obtain. 

As an example let us take chlorination of propane. The relative amounts of 
/?-propyl chloride and isopropyl chloride obtained depend upon the relative rates 
at which w-propyl radicals and isopropyl radicals are formed. If, say, isopropyl 
radicals are formed faster, then isopropyl chloride will be formed faster, and will 
make up a larger fraction of the product. As we can see, w-propyl radicals are 
formed by abstraction of primary hydrogens, and isopropyl radicals by abstraction 
of secondary hydrogens. 

H H H H H H 

abstraction 111 III 

Qf'H > H C C~C Cl i> H C- C C CI 



H H H 

C C C H 
I I 
H H 
Propane 



H H H 

/j-Propyl 
radical 



H H H 



H H H 

/i-Propyl 
chloride 



H H H 



H-i-i-i-H _a H-i-i-i-H 

'- ' -'- 



H H 

Isopropyl 



Isopronyl 



SEC. 3.21 



ORIENTATION OF HALOGENATION 



99 



Thus orientation is determined by the relative rates of competing reactions. In 
this case we are comparing the rate of abstraction of primary hydrogens with the 
rate of abstraction of secondary hydrogens. What are the factors that determine 
the rates of these two reactions, and in which of these factors may the two reac- 
tions differ? 

First of all, there is the collision frequency. This must be the same for the two 
reactions, since both involve collisions of the same particles : a propane molecule 
and a chlorine atom. 

Next, there is the probability factor. If a primary hydrogen is to be abstracted, 
the propane molecule must be so oriented at the time of collision that the chlorine 
atom strikes a primary hydrogen ; if a secondary hydrogen is to be abstracted, the 
propane must be so oriented that the chlorine collides with a secondary hydrogen. 
Since there are six primary hydrogens and only two secondary hydrogens in each 
molecule, we might estimate that the probability factor favors abstraction of 
primary hydrogens by the ratio of 6:2, or 3: 1. 

Considering only collision frequency and our guess about probability factors, 
we predict that chlorination of propane would yield w-propyl chloride and iso- 
propyl chloride in the ratio of 3 : 1 . As shown on page 95, however, the two 
chlorides are formed in roughly equal amounts, that is, in the ratio of about 1:1, 
or 3:3. The proportion of isopropyl chloride is about three times as great as 
predicted. Evidently, about three times 'as many collisions with secondary hydro- 
gens are successful as collisions with primary hydrogens. If our assumption 
about the probability factor is correct, this means that act is less for abstraction 
of a secondary hydrogen than for abstraction of a primary hydrogen. 

Chlorination of isobutane presents a similar problem. In this case, abstrac- 
tion of one of the nine primary hydrogen leads to the formation of isobutyl 
chloride, whereas abstraction of the single tertiary hydrogen leads to the formation 
of re/7-butyl chloride. We would estimate, then, that the probability factor favors 



abstraction 

of r H v 



H 

H C H 
H 

r r r . ci : 




H H H 

Isobutane 



abstraction 
of3H . 



H H H 
Isobutyl 
radical 



H C H 
H H 

H-C-C-C-H 

i i 

fer/-Butyl 
radical 



H 
H-C-H 

" I 3 

H-C-CC Ci 



Isobutyl 
chloride 




/*r/-Butyl 
chloride 



formation of isobutyl chloride by the ratio of 9: 1. The experimental results given 
on page 96 show that the ratio is roughly 2:1, or 9:4.5. Evidently, about 4.5 
times as many collisions with the tertiary hydrogen are successful as collisions wi' 



100 ALKANES CHAP. 3 

the primary hydrogens. This, in turn, probably means that /T act is less for abstrac- 
tion of a tertiary hydrogen than for abstraction of a primary hydrogen, and, in 
fact, even less than for abstraction of a secondary hydrogen. 

Study of the chlorination of a great many alkanes has shown that these are 
typical results. After allowance is made for differences in the probability factor, 
the rate of abstraction of hydrogen atoms is always found to follow the sequence 
3 > 2 > 1. At room temperature, for example, the relative rates per hvdrogen 
atom are 5.0:3.8:1.0. Using these values we can predict quite well the ratio of 
isomeric chlorination products from a given alkane. For example: 

CH 3 CH 2 CH 2 CH3 -> CH,CH,CH,CH 2 C1 and CH.CH.CHCKHi 

light, 25 ~ ' 

n-Butane //-Butyl chloride Arr-Butyl chloride 

K-butyl chloride _ no. of 1 J H reactivity of 1' H 
"ec-butyl chloride no. of 2 H reactivity of 2" H 

6 1.0 

4 X 18 

6 . , 1*% 

= YC~J equivalent to j^~ 

In spite of these differences in reactivity, chlorination rarely yields a great 
preponderance of any single isomer. In nearly every alkane, as in the example \vc 
have studied, the less reactive hydrogens are the more numerous; their lower 
reactivity is compensated for by a higher probability factor, with the result that 
appreciable amounts of every isomer are obtained. 

Problem 3.14 Predict the proportions of isomcnc products from chlorination 
at room temperature of: (a) propane; (b) isobutanc; (c) 2,3-dimcthylbutanc, 
(d) w-pentane (Note- There are three isomeric products); (c) isopentane; (f ) 2,2.3-m- 
methylbutane; (g) 2,2,4-tnmethvlpentane. For (a) and (b) check your calculations 
igainst the experimental values given on pai^cs ^ and %. 

The same sequence of reactivity, 1 > 2^ > 1', is found in bromination, but 
with enormously larger reactivity ratios. At 127 , foi example, the relative rates 
per hydrogen atom are 1600:82:1. Here, differences m reactivity are so marked 
as vastly to outweigh probability factors. 

Problem 3.15 Answer Problem 3.14 lor brommation at 127. 



3.22 Relative reactivities of alkanes toward halogenation 

The best way to measure the relative reactivities of different compounds toward 
the same reagent is by the method of competition, since this permits an exact 
quantitative comparison under identical reaction conditions. Equimolar amounts 
of two compounds to be compared are mixed together and allowed to react with a 
limited amount of a particular reagent. Since there is not enough reagent for 
both compounds, the two compete with each other. Analysis of the reaction 
products shows which compound has consumed more of the reagent and hence is 
more reactive. 



SEC. 3.23 EASE OF ABSTRACTION OF HYDROGEN ATOMS 101 

For example, if equimolar amounts of methane and ethane are allowed to 
react with a small amount of chlorine, about 400 times as much ethyl chloride as 
methyl chloride is obtained, showing that ethane is 400 times as reactive as 
methane. When allowance is made for the relative numbers of hydrogens in the 
two kinds of molecules, we see that each hydrogen of ethane is^AbonlfyBBF tifijes 
as reactive as each hydrogen of methane. 

CH 3 C1 --"- C1 2 CiHj ^ C 2 H 5 CI 

1 l.ght, 25 4(X) 

Problem 3.16 Because of the rathci large difference in reactivity belft^tfcjhane 
and methane, competition experiments have actually used mixtures containing more 
methane than ethane. If the molar ratio of methane to ethane vsere JO: 1, what ratio 
of ethyl . chloride to methyl chloride would you expect to obtain? What practical 
advantage would this experiment have over one ; nvolving a 1 : 1 ratio? 

Data obtained from similar studies of other compounds are consistent with 
this simple generalization : the reactirity of a hydrogen depends chiefly upon its 
class, and not upon the alkane to which it is attached. Each primary hydrogen of 
propane, for example, is about as easily abstracted as each primary hydrogen in 
H-butane or isobutane ; each secondary hydrogen of propane, about as easily as 
each secondary hydrogen of w-butane or w-pentane; and so on. 

The hydrogen atoms of methane, which fall into a special class, are even less 
reactive than primary hydrogens, as shown by the above competition with ethane. 

Problem 3.17 On chlonnation, an equimolar mixture of ethane and neo- 
pentane yields neopentyl chloride and ethyl chloride in the ratio of 2.3:1. How does 
the reactivity of a primary hydrogen in neopentane compare with that of a primary 
hydrogen in ethane ? 



3.23 Ease of abstraction of hydrogen atoms. Energy of activation 

At this stage we can summarize the effect of structure on halogenation of 
alkanes in the following way. The controlling step in halogenation is abstraction 
of hydrogen by a halogen atom : 

R H + X- > H-X + R- 

The relative ease with which the different classes of hydrogen atoms are abstracted 
is: 

Ease of abstraction 10 ^ *>* ^ , ^ rv , 

-. . . 3 > 2 > 1 > i~rl4 

of hydrogen atoms 

This sequence applies (a) to the various hydrogens within a single alkane and hence 
governs orientation of reaction, and (b) to the hydrogens of different alkanes and 
hence governs relative reactivities, 

Earlier, we concluded that these differences in ease of abstraction like most 
differences in rate between closely related reactions (Sec. 2.19) are probably due 
to differences in act . By study of halogenation at a series of temperatures (Sec. 
2.18), the values of ftct listed in Table 3.5 were ireasured. In agreement with our 
tentative conclusions, the increasing rate of reaction along the series, methyl, 1, 



102 ALKANES CHAP. 3 

Table 3.5 ENERGIES OF ACTIVATION, KCAL/MOLE 

R-H + X- * R- -f H-X 
R X = Cl X = Bi 



CH 3 


4 


18 


1 


1 


13 


2 


0.5 


10 


3 


0.1 


7.5 



2, 3, is paralleled by a decreasing act . In chlorination the differences in act , 
like the differences in rate, are small; in bromination both differences are large. 

We have seen (Sec. 2.18) that the larger the E* ct of a reaction, the larger the increase 
in rate brought about by a given rise in temperature. We have just found that the dif- 
ferences in rate of abstraction among primary, secondary, and tertiary hydrogens are 
due to differences in aet . We predict, therefore, that a rise in temperature should speed 
up abstraction of primary hydrogens (with the largest act ) most, and abstraction of 
tertiary hydrogens (with the smallest act ) least; the three classes of hydrogen should then 
display more nearly the same reactivity. 

This leveling-out effect has indeed been observed: as the temperature is raised, the 
relative rates per hydrogen atom change from 5.0:3.8:1.0 toward 1:1:1. At very high 
temperatures virtually every collision has enough energy for abstraction of even primary 
hydrogens. It is generally true that as the temperature is raised a given reagent becomes 
less selective in the position of its attack ; conversely, as the temperature is lowered it 
becomes more selective. 

How can we account for the effect of structure on ease of abstraction of 
hydrogen atoms ? Since this is a matter of act , we must look for our answer, as 
always, in the transition state. To do this, however, we must first shift our focus 
from the hydrogen atom being abstracted to the radical being formed. 



3.24 Stability of free radicals 

In Table 1.2 (p. 21) we find the dissociation energies of the bonds that hold 
hydrogen atoms to a number of groups. These values are the A//'s of the following 
reactions: 

CH 3 -H > CH 3 - + H- A//-104kcal 

CH 3 CH 2 -H > CH 3 CH 2 - + H- A// = 98 

A 1 radical 

CH 3 CH 2 CH 2 -H -> CH 3 CH 2 CH 2 - -I- H- A// . 98 
A 1 radical 

CH 3 CHCH 3 > CH 3 CHCH 3 + H- A// = 95 

H A 2 radical 

CH 3 CH 3 

CH 3 CCH 3 > CH 3 CCH 3 + H- A// 91 

H A3 radical 

By definition, bond dissociation energy is the amount of energy that must be 



SEC. 3.25 



EASE OF FORMATION OF FREE RADICALS 



103 



supplied to convert a mole of alkane into radicals and hydrogen atoms. As we 
can see, the amount of energy needed to form the various classes of radicals 
decreases in the order: CH 3 - > 1 > 2 > 3. 



R-H 



R- + H- A// = bond dissociation energy 



If less energy is needed to form one radical than another, it can only mean 
that, relative to the alkane from which it is formed, the one radical contains less 
energy than the other, that is to say, is more stable (see Fig. 3.6). 



CH 3 - (104) 



RH 



R- +H- 



o- (98) 



CH 3 CHCH 3 (95) 



CH 3 
CHiC-CHj (91) 



CH 4 



CH, 

CHr-C-CHa 
H 



Figure 3.6. Relative stabilities of free radicals. (Plots aligned with each 
other for easy comparison.) 

We are not attempting to compare the absolute energy contents of, say, 
methyl and ethyl radicals; 'we are simply saying that the difference in energy 
between methane and methyl radicals is greater than the difference between ethane 
and ethyl radicals. When we compare stabilities of free radicals, it must be under- 
stood that our standard for each radical is the alkane from which it is formed. As 
we shall see, this is precisely the kind of stability that we are interested in. 

Relative to the alkane from which each is formed, then, the order of stability 
of free radicals is: 

Stability of 
free radicals 



3 > 2 > 1 > CH r 



3.25 Ease of formation of free radicals 

Let us return to the halogenation of alkatfes. Orientation and reactivity, 
we have seen (Sec. 3.23), are governed by the relative ease with which the different 
classes of hydrogen, atoms are abstracted. Jut by definition, the hydrogen being 
abstracted and the radical being formed belong to the same class. Abstraction of 
a primary hydrogen yields a primary radical, abstraction of a secondary hydrogen 
yields a secondary radical, and so on. For example: 



CH 3 CH 2 CH 2 ~H + Br- 

A 1 hydrogen 



+ CH 3 CH 2 CH 2 
A 1 radical 



104 ALKANES CHAP. 3 

CH 3 CHCH 3 + Br- > H~Br + CH 3 CHCH 3 

H A 2 radical 

A 2 hydrogen 

CH 3 CH 3 

CH 3 CCH 3 -I- Br > H-Br + CH 3 CCH 3 

H A3 radical 

A 3 hydrogen 

If the ease of abstraction of hydrogen atoms follows the sequence 3 > 2 > 
1 > CH 4 , then the ease of formation of free radicals must follow the same 
sequence: 

Ease of formation 10 00 . |0 . ri * 

ff . j>z.>j.> \^ri-i* 

of free radicals 

In listing free radicals in order of their ease of formation, we find that we have 
at the same time listed them in order of their stability. The more stable the free 
radical, the more easily it is formed. 

This is an extremely useful generalization. Radical stability seems to govern 
orientation and reactivity in many reactions where radicals are formed. The addition 
of bromine atoms to alkenes (Sec. 6. 1 7), for example, is a quite different sort of 
reaction from the one we have just studied; yet, there too, orientation and reac- 
tivity are governed by radical stability. (Even in those cases where other factors 
steric hindrance, polar effects are significant or even dominant, it is convenient 
to use radical stability as a point of departure.) 

3.26 Transition state for halogenation 

Is it reasonable that the more stable radical should be formed more easily? 

We have already seen that the differences in reactivity toward halogen atoms 
are due chiefly to differences in E^ : the more stable the radical, then, the lower 
the act for its formation. This, in turn, means that the more stable the radical, 
the more stable the transition state leading to its formation both stabilities being 
measured, as they must be, against the same standard, the reactants. (Remember: 
ftct is the difference in energy content between reactants and transition state.) 

Examination of the transition state shows that this is exactly what we would 
expect As we saw before (Sec. 2.22), the hydrogen-halogen bond is partly formed 
and the carbon-hydrogen bond is partly broken. To the extent that the bond is 



C H+ -X 



H---H---X > C- + H X 
J I 



Reactants Transition state Products 

Halogen has Carbon acquiring Carbon has 

odd electron free-radical character odd electron 

broken, the alkyl group possesses character ot the free radical it will become. 
Factors that tend to stabilize the free radical tend to stabilize the incipient free 
radical in the transition state. 

We have seen that the stabilities of free radicals follow the sequence 3 > 2 > 
1 > CH 3 -. A certain factor (delocalization of the odd electron, Sec. 6.28) causes 



SEC. 3.27 



ORIENTATION AND REACTIVITY 



105 



the energy difference between isobutane and the tert-buty\ radical, for example, 
to be smaller than between propane and the isopropyl radical. It is not unreason- 
able that this same factor should cause the energy difference between isobutane 
and the incipient tert-buiyl radical in the transition state to be smaller than between 
propane and the incipient isopropyl radical in its transition state (Fig. 3.7). 



CH 3 




Stabilization of 
transition state 



CH, 

CH.,- C H Br 
I 
CH, 



Stabilization of 
L^ radical 




Progress of reaction > 

Figure 3.7. Molecular structure and rate of reaction. Stability of tran- 
sition state parallels stability of radical: more stable radical formed faster. 
(Plots aligned with each other for easy comparison.) 



3.27 Orientation and reactivity 

Throughout our study of organic chemistry, we shall approach the problems 
of orientation and reactivity in the following way. 

Both problems involve comparing the rates of closely related reactions: in the 
case of orientation, reactions at different sites in the same compound; in the case 
of reactivity, reactions with different compounds. For such closely related reac- 
tions, variations in rate are due mostly to differences in act ; by definition, E ACt 
is the difference in energy content between reactants and transition state. 

We shall examine the most likely structure for the transition state, then, to 
see what structural features affect its stability without at the same time affecting 
by an equal amount the stability of the reactants; that is, we shall look for factors 
that tend to increase or decrease the energy difference between reactants and 
transition state. Having decided what structural features affect the act , we shall 
compare the transition states for the reactions whose rates we wish to compare: 
the more stable the transition state, the faster the reaction. 

In many, if not most, reactions where a free radical is formed, as in the present 
case, tlie transition state differs from the reactants chiefly in being like the product. 
It is reasonable, ihen, that the factor most affecting the act should be the radical 
character of the transition state. Hence we find that the more stable the radical 



106 ALKANES CHAP. 3 

the more stable the transition state leading to its formation, and the faster the 
radical is formed. 



3.28 Reactivity and selectivity 

In its attack on alkanes, the bromine atom is much more selective than the 
chlorine atom (with relative rate factors of 1600:82: 1 as compared with 5.0:3.8: 1). 
It is also much less reactive than the chlorine atom (only 1/250,000 as reactive 
toward methane, for example, as we sa\s in Sec. 2.19). This is just one example of 
a general relationship: in a set of similar reactions, the less reactive the reagent, the 
more selective it is in its attack. 

To account for this relationship, we must recall what we learned in Sec. 2.23. 
In the attack by the comparatively unreactive bromine atom, the transition state 
is reached late in the reaction process, after the alkyl group has developed con- 
siderable radical character. In the attack by the highly reactive chlorine atom, the 
transition state is reached early, when the alkyl group has gained very little radical 
character. 

Bromination 



. 5. 1 

R -H-BrJ 



R _H + fir- > LR -H-BrJ > R- 4- H-Br 

Low reactivity; Transition state 

high selectivity 

Reached late: 

much radical 
character 

Chlorination 



p. 8- 1 

LR--H cij 



R-H + a- > IR-H- cij > R- + H-CI 

High reactivity; Transition state 

tow selectivity 

Reached early: 

little radical 
character 

Now, by "selectivity" we mean here the differences in rate at which the various 
classes of free radicals are formed; a more stable free radical is formed faster, we 
said, because the factor that stabilizes it delocalization of the odd electron (Sec. 
6.28) also stabilizes the incipient radical in the transition state. If this is so, 
then the more fully developed the radical character in the transition state, the more 
effective delocalization will be in stabilizing the transition state. The isopropyl 
radical, for example, is 3 kcal more stable than the w-propyl radical; if the radicals 
were completely formed in the transition state, the difference in act would be 3 
kcal. Actually, in bromination the difference in act is 3 kcal: equal, within the 
limits of experimental error, to the maximum potential stabilization, indicating, 
as we expected, a great deal of radical character. In chlorination, by contrast, 
the difference in act is only 0.5 kcal, indicating only very slight radical character. 

A similar situation exists for reactions of other kinds. Whatever the factor 
responsible for differences in stability among a set of transition states whether it is 
delocalization of an odd electron, or accommodation of a positive or negative 



SEC. 3.29 NON-REARRANGEMENT OF FREE RADICALS 107 

charge, or perhaps a change in crowding of the atoms the factor will operate more 
effectively when the transition state is more fully developed, that is, when the 
reagent is less reactive. 

3.29 Non-rearrangement of free radicals. Isotopic tracers 

Our interpretation of orientation (Sec. 3.21) was based on an assumption that 
we have not yet justified: that the relative amounts of isomeric halides we find in 
the product reflect the relative rates at which various free radicals were formed 
from the alkane. From isobutane, for example, we obtain twice as much isobutyl 
chloride as te/7-butyl chloride, and we assume from this that, by abstraction of 
hydrogen, isobutyl radicals are formed twice as fast as /erf-butyl radicals. 

Yet how do we know, in this case, that every isobutyl radical that is formed 
ultimately yields a molecule of isobutyl chloride? Suppose some isobutyl radicals 
were to change- by rearrangement of atoms into /<?r/-butyl radicals, which then 
react with chlorine to yield ter/-butyl chloride. This supposition is not so far- 

CH 3 CH 3 CH 3 

f\ ' rearrangement 

CH, C-CH, - - CHi C-CH,. CH V C-CH, 

I Docs nut w//Y>f/j 

U |j[ /<?rf-Butyl radical 

Isobutane Isohutyl radical | Ch 

CH 3 
CH 3 - -C-CH 3 

Cl 

rf/7-But>l chloride 

fetched as we, in our present innocence, might think; the doubt Jt raises is a very 
real one. We shall shortly see that another kind of reactive intermediate particle, 
the carbonium ion, is \ery prone to rearrange, with less stable ions readil> changing 
into more stable ones (See. 5.22). 

H. C. Brown (of Purdue University) and Glen Russell (no\\ of loua State 
University) decided to test the possibility that free radicals, like carbonium ions, 
might rearrange, and chose the chlorination of isobutane as a good test case, 
because of the large difference in stability between w/-butyl and isobutyl radicals. 
If rearrangement of alkyl radicals can indeed lake place, it should certainly happen 
here. 

What the problem comes down to is this: does every abstraction of primary 
hydrogen lead to isobuiyl chloride, and every abstraction of tertiary hydrogen lead 
to tert-butyl chloride? This, we might say, we could never know, because all 
hydrogen atoms are exactly alike. But are they? Actually, three isotopes of hy- 
drogen exist: 'H, prof htm, ordinary hydrogen: 2 H or D, deuterium* heavy hydro- 
gen; and 3 H or T, tritium. Protium and deuterium are distributed in nature in 
the ratio of 5000: 1. '(Tritium, the unstable, radioactive isotope, is present in 
traces, but can be made by neutron bombardment of 6 Li.) Modern methods of 
separation of isotopes have made very pure deuterium available, at moderate 
prices, in the form of deuterium oxide. D 2 O, heavy water. 



108 ALKANES CHAP. 3 

Brown and Russell prepared the deuterium-labeled isobutane I, 



DC1 i CH 3 C- CH 3 --> CH 3 -C-~CH 3 




CH 3 
CH 3 -C-CH 3 

D n 

HC1 f CH r C CH 2 ~^ CH 3 C~CH 2 CJ 

I "I 

D , D 

photochemically chlorinated it, and analyzed the products. The DClrHCl ratio 
(determined by the mass spectrometer) was found to be equal (within experimental 
error) to the tert-buiyl chloride: isobutyl chloride ratio. Clearly, every abstraction 
of a tertiary hydrogen (deuterium) gave a molecule of tert~buty\ chloride, and every 
abstraction of a primary hydrogen (protiwn) gave a molecule of isobutyl chloride. 
Rearrangement of the intermediate free radicals did not occur. 

All the existing evidence indicates quite strongly that, although rearrangement 
of free radicals occasionally happens, it is not very common and does not involve 
simple alkyl radicals. 

Problem 3.18 (a) What results would have been obtained if some isobutyl 
radicals had rearranged to tert-buiyl radicals? (b) Suppose that, instead of rearrang- 
ing, isobutyl radicals were, in effect, converted into terr-butyl radicals by the reac- 
tion 

CH 3 CH 3 CH 3 CH 3 

CH 3 -CH~-CH 7 4- CHr- C-CH 3 > CH 3 -CH-CH 3 + CH 3 -C-CH 3 

H 

What results would Brown and Russell have obtained? 

Problem 3.19 Keeping in mind the availability of D2O, suggest a way to make I 
from ter/~butyl chloride. (Hint: See Sec, 3.16.) 

The work of Brown and Russell is just one example of the way in which we can 
gain insight into a chemical reaction by using isotopically labeled compounds. We 
shall encounter many other examples in which isotopes, used either as tracers, as 
in this case, or for the detection of isotope effects (Sec. 11.15), give us information 
about reaction mechanisms that we could not get in any other way. 

Besides deuterium and tritium, isotopes commonly used in organic chemistry 
include: 14 C, available a<j 14 CH 3 OH and Ba 14 CO 3 ; 18 O, as H 2 8 O; 15 N, as 15 NH 3 , 
15 N(V, and 15 NO 2 ~ ; 36 C1, as chlorine or chloride; 131 I, as iodide. 

Problem 3.20 Bromination of methane is slowed down by the addition of 
HBr (Problem 14, p. 71); this is attributed to the reaction 

CH 3 - + HBr > CH 4 + Br- 

which, as the reverse of one of the chain-carrying steps, slows down bromination. 
How might you test whether or not this reaction actually occurs in the bromination 
mixture? 



SFC. 3.30 COMBl'SIION 109 

Problem 3.21 In Sec. 2.12 the reaction 

Cl- + CI 2 > C1 2 + C|. 

was listed as probable but unproductive. Given ordinary chlorine (made up of 35 C1 
and 37 C1) and 36 C1 2 , and a mass spectrometer, how would you go about rinding out 
whether or not the reaction actually occurs? 

3.30 Combustion 

The reaction of alkanes with oxygen to form carbon dioxide, water, and 
most important of all heat* is the chief reaction occurring in the internal com- 
bustion engine; its tremendous practical importance is obvious. 

The mechanism of this reaction is extremely complicated and is not yet fully 
understood. 'I here seems to be no doubt, however, that it is a free-radical chain 
reaction. 'I he reaction is extremely exothermic and yet requires a very high tem- 
perature, that of a flame, for its initiation. As in the case of chlorination, a great 
deal of energy is required for the bond-breaking that generates the initial reactive - 
particles; once this energy barrier is surmounted, the subsequent chain-carrying 
steps proceed readily and with the evolution of energy. 

A higher compression ratio has made the modern gasoline engine more 
efficient than earlier ones, but has at the same time created a new problem. Under 
certain conditions the smooth explosion of the fuel-air mixture in the cylinder is 
replaced by knocking, which greatly reduces the power of the engine. 

The problem of knocking has been successfully met in two general ways: 
(a) proper selection of the hydrocarbons to be used as fuel, and (b) addition of 
tetraethyllead. 

Experiments with pure compounds have shown that hydrocarbons of differing 
structures differ widely in knocking tendency. The relative antiknock tendency of 
a fuel is generally indicated by its octane number. An arbitrary scale has been set 
up, with //-heptane, which knocks ver> badly, being gi\en an octane number of 
/ero, and 2,2,4-tnmelhylpentane (*"iso-octane") being gi\en the octane number 
of 100. There are available today fuels with better antiknock qualities than "iso- 
octane." 

The gasoline fraction obtained by direct distillation of petroleum (straight-run 
gasoline) is improved by addition of compounds of higher octane number; it is 
sometimes entirely replaced by these better fuels. Branched-chain alkanes and 
alkenes, and aromatic hydrocarbons generally have excellent antiknock qualities; 
these are produced from petroleum hydrocarbons by catalytic cracking (Sec. 3.31) 
and catalytic reforming (Sec. 12.4). Highly branched alkanes are synthesized from 
alkenes and alkanes by alkylatton (Sec. 6.16). 

In 1922 T. C. Midgley, Jr., and T. A. Boyd (of the General Motors Research 
Laboratory) found that the octane number of a fuel is greatly improved by addition 
of a small amount of tetraethyllead, (C 2 H 5 )4Pb. Gasoline so treated is called 
ethyl gasoline or leaded gasoline. Nearly 50 years of research has finally shown 
that tetraethyllead probably works by producing tiny particles of lead oxides, on 
whose surface certain reaction chains are broken. 

In addition to carbon dioxide and water, however, the gasoline engine dis- 
charges other substances into the atmosphere, substances that are either smog- 



110 ALKANES CHAP. 3 

producing or downright poisonous: unburned hydrocarbons, carbon monoxide, 
nitrogen oxides, and, from leaded gasoline, various compounds of lead in the 
United States, hundreds of tons of lead a day. Growing public concern about these 
pollutants has caused a minor revolution in the petroleum and auto industries. 
Converters are being developed to clean up exhaust emissions: by catalytic oxida- 
tion of hydrocarbons and carbon monoxide, and by the breaking down of nitrogen 
oxides into nitrogen and oxygen. But most of these oxidation catalysts contain 
platinum, which is poisoned by lead; there has been a move to get the lead out of 
gasoline not, initially, to cut down on lead pollution, but to permit converters 
to function. This has, in turn, brought back the problem of knocking, which is 
being met in two ways: (a) by lowering the compression ratio of tfie new auto- 
mobiles being built: and (b) by increasing the octane number of gasoline through 
changes in hydrocarbon composition through addition of aromatics and through 
increased use of isomerization (Sec. 3.13). 

3.31 Pyrolysis: cracking 

Decomposition of a compound by the action of heat alone is known as 
pyrolysis. This word is taken from the Greek pyr, fire, and lysis, a loosing, and 
hence to chemists means "cleavage by heat"; compare hydro-lysis, "cleavage by 
water." 

The pyrolysis of alkanes, particularly when petroleum is concerned, is known 
as cracking. In thermal cracking alkanes are simply passed through a chamber 
heated to a high temperature. Large alkanes are converted into smaller alkanes, 
alkenes, and some hydrogen. This process yields predominantly ethylene (C 2 H 4 ) 
together with other small molecules. In a modification called steam cracking, 
the hydrocarbon is diluted with steam, heated for a fraction of a second to 700- 
900, and rapidly cooled. Steam cracking is of growing importance in the pro- 
duction of hydrocarbons as chemicals, including ethylene, propylene, butadiene, 
isoprene, and cyclopentadiene. Another source of smaller hydrocarbons is hydro- 
cracking, carried out in the presence of hydrogen at high pressure and at much 
lower temperatures (250-450). 

The lov\ -molecular-weight alkenes obtained from these cracking processes can 
be separated and purified, and are the most important raw materials for the large- 
scale synthesis of aliphatic compounds. 

Most cracking, however, is directed toward the production of fuels, not 
chemicals, and for this catalytic cracking is the major process. Higher boiling 
petroleum fractions (typically, gas oil), are brought into contact with a finely 
divided silica-alumina catalyst at 450-550 and under slight pressure. Catalytic 
cracking not only increases the yield of gasoline by breaking large molecules into 
smaller ones, but also improves the quality of the gasoline: this process involves 
carbonwm ions (Sec. 5.15), and yields alkanes and alkenes with the highly branched 
structures desirable in gasoline. 

Through the process of alkylation (Sec. 6.16) some of the smaller alkanes and 
alkenes are converted into high-octane synthetic fuels. 

Finally, by the process of catalytic reforming (Sec. 12.4) enormous quantities 
of the aliphatic hydrocarbons of petroleum are converted into aromatic hydro 



SIC. 3.32 DETERMINATION OF STRUCTURE 111 

carbons which arc used not only as superior fuels but as the starting materials in 
the synthesis of most aromatic compounds (Chap. 10). 



3.32 Determination of structure 

One of the commonest and most important jobs in organic chemistry is to 
determine the structural formula of a compound just synthesized or isolated from 
a natural source. 

The compound will fall into one of two groups, although at first we probably 
shall not know which group. It will be either, (a) a previously reported compound, 
which we must identify, or (b) a new compound, whose structure we must prove. 

If the compound has previously been encountered by some other chemist 
who determined its structure, then a description of its properties will be found 
somewhere in the chemical literature, together with the evidence on which its 
structure was assigned. In that case, we need only to show that our compound 
is identical with the one previously described. 

If. on the other hand, our compound is a new one that has never before been 
reported, then we must carry out a much more elaborate proof of structure. 

Let us see in a general way now, and in more detail later just how we 
would go about this job. We are confronted by a flask filled with gas, or a few 
millihters of liquid, or a tiny heap of crystals. We must find the answer to the 
question: what is if? 

First, we purify the compound and determine its physical properties: melting 
point, boiling point, density, refractive index, and solubility in various solvents. 
In the laboratory today, we would measure various spectra of the compound 
(Chap. 13), in particular the infrared spectrum and the nmr spectrum; indeed, 
because of the wealth of information to be gotten in this way, spectroscopic examina- 
tion might well be the first order of business after purification. From the mass 
spectrum we would get a very accurate molecular weight. 

We would carry out a qualitative elemental analysis to see what elements are 
present (Sec. 2.25). We might follow this with a quantitative analysis, and from this 
and the molecular weight we could calculate a molecular formula (Sec. 2.26); we 
would certainly do this if the compound is suspected of being a new one. 

Next, we study systematically the behavior of the compound toward certain 
reagents. This behavior, taken with the elemental analysis, solubility properties, 
and spectra, generally permits us to characterize the compound, that is, to decide 
what family the unknown belongs to. We might find, for example, that the com- 
pound is an alkane, or that it is an alkene, or an aldehyde, or an ester. 

Now the question is: which alkane is it? Or which alkene, or which aldehyde, 
or which ester? To find the answer, we first go to the chemical literature and look 
up compounds of the particular family to which our unknown belongs. 

If we find one described whose physical properties are identical with those of 
our unknown, then the chances are good that the two compounds are identical. 
For confirmation, we generally convert the unknown by a chemical reaction into 
a new compound called a derivative, and show that this derivative is identical with 
the product derived in the same way from the previously reported compound. 

If, on the other hand, we do not find a compound described whose physical 



112 ALKANES CHAP. 3 

properties are identical with those of our unknown, then we have a difficult job 
on our hands: we have a new compound, and must prove its structure. We may 
carry out a degradation : break the molecule apart, identify the fragments, and deduce 
what the structure must have been. To clinch any proof of structure, we attempt 
to synthesize the unknown by a method that leaves no doubt about its structure. 

Problem 3.22 The final step in the proof of structure of an unknown alkane 
\\as its synthesis by the coupling of lithium di(ft>r/-butyl)copper with w-butyl bro- 
mide. What \\as the alkane? 

In Chap. 13, after we have become familiar with more features of organic 
structure, we shall see how spectroscopy fits into the general procedure 1 outlined 
above. 



3.33 Analysis of alkanes 

An unknown compound is characterized as an alkane on the basis of negative 
evidence. 

Upon qualitative elemental analysis, an alkane gives negative tests for all 
elements except carbon and hydrogen. A quantitative combustion, if one is carried 
out, shows the absence of oxygen; taken with a molecular weight determination, 
the combustion gives the molecular formula, -C n H 2n + 2 which is that of an alkane. 

An alkane is insoluble not only in water but also in dilute acid and base and in 
concentrated sulfuric acid. (As we shall see, most kinds of organic compounds 
dissolve in one or more of these solvents.) 

An alkane is unreactive toward most chemical reagents. Its infrared spectrum 
lacks the absorption bands characteristic of groups of atoms present in other families 
of organic compounds (like OH, C~O, ,C C, etc.). 

Once the unknown has been characterized as an alkane, there remains the 
second half of the problem: finding out which alkane. 

On the basis of its physical properties boiling point, melting point, density, 
refractive index, and, most reliable of all, its infrared and mass spectra it may be 
identified as a previously studied alkane of known structure. 

If it turns out to be a new alkane, the proof of structure can be a difficult 
job. Combustion and molecular weight determination give its molecular formula. 
Clues about the arrangement of atoms are given by its infrared and nmr spectra. 
(For compounds like alkanes, it may be necessary to lean heavily on x-ray diffrac- 
tion and mass spectrometry.) 

Final proof lies in synthesis of the unknown by a method that can lead only 
to the particular structure assigned. 

(The spectroscopic analysis of alkanes will be discussed in Sees. 13.15-13.16.) 



PROBLEMS 

1. Give the structural formula of: 

(a) 2,2,3,3-tetramethylpentane (e) 2,4-dimethyl-4-ethylheptane 

(b) 2,3-dimethyIbutane (f) 2,5-dimethylhexane 

(c) 3,4,4,5-tetramethylheptane (g) 2-methyI-3-ethylpentane 

(d) 3,4-dimethyI-4-ethylheptane (h) 2,2,4-trimethylpentane 



PROBLEMS 113 

2. Draw out the structural formula and give the IUPAC name of: 

(a) (CH 3 ) 2 CHCH 2 CH 2 CH 3 (0 (CH 3 )3CCH 2 C(CH 3 )3 

(b) CH 3 CH 2 C(CH 3 ) 2 CH 2 CH 3 (g) (CH 3 ) 2 CHCH 2 CH 2 CH(C 2 H 5 ) 2 

(c) (C 2 H 5 ) 2 C(CH 3 )CH 2 CH 3 (h) (CH 3 ) 2 CHCH(CH 3 )CH 2 C(C 2 H 5 ) 2 CH 3 

(d) CH 3 CH 2 CH(C H 3 )CH(CH 3 )CH(CH 3 ) 2 (i) (CH 3 ) 2 CHC(C 2 H 5 ) 2 CH 2 CH 2 CH 3 

CH 3 CH 3 

(e) CH 3 CH 2 CHCH 2 CHCH 2 CH 3 (j) CH 3 CH 2 CHCH 2 CHCHCH 3 

CH 3 CH 2 CH 2 CH 3 CH 2 CH 2 CH 3 

3. Pick out a compound in Problem 1 or 2 that has: (a) no tertiary hydrogen; 
(b) one tertiary hydrogen; (c) two tertiary hydrogens; (d) no secondary hydrogen; 

(e) two secondary hydrogens; (f) half the number of secondary hydrogens as primary 
hydrogens. 

4. Pick out a compound (if any) in Problem 1 or 2 that contains: 

(a) one isopropyl group (g) one /*r/-butyl group 

(b) two isopropyl groups (h) two tert-buiyl groups 

(c) one isobutyl group (i) an isopropyl group and a j^r-butyl group 

(d) two isobutyl groups (j) a /m-butyl group and an isobutyl group 

(c) one sec-butyl group (k) a methyl, an ethyl, a //-propyl, and a jer-butyl 

(f ) two .w-butyl groups group 

5. What alkane^ or alkancs of molecular weight 86 have: (a) two monobromo 
derivatives? (b) three? (c) four? (d) five? (e) How many dibromo derivatives does the 
alkane in (a) have? 

6. How many mono-, di-, and trichloro derivatives are possible for cyclopentane ? 
(Structure given in Sec. 9.5.) 

7. Without referring to tables, list the following hydrocarbons in order of decreasing 
boiling points (i e., highest boiling at top, lowest at bottom): 

(a) 3.3-dimcth>lpcntane (c) 2-methylheptane (e) 2-methylhexane 

(b) //-heptane (d) //-pentane 

8. Wnte balanced equations, naming all organic products, for the following reac- 
tions: 

(a) isobutyl bromide f Mg/ether (d) product of (b) -f H 2 O 

(b) rm-butjl bromide + Mg/ether (e) product of (a) + D ; O 

(c) product of (a) -i H : O (f) .wc-butyl chloride + Li, then Cul 

(g) product of (f ) -f ethyl bromide 

9. Write equations for the preparation of //-butane from: 

(a) //-butyl bromide (d) l-butenc, CH,CH 2 CH--CH 2 

(b) .tec-butyl bromide (e) 2-butene, CH 3 CH- CHCH 3 

(c) ethyl chloride 

10. Draw structures of all products expected from monochlorination at room tem- 
perature of: 

(a) if-hexane (c) 2,2,4-trimethylpentane 

(b) isohexane (d) 2,2-dimethyl butane 

11. Predict the proportions of products in the previous problem. 

12. (a) Reaction of an aldehyde with a Grignard reagent is an important way of 
making alcohols. Why must one scrupulously dry the aldehyde before adding it to the 
Grignard reagent? (b) Why would one not prepare a Grignard reagent from 
BrCH 2 CH 2 OH? 



114 ALKANES CHAP. 3 

13. On the basis of bond strengths in Table 1.2, page 21, add the following free 
radicals to the stability sequence of Sec. 3.24: 

(a) vinyl, HiC-^CH- 

(b) ally], H 2 C=-CHCH 2 - 

(c) benzyl, C 6 H 5 CH 2 - 
Check your answer on page 211. 

14. On the basis of your answer to Problem 1 3, predict how the following would 
fit into the sequence (Sec. 3.23) that shows ease of abstraction of hydrogen atoms: 

(a) vinylic hydrogen, H 2 O- CH--H 

(b) allylic hydrogen, H 2 C- CHCH 2 -H 

(c) benzylic hydrogen, C 6 H 5 CH 2 H 
Check your answer against the facts on page 210. 

15. Free-radical chlorination of either w-propyl or isopropyl bromide gives 1-bromo- 
2-chloropropane, and of either isobutyl or ter/-butyl bromide gives l-bromo-2-chloro- 
2-methylpropane. What appears to be happening? Is there any pattern to this behavior? 

16. (a) If a rocket were fueled with kerosene and liquid oxygen, what weight of 
oxygen would be required for every I'ter of kerosene? (Assume kerosene to have the 
average composition of /!-Ci 4 H 30 .) (b) How much heat would be evolved in the com- 
bustion of one liter of kerosene? (Assume 157 kcal/mote for each -CH 2 ~ group and 
186 kcal/rnole for each CH< group.) (c) If it were to become feasible to fuel a rocket 
with free hydrogen atoms, what \\eight of fuel would be required to provide the same 
heat as a liter of kerosene and the necessary oxygen? (Assume H 2 as the sole product.) 

17. By what two quantitative methods could you show that a product isolated from 
the chlorination of propane was a monochloro or a dichloro derivative of propane? Tell 
exactly what results you would expect from each of the methods. 

18. On the basis of certain evidence, including its infrared spectrum, an unknown 
compound of formula C^H^ is suspected of being 2,7-dimethyloctane. How could 
you confirm or disprove this tentatively assigned structure? 

19. (a) A solution containing an unknown amount of methyl alcohol (CH 3 OH) 
dissolved in /?-octane is added to an excess of methylmagncsium iodide dissolved in the 
high-boiling solvent, //-butyl ether. A gas is evolved, and is collected and its volume 
measured: 1.04 cc (corrected to STP). What is the gas, and how is it formed? What 
weight of methyl alcohol was added to the Gngnard reagent? 

(b) A sample of 4.12 mg of an unknown alcohol, ROH, is added to methylmagncsium 
iodide as above; there is evolved 1.56 cc of gas (corrected to STP). What is the molecular 
weight of the alcohol? Suggest a possible structure or structures for the alcohol. 

(c) A sample of 1.79 mg of a compound of mol. wt. about 90 gave 1.34 ml of the gas 
(corrected to STP). How many "active (that is, acidic) hydrogens" are there per mole- 
cule? Assuming all these to be in OH groups, suggest a structure for the alcohol. 
(This is an example of the Zerewitinoff active hydrogen determination.) 

20. (a) /er/-Butyl peroxide is a stable, easy-to-handle liquid that serves as a con- 
venient source of free radicals: 

<CH 3 ) 3 CO- OC(CH 3 ) 3 ~ > 2(CH 3 ) 3 CO- 

A mixture of isobutane and CCI 4 is quite stable at 130 140". If a small amount of 
/T/-butyl peroxide is added, a reaction occurs, that yields (chiefly) tert-buiyl chloride 
and chloroform. A small amount of tert-bulyl alcohol ((CH 3 ) 3 COH, equivalent to the 
peroxide used) is also isolated. Give all steps in a likely mechanism for this reaction. 

(b) When irradiated with ultraviolet light, or in the presence of a small amount of 
peroxides, /er/-butyl hypochlorite, (CH 3 ) 3 C-~O- Cl, reacts with alkanes to form, in 
equimolar amounts, aJkyl chlorides and tert-buiyl alcohol. Outline all steps in a likely 
mechanism for this reaction. 



Chapter 

4 



Stereochemistry I. 
Stereoisomers 



4.1 Stereochemistry and stereoisomerism 

The science of organic chemistry, we said, is based on the relationship between 
molecular structure and properties. That part of the science which deals with 
structure in three dimensions is called stereochemistry (Gr.: stereos, solid). 

One aspect of stereochemistry is stereoisomerism. Isomers, we recall, are 
different compounds that have the same molecular formula. The particular kind 
of isomers that are different from each other only in the way the atoms are oriented 
in space (but are like one another with respect to which atoms are joined to which 
other atoms) are called Stereoisomers. 

Pairs of Stereoisomers exist that differ so little in structure and hence in 
properties that of all the physical measurements we can make, only one, involving 
a special instrument and an unusual kind of light, can distinguish between them. 
Yet, despite this close similarity, the existence of such Stereoisomers provides us 
\\ith one of our most sensitive probes into mechanisms of chemical reactions; very 
often, one of these isomers is selected for study, not because it is different from 
ordinary compounds in its three-dimensional chemistry, but because it can be made 
to reveal \vhut ordinary compounds hide. And, again despite their close similarity, 
one isomer of such a pair may serve as a nourishing food, or as an antibiotic, or as 
a powerful heart stimulant, and the other isomer ma> be useless. 

In this chapter, we shall learn how to predict the existence of the kinds of 
Stereoisomers called enantiomers and diastereomers, how to represent and designate 
their structures, and, in a general way, how their properties will compare. Then, in 
following chapters, we shall begin to use what we learn in this one. In Sees. 5.5-5.6. 
we shall learn about the kind of Stereoisomers called geometric isomers. In Chapter 7, 
the emphasis will shift from what these Stereoisomers fire, to how they are formed, 
what they do, and what they can tell us. 

We have already (Sees. 3.3 and 3.5) begun 6ur study of the branch of stereo- 
chemistry called conf or national analysis \ we shall return to it, especially in Chap. 
9, and make use of it throughout the rest of the book. 



116 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 



4.2 Isomer number and tetrahedral carbon 

Let us begin our study of stereochemistry with methane and some of its 
simple substitution products. Any compound, however complicated, that contains 
carbon bonded to four other atoms can be considered to be a derivative of methane; 
and whatever we learn about the shape of the methane molecule can be applied to 
the shapes of vastly more complicated molecules. 

The evidence of electron diifraction, X-ray diffraction, and spectroscopy shows 
that when carbon is bonded to four other atoms its bonds are directed toward the 
corners of a tetrahedron. But as early as 1874, years before the direct determina- 
tion of molecular structure was possible, the tetrahedral carbon atom was proposed 
by J. H. van't Hoff, while he was still a student at the University of Utrecht. His 
proposal was based upon the evidence of isomer number. 

For any atom Y, only one substance of formula CH 3 Y has ever been found. 
Chlorination of methane yields only one compound of formula CH 3 C1; bromina- 
tions yields only one compound of formula CH 3 Br. Similarly, only one CH 3 F is 
known, and only one CH 3 I. Indeed, the same holds true if Y represents, not just 
an atom, but a group of atoms (unless the group is so complicated that in itself 
it brings about isomerism) ; there is only one CH 3 OH, only one CH 3 COOH, only 
one CH 3 SO 3 H. 

What does this suggest about the arrangement of atoms in methane? It 
suggests that every hydrogen atom in methane is equivalent to every other hydrogen 
atom, so that replacement of any one of them gives rise to the same product. If 
the hydrogen atoms of methane were not equivalent, then replacement of one would 
yield a different compound than replacement of another, and isomeric substitution 
products would be obtained. 

In what ways can the atoms of methane be arranged so that the four hydrogen 
atoms are equivalent? There are three such arrangements: (a) a planar arrange- 
ment (I) in which carbon is at the center of a rectangle (or square) and a hydrogen 



H'/ \^H 
H' 



I II III 

atom is at each corner; (b) a pyramidal arrangement (II) in which carbon is at the 
apex of a pyramid and a hydrogen atom is at each corner of a square base; (c) a 
tetrahedral arrangement (III) in which carbon is at the center of a tetrahedron and a 
hydrogen atom is at each corner. 

How do we know that each of these arrangements could give rise to only one 
substance of formula CH 3 Y? As always for problems like this, the answer lies in 
the use of molecular models. (Gumdrops and toothpicks can be used to make 
structures like I and II, for which the bond angles of ordinary molecular models are 
not suited.) For example, we make two identical models of I. In one model we 





SEC. 4.3 OPTICAL ACTIVITY. PLANE-POLARIZED LIGHT 117 

replace, say, the upper right-hand H with a different atom Y, represented by a 
differently colored ball or gumdrop; in the other model we similarly replace, say, 
the lower right-hand H. We next see whether or not the two resulting models are 
superimposable - t that is, we see whether or not, by any manipulations except bend- 
ing or breaking bonds, we can make the models coincide in all their parts. If the 
two models are superimposable, they simply represent two molecules of the same 
compound; if the models are not superimposable, they represent molecules of 
different compounds which, since they have the same molecular formula, are by 
definition Isomers (p. 37). Whichever hydrogen we replace in I (or in II or III), 
we get the same structure. From any arrangement other than these three, we 
would get more than one structure. 

As far as compounds of the formula CH 3 Y are concerned, the evidence of 
isomer number limits the structure of methane to one of these three possibilities. 

Problem 4.1 How many isomers of formula CH 3 Y would be possible if methane 
were a pyramid with a rectangular base? What are they? (Hint: If you have trouble 
with this question now, try it again after you have studied Sec. 4.7.) 

For any atom Y and for any atom Z, only one substance of formula CH 2 YZ 
has ever been found. Halogenation of methane, for example, yields only one 
compound of formula CB 2 C1 2 , only one compound of formula CH 2 Br 2 , and only 
one compound of formula CH 2 ClBr. 

Of the three possible structures of methane, only the letrahedral one is 
consistent with this evidence. 

Problem 4*2 How many isomers of formula CH 2 YZ would be expected from 
each of the following structures for methane? (a) Structure I with carbon at the center 
of a rectangle; (b) structure I with carbon at the center of a square; (c) structure II; 
fd) structure III. 

Thus, only the tetrahedral structure for methane agrees with the evidence of 
isomer number. It is true that this is negative evidence; one might argue that 
isomers exist which have never been isolated or detected simply because the 
experimental techniques are not good enough. But, as we said before, any com- 
pound that contains carbon bonded to four other atoms can be considered to 
be a derivative of methane; in the preparation of hundreds of thousands of .com- 
pounds of this sort, the number oif isomers obtained has always been consistent 
with the concept of the tetrahedral carbon atom. 

There is additional, positive evidence for the tetrahedral carbon atom: the 
finding of just the kind of isomers enantiomers that are predicted for compounds 
of formula CWXYZ. It was the existence of enantiomers that convinced van't 
Hoff that the carbon atom is tetrahedral. But to understand what enantiomers 
are, we must first learn about the property called optical activity. 



4.3 Optical activity. Plane-polarized light 

Light possesses certain properties that are best understood by considering it 
to be a wave phenomenon in which the vibrations occur at right angles to the 
direction in which the light travels. There are an infinite number of planes passing 



118 



STEREOCHEMISTRY I. STEREOISOMERS 



CHAP. 4 



through the line of propagation, and ordinary light is vibrating in all these planes. 
If we consider that we are looking directly into the beam of a flashlight, Fig. 4.1 



Figure 4.1. Schematic representation 
of (a) ordinary light and (b) plane- 
polarized light. Light traveling per- 
pendicular to page; vibrations in 
plane of page. 




shows schematically the sort of vibrations that are taking place, ail perpendicular 
to a line between our eye and the paper (flashlight). Plane-polarized light is light 
whose vibrations take place in only one of these possible planes. Ordinary light is 
turned into plane-polarized light by passing it through a lens made of the material 
kirown as Polaroid or more traditionally through pieces of calcite (a particular 
crystalline form of CaCO 3 ) so arranged as to constitute what is called a Nicol 
prism. 

An optically active substance is one that rotates the plane of polarized light. 
When polarized light, vibrating in a certain plane, is passed through an optically 
active substance, it emerges vibrating in a different plane. 



4.4 The polarimeter 

How can this rotation of the plane of polarized light this optical activity be 
detected? It is both detected and measured by an instrument called the polari- 
meter, which is represented schematically in Fig. 4.2. It consists of a light source, 
two lenses (Polaroid or Nicol), and between the lenses a tube to hold the substance 
that is being examined for optical activity. These are arranged so that the light 



Light 
source 




Figure 4.2. Schematic representation of a polarimeter. Solid lines: before 
rotanon. Broken lines: after rotation. is angle of rotation. 



SEC. 4.5 SPECIFIC ROTATION 119 

passes through one of the lenses (polarizer), then the tube, then the second lens 
(analyzer), and finally reaches our eye. When the tube is empty, we find that the 
maximum amount of light reaches our eye whe/ the two lenses are so arranged 
that they pass light vibrating in the same plane. If we rotate the lens that is nearer 
our eye, say, we find that the light dims, and reaches a minimum when the lens 
is at right angles to its previous position. 

Let us -adjust the lenses so that a maximum amount of light is allowed to pass. 
(In practice, it is easier to detect a minimum than a maximum; the principle 
remains the same.) Now let us place the sample to he tested in the tube. If the 
substance does not afTect the plane of polarization, light transmission is still at a 
maximum and the substance is said to be optically inactive. If, on the other hand, 
the substance rotates the plane of polarization, then the lens nearer our eye must 
be rotated to conform with this new plane if light transmission is again to be a 
maximum, and the substance is said to be optically active. If the rotation of the 
plane, and hence our rotation of the lens, is to the right (clockwise), the substance 
is dextrorotatory (Latin: dexter, right); if the rotation is to the left (counterclock- 
wise), the substance is Icvorotatory (Latin: laevus. left). 

We can determine not onl> that the substance has rotated the plane, and in 
which direction, but also by how much. The amount of rotation is simply the 
number of degrees that we must rotate the lens to conform with the light. The 
symbols + and are used to indicate rotations to the right and to the left, respec- 
tively. 

The lactic acid (p. 121 ) that is extracted from muscle tissue rotates light to the 
right, and hence is known as dextrorotatory lactic acid, or ( i-)-luctic acid. The 
2-methyI-l-butanoI thai is obtained from fusel oil (a by-product of the fermentation 
of starch to ethyl alcohol) rotates light to the left, and is known as Icvorotatory 
2-methyl-l-butanol, or ( )-2-methyl-l-butanol. 

4.5 Specific rotation 

Since optical rotation of the kind we are interested in is caused by individual 
molecules of the active compound, the amount of rotation depends upon how manv 
i.wleciiics the light encounters in passing through llie lube. 

The light will encounter twice as many molecules in a tube 20 cm long as in a 
tube 10 cm long, and the rotation will be twice as large. If the active compound 
is in solution, the number of molecules encountered by the light will depend upon 
the concentration. For a given tube length, light will encounter twice as many 
molecules in a solution of 2 g per 100 cc of solvent as in a solution containing I g 
per 100 cc of solvent, and the rotation will be twice as large. When allowances are 
made for the length of tube and the concentration, it is found that the amount of 
rotation, as well as its direction, is a characteristic of each individual optically 
active compound. 

Specific rotation is the number of degrees of rotation observed if a 1 -decimeter 
tube is used, and the compound being examined is present to the extent of Ig cc. 
This is usually calculated from observations with tubes of other lengths and at 
different concentrations by means of the equation 

r i M 

WM ,_ 



120 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

.~ . ,. observed rotation (degrees) 

specific rotation = ; . ,. . i~s - 

length (dm) x g/cc 

where d represents density for a pure liquid or'concentration for a solution. 

The specific rotation is as much a property of a compound as its melting point, 
boiling point, density, or refractive index. Thus the specific rotation of the 
2-methyl-l-butanol obtained from fusel oil is 

[ a ] D 20 = -5.756' 

Here 20 is the temperature and D is the wavelength of the light used in the measure- 
ment (D line of sodium, 5893 A). 

rroDiem 4..$ me concentration ot cholesterol dissolved in chloroform is 6.1 5 g 
per 100 ml of solution, (a) A portion of this solution in a 5-cm polarimeter tube causes 
an observed rotation of 1.2\ Calculate the specific rotation of cholesterol, (b) Pre- 
dict the observed rotation if the same solution were placed in a 10-cm tube (c) Pre- 
dict the observed rotation if 10 ml of the solution \vere diluted to 20 ml and placed in 
a 5-cm tube. 

Problem 4.4 A sample of a pure liquid in a 10-ctn tube is placed in a polari- 
meter, and a reading of +45 J is made. How could you establish that [a] is really +45 
and not -315? That it is 4- 45" and not 4-405' or, for that matter, 4-765? 



4.6 Enantiomerism: the discovery 

The optical activity we have just described was discovered in 1815 at the 
College de France by the physicist Jean-Baptiste Biot. 

In 1848 at the Ecole normale in Paris the chemist Louis Pasteur made a set of 
observations which led him a few years later to make a proposal that is the foun- 
dation of stereochemistry. Pasteur, then a young man, had come to the cole 
normale from the Royal College of Besancon (where he had received his baccalau- 
reni es sciences with the rating of mediocre in chemistry), and had just won his 
docteur es sciences. To gain some experience in crystallography, he was repeating 
another chemist's earlier work on salts of tartanc acid when he saw something 
that no one had noticed before: optically inactive sodium ammonium tartrate 
existed as a mixture of two different kinds of crystals, which were mirror images of 
each other. Using a hand lens and a pair of tweezers, he carefully and laboriously 
separated the mixture into two tiny piles one of right-handed crystals and the 
other of left-handed crystalsmuch as one might separate right-handed and left- 
handed gloves lying jumbled together on a shop counter. Now, although the 
original mixture was optically inactive, each set of crystals dissolved in water was 
found to be opticalh active] Furthermore, the specific rotations of the two solu- 
tions were exactly equal, but of opposite sign: that is to say, one solution rotated 
plane-polarized light to the right, and the other solution an equal number of 
degrees to the left. In all other properties the two substances were identical. 

Since the difference in optical rotation was observed in solution, Pasteur 
concluded that it was characteristic, not of the crystals, but of the molecules. He 
proposed that, like the two sets of crystals themselves, the molecules making up 
the crystals were mirror images of each other. He was proposing the existence Of 
isomers whose structures differ only hi being mirror images of each other, and 
whose properties differ only in the direction of rotation of polarized light. 



SEC. 4.7 



ENANTIOMERISM AND TETRAHEDRAL CARBON 



121 



There remained only for van't Hoffto point out that a tetrahedral carbon atom 
would account not only for the absence of isomers of formula CH 3 Y and CH 2 YZ, 
but also for the existence of mirror-image isomers enantiomers like Pasteur's 
tartaric acids. ^ 

4.7 Enantiomerism and tetrahedral carbon 

Let us convince ourselves that such mirror-image isomers should indeed exist. 
Starting with the actual, tetrahedral arrangement for methane, let us make a model 
of a compound CWXYZ, using a ball of a different color for each different atom 
or group represented as W, X, Y, and Z. Let us then imagine that we are holding 
this model before a mirror, and construct a second model of what its mirror image 
would look like. We now have two models which look something like this: 



Blue 



Yellow 




Red 



Blue 



Red 




Green 

which are understood to stand for this: 

mirror 
X 



Green 









Not superimposable: isomers 

Are these two models superimposable? No. We may twist and turn them as 
much as we please (so long as no bonds are broken), but although two groups 
of each may coincide, the other two do not. The models are not superimposable, 
and therefore must represent two isomers of formula CWXYZ. 

As predicted, mirror-image isomers do indeed exist, and thousands of instances 
besides the tartaric acids are known. There are, for example, two isomeric lactic 




COOH 



C 2 H 5 



C 2 H 5 
-OH HO rt H HCKTH 2 ^Q H H Q- CH 2 OH 

CH 3 CH 3 CH 3 CHj 

Lactic acid 2-Methyl- 1 -butanol 

acids and two 2-methyl-l-butanols, two chloroiodomethanesulfonic acids and two 
sec-butyl chlorides. 



122 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

C,H S 





SOjH SO 3 H CH 3 

Chloroiodomethanesulfonic acid sec-Butyl chloride 

As we can see, the structures of each pair are mirror images; as we can easily 
verify by use of models, the structures of each pair are not superimposable and 
therefore represent isomers. (In fact, we have a/ready verified this, since the 
models we made for CWXYZ can, of course, stand for any of these.)* 

At this point we do not need to know the chemistry of these compounds, or 
even what structure a particular collection of letters ( COOH, say, or -CH 2 OH) 
stands for; we can tell when atoms or groups are the same or different from each 
other, and whether or not a model can be superimposed on its mirror image. Even 
two isotopes of the same element, like protfum (ordinary hydrogen, H) and deu- 
terium (heavy hydrogen, D) are different enough to permit detectable isomerism : 

C "3 CH 3 



D 

QH 5 C 6 H 5 

a -Deuteriocthylbenzene 

We must remember that everything (except, of course, a vampire) has a mirror image, 
including all molecules. Most molecules, however, are superimposable on their mirror 
images, as, for example, bromochloromethane, and do not show this mirror-image 
isomerism. 



mirror 



Cl 



r 

Br 



Cl 



^ 

Br 



Bromochloromethane 
Superimposable: no isomerism 

Mirror-image isomers are called enantiomers. Since they differ from one 
another only in the way the atoms are oriented in space, enantiomers belong to the 
general class called stereoisomers. Later on we shall encounter btereoisomers that 
are not mirror images of each other; these are called ttiastereomers. Any two stereo- 
isomers are thus classified either as enantiomers or as diastereomers, depending 
upon whether or not they are mirror images of each other. 

The non-superimposability of mirror images that brings about the existence of enantio- 
mers also, as we shall see, gives them iheir optical activity, and hence enantiomers arc 
often referred to as (one kind of) optical isomers. We shall make no use of the leim 
optical isomer, since it is hard to define- indeed, is often used undefined - and of doubtful 
usefulness. 



SEC. 4.9 PREDICTION OF ENVM MOMKRISM. CHIRALIIY 123 

4.8 Enantiomerism and optical activity 

Most compounds do not rotate the plane of polarized light. How is it that 
some do? It is not the particular chemical family that they belong to, since optic- 
ally active compounds are found in all families. To see what special structural 
feature gives rise to optical activity, let us look more closel> at what happens when 
polari/ed light is passed through a sample of a single pure compound. 

When a beam of polari/ed light passes through an individual molecule, in 
nearly every instance its plane is rotated a tiny amount by interaction with the 
charged particles of the molecule; the direction and extent of rotation varies with 
the orientation of the particular molecule in the beam. For most compounds, 
because of the random distribution of the large number of molecules that make up 
even the smallest sample of a single pure compound, for every molecule that the 
light encounters, there is another (identical) molecule oriented as (he mirror image 
of the first i \vhich exactly cancels its effect. The net result is no rotation, that is, 
optical inactivity. Thus optical inactivity is not a property of individual molecules, 
but rather of the random distribution of molecules that can serve as mirror images oj 
each o( her. 

Optical inactivity requires, then, that one molecule of a compound act as the 
mirror image of another. But in the special case of CWXY/, we have found (Sec. 
4.7) a molecule whose mirror image is not just another, identical molecule, but 
rather a molecule of a different, isomeric compound. In a pure sample of a single 
enantiorncr, no molecule can serve as the mirror image of another; there is no 
exact canceling-out of rotations, and the net result is optical activity Thus, the 
same non-superimposabilily of mirror images that gives rise to enanLiomerism 
also is responsible for optical activity. 



4.9 Prediction of enantiomerism. Chirality 

Molecules that are not superimposahle on their mirror images are chiral. 

Chirality is the necessary and sufficient condition for the existence^oT enan- 
tiomers. That is to say: a compound \\hose molecules are chiral can exist as enantio- 
mers; a compound whose molecules are achiral (vMthout chirality) cannot exist as 
enantiomers. 

When we say that a molecule and its mirror image are superimposable, we 
mean that if in our mind's eye we were to bring the image from behind the 
mirror where it seems to be, it could be made to coincide in all its parts with the 
molecule. To decide whether or not a molecule is chiral, therefore, we 
make a model of it and a model of its mirror image, and see if we can superimpose 
them. This is the safest way, since properly handled it must give us the right 
answer. It is the method that we should use until we have become quite familiar 
with the ideas involved; even then, it is the method we should use when we en- 
counter a new type of compound. 

After we have become familiar with the models themselves, we can draw 
pictures of the models, and mentally try to superimpose them. Some, we find, are 
not superimposablc, like these: 



124 



STEREOCHEMISTRY I. STEREOISOMERS 

mirror 
H 



CHAP. 4 




SOjH 



"\ 

SOjH 



Chloroiodomethanesulforuc acid 
Not superimposable : enantiomers 

These molecules are chiral, and we know that chloroiodomethanesulfonic acid can 
exist as enantiomers, which have the structures we have just made Or drawn. 
Others, we find, are superimposable, like these : 

mirror 



H 

CH 3 ^Q_-CH 3 
CI 



CH 3 



^ 

CI 



Isopropyl chloride 
Superimposable : no enantiomers 

These molecules are achiral, and so we know that isopropyl chloride cannot exist 
as enantiomers. 

"I call any geometrical figure, or any group of points, chiral, and say it has 
chirality, if its image in a plane mirror, ideally realized, cannot be brought to coin- 
cide with itself. "Lord Kelvin, 1893. 

In 1964, Cahn, Ingold, and Prelog(see p. 130) proposed that chemists use the terms 
"chiral" and "chirality" as defined by Kelvin. Based on the Greek word for "hand" 
(cheir\ chirality means " handedness," in reference to that pair of non-superimposable 
mirror images \ve constantly -have before us: our two hands. There has been wide-spread 
acceptance of Kelvin's terms, and they have largely displaced the earlier "dissymmetric" 
and "dissymmetry" (and the still earlier and less accurate "asymmetric" and "asym- 
metry"), although one must expect to encounter the older terms in the older chemical 
literature. 

Whatever one calls it, it is non-superimposability-on-mirror-image that is the neces- 
sary and sufficient condition for enantiomerism ; it is also a necessary- but not sufficient 
condition for optical activity (see Sec. 4.13). 



4.10 The chiral center 

So far, all the chiral molecules we have talked about happen to be of the kind 
CWXYZ; that is, in each molecule there is a carbon (C*) that holds four different 
groups. 

H 



H 



H 



C 2 H 5 -C*~CH 2 OH 
CH 3 

2-MethyM-butanol 



CH 3 ~-C*-~COOH C 2 H 5 -~C* CH 3 
OH CI 




~CH 3 



Lactic acid 



.fee-Butyl chloride a-Dcuterioethyl benzene 



SEC. 4.10 THE CHIRAL CENTER 125 

A carbon atom to which four different groups are attached is a chiral center. (Some- 
times it is called chiral carbon, when it is necessary to distinguish it from chiral 
nitrogen, chiral phosphorus, etc.) 

Many but not all molecules that contain a chiral center are chiral. Many 
but not all chiral molecules contain a chiral center. There are molecules that 
contain chiral centers and yet are achiral (Sec. 4.18). There are chiral molecules 
that contain no chiral centers (see, for example, Problem 6, p. 315). 

The presence or absence of a chiral center is thus no criterion of chirality. 
However, most of the chiral molecules that we shall take up do contain chiral 
centers, and it will be useful for us to look for such centers; if we find a chiral 
center, then we should consider the possibility that the molecule is chiral, and 
hence can exist in enantiomeric forms. We shall later (Sec. 4.18) learn to recognize 
the kind of molecule that may be achiral in spite of the presence of chiral centers; 
such molecules contain more than one chiral center. 

After becoming familiar with the use of models and of pictures of models, 
the student can make use of even simpler representations of molecules containing 
chiral centers, which can be drawn much faster. This is a more dangerous method, 
however, and must be used properly to give the right answers. We simply draw a 
cross and attach to the four ends the four groups that are attached to the chiral 
center. The chiral center is understood to be located where the lines cross. Chemists 
have agreed that such a diagram stands for a particular structure: the horizontal 
lines represent bonds coming toward us out of the plane of the paper, whereas the 
vertical lines represent bonds going away from us behind the plane of the paper. 
That is to say: 

H-X-Cl 

CH 3 

can be represented by 

C 2 H 5 C 2 H 5 




H- 



-Cl Cl- 



-H 



CH 3 CH 3 



In testing the superimposability of two of these flat, two-dimensional repre- 
sentations of three-dimensional objects, we must follow a certain procedure and 
obey certain rules. First, we use these representations only for molecules that 
contain a chiral center. Second, we draw one of them, and then draw the other as 
its mirror image. (Drawing these formulas at random can lead to some interesting 
but quite wrong conclusions about isomer numbers.) Third, in our mind's eye we 
may slide these formulas or rotate them end for end, but we may not remove them 
from the plane of the paper. Used with caution, this method of representation is 
convenient; it is not foolproof, however, and in doubtful cases models or pictures 
of models should be used. 



126 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

Problem 4.5 Using cross formulas, decide which of the following compounds 
arc chiral. Check your ansv\ers by use of stick-and-ball formulas, and finally by use 
of models. 

(a) 1-chloropentane (e) 2-chloro-2-methylpentane 

(b) 2-chloropentane (f) 3-chloro-2-methylpentane 

(c) 3-chloropentane (g) 4-chloro-2-methylpentane 

(d) l-chloro-2-methylpentane (h) l-chloro-2-bromobutane 

Problem 4.6 (a) Neglecting stereoisomers for the moment, draw all isomers of 
formula C*H 6 DO. (b) Decide, as in Problem 4.5, which of these are chiral. 



4.11 Enantiomers 

Jsotners that are mirror jmagesjtfjeji^^ The two 

different Jactic acids \\hose models we made in Sec. 4.7 are enantiomers (Gr. : 
enantio-i opposite). So are the two 2-methyl-l-butanols, the two jer-butyl chlorides, 
etc. How do the properties of enantiomers compare? 

Enantiomers have identical physical properties, except for the direction of 
rotation of the plane of polarized light. The two 2-methyl-l-butanols, for example, 

( 4- )-2-Meth>M -butanol ( )2-Methyl-l-butanol 
(Fermentation Product) 



Specific rotation 


+ 5.756 


-5.756 


Boiling point 


128.9 3 


128.9 


Density 


0.8193 


0.8193 


Refractive index 


1.4107 


1.4107 



have identical melting points, boiling points, densities, refractive indices, and any 
other physical constant one might measure, except for this: one rotates plane- 
polarized light to the right, the other to tire left. This fact is not surprising, since 
the interactions of both kinds of molecule with their fellows should be the same. 
Only the direction of rotation is different; the amount of rotation is the same, the 
specific rotation of one being -f 5.756, the other 5.756. It is reasonable that 
these molecules, being so similar, can rotate light by the same amount. The 
molecules are mirror images, and so are their properties: the mirror image of a 
clockwise rotation is a counterclockwise rotation and of exactly the same 
magnitude. 

Enantiomers have identical chemical properties except toward optically active 
reagents. The two lactic acids are not only acids, but acids of exactly the same 
strength; that is, dissolved in water at the same concentration, bo'h ionize to 
exactly the same degree. The two 2-methyl-l-butanols not only form the same 
products alkenes on treatment with hot sulfuric acid, alky I bromides on treatment 
with HBr, esters on treatment with acetic acid but also form them at exactly ihe 
same rate. This is quite reasonable, since the atoms undergoing attack in each 
case are influenced in their reactivity by exactly the same combination of substi- 
tuents. The reagent approaching either kind of molecule encounters the same 
environment, except, of course, that one environment is the mirror image of the 
other. 

In the special case of a reagent that is itself optically active, on the other hand, 
the influences exerted on the reagent are not identical in the attack on the two 



SEC. 4.12 THE RACEM1C MODIFICATION 127 

enantiomers, and reaction rates will be different so different, in some cases, that 
reaction with one isomer does not take place at all. In biological systems, for 
example, such stereochemical specificity is the rule rather than the exception, since 
the all-important catalysts, enzymes, and most of the compounds they work on, 
are optically active. The sugar ( + )-glucose plays a unique role in animal metab- 
olism (Sec. 34.3) and is the basis of a multimillion-dollar fermentation industry 
(Sec. 15.5); yet ( )-glucose is neither metabolized by animals nor fermented by 
yeasts. When the mold PeniciUium glaucum feeds on a mixture of enantiomeric 
tartaric acids, it consumes only the (-f)-enantiomer and leaves ( )-tartanc acid 
behind. The hormonal activity of ( )-adrenaline is many times that of its enantio- 
mer; only one stereoisomer of chloromycetin is an antibiotic, (-f )-Ephedrine not 
only has no activity as a drug, but actually interferes with the action of its enan- 
tiomer. Among amino acids, only one asparagine and one leucine are sweet, and 
only one glutamic acid enhances the flavor of food. It is ( )-carvone that gives 
oil of spearmint its characteristic odor; yet the enantiomeric ( 4- )-carvone is the 
essence of caraway. 

Consider, as a crude analogy, a right and left hand of equal strength (the 
enantiomers) hammering a nail (an optically inactive reagent) and inserting a 
right-handed screw (an optically active reagent). Hammering requires exactly 
corresponding sets of muscles in the two hands, and can be done at identical rates. 
Inserting the screw uses different sets of muscles: the right thumb pushes, for 
example, whereas the left thumb pulls. 

Or, let us consider reactivity in the most precise way we know: by the tran- 
sition-state approach (Sec. 2.22). 

Take first the reactions of two enantiomers with an optically inactive reagent. 
The reactants in both cases are of exactly the same energy: one enantiomer plus 
the reagent, and the other enantiomer plus the same reagent. The two transition 
states for the reactions are mirror images (they are enantiomeric), and hence are 
of exactly the same energy, too. Therefore, the energy differences between reac- 
tants and transition statesthe act 's -are identical, and so are the rates of 
reaction. 

Now take the reactions of two enantiomers with an optically active reagent. 
Again the reactants are of the same energy. The two transition states, however, 
are not mirror images of each other (they are diastereomeric), and hence are of 
different energies; the ac t's are different, and so are the rates of reaction. 



4.12 The racemic modification 

Ajivxture ofjeguajj^arls of enantiomers is called a racemjfcjggodificatioii. A 
racemic modificatio^ is optically^ inactive: when enantiomers are mixe3TToge1her, 
the rotation caused by a molecule of one isomer is exactly canceled by an equal and 
opposite rotation caused by a molecule of its enantiomer. 

The prefix is used to specify the racemic nature of the particular sample, as, 
for example, ( )-lactic acid or ( )-2-methyl-l-butanol. 

It is useful to compare a racemic modification with a compound whose 
molecules are superimposable on their mirror images, that is, with an achiral 
compound. They are both optically inactive, and for exactly the same reason. 
Because of the random distribution of the large number of molecules, for every 



128 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

molecule that the light encounters there is a second molecule, a mirror image of 
the first, aligned just right to cancel the effect of the first one. In a racemic modifica- 
tion this second molecule happens to be an isomer of the first; for an achiral com- 
pound it is not an isomer, but another, identical molecule (Sec. 4.8). 

(For an optically active substance uncontaminated by its enantiomer, we have 
seen, such cancellation of rotation cannot occur since no other molecule can serve 
as the mirror image of another, no matter how random the distribution.) 

Problem 4.7 To confirm the statements of the three preceding paragraphs, make 
models of: (a) a pair of enantiomers, e.g., CHClBrI; (b) a pair of identical achiral 
molecules, e.g., CH 2 ClBr; (c) a pair of identical chiral molecules, e.g., CHClBrI. 
(d) Which pairs are mirror images? 

The identity of most physical properties of enantiomers has one consequence 
of great practical significance. They cannot be separated by ordinary methods: 
not by fractional distillation, because their boiling points are identical; not by 
fractional crystallization, because their solubilities in a given solvent are identical 
(unless the solvent is optically active); not by chromatography, because they are 
held equally strongly on a given adsorbent (unless it is optically active). The 
separation of a racemic modification into enantiomers the resolution of a racemic 
modification is therefore a special kind of job, and requires a special kind of 
approach (Sec. 7.9). 

The first resolution was, of course, the one Pasteur carried out with his hand lens and 
tweezers (Sec. 4.6). But this method can almost never be used, since racemic modifica- 
tions seldom form mixtures of crystals recognizable as mirror images. Indeed, even 
sodium ammonium tart rate does not, unless it crystallizes at a temperature below 28. 
Thus partial credit for Pasteur's discovery has been given to the cool Parisian climate 
and, of course, to the availability of tartaric acid from the winemakers of France. 

The method of resolution nearly always used one also discovered by Pasteur 
involves the use of optically active reagents, and is described in Sec. 7.9. 

Although popularly known chiefly for his great work in bacteriology and medicine, 
Pasteur was by training a chemist, and his work in chemistry alone would have earned him 
a position as an outstanding scientist. 

4.13 Optical activity: a closer look 

We have seen (Sec. 4.8) that, like enantiomerism, optical activity results from 
and only from chirality: the non-superimposability of certain molecules on their 
mirror images. Whenever we observe (molecular) optical activity, we know we are 
dealing with chiral molecules. 

Is the reverse true? Whenever we deal with chiral molecules with compounds 
that exist as enantiomers must we always observe optical activity? No. We have 
just seen that a 50:50 mixture of enantiomers is optically inactive. Clearly, if we 
are to observe optical activity, the material we are dealing with must contain an 
excess of one enantiomer: enough of an excess that the net optical rotation can 
be detected by the particular polarimeter at hand. 

Furthermore, this excess of one enantiomer must persist long enough for the 
optical activity to be measured. If the enantiomers are rapidly interconverted, 
then before we could measure the optical activity due to one enantiomer, it would 
be converted into an equilibrium mixture, which since enantiomers are of exactly 
the same stability must be a 50:50 mixture and optically inactive. 



SEC. 4.14 CONFIGURATION 129 

Even if all these conditions are met, the magnitude and hence the detectability 
of the optical rotation depends on the structure of the particular molecule concerned. 
In compound I, for example, the four groups attached to the chiral center differ only in 
chain length. 

CH 2 CH 3 



CH 2 CH 2 CH 3 

I 
Ethyl--propyl--butyl-/i-hexylmethane 

It has been calculated that this compound should have the tiny specific rotation of 0.00001 
far below the limits of detection by any existing polarimeter. In 1965, enantiomerically 
pure samples of both enantiomers of I were prepared (see Problem 19, p. 1026), and each 
was found to be optically inactive. 

At our present level of study, the matter of speed of interconversion will give 
us no particular trouble. Nearly all the chiral molecules we encounter in this book 
lie at either of two extremes, which we shall easily recognizef (a) molecules like 
those described in this chapter which owe their chiralily to chiral centers; here 
interconversion o r enantiomers (configuration^ enantiomers) is so slow because 
bonds have to be Broken that we need not concern ourselves at all about inter- 
conversion; (b) moK'cules whose enantiomeric forms (conformational enantiomers) 
are interconvertible Dimply by rotations about single bonds; here for the com- 
pounds we shall encounter interconversion is so fast that ordinarily we need not 
concern ourselves at all about the existence of the enantiomers. 



4.14 Configuration 

The arrangement of atoms that characterizes a particular stereoisomer is called 
its configuration. 

Using the test of superimposability, we conclude, for example, that there are 
two stereoisomeric sec-butyl chlorides; their configurations are I and IL Let us 

C,H 5 C 2 H 5 

a V- H 

CHj CH 3 

I " 

sec-Butyl chloride 

say that, by methods we shall take up later (Sec. 7.9), we have obtained in the 
laboratory samples of two compounds of formula C2H 5 CHC1CH 3 . We find that 
one rotates the plane of polarized light to the right, and the other to the left; we 
put them into two bottles, one labeled "(+)-sec-butyl chloride" and the other 
" (-)-^r-butyl chloride." 

We have made two models to represent the two configurations of this chloride. 
We have isolated two isomeric compounds of the proper formula. Now the 
question arises, which configuration does each isomer have? Does the (-f )-isomer, 




130 STEREOCHEMISTRY I. STERKOISOMERS CHAP. 4 

say, have configuration I or configuration II? How do we know which structural 
formula, I or II, to draw on the label of each bottle? That is to say, how do we 
assign configuration ? 

Until 1949 the question of configuration could not be answered in an absolute 
sense for any optically active compound. But in that year J. M. Bijvoet most 
fittingly Director of the van't Hoflf Laboratory at the University of Utrecht (Sec. 
4.2) reported that, using a special kind of x-ray analysis (the method of anomalous 
scattering), he had determined the actual arrangement in space of the atoms of an 
optically active compound. The compound was a salt of (4- )-tartaric acid, the 
same acid that -almost exactly 100 years before had led Pasteur to his discovery 
of optical isomerism. Over the years prior to 1949, the relationships* between the 
configuration of (-f)-tartaric acid and the configurations of hundreds of optically 
active compounds had been worked out (by methods that we shall take up later, 
Sec. 7.5); when the configuration of ( f )-tartaric acid became known, these other 
configurations, too, immediately became known. (In the case of the sec-butyl 
chlorides, for example, the ( )-isomer is known to have configuration I, and the 
(f )-isomcr configuration II.) 



4.15 Specification of configuration: R and S 

Now, a further problem arises. How can we specify a particular configuration 
in some simpler, more convenient way than by always having to draw its picture? 
The most generally useful way yet suggested is the use of the prefixes R and S. 
According to a procedure proposed by R. S. Cahn (The Chemical Society, London), 
Sir Christopher Ingold (University College, London), and V. Prelog (Eidgenossiche 
Technische Hochschule, Zurich), two steps are involved. 

Step 1. Following a set of sequence rules (Sec. 4.16), we assign a sequence of 
priority to the four atoms or groups of atoms attached to the chiral center. 

In the case of CHClBrI, for example, the four atoms attached to the chiral 
center are all different and priority depends simply on atomic number, the atom 
of higher number having higher priority. Thus I, Br, CI, H. 



Br 




I H 

Bromochloroiodomethane 

Step 2. We visualize the molecule oriented so that the group of lowest priority 
is directed away from us, and observe the arrangement of the remaining groups. 
If, in proceeding from the group of highcstj^riorityjojlie group ofsecondj>riojjty 
and thence to thejhird, our eye travels in a clockwise direction, the configuration is 
specified R (Catinl rectus, right); if counterclockwise, the configuration is specified 
S (Latin: sinister, left). 

Thus, configurations I and II are viewed like this: 



SEC. 4.16 



SEQUENCE RULES 



131 





and are specified R and S, respectively. 

A complete name for an optically active compound reveals if they are known 
both configuration and direction of rotation, as, for example, (S)-(-(-)^ec-butyl 
chloride. A raccmic modification can be specified by the prefix RS, as, for example, 
(RS)-seobutyl chloride. 

(Specification of compounds containing more than one chiral center is dis- 
cussed in Sec. 4.19.) 

We must not, of course, confuse the direction of optical rotation of a compound- 
a physical property of a real substance, like melting point or boiling point- -\\ith the 
direction in which our eve happens to travel when \ve imagine a molecule held in an arbi- 
tiary manner. So far as \\e are concerned, unless \vc happen to know what has been 
established experimentally for a specific compound, \\e have no idea \\hether (4 ) or ( ) 
rotation is associated \\ith the (R)- or the (S)-configuration. 



4.16 Sequence rules 

For case of reference and for convenience in reviewing, \\e shall set do\\n here 
those sequence rules \\e shall have need of. The student should study Rules 1 and 
2 now, and Rule 3 later \\hen the need for it arises. 

Sequence Rule 1. If the four atoms attached to the chiral center are all dif- 
ferent, priority depends on atomic number, with the atom of higher atomic num- 
ber getting higher pnont>. If two atoms are isotopes of the same element, the 
atom of higher mass number has the higher priority. 

For example, in chloroiodomcthanesulfonic acid the sequence is I, Cl, S, H; 
in -deuterioethyl bromide it is Br, C, D, H. 



Cl 
H C SOjH 



H 

H 3 C C Br 
D 



Chloroiodomethanesulfonic a-Deuierioethyl bromide 
acid 



Problem 4.8 Make models and then draw both stick-and-ball pictures and cross 
formulas for the cnunnomcrs of: (a) chloroiodomcthanesulfonic acid and (b) - 
dcutcnocthvl biomide. Label each as R or S. 

Sequence Rule 2. If the relative priority of two groups cannot be decided by 
Rule I, it shall be determined by a similar comparison of the next atoms in the 
groups (and so on, if necessary, \\orking outward from the chiral center). That is 
to say, if two atoms attached to the chiral center are the same, we compare the 
atoms attached to each of these first atoms. 



132 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

For example, take sec-butyl chloride, in which two of the atoms attached to 
the chiral center are themselves carbon. In CH 3 the second atoms are H, H, H; 

H 
CH 3 CH 2 C CH 3 

Cl 
sec-Butyl chloride 

in C 2 H 5 they are C, H, H. Since carbon has a higher atomic number than hydrogen, 
C 2 H 5 has the higher priority. A complete sequence of priority for sec-butyl 
chloride is therefore Cl, C 2 H 5 , CH 3 , H. 

In 3-chloro-2-methylpentane the C, C, H of isopropyl takes priority over the 
C, H, H of ethyl, and the complete sequence of priority is Cl, isopropyl, ethyl, H. 

CH,H CH 3 H 

CH 3 -CH-C-CH 2 -CH 3 CH 3 -CH-C-CH 2 C1 

Cl Cl 

3-Chloro-2-methylpentane 1 ,2-Dichloro-3-methylbutane 

In l,2-dichloro-3-methylbutane the Cl, H, H of CH 2 C1 takes priority over the 
C, C, H of isopropyl. Chlorine has a higher atomic number than carbon, and the 
fact that there are two C's and only one Cl does not matter. (One higher number is 
worth more than two or three of a lower number.) 

Problem 4.9 Into what sequence of priority must these alkyl groups always fall: 
CH 3 , 1,2, 3? 

Problem 4.10 Specify as R or S each of the enantiomers you drew : (a) in Prob- 
lem 4.5 (p. 126); (b) in Problem 4.6 (p. 126). 



Sequence Rule 3. (The student should defer study of this rule until he needs it.) 
Where there is a double or triple bond, both atoms are considered to be 
duplicated or triplicated. Thus, 

A C 
(J=A equals C A and C==A equals C A 

Jli li 

For example, in glyceraldehyde the OH group has the highest priority of all, 
H 



H H 

H- c OH C-O equals C O 

CH 2 OH O C 

Glyceraldehyde 

and the O, O, H of CHO takes priority over the O, H, H of CH 2 OH. The com- 
plete sequence is then OH, CHO, CH 2 OH, H. 



SEC. 4.17 DIASTERFOMERS 133 

The phenyl group, C 6 H 5 is handled as though it had one of the Kekule 
structures: 



equals [ |] equals I 

C 

In l-amino-2-methyl-l-phenylpropane, for example, the C, C, C, of phenyl takes 

H 

-CH(CH 3 ) 2 
NH 2 

priority over the C, C, H of isopropyl, but not over N, which has a higher atomic 
number. The entire sequence is then NH 2 , C 6 H 5 , C 3 H 7 , H. 

The vinyl group, CH 2 =CH , takes priority over isopropyl. 
H H H 

CH=CH 2 equals C C C takes priority over C CH 3 

C H CH 3 

Following the "senior" branch. CH 2 C, we arrive at C in vinyl as compared 
v/ith H in the CH 2 H of isopropyl. 



Probjem 4.AJ 

~3-chloro-l-pe 



Hi. A A Draw and specify as R or S the enantiomers (if any) of: 

(a)*3-chloro-l-pentene (e) methvlethyl-/i-propylisopropylmethane 

(b) 3-chloro-4-methyl-l-pentene (f) C 6 H 5 CHOHCOOH, mandelic acid 

(c) HOOCCH2CHOHCOOH, malic acid (g) CH^CH(NH 2 )C(X)H, alanine 

(d) C 6 H 5 CH(CH 3 )NH* 



4.17 Diastereomers 

Next, we must learn what stereoisomers are possible for compounds whose 
molecules contain, not just one, but more than one chiral center. (In Chapter 34, 
we shall be dealing regularly with molecules that contain five chiral centers.) 

Let us start with 2,3-dichloropentane. This compound contains two chiral 

CH 3 CH 2 -CH CH CH 3 

Cl Cl 
2,3-Dichloropentane 

centers, C-2 and C-3. (What four groups are attached to each of these carbon 
atoms?) How many stereoisomers are possible? 

Using models, let us first make structure I and its mirror image II, and see if 
these are superimposable. We find that I and II are not superimposable, and 
hence must be enantiomers. (As before, we may represent the structures by pic- 
tures, and mentally try to superimpose these. Or, we may use the simple "cross" 
representations, being careful, as before (Sec. 4.10), not to remove the drawings 
from the plane of the paper or blackboard.) 

Next, we try to interconvert I and II by rotations about carbon-carbon bonds. 
We find that they are not interconvertible in this way, and hence each of them is 
capable of retaining its identity and, if separated from its mirror image, of showing 
optical activity. 



134 



STEREOCHEMISTRY I. STEREOISOMERS 

mirror 



CHAP. 4 





CH 3 



H- 

Cl- 



CH, 



-Cl 
-H 



Cl- 
H- 



-H 



C 2 H 5 

I 



C 2 H 5 

II 



Not superimposable 
Enantiomers 



Are there any other stereoisomers of 2,3-dichloropentane? We can make struc- 
ture III, which we find to be non-superimposable on either I or II: it is not, of 



mirror 




H 



Cl 




Cl 



H 



H 



Not superimposable 
Enantiomers 

course, the mirror image of either. What is the relationship between III and I? 
Between III and II? They are stereoisomers but not enantiomers, Stereoisomers 
that are not mirror images of each other are called diastereomers. Compound III 
is a diastereomer of I, and similarly of II. 



SEC. 4.18 MESO STRUCTURES 135 

Now, is III chiral? Using models, we make its mirror image, structure IV, 
and find that this is not superimposable on (or interconvertible with) III. Struc- 
tures III and IV represent a second pair of enantiomers. Like III, compound IV 
is a diastereomcr of I and of II. 

How do the properties of diastereomers compare? 

Diastereomers have similar chemical properties, since they are members of the 
same family. Their chemical properties are not identical, however. In the reaction 
of two diastereomers with a given reagent, neither the two sets of reactants nor 
the two transition states are mirror images, and hence except by sheer coinci- 
dence- -will not be of equal energies. aft 's will be different and so will the rates 
of reaction. 

Diastereomers have different physical properties: different melting points, 
boiling points, solubilities in a given solvent, densities, refractive indexes, and 
so on. Diaslereomers differ in specific rotation; they may have the same or oppo- 
site signs of rotation, or some may be inactive. 

As a result of their differences in boiling point and in solubility, they can, in 
principle at least, be separated from each other by fractional distillation or frac- 
tional crystallization; as a result of differences in molecular shape and polarity, 
they differ in adsorption, and can be separated by chromatography. 

Given a mixture of all four stereoisomeric 2,3-dichloropentanes, we could 
separate it, by distillation, for example, into two fractions but no further. One 
fraction would be the racemic modification of I plus II; the other fraction would 
be the racemic modification of HI plus IV. Further separation would require 
resolution of the racemic modifications by use of optically active reagents (Sec. 
7.9). 

Thus the presence of two chiral centers can lead to the existence of as many 
as four stcreoisomers. For compounds containing three chiral centers, there could 
be as many as eight stcreoisomers; for compounds containing four chiral centers, 
there could be as many as sixteen stercoisomers, and so on. The maximum number 
of slcrcoisomers that can exist is equal to 2", where n is the number of chiral cen- 
ters. (In any case where meso compounds exist, as discussed in the following sec- 
lion, there will be fewer than this maximum number.) 



4.18 Meso structures 

Now let us look at 2,3-dichlorobutane, which also has two chiral centers. 
Does this compound, too, exist in four stereoisomeric forms? 



CH 3 -CH-CH-CH 3 

i i 
Cl Cl 

2,3-Dichlorobutane 

Using models as before, we arrive first at the two structures V and VI. These 
are mirror images that are not superimposable or interconvertible; they are there- 
fore enantiomers, and each should be capable of optical activity. 



136 



STEREOCHEMISTRY I. STEREOISOMERS 

mirror 



H 



CHAT 





H 



CH 3 



CH, 











- 

a.. 




11 
H 


CI 







CH 3 CH 3 

V VI 

Not superimposable 

Enantiomers 



mirror 





H 



CH 3 



CH 3 



H 
H 


CI 
CI 















CH 3 CH 3 

VII VIII 

Superimposable 
A meso compound 

Next, we make VII, which we find to be a diastereomer of V and of VI. 
We now have three stereoisomers; is there a fourth? No. If we make VIII, the 
mirror image of VII, we find the two to be superimposable; turned end-for^end, 



SEC. 4.19 



SPECIFICATION OF CONFIGURATION 



137 



VII coincides in every respect with VIII. In spite of its chiral centers, VII is not 
chiral. It cannot exist in two enantiomeric forms, and it cannot be optically active. 
It is called a meso compound. 

A meso compound is one whose molecules are superimpo sable on their mirror 
images even though they contain chiral centers. A meso compound is optically 
inactive for the same reason as any other compound whose molecules are achiral: 
the rotation caused by any one molecule is cancelled by an equal and opposite 
rotation caused by another molecule that is the mirror image of the first (Sec. 4.8). 

We can often recognize a meso structure on sight by the fact that (in at least 
one of its conformations) one half of the molecule is the mirror image of the other 
half. This can be seen for wes0-2,3-dichlorobutane by imagining the molecule to 
be cut by a plane lying where the dotted line is drawn. The molecule has a plane 
of symmetry, and cannot be chiral. (Caution: If we do not see a plane of symmetry, 
however, this does not necessarily mean that the molecule is chiral.) 




CH 3 



H- 



H- 



-Cl 



-Cl 



CH 3 



Problem 4.12 Draw stereochemical formulas for all the possible stereoisomers 
of the following compounds. Label pairs of enantiomers, and meso compounds. Tell 
which isomers, if separated from all other stereoisomers, will be optically active. Pick 
out several examples of diastereomers. 

(a) 1,2-dibromopropane (e) 1,2,3,4-tetrabromobutane 

(b) 3,4-dibromo-3,4-dimethylhexane (f) 2-bromo-3-chlorobutane 

(c) 2,4-dibromopentane (g) l-chloro-2-methylbutane 

(d) 2,3,4-tribromohexane (h) 1 ,3-dichloro-2-methyIbutane 



4.19 Specification of configuration: more than one chiral center 

Now, how do we specify the configuration of compounds which, like these, 
contain more than one chiral center? They present no special problem; we simply 
specify the configuration about each of the chiral centers, and by use of numbers 
tell which specification refers to which carbon. 

Consider, for example, the 2,3-dichloropentanes (Sec. 4.17). We take each 
of the chiral centers, C-2 and G-3, in turn ignoring for the moment the existence 



Cl Cl 
2,3-Dichloropentane 



of the other and follow the steps of Sec. 4.15 and use the Sequence Rules of 



138 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

Sec. 4.16. In order of priority, the four groups attached to C-2 are Cl, 
CH 3 CH 2 CHC1~, CH 3 , H. On C-3 they are CI, CH 3 CHC1-, CH 3 CH 2 -, H. 
(Why is CH 3 CHC1- "senior" to CH 3 CH 2 -?) 

Taking in our hands or in our mind's eye a model of the particular Stereo- 
isomer we are interested in, we focus our attention first on C-2 (ignoring C-3), and 
then on C-3 (ignoring C-2). Stereoisomer I (p. 134), for example, we specify 
(2S,3S)-2,3-dichloropentane. Similarly, II is (2R,3R), III is (2S,3R), and IV is 
(2R,3S). These specifications help us to analyze the relationships among the 
stereoisomers. As enantiomers, I and II have opposite that is, mirror-image 
configurations about both chiral centers: 2S,3S and 2R,3R. As diastereomers, I 
and III have opposite configurations about one chiral center, and thS same con- 
figuration about the other: 2S,3S and 2S,3R. 

We would handle 2,3-dichlorobutane (Sec. 4.18) in exactly the same way. 
Here it happens that the two chiral centers occupy equivalent positions along the 

CH 3 -CHCH~ CH 3 

Cl Cl 

2,3-Dichlorobutane 

chain, and so it is not necessary to use numbers in the specifications. Enaniiomers 
V and VI (p. 136) are specified (S,S)- and (R,R)-2,3-dichlorobutane ; respectively. 
The meso isomer, VII, can, of course, be specified either as (R,S)- or (S,R)-2,3- 
dichlorobutane the absence of numbers emphasizing the equivalence of the two 
specifications. The mirror-image relationship between the two ends of this mole- 
cule is consistent with the opposite designations of R and S for the two chiral cen- 
ters. (Not all (R,S)-isomers, of course, are meso structures only those whose two 
halves are chemically equivalent.) 

Problem 4.13 Give the R/S specification for each Stereoisomer you drew in 
Problem 4. 12 (p. 137). 



4.20 Conformational isomers 

In Sec. 3.5, we saw that there are several different staggered conformations of 
w-butane, each of which lies at the bottom of an energy valley at an energy 
minimum separated from the others by energy hills (see Fig. 3.4, p. 79). Different 
conformations corresponding to energy minima are called conformational isomers, 
or conformers. Since conformational isomers differ from each other only in the 
way their atoms are oriented in space, they, too, are stereoisomers. Like stereo- 
isomers of any kind, a pair of conformers can either be mirror images of each other 
or not. 

-Butane exists as three conformational isomers, one ami and two gauche (Sec. 
3.5). The gauche conformers, II and III, are mirror images of each other, and hence 
are (conformational) enantiomers. Conformers I and II (or I and III) are not mirror 
images of each other, and hence are (conformational) diastereomers. 

Although the barrier to rotation in w-butane is a little higher than in ethane, 
it is still low enough that at ordinary temperatures, at least interconversion of 
conformers is easy and rapid. Equilibrium exists, and favors a higher population 
of the more stable anti conformer; the populations of the two gauche conformers- 



SEC. 4.20 CONFORMATIONAL ISOMERS 139 

mirror images, and hence of exactly equal stability are, of course, equal. Put 
differently, any given molecule spends the greater part of its time as the ami 
conformer, and divides the smaller part equally between the two gauche conformers. 
As a result of the rapid interconversion, these isomers cannot be separated. 

Problem 4.14 Return to Problem 3.4 (p. 79) and, for each compound: (a) tell 
how many conformers there are, and label pairs of (conformational) enantiomers; 
(b) give the order of relative abundance of the various conformers. 

Easy interconversion is characteristic of nearly every set of conformational 
isomers, and is the quality in which such isomers differ most from the kind of 
stereoisofners we have encountered so far in this chapter. This difference in inter- 
convertibility is due to a difference in height of the energy barrier separating stereo- 
isomers, which is, in turn, due to a difference in origin of the barrier. By definition, 
interconversion of conformational isomers involves rotation about single bonds; 
the rotational barrier is in most cases a very low one and interconversion is 
easy and fast. The other kind of stereoisomers, configurational isomers, or inver- 
sional isomers, differ from one another in configuration about a chiral 
center. Fnterconversion here involves the breaking of a covalent bond, for which 
there is a very high barrier: 50 kcal/mole or more (Sec. 1.14). Interconversion is 
difficult, and unless one deliberately provides conditions to bring it about is 
negligibly slow. 

Interconvertibility of stereoisomers is of great prac* k cal significance because 
it limits their isolabitity. Hard-to-interconvert stereoisomers can be separated 
(with special methods, of course, for resolution of enantiomers) and studied in- 
dividually; among other things, their optical activity can be measured. Easy-to- 
interconvert isomers cannot be separated, and single isolated isomers cannot be 
studied; optical activity cannot be observed, since any chiral molecules are present 
only as non-resolvable racemic modifications. 

Our general approach to stereoisomers involves, then, two stages; first, we 
test the wperimposability of possible isomeric structures, and then we test their 
intercofwertibility. Both tests are best carried out with models. We make models 
of the two molecules and, without allowing any rotations about single bonds, we 
try to superimpose them: if they cannot be superimposed, they represent isomers. 
Next, we allow the models all possible rotations about single bonds, and repeatedly 
try to superimpose them: if they stilUrannot be superimposed, they arc non-inter- 
convertible, and represent confi^itraTional isomers \ bjj! if they can be superimposed 
after rotation, they are interconvertible and represent confornigtiwl i*nmm 

In^IeinTng with those aspects of stereochemistry that depend on isolation of 
stereoisomers isomer number or optical activity, for example, or study of the 
reactions of a single stereoisomer we can ignore the existence of easy-to-inter- 
convert isomers, which means most conformational isomers. For convenience the 
following "ground rule" will hold for discussions and problems in this book: 
unless specifically indicated otherwise, (he terms "stereoisomers" "enantiomers" 
ami " diastereomers" will refer only to configurational isomers, including geometric 
isomers (Sec. 5.6), and will exclude conformational isomers. The latter will be 
referred to as "conformational isomers/* "conformers," "conformational 
enantiomers," and "conformational diastereomers." 



140 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

There is no sharp boundary between easy-to-interconvert and hard-to-interconvert 
stereoisomers. Although we can be sure that Intel-conversion of configurational isomers 
will be hard, we cannot be sure that interconversion of conformational isomers will be 
easy. Depending upon the size and nature of substituents, the barrier to rotation about 
single bonds can be of any height, from the low one in ethane to one comparable to that 
for breaking a covalent bond. Some conformational isomers exist that are readily iso- 
lated, kept, and studied; indeed, study of such isomers (atropisomers) makes up a large 
and extremely important part of stereochemistry, one which, unfortunately, we shall 
not be able to take up in this beginning book. Other conformational isomers exist that 
can be isolated, not at ordinary temperatures, but at lower temperatures, where the 
average collision energy is lower. The conformational isomers that we shall encounter 
in this book, however, have low rotational barriers, and we may assume --until vve learn 
otherwise that when we classify stereoisomers as configurational or conformational, we 
at the same time classify them as hard-to-interconvert or easy-to-interconvert, 

Problem 4.15 At low temperatures, where collision energies are small. t*o iso- 
meric forms of the badly crowded CHBr 2 CHBr 2 have been isolated by crystallization 
<a) Give a formula or formulas (Newman projections) corresponding to each of the 
>eparable forms, (b) Which, if either, of the materials, as actually isolated at low tem- 
peratures, would be optically active? Explain. 



PROBLEMS 

1. What is meant by each of the following? 

(a) optical activity (k) meso compound 

(b) dextrorotatory (1) racemic modification 

(c) levorotatory (m) configuration 

(d) specific rotation (n) conformations 

(e) chirality (o) R 

(f) chiral molecule (p) S 

(g) chiral center (q) + 
(h) superimposable (r) 

(i) enantiomers (s) configurational isomers 

(j) diastereomers (t) conformational isomers 

2. (a) What is the necessary and sufficient condition for enantiomerism ? (b) What 
is a necessary but not a sufficient condition for optical activity? (c) What conditions 
must be met for the observation of optical activity ? (d) How can you tell from its formula 
whether or not a compound can exist as enantiomers? (e) What restrictions, if any, 
must be applied to the use of planar formulas in (d)? To the use of models in (d)? 
(f) Exactly how do you go about deciding whether a molecule should be specified as R 
orasS? 

3. Compare the dextrorotatory and levorotatory for^s of 5*r-butyl alcohol, 
CH 3 CH 2 CHOHCH 3 , with respect to: 

(a) boiling point (g) rate of reaction with HBr 

(b) melting point (h) infrared spectrum 

(c) specific gravity (i) nmr spectrum 

(d) specific rotation (j) adsorption on alumina 

(e) refractive index (k) retention time in gas chromatography 

(f) solubility in 100 g of water (1) specification as R or S 

4. Which of the following objects are chiral? 

(a) nail, screw, pair of scissors, knife, spool of thread; 

(b) glove, shoe^ sock, pullover sweater, coat sweater, scarf tied around your neck; 

(c) child's block, rubber ball, Pyramid of Cheops, helix (p. 11 57), double helix (p. 1 179); 



PROBLEMS 141 

(d) basketball, football, tennis racket, golf club, baseball bat, shotgun barrel, rifle 
barrel; 

(e) your hand, your foot, your ear, your nose, yourself. 

5. Assuming both your hands to be of equal strength and skill, which of the following 
operations could you perform with equal speed and efficiency ? 

(a) driving a screw, sawing a board, drilling a hole; 

(b) opening a door, opening a milk bottle, opening a coffee jar, turning on the hot 
water; 

(c) signing your name, sharpening a pencil, throwing a ball, shaking hands with another 
right hand, turning to* page 142. 

6. Draw and specify as R or S the enantiomers (if any) of: 

(a) 3-bromohexane (d) 1,3-dichloropentane 

(b) 3-chloro-3-methylpentane (e) 3-chloro-2,2,5-trimethylhexane 

(c) 1 ,2-dibromo-2-methylbutane (f) 1-deuterio-l-chlorobutane, 

CH 3 CH 2 CH2CHDa 

7. (a) What is the lowest molecular weight alkane that is chiral ? Draw stereochemical 
formulas of the enantiomers and specify each as R or S. (b) Is there another alkane of 
the same molecular weight that is also chiral ? If there is, give its structure and name, and 
specify the enantiomers as R or S. 

8. Draw stereochemical formulas for all the possible stereoisomers of the following 
compounds. Label pairs of enantiomers, and mesc compounds. Tell which isomcrs, if 
separated from all other stereoisomers, will be optically active. Give one isomer of each 
set its R/S specification. 

(a) CH 3 CHBrCHOHCH 3 (g) HOCH 2 (CHOH) 3 CH 2 OH 

(b) CH 3 CHBrCHBrCH 2 Br (h) C H,-CHC1 

(c) C 6 H 5 CH(CH 3 )CH(CH 3 )C 6 H 5 i " I (Make models.) 

(d) CH 3 CH 2 CH(CH 3 )CH : C^H 2 CH(CH 3 )CH 2 CH 3 CH 2 -CHC1 

(e) CH 3 CH(C 6 H 5 )CHOHCH, (i) CH 2 CHC1 

(f) CH 3 CHOHCHOHCMOHCH ? OH CHC1-CH, 

(j) methylethyl-A7-prop>i->/-butyldmmoniurn chloride, (RR / R*R / "N)*C1~ (See Sec. 

1.12.) 
(k) methylethyl-//-propyl-s^r-butylarnmonium chloride 

9. (a) In a study of chlorination oP propane, four products (A, B, C, and D) of 
formula r,H 6 C'l 2 \\cre isolated. What are their structures? 

(b) Each v\as chlorinated further, and the number of trichloro products (C 3 H 5 C1 3 ) 
obtained from each was determined by gas chromatography. A gave one trichloro 
product; B gave two: and C and D each gave three. What is the structure of A? Of B? 
Of C and D? 

(c) By another synthetic method, compound C was obtained in optically active form. 
Now what is the structure of C? Of D? 

(d) When optically active C was chlorinated, one of the trichloropropanes (E) obtained 
was optically active, and the other two were optically inactive. What is the structure of 
E? Of the other two? 

10. Draw configurational isomers (if any) of: (a) CH 2 BrCH 2 Cl ; (b) CH 3 CHBrCH 2 Cl. 
(c) For each substance of (a) and (b), draw all conformers. Label pairs of conformational 
enantiomers. 

11. The more stable conformer of >/-propyl chloride, CH 3 CH 2 CH 2 C1, is the 
gauche. What does this indicate about the interaction between Cl and CH 3 ? How do 
you account for this interaction? (Hint: See Sec. 1.19.) 

12. (a) What must be the dipole moment of the anti conformation of 1,2-dichloro- 
ethane, CH 2 C1 CH 2 C1? (b) At 32 in the gas phase, the measured dipole moment of 



142 STEREOCHEMISTRY I. STEREOISOMERS CHAP. 4 

1,2-dichloroethane is 1.12 D. What does this single fact tell you about the conformational 
make-up of the compound? (c) The dipole moment of a mixture of X and Y is given by 
the expression 



where N is the mole fraction of each kind of molecule. From bond moments, it has been 
estimated that the gauche conformation of 1,2-dichloroethane should have a dipole 
moment of about 3.2 D. Calculate the conformational composition of 1,2-dichloroethane 
at 32 in the gas phase. 



Chapter 



Alkenes I. Structure 
and Preparation 

Elimination 



5.1 Unsaturated hydrocarbons 

In our discussion of the alkanes we mentioned briefly another family of 
hydrocarbons, the alkenes, which contain less hydrogen, carbon for carbon, than 
the alkanes, and which can be converted into alkanes by addition of hydrogen. 
The alkenes were further described as being obtained from alkanes by loss of 
hydrogen in the cracking process. 

Since alkenes evidently contain less than the maximum quantity of hydrogen, 
they arc referred to as unsaturated hydrocarbons. This unsaturation can be satisfied 
by reagents other than hydrogen and gives rise to the characteristic chemical 
properties of alkenes. 

5.2 Structure of ethylene. The carbon-carbon double bond 

The simplest member of the alkene family is ethylene, C 2 H 4 . In view of the 
ready conversion of ethylene into ethane, we can reasonably expect certain structural 
similarities between the two compounds. 

To start, then, we connect the carbon atoms by a covalent bond, and then 
attach two hydrogen atoms to each carbon atom. At this stage we find that each 
carbon atom possesses only six electrons in its valence shell, instead of the required 
eight, and thai the entire molecule needs an additional pair, of electrons if it is to be 
neutral. We can solve both these problems by assuming that the carbon atoms 
can share two pairs of electrons. To describe this sharing of two pairs of electrons, 
we say that the carbon atoms are joined by a double bond. The carbon-carbon 
double bond is the distinguishing feature of the alkene structure. 



H H 
H:C::C:H 



H 



H 



Ethylene 
143 




144 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

Quantum mechanics gives a more detailed picture of ethylene and the carbon- 
carbon double bond. To form bonds with three other atoms, carbon makes use of 
three equivalent hybrid orbitals: sp 2 orbitals, formed by the mixing of one s and 
two p orbitals. As we have seen (Sec. 1.10), sp 2 orbitals lie in one plane, that of the 
carbon nucleus, and are directed toward the corners of an equilateral triangle; the 
angle between any pair of orbitals is thus 120. This trigonal arrangement (Fig. 
5.1) permits the hybrid orbitals to be as far apart as possible. Just as mutual 



Figure 5.1. Atomic orbitals: hybrid 
sp 2 orbitals. Axes directed toward 
corners of equilateral triangle. 



repulsion among orbitals gives four tetrahedral bonds, so it gives three trigonal 
bonds. 

If we arrange the two carbons and four hydrogens of ethylene to permit 
maximum overlap of orbitals, we obtain the structure shown in Fig. 5.2. Each 



Figure 5.2. Ethylene molecule: only 
a bonds shown. 



carbon atom lies at the center of a triangle, at whose corners are located the two 
hydrogen atoms and the other carbon atom. Every bend angle is 120. Although 
distributed differently about the carbon nucleus, these bonds individually are very 
similar to the bonds in ethane, being cylindrically symmetrical about a line joining 
the nuclei, and are given the same designation: a bond(sigma bond). 

The molecule is not yet complete, however. In forming the sp 2 orbitals, each 
carbon atom has used only two of its three p orbitals. The remaining p orbital 
consists of two equal lobes, one lying above and the other lying below the plane 
of the three sp 2 orbitals (Fig. 5.3); it is occupied by a single electron. If the/> orbital 





Figure 5.3. Ethylene molecule: carbon-carbon double bond. Overlap 
of/? orbitals gives IT bond; -n cloud above and below plane. 

of one carbon atom overlaps the p orbital of the other carbon atom, the electrons 
pair up and an additional bond is formed. 

Because it is formed by the overlap of p orbitals, and to distinguish it from 



SI ( . 5.4 HYBRIDIZATION AND ORBITAL SIZE 145 

the dilVcrcntly shaped a bonds, this bond is called a n bond (pi bond). It consists 
of uvo parts, one electron cloud that lies above the plane of the atoms, and another 
electron cloud that lies below. Because of less overlap, the -n bond is weaker than 
the carbon-carbon a bond. As we can see from Fig. 5.3, this overlap can occur 
only when all six atoms lie in the same plane. Ethylene, then, is aflat molecule. 

The carbon-carbon "double bond" is thus made up of a strong a bond and 
a weak -n bond. The total bond energy of 163 kcal is greater than that of the car- 
bon carbon single bond of ethane (88 kcal). Since the carbon atoms are held more 
tightly together, the CC distance in ethylene is less than the CC distance in 
ethane; that is to say, the carbon-carbon double bond is shorter than the carbon- 
carbon single bond. 

The a bond in ethylene has been estimated to have a strength of about 95 
kcal: stronger than the one in ethane because it is formed by overlap of sp 2 orbitals 
(Sec. 5.4). On this basis, we would estimate the strength of the -n bond to be 68 
kcal. 

This quantum mechanical structure of ethylene is verified by direct evidence. 
Electron diffraction and spectroscopic studies show ethylene (Fig. 5.4) to be a flat 
molecule, with bond angles very close to 120. The CC distance is 1.34 A as 
compared with the CC distance of 1.53 A in ethane. 

Figure 5.4. Ethylene molecule: shape H .7^>r> ! - 34 A ^ 

nnH ci \ ^\>^^ J C 



and size. 



In addition to these direct measurements, we shall soon see that two important 
aspects of alkene chemistry are consistent with the quantum mechanical picture of 
the double bond, and are most readily understood in terms of that picture. These 
are (a) the concept of hindered rotation and the accompanying phenomenon of 
geometric isomerism (Sec. 5.6), and (b) the kind of reactivity characteristic of the 
carbon-carbon double bond (Sec. 6.2). 

5.3 Propylene 

The next member of the alkene family is propylene, C 3 H 6 . In view of its 
great similarity to ethylene, it seems reasonable to assume that this compound also 
contains a carbon-carbon double bond. Starting with two carbons joined by a 
double bond, and attaching the other atoms according to our rule of one bond per 
hydrogen and four bonds per carbon, we arrive at the structure 

H H H 
H~C~C=C-H 

i 

Propylene 

5.4 Hybridization and orbital size 

The carbon-carbon double bond in alkenes is shorter than the carbon-carbon 
single bond in alkanes because four electrons bind more tightly than two. But, in 



146 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

addition, certain other bonds in alkenes are significantly shorter than their counter- 
parts in alkanes: for example, the C H distance is 1.103 A in ethylene compared 
with 1.112 A in ethane. To account for this and other differences in bond length, 
we must consider differences in hybridization of carbon. 

The carbon-hydrogen bonds of ethylene are single bonds just as in, say, 
ethane, but they are formed by overlap of sp 2 orbitals of carbon, instead of sp 3 
orbitals as in ethane. Now, compared with an sp 3 orbital, an sp 2 orbital has 
less p character and more s character. A p orbital extends some distance 
from the nucleus; an s orbital, on the other hand, lies close about the nucleus. 
As the s character of a hybrid orbital increases, the effective size of the 
orbital decreases and, with it, the length of the bond to a given second atom. Thus 
an sp 2 -s carbon-hydrogen bond should be shorter than an sp 3 -s carbon-hydrogen 
bond. 

Benzene, in most ways a quite different kind of molecule from ethylene 
(Sec. 10.1), also contains sp 2 -s carbon-hydrogen bonds; the C~-H bond distance is 
1.084 A, almost exactly the same as in ethylene. Acetylene (Sec. 8.2) contains 
.s/j-hybridized carbon which, in view of the even greater s character of the orbitals, 
should form even shorter bonds than in ethylene; this expectation is correct, the 
sp-s bond being only 1 .079 A. 

A consideration of hybridization and orbital size would lead one to expect 
an sp 2 -sp* bond to be shorter than an sp*-sp* bond. In agreement, the carbon- 
carbon single bond-distance in propylene is 1.501 A, as compared with the carbon- 
carbon distance of 1.534 A in ethane. The sp-sp* carbon-carbon single bond in 
methylacetylene (Sec. 8.19) is even shorter, 1.459 A. These differences in carbon- 
carbon single bond lengths are greater than the corresponding differences in 
carbon-hydrogen bond lengths; however, another factor (Sec. 8.18) besides the 
particular hybridization of carbon may be at work here. 

Consideration of hybridization arid orbital size helps us to understand other 
properties of molecules besides bond length: the relative acidities of certain hydro- 
carbons (Sec. 8.10), for example, and the relative basicities of certain amines (Sec. 
31.11), We might reasonably expect shorter bonds to be stronger bonds, and in 
agreement Table 1.2 (p. 21) shows that the C H bond dissociation energy in 
ethylene (104 kcal) is larger than that in ethane (98 kcal), and the C C (single) 
bond dissociation energy in propylene (92 kcal) is greater than that in ethane 
(88 kcal). Indeed, as will be discussed in Sec. 8.19, by affecting the stability of 
molecules, changes in hybridization may be of more fundamental importance than 
has been generally recognized. 



5.5 Th4 butylencs 

Going on to the butylenes, C 4 H a , we find that there are a number of possible 
arrangements. First of all, we may have a straight-chain skeleton as in //-butane, 
or a branched-chain structure as in isobutane. Next, even when \\e restrict our- 
selves to the straight-chain skeleton, we find that there are two possible arrange- 
ments that differ in position of the double bond in the chain. So far, then, we have 
a total of three structures; as indicated, these are given the names 1-butene, 2-butene, 
and isobutylene. 



SEC. 5.5 THE BUTYLENES 147 

HHHH HHHH HH 

H-C-C-C-C-H H-6-C-C-C-H H-C-C=C-H 

I I I I I I 

H H H H H j 

1-Butene 2-Butcne H CH 

H 

Isobutylene 

How do the facts agree with the prediction of three isomeric butylenes? 
Experiment has shown that not three but four alkenes of the formula C 4 H 8 exist; 
they have the physical properties shown in Table 5.1. 



Table 5.1 PHYSICAL PROPERTIES OF THE BUTYLENES 











Refractive 


Name 


b.p., C 


m.p., C 


Density (-20) 


Index (-12.7) 


Isobutylene 


-7 


-141 


0.640 


1.3727 


1-Butene 


-6 


<-J95 


.641 


1.3711 


r/ww-2-Butene 


4 1 


-106 


.649 


1.3778 


rw-2-Butene 


+ 4 


-139 


.667 


1.3868 



On hydrogenation, the isomer of b.p. 7 yields isobutane; this butylene 
evidently contains a branched chain, and has therefore the structure we have 
designated isobutylene. 

On hydrogenation, the other three isomers all yield the same compound, 
rt-butane; they evidently have a straight-chain skeleton. In ways that we shall 
study later (Sec. 6.29), it is possible to break an alkene molecule apart at the double 
bond, and from the fragments obtained deduce the position of the double bond in 
the molecule. When this procedure is carried out, the isomer of b.p. 6 yields 
products indicating clearly that the double bond is at the end of the chain; this 
butylene has therefore the structure we have designated 1-butene. When the same 
procedure is carried out on the two remaining isomers, both yield the same mix- 
ture of products; these products show that the double bond is in the middle of 
the chain. 

Judging from the products of hydrogenation and the products of cleavage, 
we would conclude that the butylenes of b.p. + 1 and +4 both have the structure 
we have designated 2-butene. Yet the differences in boiling point, melting point, 
and other physical properties show clearly that they are not the same compound, 
that is, that they are isomers. In what way can their structures differ? ^ 

To understand the kind of isomerism that gives rise to two 2-butenes, we must 
examine more closely the structure of alkenes and the nature of the carbon-carbon 
double bond. Ethylene is a flat molecule. We have seen that this flatness is a 
result of the geometric arrangement of the bonding orbitals, and in particular the 
overlap that gives rise to the IT orbital. For the same reasons, a portion of any 
alkene must also be flat, the two doubly-bonded carbons and the four atoms 
attached to them lying in the same plane. 

If we examine the structure of 2-butene more closely, and particularly if we 




148 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

use molecular models, we find that there are two quite different ways, I and II, in 
which the atoms can be arranged (aside from the infinite number of possibilities 
arising from rotation about the single bonds). In one of the structures the methyl 
groups lie on the same side of the molecule (I), and in the other structure they lie on 
opposite sides of the molecule (II). 






Now the question arises: can we expect to isolate two isomeric 2-butenes 
corresponding to these tv\o different structures, or are they too readily inter- 
converted like, say, the conformations of //-butane (Sec. 3.5)? 

Conversion of 1 into II involves rotation about the carbon-carbon double 
bond. The possibility of isolating isomers depends upon the energy required for 
this rotation. We have seen that the formation of the rr bond involves overlap of 
the p orbitals that lie above and below the plane of the a orbilals. To pass from 
one of these 2-butenes to the other, the molecule must be twisted so that the p 
orbitals no longer overlap; that is, the n bond must be broken (see Fig. 5.5). 



Figure 5.5. Hindered rotation about 
carbon-carbon double bond. Rota- 
tion v\ould prevent overlap of p orbi- 
tals and would break n bond. 




CH, 



Breaking the TT bond requires about 7Q kcal of energy, at room temperature an 
insignificant proportion of collisions possess this necessary energy, and hence the 
rate of this intercom ersion is extremely small. Because of this 70-kcaI energy 
barrier, then, there is hindered rotation about the carbon carbon double bond. As a 
result of this hindered rotation, two isomeric 2-butenes can be isolated. These 
are, of course, the butylenes of b.p. f- l' J and b.p. f 4. 



5.6 Geometric isomerism 

Since the isomeric 2-butencs diiTer from one another only in the way the atoms 
are oriented in space (but are like one another with respect to which atoms are 
attached to which other atoms), they belong to the general class we have called 
stereoisomers (Sec. 4.1). They are not, however, mirror images of each other, and 
hence are not enantiomers. As we have already said, stereoisomers that are not 
mirror images of each other are called diascereomers. 

The particular kind of diastereomers that owe their existence to hindered 
rotation about double bonds are called geometric isomers. The isomeric 2-butenes, 
then, are diastereomers, and more specifically, geometric isomers. 



SEC. 5.6 GEOMETRIC ISOMERISM 149 

We recall that the arrangement of atoms that characterizes a particular 
stereoisomer is called its configuration. The configurations of the isomeric 
2-butenes are the structures I and II. These configurations are differentiated in 
their names by the prefixes cis- (Latin: on this side) and trans- (Latin: across), 
which indicate that the methyl groups are on the same side or on opposite sides of 
the molecule. In a way that we are not prepared to take up at this time, the isomer 
of b.p. +4 has been assigned the cis configuration and the isomer of b.p. + 1 the 
trans configuration. 

CH 3 H CH 3 H 

C C Geometric 

II || isomers 

C C 

CH/ X H \\' X CH 3 

I II 

m-2-Butene f/ww-2-Butene 

b.p. 4-4 b.p. 4-1 

There is hindered rotation about any carbon-carbon double bond, but it 
gives rise to geometric isomerism only if there is*a certain relationship among the 
groups attached to the doubly-bonded carbons. We can look i^or this isomerism 
by drawing the possible structures (or better yet, by constructing them from molec- 
ular models), and then seeing if these are indeed isomeric, or actually identical. 
On this basis we find that propylene, 1-butene, and isobutvlene should not show 

H H H v H II H 

C C C No geometric 

II j| i| isomerism 

C C C 

/ \ / \ / \ 

/it TJ s~* ft u /"'Uf /^*L1 

v-jn 3 jn \_2-t' 5 jn V. n 3 v_jn 3 

Propylene 1-Butene Isobutylene 

isomerism; this conclusion agrees with the facts. Many higher alkenes may, of 
course, show geometric isomerism 

If we consider compounds other than hydrocarbons, we find that 1,1-dichloro- 
and 1,1-dibromoethene should not show isomerism, whereas the 1,2-dichloro- and 
1,2-dibromoethenes should. In every case these predictions have been found cor- 
rect. Isomers of the following physical properties have been isolated. 

Cl H Cl H Br H Br H 

V V V V 

II II li :! 

C C C C 

C I H H Cl Br H H Br 

ciS" trans- cis- trans- 

\ ,2-Dichloroethene 1 ,2-Dibromoethene 

b.p. 60 b.p. 48 b.p. 110 b.p. 108 

m.p. -80 m.p. -50" m.p. -53 m.p. -6 

As we soon conclude from our examination of these structures, geometric 
isomerism cannot exist if either carbon carries two identical groups. Some possible 
combinations are shown below. 



150 \ ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 



abab abab aa 



V V 



a a 



V 

c ' c c I- 

abba cddc cd 

Isomerism f Isomerism No isomerism 

The phenomenon of geometric isomerism is a general one and can be en- 
countered in any class of compounds that contain carbon-carbon double bonds 
(or even double bonds of other kinds). 

The prefixes cis and trans work very well for disubstituted ethylenes and some 
trisubstituted ethylenes. But how are we to specify configurations like these? 

CH 3 H CH 3 H CH 3 Br CH 3 Br H Cl H Cl 

V V V V V V 



B/ N C, C. X V H X X C1 a' \ cC \r B/ N C. 

Z Z Z E 

1-Bromo-l-chloropropene 2-Bromo-l-chloropropene l-Bromo-l,2-dichloroethene 

CH 3 > H Br > CH 3 Cl > H 

Br > Cl Cl > H Br > Cl 

Which groups are our reference points? Looking at each doubly-bonded carbon 
in turn, we arrange its two atoms or groups in their Cahn^ngoldUPrelog sequence. 
We then take the group of higher priority on the olne carbon and the group of 
higher priority on the other carbon, and tell whether they are on the same side of 
the molecule or on opposite sides. So that it will be clear that we are using this 
method of specification, we use the letter Z to mean on the same side, and the let- 
ter E to mean on opposite sides. (From the German: zusanfmen, together, and 
entgegen, opposite.) 

In so far as chemical and physical properties are concerned, geometric isomers 
show the same relationship to each other as do the other diastereomers we have 
encountered (Sec. 4.17). They contain the same functional groups and hence show 
similar chemical properties. Their chf "nical properties are not identical, however, 
since their structures are neither idei ical nor mirror images; they react with the 
same reagents, but at different rates. 

As the examples above illustrate, geometric isomers have different physical 
properties: different melting points, boiling points, refractive indices, solubilities, 
densities, and so on. On the basis of these different physical properties, they can 
be distinguished from each other and, once the configuration of each has been 
determined, identified. On the basis of these differences in physical properties they 
can, in principle at least, be separated. (See Sec. 4.17.) 

When we take up the physical properties of the alkenes (Sec. 5.9), we shall 
discuss one of the ways in which we can tell whether a particular substance is the 
els- or /raws-isomer, that is, one of the ways in which we assign configuratio 

A pair of geometric isomers are, then, diastereomers. Where do they fit into the 
other classification scheme, the one based on how the stereoisomcrs are intercon verted 



SEC. 5.8 NAMES OF ALKENES 151 

(Sec. 4.20)? We shall discuss this question in more detail later (Sec. 7.1), but for the 
moment we can say this. In the important quality of isolability, geometric isomers resem- 
ble configurational isomers, and for a very good reason: in both cases interconversion 
requires bond breaking a TT bond in the case of geometric isomers. 

5.7 Higher alkenes 

As we can see, the butylenes contain one carbon and two hydrogens more than 
propylene, which in turn contains one carbon and two hydrogens more than 
ethylene. The alkenes, therefore, form another homologous series, the increment 
being the same as for the alkanes: CH 2 . The general formula for this family is 

C n H 2n . 

As we ascend the series of alkenes, the number of isomeric structures for each 
member increases even more rapidly than in the case of the alkane series; in 
addition to variations in the carbon skeletons, there are variations in the position 
of the double bond for a given skeleton, and the possibility of geometric isomerism. 

Problem 5.1 Neglecting enantiomerism, draw structures of: (a) the six iso- 
meric pentylenes (C 5 H| ); (b) the four chloropropylenes (C 3 H 5 C1); (c) the eleven 
chlorobutylenes (C 4 H 7 C1). Specify as Z or E each geometric isomer. 

5.8 Names of alkenes 

Common names are seldom used except for three simple alkenes: ethylene, 
propylene, and isobutylene. The various alkenes of a given carbon number are, 
however, sometimes referred to collectively as the pentylenes (amylenes), hexylenes, 
heptylenes, and so on. (One sometimes encounters the naming of alkenes as deriva- 
tives of ethylene: as, for example, tetramethylethylene for (CH 3 ) 2 C C(CH 3 ) 2 .) 
Most alkenes are named by the IUPAC system. 

The rules of the IUPAC system are: 

1 . Select as the parent structure the longest continuous chain that contains the 
carbon-carbon double bond; then consider the compound to have been derived 
from this structure by replacement of hydrogen by various alkyl groups. The 
parent structure is known as ethene<propene,butene,pentene, and so on, depending 
upon the number of carbon atoms; each name is derived by changing the ending 
-ane of the corresponding alkane name to -ene: 

H 2 C=-CH 2 CH 3 -CH=CH 2 CH 3 CH 2 CH=- CH 2 CH 3 CH=-CHCH 3 

Ethene Propene 1-Butene 2-Butene 

(c/5- or trans-) 

CH 3 CH 3 CH 3 

CH 3 C=CH 2 CH 3 -C CH=CH 2 CH 3 C CH=CH CH 3 

2-Methylpropene I ^ 

3,3-DimethyI-l -butene 4-Methy!-2-pentene 

(cis- or trans-) 

2. Indicate by a number the position of the double bond in the parent chain. 
Although the double bond involves two carbon atoms, designate its position by 
the number of the first doubly-bonded carbon encountered when numbering from 
the end of the chain nearest the double bond; thus J -butene and 2-butene. 



152 ALKENES I. STRUCTURE AND PREPARATION CHAP. S 

3. Indicate by numbers the positions of the alkyl groups attached to the 
parent chain. 

Problem 5.2 Give the structural formula of: fa) 2,3-dimethyl-2-butene; 
(b) 3-bromo2-methylpropene; (c) m-2-methyl-3-heptene: (d) (>2-chloro-2-butene. 

Problem 5.3 Referring to your answer to Problem 5.1 (p. 151), give IUPAC 
names for: (a) the isomeric pentylenes; (b) the isomerk chloropropenes. 



5.9 Physical properties 

As a class, the alkenes possess physical properties that are essentially the same 
as those of the alkanes. They are insoluble in water, but quite soluble in non- 
polar solvents like benzene, ether, chloroform, or ligroin. They are less dense than 
water. As we can see from Table 5.2, the boiling point rises with increasing carbon 

Table 5.2 ALKENES 



Name 


Formula 


M.p., 
C 


B.P., 
C 


Density 
(at 20C) 


Ethylene 


CH 2 ^CH 2 


-169 


-102 




Propylene 


CH 2 ^CHCH 3 


-185 


- 48 




-Butene 


CH 2 ^CHCH 2 CHi 




-6.5 




-Pentene 


CH 2 ^CH(CH 2 ) 2 CH, 




30 


0.643 


-Hexene 


CH 2 CH(CH 2 ) 3 CH 3 


-138 


63.5 


.675 


-Heptene 


CH 2 -=CH(CH 2 )4CH, 


-119 


93 


.698 


-Octcne 


CH 2 ^CH(CH 2 )5CH 3 


-104 


122.5 


.716 


-Nonene 


CH 2 - CH(CH 2 ) 6 CH 3 




146 


.731 


-Decene 


CH 2 = CH(CH 2 ) 7 CH 3 


- 87 


171 


.743 


cw-2-Butene 


c/>CH 3 CH =CHCH 3 


-139 


4 




fro/js-2-Butene 


/ttim-CHjCH CHCFh 


-106 


1 




fsobutylcne 


CH 2 =C(CH 3 ) 2 


-141 


7 




cis-2- Pentene 


c/A-CH 3 CH- CHCH 2 CH 3 


-151 


37 


.655 


'ra/7y-2-Pentene 


/fwu-CHiCH CHCH 2 CH 3 




36' 


.647 


J-Methyl-1-butene 


CH 2 =CHCH(CH 3 )2 


-135 


25 


.648 


2-Methyl-2-butene 


CHiCH--C(CHi)2 


-123 


39 


.660 


Z,3-Dimethyl-2-butene 


(CH 3 ) 2 C- C(CH 3 ) 2 


- 74 


73 


.705 



number; as with the alkanes, the boiling point rise is 20 30" for each added carbon, 
except for the very small homologs. As before, branching lowers the boiling point. 
A comparison of Table 5.2 with Table 3.3 (p. 86) shows that the boiling point of 
an alkene is very nearly the same as that of the alkane with the corresponding car- 
bon skeleton. 

Like alkanes, alkenes are at most only weakly polar. Since the loosely held 
rr electrons of the double bond are easily pulled or pushed, dipole moments are 
larger than for alkanes. They are still small, however: compare the dipole moments 
shown for propylene and 1-butene, for example, with the moment of 1.83 r> for 
methyl chloride. The bond joining the alkyl group to the doubly-bonded carbon 
has a small polarity, which is believed to be in the direction shown, that is, with the 
alkyl group releasing electrons to the doubly-bonded carbon. Since this polarity 
is not canceled by a corresponding polarity in the opposite direction, it gives a 
net dipole moment to the molecule. 



SEC. 5.9 PHYSICAL PROPERTIES 153 

CH 3 H C 2 H 5 ^ H 

"V ^c 7 

V V 

^ ^>i 

/i = 0.35 D /i = 0.37 D 

cw-2-Butene, with two methyl groups on one side of the molecule and two 
hydrogens on the other, should have a small dipole moment. In mws-2-butene, 
on the other hand, with one methyl and one hydrogen on each side of the molecule, 
the bond moments should cancel out. Although the dipole moments have not 



CH \ 
I I 



m-2-Butene // ans- 2-Butenc 

expect small -\ > expect ^ = 

b.p. +4 b.p. -f 1 

m.p -139 m.p. -10<> J 

been measured directly, a small difference in polarity is reflected in the higher 
boiling point of the c/j-isomer. 

This same relationship exists for many pairs of geometric isomers. Because 
of its higher polarity the cfr-isomer is generally the higher boiling of a pair; 
because of its lower symmetry it fits into a crystalline lattice more poorly, and thus 
generally has the lower melting point. 

The differences in polarity, and hence the differences in melting point and 
boiling point, are greater for alkenes that contain elements whose electronegativities 
differ widely from that of carbon. For example: 

H Cl H Cl H Br H Br H I H I 

V V V V V V 



* 



\r Br 



cis trans cis trans cis trans 

/*=1.85D /x = p = 1.35 D /x = /i = 0.75D ^ = 

b.p. 60 C b.p. 48 b.p. 110 b.p. 108 b.p. 188 b.p. 192 

m.p. -80 m.p. -50 m.p. -53 m.p. -6 m.p. -14 m.p. +72 

The relationship between configuration and boiling point or melting point is 
only a rule of thumb, to which there are many exceptions (for example, the boil- 
ing points of the diiodoethenes). Measurement of dipole moment, on the other 
hand, frequently enables us positively to designate a particular isomer as c/$ or 
trans. 



154 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 



k 5.4 (a) Indicate the direction of the net dipole moment for each of the 
dihaloethenes. (b) Would os-2,3-dichloro-2-butene have a larger or smaller dipole 
moment than cfr-l,2-dichloroethene? (c) Indicate the direction of the net dipole mo- 
ment of <^],2-dibromo-l,2-dichloroethene. Will it be larger or smaller than the 
dipole moment of ctf-l,2-dichloroethene? Why? 



5.10 Industrial source 

Alkenes are obtained in industrial quantities chiefly by the cracking of petro- 
leum (Sec. 3.31). The smaller alkenes can be obtained in pure form by fractional 
distillation and are thus available for conversion into a large number of important 
aliphatic compounds. Higher alkenes, which cannot be separated from the compli- 
cated cracking mixture, remain as valuable components of gasoline. 

1-Alkenes of even carbon number, consumed in large quantities in the manu- 
facture of detergents, are available through controlled ionic polymerization of 
ethylene by the Ziegler-Natta method (Sec. 32-6). 

5.11 Preparation 

Alkenes containing up to five carbon atoms can be obtained in pure form 
from the petroleum industry. Pure samples of more complicated alkenes must be 
prepared by methods like those outlined below. 

The introduction of a carbon-carbon double bond into a molecule containing 
only single bonds must necessarily involve the elimination of atoms or groups from 
two adjacent carbons: 



-C-C- * C=-C~ Elimination 

I I 
Y Z 

In the cracking process already discussed, for example, the atoms eliminated are 
both hydrogen atoms: 

C- C J^ C--C + H 2 

I : 
H H 

The elimination reactions described below not only can be used to make 
simple alkenes, but also and this is much more important provide the best 
general ways to introduce carbon-carbon double bonds into molecules of all kinds. 



PREPARATION OF ALKENES 
1. Dehydrohalogenation of alkyl halides. Discussed in Sec. <U2 5.14. 

I 1 alcohol ! i Ease of dehydrohalogenation 

C-C + KOH > -C C- + KX f H 2 O of alkyl halides 

H X 3 > 2 > 1 



SEC. 5.11 PREPARATION 155 

Examples: 

CH 3 CH 2 CH 2 CH 2 C1 KOHfalc) > CHjCH 2 CH-CH> 

w-Butyl chloride 1-Butene 

CH 3 CH 2 CHCiCH 3 KQH(alc) > CH 3 CH CHCH 3 -f CH,CH 2 CH CH 2 

sec-Buiyl chloride 2-Butene 1-Butene 

W% M/ Q 

2. Dehydration of alcohols. Discussed in Sec. 5.19 5.23. 

I I acid I ' 

C-C- - > O=C- - + H 2 O Ease of dehydration of alcohols 

H OH Alkenes 3' > 2 > 1 

Alcohols 

Examples: 

H H H H 

H-C-C-H -^^ H C---C-H 4 H : 

H (^H Ethylene 

Ethyl alcohol 

CH 3 CH 2 CH 2 CH 2 OH "^^ CH 3 CH 2 CH-CH 2 + CH 3 CH=--CHCH 3 
-Butyl alcohol 1-Butene 2-Butcne 

Chief product 

CH 3 CH 2 -CH-CH 3 -^^ CH 3 CH-CHCH 3 + CH 3 CH 2 CH-CH 2 
I 2-Butene 1-Butene 

n 4 i i u i Chief product 

sec-Butyl alcohol 

3. Dehalogenation of vicinal dihalides. Discussed in Sec. 5.11. 

-C-C- + Zn * -C-C- + ZnX 2 



Example: 



CH 3 CHBrCHBrCH 3 --> CH 3 CH=CHCH 3 

2,3-Dibromobutane 2-Butene 



4. Reduction of alkynes. Discussed in Sec. 8.9. 



I R 

i An alkyne 



Pd or Ni-B (P-2) 



R W 

H" \ 



R H 

\ __ / 

(2 C Trans 

H R 



156 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

The most important of these methods of preparation since they are the most 
generally applicable are the dehydrohalogenation of alkyl halides and the dehydra- 
tion of alcohols. Both methods suffer from the disadvantage that, where the struc- 
ture permits, hydrogen can be eliminated from the carbon on either side of the 
carbon bearing the X or OH; this frequently produces isomers. Since the 
isomerism usually involves only the position of the double bond, it is not important 
in the cases where we plan to convert the alkene into an alkane. 

As we shall see later, alkyl halides are generally prepared from the correspond- 
ing alcohols, and hence both these methods ultimately involve preparation from 
alcohols; however, dehydrohalogenation generally leads to fewer complications 
and is often the preferred method despite the extra step in the sequence. * 

Dehaiogenation of vicinal (Latin: vicinalis, neighboring) dihalides is severely 
limited by the fact that these dihalides are themselves generally prepared from the 
alkenes. However, it is sometimes useful to convert an alkene to a dihalide while 
we perform some operation on another part of the molecule, and then to regenerate 
the alkene by treatment with zinc; this procedure is referred to as protecting the 
double bond. 

When a pure rw- or /ra/tf-alkene is wanted, uncontaminated with its stereoiso- 
mer, it can often be prepared by reduction of an alkyne with the proper reagent 
(Sec. 8.9). 



5,12 Dehydrohalogenation of alkyl halides 

Alkyl halides are converted into alkenes by dehydrohalogenation: elimination 
of the elements of hydrogen halide. Dehydrohalogenation involves removal of the 
halogen atom together with a hydrogen atom from a carbon adjacent to the one 

Dehydrohalogenation: elimination of HX 

i I II 

C C + KOH (alcoholic) - > C=C - + KX + H 2 O 

' I Alkene 

H A. 

Alky] halide 

bearing the halogen. It is not surprising that the reagent required for the elimina- 
tion of what amounts to a molecule of acid is a strong base. 

The alkene is prepared by simply heating together the alkyl halide and a 
solution of potassium hydroxide in alcohol. For example: 



CH 3 CH 2 CH 2 CI -* CH 3 CH^CH 2 ' CH 3 CHCH 3 

w-Propyl chloride Propylene ' 

Isopropyl chloride 



CH 3 CH2CH 2 CH 2 C1 -> CH 3 CH 2 CH - 
/i-Butyl chloride 1-Butene 



CH 3 CH 2 CHCH 3 ~~-- CHjCH-CHCH, + CH,CH 2 CH--=CH 2 
I 2-Butene 1-Butcnc 

0/)0 , <V}<> ' 

*?c-Butyl chloride /0 ^ /w 



SEC. 5.13 MECHANISM OF DEHYDROHALOGENAT1ON 157 

As we can see, in some cases this reaction yields a single alkene. and in other cases 
yields a mixture. w-Butyl chloride, for example, can eliminate hydrogen only from 
C-2 and hence yields only 1-butene. sec-Butyl chloride, on the other hand, can 
eliminate hydrogen from either C-l or C-3 and hence yields both 1-butene and 
2-butene. Where the two alkenes can be formed, 2-butene is the chief product; 
this fact fits into a general pattern for dehydrohalogenation which is discussed in 
Sec. 5.14. 

Problem 5.5 Give structures of all alkenes expected from dehydrohalogenation 
of: (a) 1-chloropentane, (b) 2-chloropentane, (c) 3-chloropentane, (d) 2-chloro-2-methyl- 
butane, (e) 3-chloro-2-methylbutane, (f) 2-chloro-2,3-dimethylbutane, (g) 1-chloro- 
2,2-dimcthylpropane. 

Problem 5.6 What alkyl halidc (if any) would yield each of the following pure 
alkenes upon dehydrohalogenation? (a) isobutylene, (b) 1-pentene, (c) 2-pentene, 
(d) 2-methyM-butene, (e) 2-methyl-2-butene, (f) 3-methyl-l-butene. 



5.11 Mechanism of dehydrohalogenation 

The function of hydroxide ion is to pull a hydrogen ion away from carbon; 
simultaneously a halide ion separates and the double bond forms. We should 



I I 
C::C + :X:~ -* H 2 O 




represented as 



t? 1 

f* ^-1 

H 


-->\:==c^ + x- 


+ H 2 



where arrows show the 
direction of electron shift 

notice that, in contrast to free radical reactions, the breaking of the CH and 
C X bonds occurs in an unsymmetrical fashion: hydrogen relinquishes both 
electrons to carbon, and halogen retains both electrons. The electrons left behind 
by hydrogen are now available for formation of the second bond (the TT bond) 
between the carbon atoms. 

What supplies the energy for the breaking of the carbon-hydrogen and 
carbon-halogen bonds? 

(a) First, there is formation of the bond between the hydrogen ion and the 
very strong base, hydroxide ion. 

(b) Next, there is formation of the TT bond which, although weak, does supply 
about 70 kcal/mole of energy. 



158 ALKENES I. STRUCTURE AND PREPARATION CHAP, 5 

(c) Finallyand this is extremely important there is the energy of solvation 
of the halide ions. Alcohol, like water, is a polar solvent. A liberated halide ion 
is surrounded by a cluster of these polar molecules; each solvent molecule is 
oriented so that the positive end of its dipole is near the negative ion (Fig. 5.6). 




Figure 5.6. Ion dipole interaction: 
solvated halide ion. 



Although each of these ion- dipole bonds (Sec. 1.21) is weak, in the aggregate they 
supply a great deal of energy. (We should recall that the ion dipole bonds in 
hydrated sodium and chloride ions provide the energy for the breaking down of the 
sodium chloride crystalline lattice, a process which in the absence of water requires 
a temperature of 801 \) Just as a hydrogen ion is pulled out oj the molecule by a 
hydroxide ion* so a halide ion is pulled out by solvent molecules. 

The free-radical reactions of the alkanes, which we studied in Chap. 3, are 
chiefly gas phase reactions. It is significant that ionic reactions (like the one just 
discussed) occur chiefly in solution. 

(We shall return to dehydrohalogenation after we have learned a little more 
chemistry (Sec. 14.20), and have a look at the evidence for this mechanism.) 

5.14 Orientation and reactivity in dehydrohalogenation 

In cases where a mixture of isomeric alkenes can be formed, which isomer, if 
any, will predominate? Study of many dehydrohalogenation reactions has shown 
that one isomer generally does predominate, and that it is possible to predict which 
isomer this v\ill be- that is, to predict the orientation of elimination -on the basis 
of molecular structure. 



CH 3 CH 2 CHBKH 3 KOH <' tlc >^ CH 3 CH-CHCH 3 and CH 3 CH 2 CH=CH 2 



CH.,CH 2 CH 2 CHBrCH, KO1Ua!c) > CH 3 CH 2 CH-CHCH 3 

71% 

and CH 3 CH 2 CH 2 CH=CH 2 

CHj CH 3 CH 3 

CH 3 CH 2 CBKH< K0>f(jk) > CH 3 CH=-CCH 3 and CH 3 CH 2 C- CH 2 

71% 29% 

Once more, orientation is determined by the relative rates of competing 
reactions. For .w-butyl bromide, attack by base at any one of three hydrogens 



SEC. 5.14 ORIENTATION AND REACTIVITY IN DEHYDROHALOGENATION 



159 



(those on C-l) can lead to the formation of 1-butene; attack at either of two hydro- 
gens (on C-3) can lead to the formation of 2-butene. We see that 2-butene is the 
preferred product that is, is formed faster despite a probability factor of 3:2 
working against its formation. The other examples fit the same pattern: the pre- 
ferred product is the alkene that has the greater number of alkyl groups attached 
to the doubly-bonded carbon atoms. 

Ease of formation of alkenes 
R 2 C=CR 2 > R 2 O=CHR > R 2 O=CH 2 , RCHCHR > RCH=-CH 2 

In Sec. 6.4 we shall find evidence that the stability of alkenes follows exactly the 
same sequence. 

Stability of alkenes 

R 2 C=CR 2 > R 2 O=CHR > R 2 O=CH 2 , RCH=CHR 



In dehydrohalogenation, the more stable the alkene the more easily it is formed. 

Examination of the transition state involved shows that it is reasonable that 
the more stable alkene should be formed faster: 



X 

I I 

-c-c 

1 



OH- 










X- + H 2 



8- 
H--OH 

Transition state : 

partly formed 
double bond 



The double bond is partly formed, and the transition state has thus acquired alkene 
character. Factors that stabilize an alkene also stabilize an incipient alkene in the 
transition state. 

Alkene stability not only determines orientation of dehydrohalogenation, but 
also is an important factor in determining the reactivity of an alkyl halide toward 
elimination, as shown at the top of the next page. 



Reactant 

CH 3 CH 2 Br 

CH 3 CH 2 CH 2 Br 

CH 3 CHBrCH 3 

(CH 3 ) 3 CBr 



Product 

CH 2 -CH 2 

CH 3 CH-CH ? 

CH 3 CH=-CH 2 

(CH 3 ) 3 C-CH 2 



Relative rates 
1.0 
3.3 
9.4 
120 



Relative rates per H 
1.0 
5.0 
4.7 
40 



As one proceeds along a series of alkyl halides from 1 to 2 to 3, the structure 
by definition becomes more branched at the carbon carrying the halogen. This 
increased branching has two results: it provides a greater number of hydrogens for 
attack by base, and hence a more favorable probability factor toward elimination; 
and it leads to a more highly branched, more stable alkene, and hence a more 
stable transition state and lower act . As a result of this combination of factors, 
in dehydrohalogenation the order of reactivity of RX is 3 > 2 > 1. 



160 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

Problem 5.7 Predict the major product of each dehydrohalogenation in Problem 
5.5, page 157. 



5.15 Carbonium ions 

To account for the observed facts, we saw earlier, a certain mechanism was 
advanced for the halogenation of alkanes; the heart of this mechanism is the fleet- 
ing existence of free radicals, highly reactive neutral particles bearing an odd 
electron. 

Before we can discuss the preparation of alkenes by dehydration of alcohols, 
we must first learn something about another kind of reactive particle: the car- 
bonium ion, a group of atoms that contains a carbon atom bearing only six electrons. 
Carbonium ions are classified as primary, secondary, or tertiary after the carbon 
bearing the positive charge. They are named by use of the word cation. For exam- 
pie: 

H H H CH 3 

H:C CH,:Co CH 3 :C:CH 3 CH 3 :C:CH 3 

H H ' @ 

Methyl cation Ethyl cation Isopropyl cation /e/7-Butyl cation 

(primary, 1") (secondary, 2) (tertiary, 3) 

Like the free radical, the carbonium ion is an exceedingly reactive particle, 
and for the same reason: the tendency to complete the octet of carbon. Unlike 
the free radical, the carbonium ion carries a positive charge. 

One kind of unusually stable carbonium ion (Sec. 12.19) was recognized as 
early as 1902 by the salt-like character of certain organic compounds. But direct 
observation of simple alkyl cations should be exceedingly difficult, by virtue of 
the very reactivity and hence short life that we attribute to them. Nevertheless, 
during the 1920's and 193CTs, alkyl cations were proposed as intermediates in many 
organic reactions, and their existence was generally accepted, due largely to the 
work of three chemists: Hans Meerwein of Germany, "the father of modern car- 
bonium ion chemistry;" Sir Christopher Ingold of England; and Frank Whitmore 
of the United States. The evidence consisted of a wide variety of observations 
made in studying the chemistry of alkenes, alcohols, alkyl halides, and many 
other kinds of organic compounds: observations that revealed a basically similar 
pattern of behavior most logically attributed to intermediate carbonium ions. A 
sizable part of this book will be devoted to seeing what that pattern is. 

In 1963, George Olah (now at Case Western Reserve University) reported the 
direct observation of simple alkyl cations. Dissolved in the extremely powerful 
Lewis acid SbF 5 , alkyl fluorides (and, later, other halides) were found to undergo 
ionization to form the cation, which could be studied at leisure. There was a 

RF + SbF 5 > R+ SbF 6 - 

dramatic change in the nmr spectrum (Chap. 13), from the spectrum of the 
alkyl fluoride to the spectrum of a molecule that contained no fluorine but instead 
j/? 2 -hybridized carbon with a very low electron density. Figure 5.7 shows what 
was observed for the terf-butyl fluoride system : a simple spectrum but, by its very 



SEC. 5.16 



STRUCTURE OF CARBONIUM IONS 



161 



simplicity, enormously significant. Although potentially very reactive, the to/- 
butyl cation can do little in this environment except try to regain the fluoride ion 
and the SbF 5 is an even stronger Lewis acid than the cation. 



a (CHjhC'Fneat 
b (CHabCF in SbF 5 



K 20 




3.0 



2.0 



1.0 



0.0 6 



8.0 7.0 60 5.0 4.0 

Courtesy of The Journal jf American Chemkal Society 

Figure 5.7. Proton nmr spectrum of (a) tert-buty\ fluoride and (b) tert-bulyl 
cation. Jn (#), proton signal split imo two peaks by coupling with nearby 
fluorine. In (6), single peak, shifted far downfield; strong dcshielding due to 
low electron density on positive carbon. 

By methods like this, Olah has opened the door to the study not just of the 
existence of organic cations of many kinds, but of intimate details of their struc- 
ture. 

5.16 Structure of carbunium ions 

In a carbonium ion, the electron-deficient carbon is bonded to three other 
atoms, and for this bonding uses v/?- orbitals; the bonds are trigonal, directed to 
the corners of an equilateral triangle. This part of a carbonium ion is therefore 





(a) 

Figure 5.8. A carbonium ion. (a) Only a bonds shown, (b) Empty p orbi- 
tal above and below plane of or bonds. 



flat, the electron-deficient carbon and the three atoms attached to it lying in the 
same plane (Fig. 5.8a). 



162 ALKENES I. SIRLCTLRE AND PREPARATION CHAP. 5 

But our description of the molecule is not yet quite complete. Carbon has 
left a /> orbital, with its t\\o lobes lying above and below the plane of the <r bonds 
(Fig. 5.X/j); in a carbonium ion, the/; orbital is empty. Although formally empty, 
this /> orbital, we shall find, is intimately involved in the chemistry of carbojiium 
ions: in their stability, and the stability of various transition states leading to their 
formation. 

'Fhere can be little doubt that carbonium ions actually are flat. The quantum 
mechanical picture of a carbonium ion is exactly the same as that of boron tri- 
tluoridc (Sec. 1JO). a molecule whose flatness is firmly established. Nmr and infra- 
icd spectra of the stabilized carbonium ions studied by Olah are consistent with 
A/> : Inbndi/alion and flatness: in particular, infrared and Raman spectra of the 
/IT/- bmy I cation are strikingly similar to those of trimethylboron, known to be 
flat. 

F- \idence of another kind indicates that carbonium ions not only normally 
arc flat, but ha\c a stiong necJ to 'he flat. Consider the three tertiary alkyl bromides: 
/</7-but\l bromide: and I and II, which are bicyclic (t,vo-ringed) compounds with 

B 

H ' ( 

\\)C k 

1 II 

/r/-Bul>l bromide 1-Bromobicyclo- 1 -Bromobicyclo- 

[222]octane [2.2.1]heptanc 

bromine at the bridgehead. The impact of a high-energy electron can remove 
bromine from an alkyl bromide and generate a carbonium ion; the energy of the 
electron required to do the job can be measured. On electron impact, I requires 
5 kcal mole more energy to form the carbonium ion than does /e/7-butyl bromide, 
and II requires 20 kcal mole more energy. 

Mo\\ arc we to interpret these facts? On conversion into a carbonium ion, 
three carbons must move into the plane of the electron-deficient carbon: easy for 

fir 

I f* as 



C 

Alkyl bromide Carbonium ion 

Teirahetlral Trigonal 

the open-chum /cry-butyl group; but difficult for F, where the three carbons are 
tied back by the ring system; and still more difficult for II, where they are tied 
back more tightly by the smaller ring. 

Imagine* or, better, make a model of I or II. You could squash the top of the 
molecule flat, but only by distorting the angles of the other bonds away from their nor- 
mal tetrahedral angle, and thus introducing angle strain (Sec. 9.7). 

Now, why is there this need to be flat? Partly, to permit formation of the 
strongest possible a bonds through sp 2 hybridization. But there is a second ad- 



SEC. 5.17 STABILITY OF CARBQNILM IONS 163 

vantage of flatness, one which is related to the major factor determining carbonium 
ion stability, accommodation of charge. 



5.17 Stability of carbonium ions. Accommodation of charge 

The characteristic feature of a carbonium ion is, by definition, the electron- 
deficient carbon and the attendant positive charge. The relative stability of a 
carbonium ion is determined chiefly by how well it accommodates that charge. 

According to the laws of electrostatics, the stability of a charged system is 
increased by dispersal of the charge. Any factor, therefore, that tends to spread out 
the positive charge of the electron-deficient carbon and distribute it over the rest 
of the ion must stabilize a carbonium ion. 

Consider a substituent, G, attached to an electron-deficient carbon in place 
of a hydrogen atom. Compared with hydrogen, G may either release electrons 
or withdraw electrons (Sec. 1.23). 



G->C 



Carbonium Ion Stability 





I I 

G releases electrons: G withdraws electrons 
disperses charge, intensifies charge, 

stabilizes cation destabilizes cation 

An electron-releasing substituent tends to reduce the positive charge at the 
electron-deficient carbon; in doing this, the substituent itself becomes somewhat 
positive. This dispersal of the charge stabilizes the carbonium ion. 

An electron-withdrawing substituent tends to intensify the positive charge 
on the electron-deficient carbon, and hence makes the carbonium ion less stable. 

We consider (Sec. 1.23) electronic effects to be of two kinds: inductive effects, 
related to the electronegativity of substituents ; and resonance effects. In the case 
of carbonium ions, we shall see (Sec. 8.21), a resonance effect involves overlap of 
the "empty" p orbital of the electron-deficient carbon with orbitals on other, 
nearby atoms; the result is, of course, that the p orbital is no longer empty, and 
the electron-deficient carbon no longer so positive. Maximum overlap depends on 
coplanarity in this part of the molecule, and it is here that we find the second ad- 
vantage of flatness in a carbonium ion. 

So far, we, have discussed only factors operating within a carbonium ion to 
make it more or less stable than another carbonium ion. But what is outside the 
carbonium ion proper its environment can be even more important in deter- 
mining how fast a carbonium ion is formed, how long it lasts, and what happens 
to it. There are onions, one of which may stay close by to form an ion pair. There 
is the solvent: a cluster of solvent molecules, each with the positive end of its di- 
pole turned toward the cation; possibly one solvent molecule or two playing a 
special role through overlap of one or both lobes of the p orbital. There may be a 
neighboring group effect (Chap. 28), in which a substituent on a neighboring 
carbon approaches closely enough to share its electrons and form a covalent bond: 
an internal factor, actually, but in its operation much like an external factor. 

In all this we see the characteristic of carbonium ions that underlies their 
whole pattern of behavior: a need for electrons to complete the octet of carbon. 



164 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

5.18 Relative stabilities of alkyl cations 

The amount of energy required to remove an electron from a molecule or 
atom is called the ionization potential. (It is really the ionization energy.) The 
ionization potential of a free radical is, by definition, the A// for the conversion of 
the radical into a carbonium ion: 

R. ^ R+ + e - A H ~ ionization potential 

In ways that we cannot go into, the ionization potentials of many free radicals 
have been measured. For example: 

CH 3 - > CH 3 f 4- e" A// = 229 kcal/mole 

CH 3 CH 2 - > CH 3 CH 2 * + e~ A// = 202 

CH 3 CHCH 3 CH 3 CHCH 3 + e" A// - 182 

CH 3 CH 3 

CH 3 -C-CH 3 > CH 3 ~C-CH 3 + e A// = 171 

As we can see, the values decrease in the order: CH 3 - > 1 > 2 > 3. 

Bond dissociation energies have already shown (Sec. 3.24) that the amount 
of energy required to form free radicals from alkanes decreases in the same order: 
CH 3 - > 1 > 2 > 3. If we combine these two sets of data ionization poten- 
tials and bond dissociation energies we see (Fig. 5.9) that, relative to the various 
alkanes concerned, the order of stability of carbonium ions is: 

Stability of carbonium ions 3 > 2 > 1 > CH 3 4 

Differences in stability between carbonium ions are much larger than between 
free radicals. The ter/-butyl free radical, for example, is 13 kcal more stable than 
the methyl free radical; the tert-butyl cation is 71 kcal more stable than the methyl 
cation. 

What we really want as standards for stability of carbonium ions are, of course, the 
kinds of compounds they are generated from: alcohols at this particular point or, later, 
alkyl halides (Chap. 14). However, the relative stabilities of most ordinary neutral 
molecules closely parallel the relative stabilities of the alkanes, so that the relative order 
of stabilities that we have arrived at is certainly valid whatever the source of the car- 
bonium ions. To take an extreme example, the difference in stability between methyl and 
ferf-butyl cations relative to the alkanes, as we have just calculated it, is 71 kcal. Relative 
to other standards, the difference in stability is: alcohols, 57 kcal; chlorides, 74 kcal; 
bromides, 78 kcal; and iodides, 76 kcal. 

Now, by definition, the distinction among primary, secondary, and tertiary 
cations is the number of alkyl groups attached to the electron-deficient carbon. 
The facts are, then, that the greater the number of alkyl groups, the more stable 
the carbonium ion. 

H H R R 

H-C8 R->C0 R->C0 R->C 

H H H R 

Methyl cation Primary cation Secondary cation Tei tiary cation 

Electron release: Disperses charge, stabilizes ion 



SEC. 5.18 



RELATIVE STABILITIES OF ALKYL CATIONS 



165 



0> 

1 

8 

J* 

1 

V 

1 
C 



CHa(333) 



T 
R-H 



CH,,CH 2 '(.K)) 



CHUCHCH.M277) 



CH 3 
CH 3 C-CH 3 (262) 



lonization 
potential 



229 



202 



182 



171 



CH 3 - 



CHiCHs 



CH.<CHCH 3 



CH, 
CHa-C-CH. 



T 
R- 



104 



98 



95 



91 



Bond 

dissociation 
energy 

R- + H- 

T 

R-H 



CH 4 



CH;iCH,, 



CH 3 CH : CH 3 



CH 8 

CHj C CHs 

i 
H 



Figure 5.9. Relative stabilities of alkyl cations. (Plots aligned with each 
other for easy comparison.) 



If our generalization about dispersal of charge applies in this case, alkyl 
groups must release electrons here: possibly through an inductive effect, possibly 
through resonance (hyper conjugation, Sec. 8.21). 



The electronic effects of alkyl groups are the most poorly understood of all such 
effects. Dipole moments indicate that they are electron-releasing when attached to a ir 
electron system, but weakly electron-withdrawing in saturated hydrocarbons. Ingold 
(p. 160) has suggested that alkyl groups can do pretty much what is demanded of them by 
other groups in the molecule. This much seems clear: (a) in general, carbonium ions and 
incipient carbonium ions are stabilized by electron release; (b) alkyl groups stabilize 
carbonium ions; (c) alkyl groups affect a wide variety of reactions in a manner consistent 
with their being electron-releasing. 



\LKENES I. STRl CTl'RF. AND PREPARATION 



CHAP. 5 



5.19 Dehydration of alcohols 

Alcohols are compounds of the general formula, ROH, where R is any alkyl 
group: the hydroxyl group, OH, is characteristic of alcohols, just as the carbon- 
carbon double bond is characteristic of alkenes. An alcohol is named simply by 
naming the alkyl group that holds the hydroxyl group and following this by the 
word alcohol. It is classified as primary (\ ). secondary (2 ), or tertiary (3), depend- 
ing upon the nature of the carbon atom holding the hydroxyl group (Sec. 3.11). 
For example: 



CH 3 CH 2 OH 

Ethyl alcohol 
A pntnaty alcohol 


CH V 
VHCH.OH 

CM/ 

Isobutyl alcohol 
A primaiy alcohol 


CH 3 CHCH 3 

OH 

Isopropyl alcohol 
A secondary alcohol 


CH 3 % 
CH 3 -C-CH 3 

OH 

/T/-Butyl alcohol 
A tertiary alcohol 



An alcohol is converted into an alkene by dehydration: elimination of a 
molecule of water. Dehydration requires the presence of an acid and the applica- 
tion of heat. It is generally carried out in either of two ways: (a) heating the 

r r > rwJ + 

" C " C > -~ c = c - + 



H OH 

Alcohol 



4Cld > rw 
Teat > -~ c = c - 

Alkcnc 



Dehydration: 
elimination of H 2 O 



alcohol with sulfuric or phosphoric acid to temperatures as high as 200, or 
(b) passing the alcohol vapor over alumina, A1 2 O 3 , at 350-400, alumina here serving 
as a Lewis acid (Sec. 1.22). 

The various classes of alcohols differ widely in ease of dehydration, the order 
of reactivity being 

Ease of dehydration of alcohols 3 3 > 2 > 1 

The following examples show how these differences in reactivity affect the experi- 
mental conditions of the dehydration. (Certain tertiary alcohols are so prone to 
dehydration that they can be distilled only if precautions are taken to protect the 
system from the acid fumes in the ordinary laboratory.) 



Ethyl alcohol 

CH 3 CH2CH 2 CH 2 OH 

- Butyl alcohol 

CH 3 CH 2 CHOHCH 3 

sec- Butyl alcohol 

C 3 

CM* C CH* 


170 
15% H 2 S0 4 ^ 


Ethylene 

CH 3 CH==CHCH 3 

2-Butcne 
Chief product 

CH 3 CH==CHCH 3 

2-Butene 
Chiej product 

CH 3 

C\4 , r 1 PH 


140" * 
60% H 2 S0 4 


100 
20% H : SO< 


OH 

tort-Butyl .alcohol 


85-90" 


Isobutylene 



SEC. 5.20 MECHANISM OF DEHYDRATION OF ALCOHOLS 167 

Where isomeric alkenes can be formed, we again find the tendency for one 
isomer to predominate. Thus, sec-butyl alcohol, which might yield both 2-butene 
and 1-butene, actually yields almost exclusively the 2-isomer (see Sec. 5.23). 

The formation of 2-butene from //-butyl alcohol illustrates a characteristic of 
dehydration that is not shared by dehydrohalogenalion: the double bond can be 
formed at a position remote from the carbon originally holding the OH group. 
This characteristic is accounted for later (Sec. 5.22). It is chiefly because of the 
greater certainty as to where the double bond will appear that dehydrohalogenation 
is often preferred over dehydration as> a method of making alkenes. 

5.20 Mechanism of dehydration of alcohols 

The generally accepted mechanism for the dehydration of alcohols is sum- 
marized in the following equations; for the sake of simplicity, ethyl alcohol is used 
as the example. The alcohol unites (step 1) with a hydrogen ion to form the pro- 
tonated alcohol, which dissociates (step 2) into water and a carbonium ion; the 
carbonium ion then loses (step 3) a hydrogen ion to form the alkene. 

H H H H 

II II 

(1) H C : C H J H+ < > H C : C H 

H :0:H H :O^H 

Alcohol H 

Protonated alcohol 



(2) _ 

H 
Carbonium ion 



(3) 



The double bond is thus formed in two stages, OH being lost (as H 2 O) in 
step (2) and H being lost in step (3). This is in contrast to dehydrohalogenation 
(Sec. 5.13), where the halogen and hydrogen are lost simultaneously. 

The first step is simply an acid-base equilibrium in the Lowry-Bronsted sense 
(Sec. 1.22). When sulfuric acid, for example, is dissolved in water, the following 
reaction occurs: 






HS0 4 ~ 
Stronger base 6 " Weaker base 

The hydrogen ion is transferred from the very weak base, HSO 4 ~, to the stronger 



168 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

base, H 2 O, to form the oxonium ion, H 3 O+ ; the basic properties of each are due 
of course, to the unshared electrons that are available for sharing with the hydro- 
gen ion. An alcohol also contains an oxygen atom with unshared electrons and 
hence displays basicity comparable to that of water. The first step of the mechan- 
ism is more properly represented as 



CH 3 CH 2 :O: + HO-S-OH <=* CH 3 CH 2 :O:H f HSO 4 ~ 
Stronger base Q Weaker base 

where the hydrogen ion is transferred from the bisulfate ion to the stronger base, 
ethyl alcohol, to yield the substituted oxonium ion, C 2 H 5 OH 2 + , the protonated 
alcohol. 

In a similar way, step (3) does not actually involve the expulsion of a naked 
hydrogen ion, but rather a transfer of the hydrogen ion to a base, the strongest one 
around, C 2 H 5 OH. 

H H 



C 2 H 5 OH I HC : C-H <=n C 2 H 5 OH 2 + +H-C::C-H 




H 



For convenience we shall frequently show the addition or expulsion of a hydrogen 
ion, H + , but it should be understood that in all cases this actually involves the 
transfer of a proton from one base to another. 

In summary, the mechanism for dehydration is the following: 



1 i 

p f> _|_ Li 

II 

H OH 


Dehydration* 
H OH 2 + 




(2) -<-<- = -<- + HP 



' * C~rC < C C ~H H : B 

\^j / \ 

H 

>^ 

All three reactions are shown as equilibria, since each step is readily reversible; 
as we shall soon see, the exact reverse of this reaction sequence is involved in the 
formation of alcohols from alkenes (Sec. 6.10). Equilibrium (I) lies very far to the 



SEC. 5.20 MECHANISM OF DEHYDRATION OF ALCOHOLS 169 

right; sulfuric acid, for example, is known to be nearly completely ionized in 
alcohol solution. Since there is a very low concentration of carbonium ions 
present at any time, equilibrium (2) undoubtedly lies very far to the left. Occasion- 
ally one of these few carbonium ions undergoes reaction (3) to form the alkene. 
Under the conditions of dehydration the alkene, being quite volatile, is generally 
driven from the reaction mixture, and thus equilibrium (3) is shifted to the right. 
As a consequence the entire reaction system is forced toward completion. 

The carbonium ion is formed by dissociation of the protonated alcohol; this 
involves separation of a charged particle, R + , from a neutral particle, H 2 O. It is 
obvious that this process requires much less energy than would formation of a 
carbonium ion from the alcohol itself, since the latter process involves separation 
of a positive particle from a negative particle. Viewed in another way, the car- 
bonium ion (a Lewis acid) releases the weak base, water, much more readily than it 

ROH 2 <S > R + H 2 O Easy 

Weak base: 
good leaving group 

ROM > R + OH- Difficult 

Strong base: 
poor leaving group 

releases the extremely strong base, hydroxide ion ; that is to say, water is a much 
better leaving group than hydroxide ion. Indeed, the evidence indicates that 
separation of a hydroxide ion from an alcohol almost never occurs; reactions 
involving cleavage of the C O bond of an alcohol seem in nearly every case to 
require an acidic catalyst, the function of which, as in the present case, is to form 
the protonated alcohol. 

Finally, we must realize that even dissociation of the protonated alcohol is 
made possible only by solvation of the carbonium ion (compare Sec. 5.13). Energy 
for the breaking of the carbon-oxygen bond is supplied by the formation of many 
ion-dipole bonds between the carbonium ion and the polar solvent. 

As we shall see, a carbonium ion can undergo a number of different reactions; 
just which one occurs depends upon experimental conditions. All reactions of a 
carbonium ion have a common end : they provide a pair of electrons to complete the 
octet of the positively charged carbon. In the present case a hydrogen ion is 
eliminated from the carbon adjacent to the positive, electron-deficient carbon; 
the pair of electrons formerly shared by this hydrogen are available for formation 
of a TT bond. 

ii / 

C -f H:B 




We can see how the mechanism accounts for the fact that dehydration is 
catalyzed by acids. Now let us see how the mechanism also accounts for the fact 
that the ease with which alcohols undergo dehydration follows the sequence 
3 > 2 > 1. 



170 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

Problem 5.8 According to the principle of microscopic reversibility, a reaction 
and its reverse follow exactly the same path but in opposite directions. On this basis 
wfite a detailed mechanism for the hydration of alkenes* a reaction that is the exact 
reverse of the dehydration of alcohols. (Check your answer in Sec. 6.10.) 



5*21 Ease of formation of carbonium ions 

The ease with which alcohols undergo dehydration follows the sequence 
3 > 2 > 1. There is evidence that a controlling factor in dehydration is ihe 
formation of the carbonium ion, and that one alcohol is dehydrated more easily 
than another chiefly because it forms a carbonium ion more easily. 

Carbonium ions can be formed from compounds other than alcohols, and in 
reactions other than elimination. In all these cases the evidence indicates that the 
ease of formation of carbonium ions follows the same sequence: 

Ease of formation of carbonium ions 3 > 2 > 1 > CH 3 + 

In listing carbonium ions in order of their ease of formation, we find that we 
have at the same time listed them in order of their stability. The more stable the 
carbonium ion, the more easily it is formed. 

Is it reasonable that the more stable carbonium ion should be formed more 
easily? To answer this question, we must look at a reaction in which a carbonium 
ion is formed, and consider the nature of the transition state. 

In the dehydration of an alcohol, the carbonium ion is formed by loss of 
water from the protonated alcohol, ROH 2 *, that is, by breaking of the carbon- 
oxygen bond. In the reactant the positive charge is mostly on oxygen, and in the 
product it is on carbon. In the transition state the C O bond must be partly 
broken, oxygen having partly pulled the electron pair away from carbon. The posi- 
tive charge originally on oxygen is now divided between carbon and oxygen. Car- 
bon has partly gained the positive charge it is to carry in the final carbonium ion. 



F8+ 5+ 1 
lR-:OH 2 j 



R:OH 2 + * R-:OH 2 R+ + :OH 2 
Reactant Transition state Products 

Oxygen Carbon and oxygen Carbon 

has full have partial has full 

positive charge positive charges positive charge 

Electron-releasing groups tend to disperse the partial positive charge (8+) 
developing on carbon, and in this way stabilize the transition state. Stabilization 
of the transition state lowers ac t and permits a faster reaction (see Fig. 5.10). 

Thus the same factor, electron release, that stabilizes the carbonium ion also 
stabilizes the incipient carbonium ion in the transition state. The more stable 
carbonium ion is formed faster. 

We shall return again and again to the relationship between electronic effects 
and dispersal of charge, and between dispersal of charge and stability. We shall find 
that these relationships will help us to understand carbonium ion reactions of many 
kinds, and, in fact, all reactions in which a charge positive or negativedevelops 
or disappears. These will include reactions as seemingly different from dehydra- 



SEC. 5.22 



REARRANGEMENT OF CARBONIUM IONS 



171 



RCHr-OHs 




8+ *+ 
R>CH OH> 



*+ + 
RsC OH 2 




R 2 CHOH 2 

Secondary 



Progress of reaction > 

Figure 5.10. Molecular structure and rate of reaction. Stability of tran- 
sition state parallels stability of carbonium ion: more stable carbonium ion 
formed faster. (Plots aligned with each other for easy comparison.) 

tion of alcohols as: addition to alkenes, aromatic and aliphatic substitution, 
rearrangements, acidity and basicity. 



5.22 Rearrangement of carbonium ions 

Very often, dehydration gives alkenes that do not fit the mechanism as we have 
so far seen it. The double bond appears in unexpected places; sometimes the 
carbon skeleton is even changed. For example: 



CH 3 CH2CH 2 CH 2 OH 

H-Butyl alcohol 



CH 3 

CH 3 CH 2 CHCH 2 OH 

2-Methyl-l-bjLitanol 



CHjCCHOHCHj - 

CH, 

3,3-Dimethyl-2-butanol 



CH 3 CH-CHCH 3 

2-Butene 
Chief product 



-v CH 3 CH CCH 3 

2-Methyl-2-butene 
Chief product 

CH, 
-^ CH 3 C=-CCH 3 

CH 3 

2,3-Dimethyl-2-butene 
Chief product 



CH 3 
and CH 3 CHC=CH 2 

CH 3 

2,3-Dimethyl-l-butene 



Take the formation of 2-butene from w-butyl alcohol. Loss of water from the 
protonated alcohol gives the -butyl carbonium ion. Loss of the proton from the 



172 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

carbon adjacent to the positive carbon could give 1-butene but not the 2-butene 
that is the major product. 

CH 3 CH 2 CH 2 CH 2 OH 2 > H 2 O + CH 3 CH 2 CH 2 CH 2 <# > CH 3 CH 2 CH==CH 2 

The other examples are similar. In each case we conclude that if, indeed, the 
alkene is formed from a carbonium ion, it is not the same carbonium ion that is 
initially formed from the alcohol 

A similar situation exists for many reactions besides dehydration. The idea 
of intermediate carbonium ions accounts for the facts only if we add this to the 
theory: a carbonium ion can rearrange to form a more stable carbonium ion. 

w-Butyl alcohol, for example, yields the w-butyl cation; this rearranges to 
the sec-butyl cation, which loses a hydrogen ion to give (predominantly) 
2-butene: 

CH 3 CH 2 CH 2 CH 2 OH 2 > CH 3 CH 2 CH 2 CH 2 + H 2 O 

CH 3 Ch 2 CH 2 CH 2 ci > CH 3 CH 2 CHCH 3 Rearrangement 


(7) (2) 

CH 3 CH 2 CHCH 3 > CH 3 CH-CHCH 3 + H+ 
2-Butene 

Chief product 

In a similar way, the 2-methyl-l -butyl cation rearranges to the 2-methyl-2-butyl 
cation, 

CH 3 CH, 

CH 3 CH 2 CHCH 2 CB > CH 3 CH 2 CCH 3 Rearrangement 

e 

(/) (3) 

and the 3,3-dimethyl-2-butyl cation rearranges to the 2,3-dimethyl-2-butyl cation. 

CH 3 CH 3 CH 3 
CH 3 C CHCH 3 > CH 3 C CCH 3 Rearrangement 

I *e | 

CH 3 H 

(2) (3) 



We notice that in each case rearrangement occurs in the way that yields the more 
stable carbonium ion: primary to a secondary, primary to a tertiary, or secondary 
to a tertiary. 

Just how does this rearrangement occur? Frank Whitmore (of The Penn- 
sylvania State University) pictured rearrangement as taking place in this way: a 
hydrogen atom or alkyl group migrates with a pair of electrons from an adjacent 
carbon to the carbon bearing the positive charge. The carbon that loses the 
migrating group acquires the positive charge. A migration of hydrogen with a 
pair of electrons is known* as a hydride shift; a similar migration of an alkyl group 
is known as an alkyl shift. These are just two examples of the most common kind 




SEC. 5.22 REARRANGEMENT OF CARBONIUM IONS - 173 

of rearrangement, the 1,2-shifts: rearrangements in which the migrating group moves 
from one atom to the very next atom. 

I I 

A hydride shift 

1,2 Shifts 

C C An alkyl shift 
R 

We can account for rearrangements in dehydration in the following way. 
A carbonium ion is formed by the loss of water from the protonated alcohol. 
If a 1,2-shift of hydrogen or alkyl can form a more stable carbonium ion, then such a 
rearrangement takes place. The new carbonium ion now loses a proton to yield 
an alkene. 

In the case of the w-butyl cation, a shift of hydrogen yields the more stable 
m'-butyl cation; migration of an ethyl group would simply form a different 
w-butyl cation. In the case of the 2-methyl-l-butyl cation, a hydride shift yields a 
tertiary cation, and hence is preferred over a methyl shift, which would only 
yield a secondary cation. In the case of the 3,3-dimethyl-2-butyl cation, on the 
other hand, a methyl shift can yield a tertiary cation and is the rearrangement 
that takes place. 

H H H H 

II II 

CH 3 CH 2 "C~C-H > CH 3 CH 2 -C-C-H 

rl^ I 

w H 

/i-Butyl wr-Butyl 

cn (^ u ) 



CH 3 H 

CH 3 CH 2 C -- C-H 

I 
H 

2-MethyI-l -butyl 2-Methyl-2-butyl 

(/<) (^ e ) 

CH 3 H CH 3 H 





CH 3 

3,3-Dimethyl-2-butyl 2,3-Dimethyl-2-butyl 

(2) (3) 

Historically, it was the occurrence of rearrangements that was chiefly respon- 
sible for the development of the carbonium ion theory. Reactions of seemingly 



174 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

quite different kinds involved rearrangements that followed the same general 
pattern; the search for a common basis led to the concept of the carbonium ion. 
Today, the occurrence (or non-occurrence) of rearrangements of the kind we have 
seen here is the best and sometimes the only evidence for (or against) the 
intermediate formation of carbonium ions. 

In our short acquaintance with the carbonium ion, we have encountered two 
of its reactions. A carbonium ion may: 

(a) eliminate a hydrogen ion to form an alkene; 

(b) rearrange to a more stable carbonium ion. 

This list will grow rapidly. 

In rearrangement, as in every other reaction of a carbonium ion, the electron- 
deficient carbon atom gains a pair of electrons, this time at the expense of a neigh- 
boring carbon atom, one that can better accommodate the positive charge. 



5.23 Orientation and reactivity in dehydration 

At this point, we know this much about dehydration of alcohols. 

(a) It involves the formation of a carbonium ion. How fast dehydration 
takes place depends chiefly upon how fast this carbonium ion is formed, which, in 
turn, depends upon how Stable the carbonium ion is. The stability of the car- 
bonium ion depends upon the dispersal of the positive charge, which is determined 
by electron-release or electron-withdrawal by the attached groups. 

(b) If this initially formed carbonium ion can rearrange via a 1,2-shift to form 
a more stable carbonium ion, it will do so. 

This brings us to the last step of dehydration, (c) The carbonium ion either 
the original one or the one formed by rearrangement loses a proton to form an 
alkene. Now, if isomeric alkenes can be formed in this step, which, if any, will 
predominate? The examples we have already encountered give us the answer: 



CH 3 CH 2 CHCH 3 > CH 3 CH=HCH 3 and CH 3 CH 2 CH=CH 2 

2-Butene 
Preferred product 



2-Butene 1-Butene 



CH 3 CH 3 CH 3 

CH 3 CH 2 CCH 3 CH 3 CH=CCH 3 and CH 3 CH 2 O=CH 2 

2-Methyl-2-butene 2-Methyl-l -butene 
Preferred product 

CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 

CH 3 C CCH 3 > CH 3 C==CCH 3 and CH 3 C ^CH 2 

I I 

U IJT 

n n 

2,3-Dimethyl-2-butene 2,3-Dimethyl-l -butene 
Preferred product 

Here, as in dehydrohalogenation, the preferred alkene is the more highly sub- 
stituted one, that is, the more stable one (Sec. 6.4). In dehydration, die more stable 
alkene is the preferred product. 



PROBLEMS 175 

Once more, examination of the transition state involved shows that it is 
reasonable that the more stable alkene should be formed faster: 



-cU- 

| 



ROH 



\ 



ROH 2 



H 18. 

H-OR 
H J 

Transition state": 

partly formed 
double bond 

As the proton is pulled away by the base (the solvent), the electrons it leaves behind 
become shared by the two carbons, and the carbon-carbon bond acquires double- 
bond character. Factors that stabilize an alkene also stabilize an incipient alkene 
in the transition state. 

Problem 5.9 Predict the major product of dehydration of each of the following: 
(a) (CH 3 ) 2 C(OH)CH 2 CH 3 , (b) (CH 3 ) 2 CHCHOHCH 3 , (c) (CH 3 ) 2 C(OH)CH(CHj) 2 . 

PROBLEMS 

1. Give the structural formula of: 

(a) 3,6-dimethyl-l-octene (e) (Z)-3-chIoro-4-methyl-3-hexene ' 

(b) 3-chloropropene (f) (E)-l-deuterio-2-chloropropene 

(c) 2,4,4-trimethyl-2-pentene (g) (R)-3-bromo- 1-butene 

(d) //wi.s-3,4-dimethyl-3-hexene (h) (S)-/ra/w-4-methyl-2-hexene 

2. Draw out the structural formula and give the IUPAC name of: 

(a) isobutylene (d) /r^-(CH 3 ) 2 CHCH-CHCH(CH 3 ) 2 

(b) m-CH 3 CH 2 CH-CHCH,CH 3 (e) (CH 3 ) 2 CHCH 2 CH~C(CH 3 ) 2 

(c) (CH 3 ) 3 CCH-CH 2 (f) (CH 3 CH 2 ) 2 C=-CH 2 

3. Indicate which of the following compounds show geometric (cis-trans) isomerism, 
draw the isomeric structures, and specify each as Z or E. 

(a) 1-butene (g) 2-pentene 

(b) 2-butene (h) 1-chloropropene 

(c) 1,1-dichloroethene (i) l-chloro-2-methyl-2-butene 

(d) 1,2-dichloroethene (j) 3-methyl-4-ethyl-3-hexene 

(e) 2-methyl-2-butene (k) 2,4-hexadiene 

(f) 1-pentene (CH 3 CH--CHCH-CHCH 3 ) 

4. There are 13 isomeric hexylenes (C 6 H I2 ) disregarding geometric isomerism. 
(a) Draw the structure and give the IUPAC name for each, (b) Indicate which ones show 
geometric isomerism, draw the isomeric structures, and specify each as Z or E. (c) One 
of the hexylenes is chiral. Which one is it? Draw structures of the enantiomers, and 
specify each as R or S. 

5. In which of the following will c/s-3-hexene differ from msrw.y-3-hexene? 

(a) b.p. (g) rate cf hydrogenation 

(b) m.p. (h) product of hydrogenation 

(c) adsorption on alumina (i) solubility in ethyl alcohol 

(d) infrared spectrum (j) density 

(e) dipolc moment (k) retention time in gas chromatography 

(f ) refractive index 

(1) Which one of the above would absolutely prove the configuration of each isomer? 



176 ALKENES I. STRUCTURE AND PREPARATION CHAP. 5 

6. Write balanced equations for preparation of propylene from: 

(a) CH 3 CH 2 CH 2 OH (/i-propyl alcohol) (c) isopropyl chloride 

(b) CH 3 CHOHCH 3 (isopropyl alcohol) (d) the alkyne, CH 3 O=CH 

(e) propylene bromide (1,2-dibromopropane) 

7. Give structures of the products expected from dehydrohalogenation of: 

(a) 1-bromohexane (e) 3-bromo-2-methylpentane 

(b) 2-bromohexane (f) 4-bromo-2-methylpentane 

(c) l-bromo-2-methylpentane (g) l-bromo-4-methylpentane 

(d) 2-bromo-2-methylpentane (h) 3-bromo-2,3-dimethylpentane 

8. In those cases in Problem 7 where more than one product can be formed, predict 
the major product. 

9. Which alcohol of each pair would you expect to be more easily dehydrated? 

(a) CH 3 CH 2 CH 2 CH 2 CH20H or CH 3 CH 2 CH 2 CHOHCH 3 

(b) (CH 3 ) 2 C(OH)CH 2 CH 3 or (CH 3 ) 2 CHCHOHCH 3 

(c) (CH 3 ) 2 CHC(OH)(CH 3 ) 2 or (CH 3 ) 2 CHCH(CH 3 )CH 2 OH 

10. (a) Show all steps in the synthesis of propylene from propane by ordinary 
laboratory methods (not cracking), (b) If the steps in (a) were carried out starting with 
it-butane, would a single product or a mixture be expected ? 

11. When dissolved in SbF 5 , w-propyl fluoride and isopropyl fluoride give solutions 
with identical nmr spectra, indicating that identical species are present. How do you 
account for this ? 

12. (a) When neopentyl alcohol, (CH 3 ) 3 CCH 2 OH, is heated with acid, it is slowly 
converted into an 85 : 15 mixture of two alkenes of formula C 5 H 10 . What are these aikenes, 
and how are they formed? Which one would you think is the major product, and why? 

(b) Would you expect neopentyl bromide, (CH 3 )jCCH 2 Br, to undergo the kind of 
dehydrohalogenaticn described in Sec. 5.13? Actually, when heated in aqueous alcohol, 
neopentyl bromide slowly reacts to yield, among other products, the same alkenes as those 
in (a). Suggest a mechanism for this particular kind of dehydrohalogenation. Why does 
this reaction, unlike that in (a), not require acid catalysis? 

13. When 3,3-dimethyl-l-butene is treated with hydrogen chloride there is obtained 
a mixture of 3-chloro-2,2-dimethylbutane and 2-chIuro-2,3-dimethylbutane. What does 
the formation of the second product suggest to you? Propose a likely mechanism for 
this reaction, which is an example of electrophilic addition. Check your answer in Sees. 
6. 10 and 6. 12. 



Chapter 



Alkenes II. Reactions of the 
Carbon Carbon Double Bond 

Electrophilic and Free-Radical Addition 



6.1 The functional group 

The characteristic feature of the alkene structure is the carbon-carbon double 
bond. The characteristic reactions of an alkene are those that take place at the 
double bond. The atom or group of atoms that defines the structure of a particular 
family of organic compounds and, at the same time, determines their properties is 
called the functional group. 

In alkyl halides the functional group is the halogen atom, and in alcohols the 
OH group; in alkcnes it is the carbon- carbon double bond. We must not forget 
that an alkyl halide, alcohol, or alkene has alkyl groups attached to these functional 
groups; under the proper conditions, the alkyl portions of these molecules undergo 
the reactions typical of alkanes. However, the reactions that are characteristic of 
each of these compounds are those that occur at the halogen atom or the hydroxyl 
group or the carbon-carbon double bond. 

A large part of organic chemistry is therefore the chemistry of the various 
functional groups. We shall learn to associate a particular set of properties with a 
particular group wherever we may find it. When we encounter a complicated 
molecule, which contains a number of different functional groups, we may expect 
the properties of this molecule to be roughly a composite of the properties of the 
various functional groups. The properties of a particular group may be modified, 
of course, by the presence of another group and it is important for us to understand 
these modifications, but our point of departure is the chemistry of individual 
functional groups. 

6.2 Reactions of the carbon -carbon double bond: addition 

Alkene chemistry is the chemistry of the carbon -carbon double bond. 
What kind of reaction may we expect of the double bond ? The double* bond 
consists of a strong or bond and a weak IT bond; we might expect, therefore, that 

177 



178 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

reaction would involve the breaking of this weaker bond. This expectation is 
correct; the typical reactions of the double bond are of the sort, 



O=C + YZ > C C Addition 

Y 



J -A 




where the -n bond is broken and two strong a bonds are formed in its place. 

A reaction in which two molecules combine to yield a single molecule of product 
is called an addition reaction. The reagent is simply added to the organic molecule, 
in contrast to a substitution reaction where part of the reagent is substituted for a 
portion of the organic molecule. Addition reactions are necessarily limited to 
compounds that contain atoms sharing more than one pair of electrons, that is, 
to compounds that contain multiply-bonded atoms. 

What kind of reagent may we expect to add to the carbon-carbon double 
bond? In our structure of the bond there is a cloud of -n electrons above and below 
the plane of the atoms (see Fig. 6.1). These rr electrons are less involved than the 



Figure 6.1. Carbon-carbon double 
bond: TT bond is source of electrons. 



a electrons in holding together the carbon nuclei. As a result, they are themselves 
held less tightly. These loosely held IT electrons are particularly available to a 
reagent that is seeking electrons. It is not surprising, then, that in many of its 
reactions the carbon-carbon double bond serves as a source of electrons: that is, it 
acts as a base. The compounds with which it reacts are those that are deficient in 
electrons, that is, are acids. These acidic reagents that are seeking a pair of electrons 
are called electrophilic reagents (Greek: electron-loving). The typical reaction of an 
alkene is electrophilic addition, or, in other words, addition of acidic reagents. 

Reagents of another kind, free radicals, seek electrons or, rather, seek an 
electron. And so we find that alkenes also undergo free-radical addition. 

Most alkenes contain not only the carbon-carbon double bond but also alkyl 
groups, which have essentially the alkane structure. Besides the addition reactions 
characteristic of the carbon-carbon double bond, therefore, alkenes may undergo 
the free-radical substitution characteristic of alkanes. The most important of these 
addition and substitution reactions are summarized below, and will be discussed in 
detail in following sections. 

There are reagents that can add either as acids or as free radicals, and with 
strikingly different results ; there are reagents that are capable both of adding to the 
double bond and of bringing about substitution. We shall see how, by our choice 
of conditions, we can lead these reagents along the particular reaction path 
electrophilic or free-radical, addition or substitution we want them to follow. 

The alkyl groups attached to the doubly-bonded carbons modify the reactions 
of the double bond; the double bond modifies the reactions of the alkyl groups. 



SEC. 6.2 REACTIONS OF CARBON-CARBON DOUBLE BOND: ADDITION 179 

We shall be concerned with seeing what these modifications are and, where possible, 
how they can be accounted for. 



REACTIONS OF ALKENES 
Addition Reactions 

II II 

-O=C- + YZ > -C-C- 

I I 
Y Z 

1. Addition of hydrogen. Catalytic hydrogenation. Discussed in Sec. 6.3. 

I I P Pit rtr Nil I' 



Example: 

CHjCH CH 2 * CHjCt^CH} 

Propene- Propane 

(Propylene) 

2. Addition of halogens. Discussed in Sees. 6.5, 6.13, and 7.11-7.12. 

II II 

-C-C - 4- X 2 > -C-C- X 2 = CI 2 , Br 2 

X X 

Example: 

CH 3 CH=CH 2 Hr> ' nCCIt - CH 3 CHBrCH 2 Br 

Propene 1,2-Dibromopropane 

(Propylenc) (Propylene bromide) 

3. Addition of hydrogen halides. Discussed in Sees. 6.6-6.7 and 6.17. 

II ! I 

-C-C- + HX - -C-C- HX - HCI, HBr, HI 

! I 
H X 

Examples: 

CH,CH-CH : -- > CH 3 CHICH 3 

Propene 2-Iodopropane 

(Isopropyl iodide) 

no peroxides 

-> CH 3 CHBrCHi Markovniko\ addition 

2-Bromopropane 
(Isopropyl bromide) 

per xldes > CH 3 CH 2 CH 2 Br Anti-Markovnikov addition 

NBromopropane 
(w-Propyl bromide) 



CH 3 CH=CH 2 



HBr 



180 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP 6 

4. Addition of sulfuric acid. Discussed in Sec. 6.8. 

-O=C- + H 2 S0 4 > -C-C- 

H OSO 3 H 
Example: 



CH 3 CH=CH 2 ' 4 > CH 3 CHCH 3 
Propene 



Isopropyl hydrogen sulfate 
5. Addition of water. Hydration. Discussed in Sec. 6.9. 

-C=C + HOH -^4 C-C- 

UH 

Example: 

CH 3 CH=CH 2 H2 >H *> CH 3 CHCH 3 



Propene 



Isopropyl alcohol 
(2-Propanol) 



6. Halohydrin formation. Discussed in Sec. 6.14. 



-C=C- + X 2 + H 2 O > C-C + HX X 2 = C1 2 , Br 2 

X OH 
Example: 

CH 3 CH=CH 2 Cb>H2 > CH 3 CH-CH 2 
Propylene ^ L 

Propylene chlorohydrin 
(1 -Chloro-2-propanol) 

7. Dimerization. Discussed in Sec. 6.15. 
Example: 

CH 3 CH 3 CH 3 CH 3 

CH 3 ~O=CH 2 + CH 3 -0=CH 2 -^-+ CH 3 -C~CH=C-CH 3 
Isobutylene I 

2,4,4-Trimethyl-2-pentene 

CH 3 CH 3 
and CH 3 ~C-CH 2 -C=CH 2 

CH 3 

2,4,4-Trimethyl-l -pentene 



SEC. 6.2 REACTIONS OF CARBON-CARBON DOUBLE BOND: ADDITION 181 

8. Alkylation. Discussed in Sec. 6.16. 



Example: 

CH 3 CH 3 CH 3 CH 3 

CH 3 -0==CH 2 + CHy-C-H H2S * > CH 3 C-CH 2 -C-CH 3 

Isobutylene ! JT ! 

Isobutane 2,2,4-Trimethylpentane 
9. Oxymercuration-demercuration. Discussed in Sec. 15.8. 

9r . H 2 + Hg(OAc) 2 



_^_<i_ N.B! 


SU-A4- 


HO HgOAc 


HOH 




Markovnikov 
orientation 



10. Hydroboration-oxidation. Discussed in Sees. 15.9-15.11. 



Dibonme 



A nti'Markovnikov 
orientation 



11. Addition of free radicals. Discussed in Sees. 6.17 and 6.18. 

I I 
C=C- + Y-Z 



peroxides J, ' 

or light I I 

Y Z 



Example: 



/i-C 6 H 13 CH==CH 2 + BrCCI 3 "*""" > /7-C 6 H 13 CH-CH 2 -CCl 3 
1-Octene Bromotrichloromethane 1 

3-Bromo-l , 1 ,1 -trichlorononane 

12. Polymerization. Discussed in Sees. 6.19 and 32.3-32.6. 

13. Addition of carbenes. Discussed in Sees. 9.15-9.16. 

14. Hydroxylation. Glycol formation. Discussed in Sees. 6.20 and 17.12. 

I I - II 
-C=C- + KMnO 4 or HCO 2 OH > -C C- 



OH 



OH 



182 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

V 

Example: 

CHCH=CH 2 KMn orHCQ * H > CH 3 -CH-CH 2 
Propylene ^ ^ 

(Propene) Propylene glycol 

(1,2-Propanediol) 

Substitution Reactions 
15. Halogenation. Allylic substitution. Discussed in Sec. 6.21. 



Examples: 



;- + X 2 -+ X-C--C=C- X 2 = Cl 2 ,Br 2 

Low I 

concentration 

C* I flfi c 

CH 3 CH=CH 2 2> ^ > C1~CH 2 CH=CH 2 
Propylene Allyl chloride 

(Propene) (3-Chloro-l-propene) 

Q + NBS > Q 

Cyclohexene 3-Bromocyclohexene 

N-Bromosuccinimide 

Cleavage Reactions 
16. Ozonolysis. Discussed in Sec.' 6.29. 

^ ->c x x c~ HAZ ? -c^o + o=c- 

3 \ / structure 

Ozone ^ ' Aldehydes and ketones 

Ozonide 

Examples: 

H H 

CH 3 CH 2 CH-CH 2 -5i> H ^ >Zn > CH 3 CH 2 D-0 + O-CH 
1-Butene 

CH, CH 3 H 

Isobutylene 



6.3 Hydrogenation. Heat of hydrogenation 

We have already encountered hydrogenation as the most useful method for 
preparing alkanes (Sec. 3.15). It is not limited to the synthesis of alkanes, but is a 
general method for the conversion of a carbon-carbon double bond into a carbon- 
carbon single bond: using the same apparatus, the same catalyst, and very nearly 



SEC. 6.3 HYDROGENATION. HEAT OF HYDROGENATION 183 

* 

the same conditions, we can convert an alkene into an alkanc. an unsaturated 
alcohol into a saturated alcohol, or an unsaturated ester into a saturated ester. 
Since the reaction is generally_quantitative,_and since truTvoiume of hydrogen 
consumed can be easily measured, hydrogenation is frequently used as an analytical 
tool; it can, for example, tell us the number of double bonds in a compound. 

C=C + H H > C C A# = heat of hydrogenation 
H H 

Hydrogenation is exothermic: the two a bonds (C H) being formed are, 
together, stronger than the a bond (H H) and -n bond being broken. The quantity 
of heat evolved when one mole of an unsaturated compound is hydrogenated is called 
the heat of hydrogenation; it is simply A// of the reaction, but the minus sign is 
not included. The heat of hydrogenation of nearly every alkene is fairly close to 
an approximate value of 30 kcal for each double bond in the compound (see 
Table 6.1). 

Table 6.1 HEATS OF HYDROGENATION OF ALKENES 

Heat of hydrogenation, 
Alkene kca I/mole 



Ethylene 32.8 



Propylenc 


30.1 


1-Butene 


30.3 


1-Pentene 


30.1 


1-Heptene 


30.1 


3-Methyl-l-butene 


30.3 


3,3-Dimethyl-l -butene 


30.3 


4,4-DimethyM-pentene 


29.5 


m-2-Butene 


28.6 


tranv-2- Butene 


27.6 


Isobutylcne 


28.4 


m-2-Pentene 


28.6 


//fl//.y-2-Pentcne 


27.6 


2-Methyl-l -butene 


28.5 


2,3-Dimethyl-l-butene 


28.0 


2-Methyl-2-butene 


26.9 


2,3-Dimethyl-2-butene 


26.6 



Although hydrogenation is an exothermic reaction, it proceeds at a negligible 
rate in the absence of a catalyst, even at elevated temperatures. The uncatalyzed 
reaction must have, therefore, a very large energy of activation. The function of 
the catalysfis to lower the energy of activation ( act ) so that the reaction can proceed 
rapidly at room temperature. The catalyst does not, of course, affect the net 
energy change of the overall reaction; it simply lowers the energy hill between the 
reactants and products (see Fig. 6.2). 

A catalyst lowers act by permitting reaction to take place in, a different way, 
that is, by a different mechanism. In this case, the reactants are adsorbed on the. 



184 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



enormous surface of the finely divided metal, where reaction actually occurs. 
Reaction between the adsorbed molecules is very different from the reaction that 
would have to take place otherwise; it is believed, for example, that the catalytic 
surface breaks the * bond of the alkene prior to reaction with hydrogen. 

Lowering the energy hill, as we can see, decreases the energy of activation of 
the reverse reaction as well, and thus increases the rate of e/ehydrogenation. We 
might expect, therefore, that platinum, palladium, and nickel, under the proper 
conditions, should serve as dehydrogenation catalysts; this is indeed the case. 
We are familiar with the fact that, although a catalyst speeds up a reaction, it does 



/ \^~ No catalyst 




-C=C 



H-H 2 



V. 



Progress of reaction > 

Figure 6.2. Potential energy changes during progress of reaction : effect of 
catalyst. 

not shift the position of equilibrium ; this is, of course, because it speeds up both 
the forward and reverse reactions (See Sec. 30.7). 

Like hydrogenation, the addition of other reagents to the double bond is 
generally exothermic. The energy consumed by the breaking of the Y Z and IT 
bonds is almost always less than that liberated by formation of the C Y and C Z 
bonds. 

-G=C- + Y-Z > --C--C-- + heat 



6.4 Heat of hydrogenation and stability of alkenes 

Heats of hydrogenation can often give us valuable information about the 
relative stabilities of unsaturated compounds. For example, of the isomeric 
2-butenes, the cw-isomer has a heat of hydrogenation of 28.6 kcal, the trans- 
homer one of 27.6 kcal. Both reactions consume one mole of hydrogen and yield 
the same product, n-butane. Therefore, if the mww-isomer evolves 1 kcal less 



SEC. 6.4 HEAT OF HYDROGENATION AND STABILITY OF ALKENES 185 

energy than the cfr-isomer, it can only mean that it contains 1 kcal less energy; in 
other words, the /ronf-isomer is more stable by 1 kcal than the c/5-isomer (see 
Fig. 6.3). In a similar way, frdw$-2-pentene (heat of hydrogenation = 27.6 kcal) 
must be more stable by 1.0 kcal than c/.s-2-pentene (heat of hydrogenation = 28.6 
kcal). 



c/s-CH 3 CH=CHCH 3 

frfl/is-CH 3 CH=CHCH 3 } 1 kcal 



1 





-CH 3 CH 2 CH a CH 3 



Figure 6,3. Heats of hydrogenation and stability: cis- and mww-2-butene. 

Of simple disubstituted ethylenes, it is usually the fraws-isomer that is the more 
stable. The two larger substituents are located farther apart than in the cfa-isomer; 
there is less crowding, and less van der Waals strain (Sec. 3.5). 

Heats of h>drogenation show that the stability of an alkene also depends upon 
the position of the double bond. The following examples are typical: 

CH 3 CH 2 CH=CH 2 CH 3 CH=CHCH 3 

30.3 kcal cis 28.6; trans 27.6 

CH 3 CH 2 CH 2 CH=CH 2 CH 3 CH 2 CH=CHCH 3 

30. 1 kcal cis 28.6 ; trans 27.6 

CH 3 CH 3 CH 3 

CH 3 CHCH=CH 2 CH 2 =CCH 2 CH 3 CH 3 C=CHCH 3 

30.3 kcal 28.5 26.9 

Each set of isomeric alkenes yields the same alkane. The differences in heat of 
hydrogenation must therefore be due to differences in stability. In each case, 
the greater the number of alkyl groups attached to the doubly-bonded carbon atoms, 
the more stable the alkene. 

Stability of alkenes 
R 2 O=CR 2 > R 2 C=CHR > R*C=CH 2 , RCH=CHR > RCH=CH 2 > CH 2 =CH 2 

We Have seen (Sees. 5.14, 5.23) that the stability of alkenes determines 
orientation in dehydrohalogenation and dehydration. 



186 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

Problem 6.1 (a) Write a balanced equation for combustion of 1-butene. 
(b) How does this equation compare with the corresponding one for cw-2-butene? For 
//ww-2-butene? (c) The following heats of combustion have been measured for these 
three butenes: 648.1, 647.1, 649.8 kcal. Which heat of combustion do you think ap- 
plies to each butene? (d) Assign the following heats of combustion to 1-pentene, and 
as- and f/ww-2-pentene: 804.3, 806.9, 805.3. 

6.5 Addition of halogens 

Alkenes are readily converted by chlorine or bromine into saturated com- 
pounds that contain two atoms of halogen attached to adjacent carbons; iodine 
generally fails to react. 

II II 

C=C + X 2 > C-C 

Alkene (X 2 = Cl 2 ,Br 2 ) ^ ^ 

Vicinal dihalide 

The reaction is carried out simply by mixing together the two reactants, usually 
in an inert solvent like carbon tetrachloride. The addition proceeds rapidly at 
room temperature or below, and does nof require exposure to ultraviolet light; in 
fact, we deliberately avoid higher temperatures and undue exposure to light, as well 
as the presence of excess halogen, since under those conditions substitution might 
become an important side reaction. 

This reaction is by far the best method of preparing vicinal dihalides. For 
example: 

CH 2 =CH 2 + Br 2 -^* CH 2 -CH 2 

Ethene ^ ^ 

(Ethylene) i, 2 -Dibromoethane 

(Ethylene bromide) 

CH 3 CH=CH 2 + Br 2 -^> CH 3 CH CH 2 

Propene ^ ^ r 

(Propylene) 1,2-Dibromopropane 

(Propylene bromide) 
CH 3 CH 3 

CH 3 -O=CH 2 + Br 2 ^^> CH 3 -C-CH 2 

2-Methylpropene 1 jl 

(Isobutylene) l,2-Dibromo-2-methylpropane 

(Isobutylene bromide) 

Addition of bromine is extremely useful for detection of the carbon-carbon 
double bond. A solution of bromine in carbon tetrachloride is red; the dihalide, 
like the alkene, is colorless. Rapid decolorization of a bromine solution is charac- 
teristic of compounds containing the carbon-carbon double bond. -{However, see 
Sec. 6.30.) 

A common method of naming alkene derivatives is illustrated here. As we 
see, the product of the reaction between ethylene and bromine has the IUPAC 
name of 1,2-dibromoethane. It is also frequently called ethylene bromide, the 
word ethylene forming part of the name even though the compound is actually 



SEC. 6.6 ADDITION OF HYDROGEN HALIDES. MARKOVNIKOV'S RULE 187 

saturated. This is an old-fashioned name, and is meant to indicate the product of 
the reaction between ethylene and bromine, just as, for example, sodium bromide 
would indicate the product of the reaction between sodium and bromine. It 
should not be confused with the different compound, 1,2-dibromoethene, 
BrCH=CHBr. In a similar way, we have propylene bromide, isobutylene bromide, 
and so on. 

We shall shortly encounter other saturated compounds that are named in a 
similar way, as, for example, ethylene bromohydrin and ethylene glycoL These 
names have in common the use of two words, the first of which is the name of the 
alkene; in this way they can be recognized as applying to compounds no longer 
containing the double bond. 
* 

6.6 Addition of hydrogen halidcs. Markovnikov's rule 

An alkene is converted by hydrogen chloride, hydrogen bromide, or hydrogen 
iodide into the corresponding alkyl halide. 

-C=C + HX 
Alkene (HX = HC1, HBr, HI) 

Alkyl halide 

The reaction is frequently carried out by passing the dry gaseous hydrogen 
halide directly into the alkene. Sometimes the moderately polar solvent, acetic 
acid, which will dissolve both the polar hydrogen halide and the non-polar alkene, 
is used. The familiar aqueous solutions of the hydrogen hatides are not generally 
used; in part, this is to avoid the addition of water to the alkene (Sec. 6.9). 

Problem 6.2 (a) What is the acid in an aqueous solution of HBr? In dry HBr? 
(b) Which is the stronger acid? (c) Which can better transfer a hydrogen ion to an 
alkene? 

In this way, ethylene is converted into an ethyl halide, the hydrogen becoming 
attached to one doubly-bonded carbon and the halogen to the other. 

CH 2 =CH 2 + HI - > CH 3 CH 2 I 
Ethylene Elhyl iodide 

Propylene could yield either of two products, the //-propyl halide or the iso- 
propyl halide, depending upon the orientation of addition, that is, depending upon 
which carbon atoms the hydrogen and halogen become attached to. Actually, it is 
found that the isopropyl halide greatly predominates. 




CH 3 -~CH=CH 2 ~X-> CH 3 CH CH 2 

-Sw .'' I I 

H-l H I 

n-Propyl iodide 
CH 3 -CH=CH 2 - CH 3 ~CH-CH 2 Actual product 

I-H I H 

Isopropyl iodide 



188 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



In the same way, isobutylene could yield either of two products, isobutyl halide or 
ter/-butyl halide; here the orientation of addition is such that the tert-butyl halide 
greatly predominates. 

CH 3 
CH 3 -C=C,H 2 



CH 3 

CHr-C-CH 2 
H 



CH 3 

CH 3 C==CH 2 
I-H 



I 

Isobutyl iodide 

CH 3 
CH 3 C CH 2 Actual product 

I H 

ten-Butyl iodide 



Orientation in alkane substitutions (Sec. 3.21) depends upon which hydrogen 
is replaced; orientation in alkene additions depends upon which doubly-bonded 
carbon accepts Y and which accepts Z of a reagent YZ. 

Examination of a large number of such additions showed the Russian chemist 
Vladimir Markovnikov (of the University of Kazan) that where two isomeric 
products are possible, one product usually predominates. He pointed out in 1869 
that the orientation of addition follows a pattern which we can summarize as: 
In the ionic addition of an acid to the carbon-carbon double bond of an alkene, the 
hydrogen of the acid attaches itself to the carbon atom that already holds the greater 
number of hydrogens. This statement is generally known as Markovnikov's rule. 
Thus: "Unto everyone that hath shall be given," or "Them as has, gits." 

Thus, in the addition to propylene we see that the hydrogen goes to the carbon 
bearing two hydrogen atoms rather than to the carbon bearing one. In the addition 
to isobutylene, the hydrogen goes to the carbon bearing two hydrogens rather than 
to the carbon bearing none. 

Using Markovnikov's rule, we can correctly predict the principal product of 
many reactions. For example: 

* CH 3 CH 2 CHICH 3 

sec-Butyl iodide 
(2-Iodobutane) 

CH 3 

> CH 3 -~C-CH 2 --CH 3 

I 

ferf-Pentyl iodide 
(2-Iodo-2-methylbutane) 

> CH 3 CHICH 2 CH 3 

sec-Butyl iodide 
(2-Iodobutane) 

> CH 3 CHIC1 
1-Chloro-l-iodoethane 

* CH 3 CH 2 CHICH 2 CH 3 + CH 3 CH 2 CH 2 CHICH 3 

3-Iodopentane 2-Iodopentane 



CH 3 CH 2 CH=CH 2 + HI 
1-Butene 



CH 3 

CH 3 C=CH-CH 3 + HI 

2-Methyl-2-butene 



CH 3 CH=CHCH 3 + HI 
2-Butene 

CH 2 =CHC1 + HI 

Vinyl chloride 
(Chloroethene) 

CH 3 CH 2 CH=CHCH 3 4- HI 

2-Pentene 



SEC. 6.7 ADDITION OF HYDROGEN BROMIDE. PEROXIDE EFFECT 189 

In 2-pentene each of the doubly-bonded carbons holds one hydrogen, so that 
according to the rule we should expect neither product to predominate. Here again 
the prediction is essentially correct, roughly eqiial quantities of the two isomers 
actually being obtained. 

The examples have involved the addition of hydrogen iodide; exactly similar 
results are obtained in the addition of hydrogen chloride and, except for special 
conditions indicated in the following section, of hydrogen bromide. 

Reactions that, from the standpoint of orientation, give exclusively or nearly 
exclusively one of several possible isomeric products are called regiospecific. 
(From the Latin regio, direction, and pronounced "reejio.") 

Addition of hydrogen halides to alkenes can be used to make alkyl halides. 
The fact that addition occurs with a specific orientation, as summarized by Markov- 
nikov's rule, rather than at random, is an advantage since a fairly pure product 
can generally be obtained. At the same time, the synthesis is, of course, limited to 
those products that are formed in agreement with Markovnikov's rule; for example, 
we can make isopropyl iodide in this way, but not w-propyl iodide. (As we shall 
see later, there are other, more important ways to prepare alkyl halides.) 

6.7 Addition of hydrogen bromide. Peroxide effect 

Addition of hydrogen chloride and hydrogen iodide to alkenes follows 
Markovnikov's rule. Until 1933 the situation with respect to hydrogen bromide 
Was exceedingly confused. It had been reported by some workers that addition of 
hydrogen bromide to a particular alkene yields a product in agreement with 
Markovnikov's rule; by others, a product in contradiction to Markovnikov's rule; 
and by still others, a mixture of both products. It had been variously reported 
that the product obtained depended upon the presence or absence of water, or of 
light, or of certain metallic halides; it had been reported that the product obtained 
depended upon the solvent used, or upon the nature of the surface of the reaction 
vessel. 

In 1933, M. S. Kharasch and F. W. Mayo at the University of Chicago 
brought order to this chemical chaos by discovering that the orientation of addition 
of hydrogen bromide to the carbon-carbon double bond is determined solely by 
the presence or absence of peroxides. 

Organic peroxides are compounds containing the O O linkage. They are en- 
countered, generally in only very small amounts, as impurities in many organic com- 
pounds, where they have been slowly formed by the action of oxygen. Certain peroxides 
are deliberately synthesized, and used as reagents. 

Kharasch and Mayo found that if one carefully excludes peroxides from the 
reaction system, or if one adds certain inhibitors hydroquinone (p. 878), for ex- 
ample, or diphenylamine (p. 728) the addition of HBr to alkenes follows Markov- 
nikov's rule. On the other hand, if one does not exclude peroxides, or if one 

no peroxides > C H 3CHBrC H 3 Markovnikov addition 

Isopropyl bromide 



CH 3 CH=CH 2 

* > 

rt-Propyl bromide 



Propylene I f an ^ CH 3 CH 2 CH 2 Br Anti-Markovnikoy addition 



190 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

CH 3 
CH 3 C CH 3 Markovnikov addition 

Br 

/erf-Butyl bromide 
CH 3 -C=CH 2 " 

Isobutylene 

CH 3 

-> CH 3 ~C CH 2 Br Anti-MarkovniHov addition 



H 
Isobutyl bromide 

deliberately puts peroxides into the reaction system, HBr adds to alkenes in exactly 
the reverse direction. 

This reversal of the orientation of addition caused by the presence of peroxides 
is known as the peroxide effect. Of the reactions we are studying, only the addition 
of hydrogen bromide shows the peroxide effect. The presence or absence of 
peroxides has no effect on the orientation of addition of hydrogen chloride, hydro- 
gen iodide, sulfuric acid, water, etc. As we shall see (Sees. 6.11 and 6.17), both 
Markovnikov's rule and the peroxide effect can readily be accounted for in ways 
that are quite consistent with the chemistry we have learned so far. 

6.8 Addition of sulfuric acid 

Alkenes react with cold, concentrated sulfuric acid to form compounds of the 
genera] formula ROSO 3 H, known as alkyl hydrogen sulfates. These products are 
formed by addition of hydrogen ion to one side of the double bond and bisulfate 

O O 

-C=O- + H-O-S--O--H > -C~C~0-~S-0-H 
Alkene i 1 I 



Sulfuric acid Alkyl hydrogen sulfate 

ion to the other. It is important to notice that carbon is bonded to oxygen and 
not to sulfur. 

Reaction is carried out simply by bringing the reactants into contact: a gaseous 
alkene is bubbled through the acid, and a liquid alkene is stirred or shaken with the 
acid. Since alkyl hydrogen sulfates are soluble in sulfuric acid, a clear solution 
results. The alkyl hydrogen sulfates are deliquescent solids, and are difficult to 
isolate. As the examples below show, the concentration of sulfuric acid required 
for reaction depends upon the particular alkene involved ; we shall later account for 
this in a reasonable way (Sec. 6.11). 

If the sulfuric acid solution of the alkyl hydrogen sulfate is diluted with water 
and heated, there is obtained an alcohol bearing the same alkyl group as the 
original alkyl hydrogen sulfate. The alkyl hydrogen sulfate has been cleaved by 
water to form the alcohol and sulfuric acid, and is said to have been hydrolyzed. 
This sequence of reactions affords a route to the alcohols, and it is for this purpose 



SEC. 6.10 ELECTROPHILIC ADDITION: MECHANISM 191 

that addition of sulfuric acid to alkenes is generally carried out. This is an excel- 
lent method for the large-scale manufacture of alcohols, since alkenes are readily 

CH 2 =CH 2 98 /oH * S04 > CH 3 CH 2 OS0 3 H H2 ' hcat > CH 3 CH 2 OH 4- H 2 SO 4 

Ethylene Ethyl hydrogen sulfate Ethyl alcohol 

CH 3 CH=CH 2 80 /oH2S04 > CH 3 CHCH 3 H2 theai > CH 3 CHCH 3 



Propylene ^ 



>S0 3 H OH 

Iscpropyl hydrogen sulfate Isopropyl alcohol 



?*' 

pu /~< PH. 63/ 


CH 3 

^ HbSO* pjr p p|jr 


H 2 O, heat ^ 


CH 3 
CH-. C CH, 


Isobutybne 


OSO 3 H 

ferf-Butyl hydrogen sulfate 


OH 

terf-Butyl alcohol 



obtained by the cracking of petroleum. Because the addition of sulfuric acid fol- 
lows Markovnikov's rule, certain alcohols cannot be obtained by this method. 
For example, isopropyl alcohol can be made but not w-propyl alcohol; ferf-butyl 
alcohol, but not isobutyl alcohol. 

The fact that alkenes dissolve in cold, concentrated sulfuric acid to form the 
alkyl hydrogen sulfates is made use of in the purification of certain other kinds of 
compounds. Alkanes or alkyl halides, for example, which are insoluble in sulfuric 
acid, can be freed from alkene impurities by washing with sulfuric acid. A gaseous 
alkane is bubbled through several bottles of sulfuric acid, and a liquid alkane is 
shaken with sulfuric acid in a separatory funnel. 

6.9 Addition of water. Hydration 

Water adds to the more reactive alkenes in the presence of acids to yield 
alcohols. Since this addition, too, follows Markovnikov's rule, the alcohols are 




-C=C + H 2 
Alkene 



the same as those obtained by the two-step synthesis just described; this direct 
hydration is, of course, the simpler and cheaper of the two processes. Hydration 
of alkenes is the principal industrial source of those lower alcohols whose formation 
is consistent with Markovnikov's rule. 

CH 3 CH 3 

CH 3 -C=CH 2 H2 * H *> CH 3 -C-CH 3 

Isobutylene I 

OH 

tert-Butyl alcohol 

6.10 Electrophilic addition: mechanism 

Before we consider other reactions of alkenes, it will be helpful to examine the 
mechanism of some of the reactions we have already discussed. After we have 



192 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

done this, we shall return to our systematic consideration of alkene reactions, 
prepared to understand them better in terms of these earlier reactions. 

We shall take up first the addition of those reagents which contain ionizable 
hydrogen: the hydrogen halides, sulfuric acid, and water. The generally accepted 
mechanism will be outlined, and then we shall see how this mechanism accounts 
for certain facts. Like dehydration of alcohols, addition is pictured as involving 
carbonium ions. We shall notice certain resemblances between these two kinds of 
reaction; these resemblances are evidence that a common intermediate is involved. 

Addition of the acidic reagent, HZ, is believed to proceed by two steps: 



(1) -0=C-- + H:Z > -C-C- + :Z HZ - HC1, HBr, HI, 

e H 2 S0 4 ,H 3 0+ 



-0=C-- + H:Z > -C-C- + 
^ e 

(2) -C--C--+ :Z > -C-C- :Z C1-, Br~, I-, 

i i i HS0 4 -,H 2 

Step (1) involves transfer of hydrogen ion from : Z to the alkene to form a carbonium 
ion; this is a transfer of a proton from one base to another. 

Electrophilic addition 




U 

represented as 




Step (2) is the union of the carbonium ion with the base :Z. 

Step (1) is the difficult step, and its rate largely or entirely controls the overall 
rate of addition. This step involves attack by an acidic, electron-seeking reagent 
that is, an electrophilic reagent and hence the reaction is called electrophilic 
addition. The electrophile need not necessarily be a Lowry-Bronsted acid trans- 
ferring a proton, as shown here, but, as we shall see, can be almost any kind of 
electron-deficient molecule (Lewis acid). 

On the basis of step (2), we can add another reaction to our list of Sec. 5.22. 
A carbonium ion may: 

(c) combine with a negative ion or other basic molecule to form a halide, a 
bisulfate, an alcohol, etc. 

This reaction, like the. earlier ones, provides the electron-deficient carbon with a 
pair of electrons. 

The general mechanism is illustrated by specific examples: addition of hy- 
drogen chloride, 

(1) CH 3 --CH==CH 2 + H:C1: CH 3 -CH-CH 3 + :Clr 



SEC. 6.10 ELECTROPHILIC ADDITION: MECHANISM 193 

(2) CH 3 CH CH 3 + :Clr > CH 3 CH CH 3 

^ 

of sulfuric acid, 

(1) CH 3 CH=CH 2 + H:OSO 3 H CH 3 CH CH 3 + :OSO 3 H- 





(2) CH 3 CH CH 3 + :OSO 3 H~ > CH 3 CH CH 3 

I 

OSO 3 H 

and of water. 



(1) CH 3 -CH=CH 2 + H:OH 2 + 7 CH 3 -CH-CH 3 + :OH 2 



(2a) CH 3 -CH-CH 3 + :OH 2 CH 3 -CH-CH 3 

* I 

OH 2 

(2b) CH 3 CH CH 3 + :OH 2 ^ > CH 3 CH CH 3 + H:OH 2 + 

OH 2 OH 

We notice that the carbonium ion combines with water to form not the alcohol 
but the protonated alcohol; in a subsequent reaction this protonated alcohol re- 
leases a hydrogen ion to another base to form the alcohol. This sequence of reac- 
tions, we can see, is just the reverse of that proposed for the dehydration of alcohols 
(Sec. 5.20). In dehydration, the equilibria are shifted in favor of the alkene chiefly 
by the removal of the alkene from the reaction mixture by distillation : in hydra- 
tion, the equilibria are shifted in favor of the alcohol partly by the high concentra- 
tion of water. 

Let us see how this mechanism accounts for some of the facts. 

First, the mechanism is consistent with (a) the acidic nature of the reagents. 
According to the mechanism, the first step in all these reactions is the transfer of 
a hydrogen ion to the alkene. This agrees with the fact that all these reagents 
except water are strong acids in the classical sense ; that is, they can readily supply 
hydrogen ions. The exception, water, requires the presence of a strong acid for 
reaction to occur. 

Next, the mechanism is consistent with (b) the basic nature of alkenes. The 
mechanism pictures the alkene as a base, supplying electrons to an attacking acid. 
This agrees with the structure of the carbon-carbon double bond : basicity is due 
to the loosely held, mobile IT electrons. 

In the following sections we shall see that the mechanism is also consistent 
with (c) the orientation of addition, (d) the relative reactivities ofalkenes, and (e) the 
occurrence of rearrangements. 

Problem 6.3 Addition of D 2 O to 2-methyl-2-butene (in the presence of D+) 
was found (as we might expect) to yield the alcohol (CH 3 ) 2 C(OD)CHDCH 3 . When 
the reaction was about half over, it was interrupted and the unconsumed alkene was 
isolated; mass spectrometric analysis showed that it contained almost no deuterium. 
This fact is considered to be evidence that formation of the carbonium ion is rate- 
determining: that as soon as a carbonium ion is formed, it rapidly reacts with water 
to yield the alcohol. Show how this conclusion is justified. (Hint: What results would 



194 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



you expect if the carbonium ions were formed rapidly and reversibly, and only every 
so often combined with water?) 

6.11 Electrophilic addition: orientadon and reactivity 

The mechanism is consistent with the orientation of addition of acidic reagents, 
and with the effect of structure on relative reactivities. 

Addition of hydrogen chloride to three typical alkenes is outlined below, with 
the two steps of the mechanism shown. In accord with Markovnikov's rule, 
propylene yields isopropyl chloride, isobutylene yields terf-butyl chloride, and 
2-methyl-2-butene yields /er/-pentyl chloride. 



CH 3 -CH=CH 2 

Propylene 



CH 3 -CH-CH 3 

A 2 cation 



CH 3 -CH 2 -CH 2 e 

A 1 cation 



CH 3 

CH 3 -C=CH 2 

Isobutylene 



CH 3 ~C-CH 3 


A 3 cation 



CH3 
CHr-C-CH 2 

H 

A 1 cation 



CH 3 -CH-CH 3 

Cl 
Isopropyl chloride 



CH 3 
CH 3 -C-CH 3 

Cl 
/erf-Butyl chloride 



Actual 
product 



2-Mcthyl-2-butcnc 




A 3 C cation 



CH 3 
CH 3 -CH 2 -C-CH 3 

Cl 

tcrt-Pcnlyl chloride 

(2-Chloro-2-methylbutane) 

Actual product 



CH 3 

CH 3 -CH-C~CH 3 

I 
H 

A 2 cation 

Which alkyl halide is obtained depends upon which intermediate carbonium 
ion is formed. This in turn depends upon the alkene and upon which carbon of 
the double bond hydrogen goes to. Propylene, for example, could yield an /i-propyl 
cation if hydrogen went to C-2 or an isopropyl cation if hydrogen went to C-l. 

Orientation is thus determined by the relative rates of two competing reactions : 
formation of one carbonium ion or another. The fact that propylene is converted 



SEC. 6.11 ELECTROPHILIC ADDITION: ORIENTATION AND REACTIVITY 195 

into the isopropyl cation instead of the w-propyl cation means that the isopropyl 
cation is formed faster than the w-propyl cation. 

In each of the examples given above, the product obtained shows that in the 
initial step a secondary cation is formed faster than* a primary, or a tertiary faster 
than a primary, or a tertiary faster than a secondary. Examination of many cases 
of addition of acids to alkenes shows that this is a general rule: orientation is 
governed by the ease of formation of carbonium ions, which follows the sequence 
3 > 2 > 1. 

In listing carbonium ions in order of their ease of formation from alkenes, 
we find that once more (compare Sec. 5.21) we have listed them in order of their 
stability (Sec. 5.18). 

Stability of carbonium ions 3 > 2 > 1 > CH 3 + 

We can now replace Markovnikov's rule by a more general rule: electrophilic 
addition to a carbon-carbon double bond involves the intermediate formation of the 
more stable carbonium ion. 

Is it reasonable that the more stable carbonium ion should be formed more 
easily? We answered this question in Sec. 5.21 by considering the transition state 
leading to a carbonium ion ; let us do the same here. 

In addition reactions, the carbonium ion is formed by attachment of hydrogen 
ion to one of the doubly-bonded carbons. In the reactant the positive charge is 
entirely on the hydrogen ion; in the product it is on the carbon atom. In the 
transition state, the C H bond must be partly formed, and the double bond partly 
broken. As a result the positive charge is divided between hydrogen and carbon. 

C=C + H + > -C^C > C C Electrophilic 

S + |0 addition 

L H8+ J H 

Reactants Transition state Product 

Hydrogen Carbon and hydrogen Carbon 

has full have partial has full 

positive charge positive charges positive charge 

Electron-releasing groups tend to disperse the partial positive charge (8+) 
developing on carbon and in this way stabilize the transition state. Stabilization 
of the transition state lowers act and permits a faster reaction (see Fig. 6.4). 
As before, the electron release that stabilizes the carbonium ion also stabilizes the 
incipient carbonium ion in the transition state. The more stable carbonium ion is 
formed faster. 

Thus, the rate of addition of a hydrogen ion to a double bond depends upon 
the stability of the carbonium ion being formed. As we might expect, this factor 
determines not only the orientation of addition to a simple alkene, but also the 
relative reactivities of different alkenes. 

Alkenes generally show the following order of reactivity toward addition of 
acids: 

Reactivity of alkenes toward acids 
CH 3 

G=CH 2 > CH 3 CH=CHCH 3 , CH 3 CH 2 CH=CH2, CH 3 CH=CH 2 > 

CH 3 X 

CH 2 -=CH 2 > CH 2 =CHC1 



196 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 











CH;i-CH-CH 2 e 



CH 3 -CH-CH 2 

e 



CH 3 -CH=CH 2 +He 



Progress of reaction 



Progress of reaction 



Figure 6.4. Molecular structure and orientation of reaction. Stability 
of transition state parallels stability of carbonium ion: more stable carbon- 
ium ion formed faster. 



Isobutylene, which forms a tertiary cation, reacts faster than 2-butene, which forms 
a secondary cation. 1-Butene, 2-butene, and propylene, which form secondary 
cations, react faster than ethylene, which forms a primary cation. 



CH, 

CH 3 -C=CH 2 + H + 
Isobutylene 

CH 3 CH=HCH 3 + H+ 
2-Butene 

CH 3 CH 2 CH==CH 2 + H+ 
1-Butene 

CH 3 CH=CH 2 + H + 
Propylene 

CH 2 =CH 2 + H+ 

Ethylene 



CH 3 
CH 3 C-CH 3 

e 

A 3 cation 
CH 3 CH 2 CHCH 3 

e 

A 2 cation 
CH 3 CH 2 CHCH 3 

e 

A 2 cation 
CH 3 CHCH 3 

I/ 

A 2 cation 

CH 3 CH 2 8 
A 1 cation 



Halogens, like other elements in the upper right-hand corner of the Periodic 
Table, tend to attract electrons. Just as electron release by alkyl groups disperses 
the positive charge and stabilizes a carbonium ion, so electron withdrawal by 
halogens intensifies the positive charge and destabilizes the carbonium ion. It is 
not surprising that vinyl chloride, CH 2 ^CHC1, is less reactive than ethylene. 

We can begin to see what a powerful weapon we have for attacking the 
problems that arise in connection with a wide variety of reactions that involve 
carbonium ions. We know that the more stable the carbonium ion, the faster it 
is formed; that its stability depends upon dispersal of the charge; and that dispersal 



SEC. 6.13 MECHANISM OF ADDITION OF HALOGENS 197 

of charge is determined by the electronic effects of the attached groups. We have 
already found that this same approach enables us to deal with such seemingly 
different facts as (a) the relative ease of dehydration of alcohols; (b) the relative 
reactivities of alkenes toward addition of acids; and (c) the orientation of addition 
of acids to alkenes. 

6.12 Electrophilic addition: rearrangement 

The mechanism of electrophilic addition is consistent with the occurrence of 
rearrangements. 

If carbonium ions are intermediates in electrophilic addition, then we should 
expect the reaction to be accompanied by the kind of rearrangement that we said 
earlier is highly characteristic of carbonium ions (Sec. 5.22). Rearrangements are 
not only observed, but they occur according, to just the pattern that would be 
predicted. 

For example, addition of hydrogen chloride to 3,3-dimethyl-l-butene yields 
not only 2-chloro-3,3-dimethylbutane, but also 2-chloro-2,3-dimethylbutane: 

CH 3 CH 3 CH 3 
CHj-C 1 : CH==CH 2 2!E> CH 3 -C CH-CH 3 2^> CH 3 -C CH-CH 3 

<!:H 3 CH 3 CH 3 Cl 

3,3 Dimethyl-1-butene 2-Chloro-3,3-dimethylbutanc 

rearrangement 

CH 3 CH 3 
CH 3 -C- CH-CH 3 ^> CH 3 -C CH-CH 3 

CH 3 Cl CH 3 

2-Chloro-2,3-dimethylbutane 

Since a 1,2-shift of a methyl group can convert the initially formed secondary 
cation into the more stable tertiary cation, such a rearrangement does occur, and 
much of the product is derived from this new ion. (If we compare this change in 
carbon skeleton with the one accompanying dehydration of 3,3-dimethyl-2-butanol 
(p. 171), we can begin to see how the idea arose that these apparently unrelated 
reactions proceed through the same intermediate.) 

Problem 6.4 Addition of HC1 to 3-methyl-l-butene yields a mixture of t\\o alkyl 
chlorides. What are they likely to be, and how is each formed? Give detailed equa- 
tions. < 

Problem 6.5 The reaction of aqueous HC1 with 3,3-dimethyl-2-butanol yields 
2,3-dimethyl-2-chlorobutane. Using only reaction steps that you have already encoun- 
tered, propose a detailed mechanism for this reaction. (Check your answer in Sec. 
16.5.) 

+*/ 

6.13 Mechanism of addition of halogens 

Electrophilic addition of acids to alkenes involves two steps, the first being 
attachment of hydrogen ion to form the carbonium ion. What is the iticchanism 
of the addition of chlorine and bromine? 



198 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

From the structure of the double bond we might expect that here again it is 
an electron source, a base, and hence that the halogen acts as an electrophilic 
reagent, an acid. This idea is supported by the fact that alkenes usually show the 
same order of reactivity toward halogens as toward the acids already studied: 
electron-releasing substituents activate an alkene, and electron-withdrawing sub- 
stituents deactivate an alkene. 

The commonly accepted mechanism for addition of halogens to alkenes has 
two steps, and is quite analogous to the mechanism for addition of hydrogen- 
containing acids (protic acids). In step (1) halogen adds as a positive halogen ion 



-C=C- + :X:X: > -O-C- + :Xr 



(1) 



(2) 



to the double bond to form a carbonium ion. In step (2) the carbonium ion 
combines with a negative halide ion. (The mechanism is somewhat simplified for 
our present purpose, and will be modified in Sec. 7.12.) 

It seems reasonable that an alkene should abstract hydrogen ion from the 
very polar hydrogen halide molecule. Is it reasonable that an alkene should 
abstract a positive halogen ion from the non-polar halogen molecule? Let us 
look at this problem more closely. 

It is true that a halogen molecule is non-polar, since the two identical atoms 
share electrons equally. This is certainly not true, however, for a halogen mole- 
cule while it is under the influence of the powerful electric field of a nearby carbon- 
carbon double bond. The dense electron cloud of the double bond tends to repel 
the similarly charged electron cloud of the halogen molecule; this repulsion makes 
the halogen atom that is nearer the double bond relatively positive and its partner 



V 

4 

/ \ 



8+ S_ 

Br Br Polarization of Br 2 by a double bond 



relatively negative. The distortion of the electron distribution in one molecule 
caused by another molecule is called polarization. Here, we would say that the 
alkene has polarized the halogen molecule. 

The more positive halogen of this polarized molecule is then abstracted by 
the alkene to form a carbonium ion, leaving a negative halide ion. This halide 
ion, or more probably another just like it, finally collides with the carbonium ion 
to yield the product, a dihalide. 

Let us look at some of the evidence for this mechanism. If a carbonium 
ion is the intermediate, we might expect it to react with almost any negative ion or 
basic molecule that we care to provide. For example, the carbonium ion formed 
in the reaction between ethylene and bromine should be able to react not only 
with bromide ion but also if these are present with chloride ion, iodide ion, 
nitrate ion, or water. 



SEC. 6.14 



HALOHYDRIN FORMATION 



199 



The facts are in complete agreement with this expectation. When ethylene 
is bubbled into an aqueous solution of bromine and sodium chloride, there is 
formed not only the dibromo compound but also the bromochloro compound and 
the bromoalcohol. Aqueous sodium chloride alone is completely inert toward 
ethylene; chloride ion or water can react only after the carbonium ion has been 
formed by the action of bromine. In a similar way bromine and aqueous sodium 
iodide or sodium nitrate convert ethylene into the bromoiodo compound or the 
bromonitrate, as well as into the dibromo compound and the bromoalcohol. 

Br "> CH 2 Br-CH 2 Br 
1,2-Dibromoethane 

> CH 2 Br-CH 2 Cl 
2-Bromo- 1 -chloroethane 

CH 2 =CH 2 -^2-> CH 2 Br-CH 2 [ ! " > CH 2 Br~CH 2 I 

2-Bromo- 1 -iodoethane 

CH 2 Br CH 2 ONO 2 
2-Bromoethyl nitrate 

CH 2 Br-CH 2 OH 2 

1=^ CH 2 Br-CH 2 OH 
2-Bromoethanol 

Bromine in water with no added ions yields the dibromo compound and the 
bromoalcohol. 

In addition to the elegant work just described, the stereochemistry of the 
reaction provides powerful support for a two-step addition of halogen. At the 
same time, as we shall see in Sec. 7.12, it requires a modification in the mechanism. 



ci- 



NOj- 



6.14 Halohydrin formation 

As we hav* just seen, addition of chlorine or bromine in the presence of water 
can yield compounds containing halogen and hydroxyl groups on adjacent carbon 
atoms. These compounds are commonly referred to as halohydrins. Under proper 
conditions, they can be made the major products. For example: 



CH 2 -CH 2 

Ethylene 



CH 3 -CH=CH 2 

Propylene 



Br; . H 2 



CH 2 -CH 2 

OH Br 

Ethylene bromohydnn 
(2-Bromoethanol) 



Cl: . H : ^ 



CH 3 -CH-CH 2 

OH Cl 

Propylene chlorohydrin 
( 1 -Chloro-2-propanol) 



200 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



There is evidence, of a kind we are not prepared to go into here, that these 
compounds are formed by reaction of halogen and water (as shown in Sec. 6.13) 
rather than by addition of preformed hypohalous acid, HOX. Whatever the 
mechanism, the result is addition of the elements of hypohalous acid (HO and 
X), and the reaction is often referred to in that way. 

We notice that in propylene chlorohydrin chlorine is attached to the terminal 
carbon. This orientation is, we say, quite understandable in light of the mechanism 
and what we know about formation of carbonium ions: the initial addition of 
chlorine occurs in the way that yields the more stable secondary cation. However, 
we shall have to modify (Sec. 17.15) this interpretation of the orientation to fit 
the modified mechanism of Sec. 7.12. 



6.15 Addition of alkenes. Dimerization 

Under proper conditions, isobutylene is converted by sulfuric or phosphoric 
acid into a mixture of two alkenes of molecular formula C 8 H 16 . Hydrogenation 
of either of these alkenes produces the same alkane, 2,2,4-trimethyIpentane (Sec. 
3.30). The two alkenes are isomers, then, and differ only in position of the double 
bond. {Problem: Could they, instead, be cis-trans isomers?) When studied by 
the methods discussed at the end of this chapter (Sec. 6.29), these two alkenes 
are found to have the structures shown: 



CH 3 



CH 3 

2CH 3 -C= 

Isobutylene 



CH2=C-CH 2 -C- CH 3 -i 

CH 3 

2,4,4-Trimethyl- 1 -pentene 



CH 3 CH 3 

I 1 

-C- 



CH 3 CH 3 



H 



CH 3 

2,2,4-Trimethylpentane 
("Iso-octane") 



CH 3 

2 f 4,4-Trimethyl-2-pentene 



Since the alkenes produced contain exactly twice the number of carbon and 
hydrogen atoms as the original isobutylene, they are known as (timers (di = two, 
mer = part) of isobutylene, and the reaction is called dimerization. Other alkenes 
undergo analogous dimerizations. 

Let us see if we can devise an acceptable mechanism for this dimerization. 
There are a great many isomeric octenes; if our mechanism should lead us to just 
the two that are actually formed, this in itself would provide considerable support 
for the mechanism. 

Since the reaction is catalyzed by acid, let us write as step (1) addition of a 
hydrogen ion to isobutylene to form the carbonium ion; the tertiary cation would, 
of course, be the preferred ion. 



0) 



CH 3 
CH 3 O=CH 2 



CH 3 
CH 3 C CH 3 



SEC. 6.16 ADDITION OF ALKANES. ALKYLATION 201 

A carbonium ion undergoes reactions that provide electrons to complete the 
octet of the positively charged carbon atom. But a carbon-carbon double bond 
is an excellent electron source, and a carbonium ion might well go there in its quest 
for electrons. Let us write as step (2), then, addition of the /erf-butyl cation to 
isobutylene; again, the orientation of addition is such as to yield the more stable 

CH 3 CH 3 CH 3 CH 3 

(2) CH 3 C=CH 2 + 0C CK 3 > CH 3 -C CH 2 -C CH 3 

I & I 

CH 3 CH 3 

tertiary cation. Step (2) brings about the union of two isobutylene units, which is, 
of course, necessary to account for the products. 

What is this new carbonium ion likely to do? We might expect that it could 
add to another molecule of alkene and thus make an even larger molecule; under 
certain conditions this does indeed happen. Under the present conditions, how- 
ever, we know that this reaction stops at eight-carbon compounds, and that these 
compounds are alkenes. Evidently, the carbonium ion undergoes a reaction 
familiar to us: loss of a hydrogen ion (step 3). Since the hydrogen ion can be 
lost from a carbon on either side of the positively charged carbon, two products 
should be possible. 

CH 3 CH 3 
I I 

> H+ + CH 2 =C-CH 2 --C--CH 3 



CH 3 CH 3 
I I 

(3) CH 3 C CH 2 -C-CH 3 



CH 3 



CH 3 



H 3 CH 3 



=CH C 



> H+ 4- CH 3 O=CH C CH 3 

CH 3 

We find that the products expected on the basis of our mechanism are just 
the ones that are actually obtained. The fact that we can make this prediction 
simply on the basis of the fundamental properties of carbonium ions as we under- 
stand them is, of course, powerful support for the entire carbonium ion theory. 

From what we have seen here, we can add one more reaction to those under- 
gone by carbonium ions. A carbonium ion may: 

(d) add to an alkene to form a larger carbonium ion. 

6.16 Addition of alkanes. Alkyiation 

The large amounts of 2,2,4-trimethylpentane consumed as high-test gasoline 
are not made today by the dimerization reaction just described, but in another, 
cheaper way. Isobutylene and isobutane are allowed to react in the presence of an 



CH 3 

CH 3 -(UcH 2 + H-d:-CH3 "*>. " 
Isobutylene 



Isobutane 2,2,4-Trimethylpentane 



202 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

acidic catalyst, to form directly 2,2,4-trimethylpentane, or "iso-octane." This 
reaction is, in effect, addition of an alkane to an alkene. 

The commonly accepted mechanism of this alkylation is based on the study 
of many related reactions and involves in step (3) a reaction of carbonium ions 
that we have not previously encountered. 

CH 3 CH 3 

(1) CH 3 -G=CH 2 + H + > CH 3 -C-CH 3 



CH 3 CH 3 CH 3 CH 3 

(2) CH 3 -O=CH 2 + 0C-CH 3 > CH 3 -C~CH 2 -C-CH 3 

I e I 

CH 3 CH 3 



CH 3 CH 3 CH 3 CH 3 CH 3 CH 3 

(3) CH 3 -C~CH 2 -C~-CH 3 r H^-C--CH 3 > CH 3 ~C~CH 2 ~C-CH 3 r 0C-CH 3 

I / I I I 

:H 3 ./ CH 3 H CH 3 CH 3 



then (2), (3), (2), (3), etc. 



The first two steps are identical with those of the dimerization reaction. 
In step (3) a carbonium ion abstracts a hydrogen atom with its pair of electrons 
(a hydride ion, essentially) from a molecule of alkane. This abstraction of hydride 
ion yields an alkane of eight carbons, and a new carbonium ion to continue the 
chain. As we might expect, abstraction occurs in the way that yields the tert- 
butyl cation rather than the less stable (l c ) isobutyl cation. 

This is not our first encounter with the transfer of hydride ion to an electron- 
deficient carbon; we saw much the same thing in the 1,2-shifts accompanying the 
rearrangement of carbonium ions (Sec. 5.72). There, transfer was intramolecular 
(within a molecule); here, it is intermolecular (between molecules). We shall find 
hydride transfer playing an important part in the chemistry of carbonyl compounds 
(Chap. 19). 

Let us now bring our list of carbonium ion reactions up to date, A carbonium 
ion may: 

(a) eliminate a hydrogen ion to form an alkene; 

(b) rearrange to a more stable carbonium ion; 

(c) combine with a negative ion or other basic molecule; 

(d) add to an alkene to form a larger carbonium ion; 

(e) abstract a hydride ion from an aikane. 

A carbonium ion formed by (b) or (d) can subsequently undergo any of the reac- 
tions. 

As we see, all reactions of a carbonium ion have a common end: they provide 
a pair of electrons to complete the octet of the positively charged carbon. 

Problem 6.6 When ethylene is alkylated by isobutane in the presence of acid, 
there is obtained, not neohexane, (CH 3 ) 3 CCH 2 CH 3 , but chiefly 2,3-dimethylbutane. 
Account in detail for the formation of this product. 



SEC. 6.17 FREE-RADICAL ADDITION 203 



6.17 Free-radical addition. Mechanism of the peroxide-initiated addition of 
HBr 

In the absence of peroxides, hydrogen bromide 'adds to alkenes in agreement 
with Markovnikov's rule; in the presence of peroxides, the direction of addition is 
exactly reversed (see Sec. 6.7). 

To account for this peroxide effect, Kharasch and Mayo proposed that addition 
can take place by two entirely different mechanisms: Markovnikov addition by 
the ionic mechanism that we have just discussed, and anti-Markovnikov addition 
by a free-radical mechanism. Peroxides initiate the free-radical reaction; in their 
absence (or if an inhibitor, p. 189, is added), addition follows the usual ionic path. 

The essence of the mechanism is that hydrogen and bromine add to the double 
bond as atoms rather than as ions; the intermediate is a free radical rather than a 

(1) peroxides > Rad- *) 

V Chain-initiating steps 

(2) Rad- 4- H:Br > Rad:H + fir- J 



(3) 



Br- + -C=C- > -C-C- 

llr' 



/ Chain-propagating steps 
(4) -C-0- + HrBr > -C-C- + fir- 



JJr 



lU 



then (3), (4), (3), (4), etc. 




carbonium ion. Like halogenation of alkanes, this is a chain reaction, this time 
involving addition rather than substitution. 

Decomposition of the peroxide (step 1) to yield free radicals is a well-known 
reaction. The free radical thus formed abstracts hydrogen from hydrogen bromide 
(step 2) to form a bromine atom. The bromine atom adds to the double bond 
(step 3), and, in doing so, converts the alkene into a free radical. 

Free-radical 

addition 



This free radical, like the free radical initially generated from the peroxide, abstracts 
hydrogen from hydrogen bromide (step 4). Addition is now complete, and a new 
bromine atom has been generated to continue the chain. As in halogenation of 
alkanes, every so often a reactive particle combines with another one, or is captured 
by the wall of the 1 reaction vessel, and a chain is terminated. 

The mechanism is well supported by the facts. The fact that a very few mole- 
cules of peroxide can change the orientation of addition of many molecules of 
hydrogen bromide strongly indicates a chain reaction. So, too, does the fact that 
a very few molecules of inhibitor can prevent this change in orientation. It is not 
surprising to find that these same compounds are efficient inhibitors of many other 
chain reactions. Although their exact mode of action is not understood, it seems 
clear that they break the chain, presumably by forming unreactive radicals. 



204 



ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



We must not confuse the effects of peroxides, which may have been formed by the 
action of oxygen, with the effects of oxygen itself. Peroxides initiate free-radical reac- 
tions; oxygen inhibits free-radical reactions (see Sec. 2.14). 

The mechanism involves addition of a bromine atom to the double bond. 
It is supported, therefore, by the fact that anti-Markovnikov addition is caused 
not only by the presence of peroxides but also by irradiation with light of a wave- 
length known to dissociate hydrogen bromide into hydrogen and bromine atoms. 

Recently, the light-catalyzed addition of hydrogen bromide to several alkenes 
was studied by means of esr (electron spin resonance) spectroscopy, which not 
only can detect the presence of free radicals at extremely low concentrations, but 
also can tell something about their structure (see Sec. 13.14). Organic free radicals 
were shown to be present at appreciable concentration, in agreement with the 
mechanism. 

Is it reasonable that free-radical addition of hydrogen bromide should occur 
with orientation opposite to that of ionic addition? Let us compare the two kinds 
of addition to propylene. 

Ionic addition: Markovnikov orientation 



CH 3 CH=CH 2 
Propylene 



HBr 



CH 3 -CH-CH 3 

A 2 cation 



CH 3 -CH 2 -CH 2 

A 1 cation 



CH 3 -CH-CH 3 

Br 

Isopropyl bromide 



Free-radical addition : Anti-Markovnikov orientation 



HBr 



CH 3 CH=CH 2 

Propylene 



Br 



CH 3 -CH 2 -CH 2 Br 

fl-Propyl bromide 



CH 3 ~CH~CH 2 Br 
A 2 free radical 



CH 3 -CHCH 2 

Br 
A 1 free radical 



Ionic addition yields isopropyl bromide because a secondary cation 
is formed faster than a primary. Free-radical addition yields w-propyl bromide 
because a secondary free radical is formed faster than a primary. Examination of 
many cases of anti-Markovnikov addition shows that orientation is governed by 
the ease of formation of free radicals, which follows the sequence 3 > 2 > 1. 

In listing free radicals in order of their ease of formation from alkenes, we 
find that once more (compare Sec. 3.25) we have listed them in order of their 
stability (Sec. 3.24): 



Stability of free radicals 



3 > 2 > 1 > CH 3 



SEC. 6.18 OTHER FREE-RADICAL ADDITIONS 205 

Free-radical addition to a carbon-carbon double bond involves the intermediate 
formation of the more stable free radical. 

Thus we find the chemistry of free radicals and the chemistry of carbonium 
ions following much the same pattern: the more stable particle is formed more 
easily, whether by abstraction or dissociation, or by addition to a double bortd. 
Even the order of stability of the two kinds of particle is the same: 3 > 2 > 1 > 
CH 3 . In this particular case orientation is reversed simply because the hydrogen 
adds first in the ionic reaction, and bromine adds first in the radical reaction. 



6.18 Other free-radical additions 

In the years since the discovery of the peroxide effect, dozens of reagents 
besides HBr have been found (mostly by Kharasch) to add to alkenes in the 
presence of peroxides or light. Exactly analogous free-radical mechanisms are 
generally accepted for these reactions, too. 

For the addition of carbon tetrachloride to an alkene, for example, 

RCH=CH 2 + CC1 4 pcroxidci > RCH-CH 2 -CC1 3 

CJ 
the following mechanism has been proposed: 

(1) peroxide > Rad- 

(2) Rad + C1:CC1 3 Rad:Cl + CC1 3 

(3) CC1 3 + RCH=CH 2 > RCH-CH 2 -CC1 3 

(4) RCH-CH 2 -CC1 3 + C1:CC1 3 RCH-CH 2 -CC1 3 + -CC1 3 

Ci 
then (3), (4), (3), (4), etc. 

In the next section, we shall encounter another example of free-radical addi- 
tion polymerization which has played a key part in the creation of this age of 
plastics. 

Problem 6.7 In the presence of a trace of peroxide or under the influence of 
ultraviolet light, 1-octene reacts: 

(a) with CHCJ 3 to form 1,1,1-trichlorononane; 

(b) with CHBr 3 to form 1,1,3-tribromononane; 

(c) with CBrCl 3 to form l,l,l-trichloro-3-bromononane; 

(d) with H-S-CH 2 COOH (thioglycoiic acid) to yield /7-C 8 H, 7 S-CH 2 COOH; 

(e) with aldehydes, R C=O, to yield ketches, w-C 8 H 17 C R. 

H (!) 

Show all steps of a likely rrochanism for these reactions. 

Problem 6.8 From the addition of CC1 4 to alkenes, RCH=CH 2 , there is ob- 
tained not only RCHCICH 2 CC1 3 , but also RCHC1CH 2 CHCH 2 CC1 3 . Using only the 

R 

kinds of reactions you have already encountered, suggest a mechanism for the forma- 
tion of this second product. 



206 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

Problem 6.9 In the dark at room temperature, a solution of chlorine in 
tetrachloroethylene can be kept for long periods with no sign of reaction. When 
irradiated with ultraviolet light, however, the chlorine is rapidly consumed, with the 
formation of hexachloroethane; many molecules of product are formed for each photon 
of light absorbed; this reaction is slowed down markedly when oxygen is bubbled 
through the solution. 

(a) How do you account for the absence of reaction in the dark? (b) Outline all 
steps in the most likely mechanism for the photochemical reaction. Show how it ac- 
counts for the facts, including the effect of oxygen. 

Free-radical addition is probably even commoner than has been suspected. Recent 
work indicates that free-radical chains do not always require light or decomposition of 
highly unstable compounds like peroxides for their initiation. Sometimes a change from 
a polar solvent which can stabilize a polar transition state to a non-polar solvent 
causes a change from a heterolytic reaction to a homolytic one (Sec. 1.14). In some 
cases, it may even be that chains are started by concerted homolysis, in which cleavage of 
comparatively stable molecules (halogens, for example) is aided by the simultaneous 
breaking and making of other bonds. In the absence of the clue usually given by the 
method of initiation, the free-radical nature of such Reactions is harder to detect; one 
depends upon inhibition by oxygen, detailed analysis of reaction kinetics, or a change in 
orientation or stereochemistry. 

6.19 Free-radical polymerization of alkenes 

When ethylene is heated under pressure with oxygen, there is obtained a" 
compound of high molecular weight (about 20,000), which is essentially an alkane 
with a very long chain. This compound is made up of many ethylene units and 



CH 2 ~CH 2 CH 2 -CH 2 -CHr-CH 2 ~ 

or (~CH 2 CH 2 --) 
Polyethylene 

hence is called polyethylene (poly = many). It is familiar to most of us as the plastic 
material of packaging films. 

The formation of polyethylene is a simple example of the process called 
polymerization : the joining together of many small molecules Jo make very large 
molecules. The compound composed of these very large molecules is called a 
polymer (Greek: poly + meros, many parts). The simple compounds from which 
polymers are made are called monomers (mono one). 

Polymerization of substituted ethylenes yields compounds whose structures 
contain the long chain of polyethylene, with substituents attached at more or less 
regular intervals. For example, vinyl chloride yields poly(vinyl chloride), used to 

per " dcs 

Cl 
Vinyl chloride 

or (-CH 2 -CH-) n 

Cl 
Poly(vinyl chloride) 

make phonograph records, plastic pipe, and when plasticized with high-boiling 
esters raincoats, shower curtains, and coatings for metals and upholstery fabrics. 



SEC. 6.20 HYDROXYLATION. GLYCOL FORMATION 207 

Many other groups (e.g., COOCH 3 , CN, C 6 H 5 ) may be attached to the 
doubly-bonded carbons. These substituted ethylenes polymerize more or less 
readily, and yield plastics of widely differing physical properties and uses, but the 
polymerization process and the structure of the polymer are basically the same as 
for ethylene or vinyl chloride. 

Polymerization requires the presence of a small amount of an initiator. 
Among the commonest of these initiators are peroxides, which function by break- 
ing down to form a free radical. This radical adds to a molecule of alkene, and in 
doing so generates another free radical. This radical adds to another molecule of 
alkene to generate a still larger radical, which in curn adds to another molecule of 
alkene, and so on. Eventually the chain is terminated by steps, such as union 
of two radicals, that consume but do not generate radicals. 



Peroxide > Rad- -\ 

Rad- + CH 2 -CH RadCH 2 -CH- I Chain-initiating steps 

R R J 

RadCH 2 -CH + CH 2 =CH * RadCH 2 -CH CH 2 CH- etc. Chain- 

II II propagating 

R R R R step 



This kind of polymerization, each step of which consumes a reactive particle 
and produces another, similar particle, is an example of chain-reaction polymeri- 
zation. In Chap. 32, we shall encounter chain-reaction polymerization that takes 
place, not by way of free radicals, but by way of organic ions. We shall also en- 
counter step-reaction polymerization, which involves a series of reactions each of 
which is essentially independent of the others. 

Problem 6.10 Give the structure of the monomer from which each of the follow- 
ing polymers would most likely be made: 

(a) Orion (fibers, fabrics), ~CH 2 CH(CN)CH 2 CH(CN)~; 

(b) Saran (packaging film, seat covers), ~ CH 2 CCl 2 CH 2 CCl 2 -~; 

(c) Teflon (chemically resistant articles), ~ CF 2 CF 2 CF 2 CF 2 -~. 

Problem 6.11 Can you suggest a reason why polymerization should take place 
in a way C'head-to-taiP) that yields a polymer with regularly alternating groups? 



6.20 Hydroxylation. Glycol formation 

Certain oxidizing agents convert alkenes into compounds known as glycols. 
Glycols-are simply dihydroxy alcohols; their formation amounts to l 
of two hydroxyl groups to the double bond. 



l __l __ cold alkaline KMnO 4 __i _ r __ 
^" C or HCOjOH > | 7 , 

OH OHL 

A glycol 



Of the numerous oxidizing agents that cause hydroxylation, two of the most 
commonly used are (a) cold alkaline KMnO 4 , and (b) peroxyformic acid, 
HC0 2 OH. 



208 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

Hydroxylation with permanganate is carried out by stirring together at room 
temperature the alkene and the aqueous permanganate solution: either neutral 
the reaction produces OH~ or, better, slightly alkaline. Heat and the addition 
of acid are avoided, since these more vigorous conditions promote further oxida- 
tion of the glycol, with cleavage of the carbon-carbon double bond (Sec. 6.29). 

Hydroxylation with peroxyformic acid is carried out by allowing the alkene 
to stand with a mixture of hydrogen peroxide and formic acid, HCOOH, for a few 
hours, and then heating the product with water to hydrolyze certain intermediate 
compounds. 

A glycol is frequently named by adding the word glycol to the name of the 
alkene from which it is formed. For example: 

3CH 2 =CH 2 + 2KMnO 4 + 4H 2 O - > 3CH 2 -CH 2 + 2MnO 2 + 2KOH 
Ethylene 



Ethylene glycol 

H 2 HC 2 H > - 
Propylcne 



CH 3 -CH=CH 2 HC 2 H > -^> CH 3 -CH-CH 2 



Propylene glycol 

Hydroxylation of alkenes is the most important method for the synthesis of 
glycols. Moreover, oxidation by permanganate is the basis of a very useful analy- 
tical test known as the Baeyer test (Sec. 6.30). 

(We shall discuss the stereochemistry and mechanism of glycol formation in 
Sec. 17.12.) 

6.21 Substitution by halogen. Allylic hydrogen 

So far in our discussion of alkenes, we have concentrated on the carbon- 
carbon double bond, and on the addition reactions that take place there. Now 
let us turn to the alkyl groups that are present in most alkene molecules. 

Since these alkyl groups have the alkane structure, they should undergo 
alkane Reactions, for example, substitution by halogen. But an alkene molecule 
presents two sites where halogen can attack, the double bond and the alkyl groups. 
Can we direct the attack to just one of these sites? The answer is yes, by our 
choice of experimental conditions. 

We know that alkanes undergo substitution by halogen at high temperatures 
or under the influence of ultraviolet light, and generally in the gas phase: conditions 
that favor formation of free radicals. We know that alkenes undergo addition of 
halogen at low temperatures and in the absence of light, and generally in the liquid 
phase: conditions that favor ionic reactions, or at least do not aid formation of 
radicals. 

I I I 

-c=c~c- 



Ionic Free-radical 

attack attack 

Addition Substitution 



SEC. 6.21 



SUBSTITUTION BY HALOGEN. ALLYLIC HYDROGEN 



209 



If we wish to direct the attack of halogen to the alkyl portion of an alkene 
molecule, then, we choose conditions that are favorable for the free-radical reaction 
and unfavorable for the ionic reaction. Chemists of the Shell Development Com- 
pany found that, at a temperature of 500-600, a mixture of gaseous propylene 
and chlorine yields chiefly the substitution product, 3-chloro-l-propene, known as 
allyl chloride (CH 2 =CH CH 2 = allyl). Bromine behaves similarly. 



CH 3 -CH=CH 2 

Propylene 



low temp. 
CC1 4 soln. 



500-600 
gas phase 



CH 3 CH-CH 2 

Cl Cl 

1 ,2-Dichloropropane 
Propylene chloride 

C1~CH 2 --CH=CH 2 

3-ChIoro- 1 -propene 
Allyl chloride 



HC1 



Ionic: ~" 
addition 



Free-radical: 

substitution 



In view of Sees. 6.17-6.18, we might wonder why a halogen atom does not add 
to a double bond, instead of abstracting a hydrogen atom. H. C. Brown (of Purdue 
University) has suggested that the halogen atom does add but, at high temperatures, 
is expelled before the second step of free-radical addition can occur. 



X- + CHj CH-CH 2 



CH 3 -CH-CH 2 X 
I 



Free-radical addition 
CH 3 -CH CH 2 X + X- 



X 



Free-radical substitution 



HX + CH 2 -CH=-CH 2 
Allyl radical 



> X CH 2 -CH=CH 2 + X- 
Allyl halide 

Actual product at 

high temperature or 

low halogen concentration 

(X Cl, Br) 

Consistent with Brown's explanation is the finding that low concentration of 
halogen can be used instead of high temperature to favor substitution over (free- 
radical) addition. Addition of the halogen atom gives radical I, which falls apart 
(to regenerate the starting material) if the temperature is high or if it does not 
soon encounter a halogen molecule to complete the addition. The allyl radical, 
on the other hand, once formed, has little option but to wait for a halogen molecule, 
whatever the temperature or however low the halogen concentration. 

Problem 6.12 (a) What would the allyl radical have to do to return to the start- 
ing material ? (b) From bond dissociation energies, calculate the minimum Ut for 
this reaction. 



The compound N-bromosuccinimide (NBS) is a reagent used for the specific 
purpose of brominating alkenes at the allylic position; NBS functions simply by 



210 ALKENES II, REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

providing a constant, low concentration of bromine. As each molecule of HBr is 
formed by the halogenation, NBS converts it into a molecule of Br 2 . 

O O . 



. 

N-Bromosuccinimifle Succinimide 

(NBS) 

6.22 Orientation and reactivity in substitution 

Thus alkenes undergo substitution by halogen in exactly the same way as do 
alkanes. Furthermore, just as the alkyl groups affect the reactivity of the double 
bond toward addition, so the double bond affects the reactivity of the alkyl groups 
toward- substitution. 

Halogenation of many alkenes has shown that: (a) hydrogens attached to 
doubly-bonded carbons undergo very little substitution; and (b) hydrogens at- 
tached to carbons adjacent to doubly-bonded carbons are particularly reactive 
toward substitution. Examination of reactions which involve attack not only 
by halogen atoms but by other free radicals as well has shown that this is a general 
rule: hydrogens attached to doubly-bonded carbons, known as vinylic hydrogens, 
are harder to abstract than ordinary primary hydrogens; hydrogens attached to a 
carbon atom adjacent to a double bond, known as allylic hydrogens, are even 
easier to abstract than tertiary hydrogens. 



Vinylic hydrogen: hard to abstract 
Allylic hydrogen: easy to abstract 



^r 
-<u 



We ^an now expand the reactivity sequence of Sec. 3.23. 
*** > 3" > 2" > 1- > CH 4> vinyiic 

Substitution in alkenes seems to proceed by the same mechanism as substitu- 
tion in alkanes. For example: 



CH 2 -CH-H -> CH 2 =CH- -i CH 2 =CH Cl 
Ethylene Vinyl radical Vinyl chloride 

CH 2 -CH-CH 2 -H -U CH 2 =CH-CH 2 . -^ CH 2 =CH-~CH 2 C1 
Propylene Allyl radical Allyl chloride 

Evidently the vinyl radical is formed very slowly and the allyl radical is formed very 
rapidly. We can now expand the sequence of Sec. 3.25. 



SEC. 6.23 RESONANCE THEORY 211 

Ease of formation CH 

of free radicals ' J J 

Are these findings in accord with our rule that the more stable the radical, the 
more rapidly it is formed'? Is the slowly formed vinyl radical relatively unstable, 
and the rapidly formed allyl radical relatively stable? 

The bond dissociation energies in Table 1.2 (p. 21) show that 104 kcal of 
energy is needed to form vinyl radicals from a mole of ethylene, as compared with 
98 kcal for formation of ethyl radicals from ethane. Relative to the hydrocarbon 
from which each is formed, then, the vinyl radical contains more energy and is 
less stable than a primary radical, and about the same as a methyl radical. 

On the other hand, bond dissociation energies show that only 88 kcal is needed 
for formation of allyl radicals from propylene, as compared with 91 kcal for 
formation of tert-ouiyi radicals. Relative to the hydrocarbon from which each is 
formed, the allyl radical contains less energy and is more stable than the tert- 
butyl radical. 

We can now expand the sequence of Sec. 3.24; relative to the hydrocarbon 
from which each is formed, the order of stability of free radicals is: 



In some way, then, the double bcmd affects the stability of certain free radicals; 
it exerts a similar effect on the incipient radicals of the transition state, and thus 
affects the rate of their formation. We have already seen (Sec. 5.4) a possible 
explanation for the unusually strong bond to vinylic hydrogen. The high stability 
of the allyi radical is readily accounted for by the structural theory: specifically, 
by the concept of resonance. 

6.23 Resonance theory 

It will be helpful first to list some of the general principles of the concept of 
resonance, and then to discuss these principles in terms of a specific example, 
the structure of the allyl radical. 

(a) Whenever a molecule can be represented by two or more structures that 
differ only in the arrangement of electrons that is, by structures that have the same 
arrangement of atomic nuclei there is resonance. The molecule is a hybrid of all 
these structures, and cannot be represented satisfactorily by any one of them. 
Each of these structures is said to contribute to the hybrid. 

(b) When these contributing structures are of about the same stability (that is, 
have about the same energy content), then resonance is important. The contribution 
of each structure to the hybrid depends upon the relative stability of that struc- 
ture: the more stable structures make the larger contribution. 

(c) The resonance hybrid is more stable than any of the contributing structures. 
This increase in stability is called the resonance energy. The more nearly equal in 
stability the contributing structures, the greater the resonance energy. 

There can be resonance only between structures that contain the same number of 
odd electrons. We need concern ourselves about this restriction only in dealing with 



212 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

{//-radicals: molecules that contain two unpaired electrons. There cannot be resonance 
between a diradical structure and a structure with all electrons paired. 



6.24 Resonance structure of the allyl radical 

In the language of the resonance theory, then, the allyl radical is a resonance 
hybrid of the two structures, I and II. 

CH 2 ==CH--CH 2 . CH 2 -CH=CH 2 

I H 

This simply means that the allyl radical does not correspond to either I or 
II, but rather to a structure intermediate between I and II. Furthermore, since I 
and II are exactly equivalent, and hence have exactly the same stability, the 
resonance hybrid is equally related to I and to II; that is, I and II are said to 
make equal contributions to the hybrid. 

This does not mean that the allyl radical consists of molecules half of which 
correspond to I and half to II, nor does it mean that an individual molecule 
changes back and forth between I and II. All molecules are the same; each one 
has a structure intermediate between I and II. 

An analogy to biological hybrids that was suggested by Professor G. W. 
Wheland of the University of Chicago is helpful. When we refer to a mule as a 
hybrid of a horse and a donkey, we do not mean that some mules are horses and 
some mules are donkeys; nor do we mean that an individual mule is a horse part of 
the time and a donkey part of the time. We mean simply that a mule is an animal 
that is related to both a horse and a donkey, and that can be conveniently defined in 
terms of those familiar animals. 

An analogy used by Professor John D. Roberts of the California Institute of 
Technology is even more apt. A medieval European traveler returns home from a 
journey to India, and describes a rhinoceros as a sort of cross between a dragon and 
a unicorn a quite satisfactory description of a real animal in terms of two familiar 
but entirely imaginary animals. 

// must be understood that our drawing of two structures to represent the allyl 
radical does not imply that either of these structures (or the molecules each would 
singly represent) has any existence. The two pictures are necessary because of the 
limitations of our rather crude methods of representing molecules. We draw two 
pictures because no single one would suffice. It is not surprising that certain 
molecules cannot be represented by one structure of the sort we have employed; 
on the contrary, the surprising fact is that the crude dot-and-dash representation 
used by organic chemists has worked out to the extent that it has. 

The resonance theory further tells us that the allyl radical does not contain 
one carbon-carbon t single bond and one carbon-carbon double bond (as in 1 
or II), but rather contains two identical bonds, each one intermediate between a 
single and a double bond. This new type of bond this hybrid bond - has Kvn 
described as a one-and-a-half bond. It is said to possess one-half single-bond ehaw- 
ter and one-half double-bond character. 

[CH 2 =CH~CH 2 . .CH 2 -CH=CH 2 ] equivalent to CH^CH^CH; 

I H ' * ' 

ill 



SEC. 6.25 STABILITY OF THE ALLYL RADICAL 213 

The odd electron is not localized on one carbon or the other but is delocaliz^d, 
being equally distributed over both terminal carbons. We might represent this 
symmetrical hybrid molecule as in III, where the broken lines represent half bonds. 

Problem 6.13 The nitro group, NO 2 , is usually represented as 

v 

x o 

Actual measurement shows that the two nitrogen -oxygen bonds of a nitro compound 
have exactly the same length. In nitromethane, CH 3 NO 2 , for example, the two nitro- 
gen-oxygen bond lengths are each 1.21 A, as compared with a usual length of 1.36 A 
for a nitrogen -oxygen single bond and 1.18 A for a nitrogen-oxygen double bond, 
What is a better representation of the N(>2 group? 

Problem 6.14 The carbonate ion, CO 3 ~ ~, might be represented as 

o=c / 



Actual measurement shows that all the carbon-oxygen bonds in CaCOa have the same 
length, 1.31 A, as compared with a usual length of about 1.36 A for a carbon-oxygen 
single bond and about 1,23 A for a carbon-oxygen double bond. What is a better 
representation of the CO 3 ~ ~ ion? 



6.25 Stability of the allyl radical 

A further, most important outcome of the resonance theory is this: as a 
resonance hybrid, the allyl radical is more stable (i.e., contains less energy) than 
either of the contributing structures. This additional stability possessed by the mole- 
cule is referred to as resonance energy. Since these particular contributing struc- 
tures are exactly equivalent and hence of the same stability, we expect stabilization 
due to resonance to be large. 

Just how large is the resonance energy of the allyl radical ? To know the exact 
value, we would have to compare the actual, hybrid allyl radical with a non-existent 
radical of structure I or II something we cannot do, experimentally. We can, 
however, estimate the resonance energy by comparing two reactions: dissociation 
of propane to form a w-propyl radical, and dissociation of propylene to form an 
allyl radical. 

CH 3 CH 2 CHj > CH 3 CH 2 CH r + H- A// = + 98 kcal 

Propane w-Propyl radical 

CH 2 ^CH -CH 3 > CH 2 =CH -CH 2 - + H- A// = +85 

Propylene Allyl radical 

Propane, the /j-propyl radical, and propylene are each fairly satisfactorily repre- 
sented by a single structure; the allyl radical, on the other hand, is a resonance 
bvbrkl, WV see that the energy difference between propylene and the allyl radical 
is 10 kiai/niolc less (98 88) than the energy difference between propane and the 



214 



ALKENES H. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 



n-propyl radical; we attribute the lower dissociation energy entirely to resonance 
stabilization of the allyl radical, and estimate the resonance energy to be 10 kcal/ 
mole. 

6.26 Orbital picture of the allyl radical 

To get a clearer picture of what a resonance hybrid is and, especially, to 
understand how resonance stabilization arises let us consider the bond orbitals 
in the allyl radical. 

Since each carbon is bonded to three other atoms, it uses sp 2 orbitals (as in 
ethylene, Sec. 5.2). Overlap of these orbitals with each other and with the 5 orbitals 
of five hydrogen atoms gives the molecular skeleton shown in Fig. 6.5, with all 
bond angles 120. In addition, each carbon atom has a p orbital which, as we 
know, consists of two equal lobes, one lying above and the other lying below the 
plane of the a bonds; it is occupied by a single electron. 




equivalent to 




[CH 2 -CH CH 2 * -CH 2 CH=CH 2 ] equivalent to 



Figure 6.5. Ally! radical. The p orbital of the middle carbon overlaps 
p orbitals on both sides to permit delocalization of electrons. 



As in the case of ethylene, the p orbital of one carbon can overlap the 
p orbital of an adjacent carbon atom, permitting the electrons to pair and a bond 
to be formed. In this way we would arrive at either of the contributing structures, 
I or II, with the odd electron occupying the/? orbital of the remaining carbon atom. 
But the overlap is not limited to a pair of p orbitals as it was in ethylene; the 
p orbital of the middle carbon atom overlaps equally well the p orbitals of both 
the carbon atoms to which it is bonded. The result is two continuous IT electron 
clouds, one lying above and one lying below the plane of the atoms. 

Since no more than two electrons may occupy the same orbital (Pauli exclusion 
principle), these ir clouds are actually made up of two orbitals (See 29.5). One of these, 
containing two n electrons, encompasses all three carbon atoms; the other, containing 
the third (odd) -n electron, is divided equally between the terminal carbons. 

The overlap of the p orbitals in both directions, and the resulting participation 
of each electron in two bonds, is equivalent to our earlier description of the allyl 
radical as a resonance hybrid of two structures. These two methods of representa- 
tion, the drawing of several resonance structures and the drawing of an electron 
cloud, are merely our crude attempts to convey by means of pictures the idea that 
a given pair of electrons may serve to bind together more than two nuclei. It is this 



SEC. 6.27 USING THE RESONANCE THEORY 215 

ability of n electrons to participate in several bonds, this delocalization of elec- 
trons, that results in stronger bonds and a more stable molecule. For this reason 
the term delocalization energy is frequently used instead of resonance energy. 

The covalcnt bond owes its strength to the fact that an electron is attracted more 
strongly by two nuclei than by one. In the same way an electron is more strongly attrac- 
ted by three nuclei than by two. 

We saw earlier (Sec. 2.21) that the methyl radical may not be quite flat: that 
hybridization of carbon may be intermediate between sp 2 and sp*. For the allyl 
radical, on the other hand and for many other free radicals flatness is clearly 
required to permit the overlap of/? orbitals that leads to stabilization of the radical. 

In terms of the conventional valence-bond structures we employ, it is difficult 
to visualize a single structure that is intermediate between the two structures, 1 
and II. The orbital approach, on the other hand, gives us a rather clear picture of 
the allyl radical: the density of electrons holding the central carbon to each of the 
others is intermediate between that of a single bond and that of a double bond. 

6.27 Using the resonance theory 

The great usefulness, and hence the great value, of the resonance theory lies in 
the fact that it retains the simple though crude type of structural representation 
which we have used so far in this book. Particularly helpful is the fact that the 
stability of a structure can often be roughly estimated from its reasonableness. If 
only one reasonable structure can be drawn for a molecule, the chances are good 
that this one structure adequately describes the molecule. 

The criterion of reasonableness' is not so vague as it might appear. The fact 
hat a particular structure seems reasonable to us means that we ha\e previously 
encountered a compound whose properties are pretty well accounted for by a 
structure of that type; the structure must, therefore, represent a fairly stable kind 
of arrangement of atoms and electrons. For example, each of the contributing 
structures for the allyl radical appears quite reasonable because we have en- 
countered compounds, alkenes and free radicals, that possess the features of this 
structure. 

There are a number of other criteria that we can use to estimate relative stabili- 
ties, and hence relative importance, of contributing structures. One of these has 
to do with (a) electronegativity ami location of charge. 

For example, a convenient way of indicating the polarity (ionic character) of 
the hydrogen- chlorine bond is to represent HC1 as a hybrid c-f structures I and II. 
We judge that II is appreciably stable and hence makes significant contribution, 
because in it a negative charge is located on a highly electronegative atom, chlorine. 

H-C1 H + C1- 

I II 

On the other hand, we consider methane to be represented adequately by the 
single structure HI. 

H 

H-C-H 

H 

III 



216 ALKENES II. REACTIONS OF CARBON-CARBON DOLBLE BOND CHAP. 6 

Although it is possible to draw additional, ionic structures like IV and V, we judge 
these to be unstable since in them a negative charge is located on an atom of low 

H + H 

H- C-H H-C- H + etc. 

I ! 

H H 

IV V 

elcctroncgativity, carbon. We expect TV and V to make negligible contribution to 
the hybrid and hence we ignore them. 

In later sections we shall use certain other criteria to help us estimate stabilities 
of possible contributing structures: (b) number of bonds (Sec. 8.17); (c) dispersal 
of charge (Sec. 11.19); (d) complete vs. incomplete octet (Sec. 1 1.20); (e) separation 
of charge (Sec. 18.12). 

Finally, we shall find certain cases where the overwhelming weight of evidence 
bond lengths, dipole moments, reactivity- indicate that an accurate description 
of a given molecule requires contribution from structures of a sort that may appear 
quite unreasonable to us (Sees. 6.28 and 8.18); this simply reminds us that, after 
all, we know very little about the structure of molecules, and must be prepared to 
change our ideas of what is reasonable to conform with e\idence provided by 
experimental facts. 

In the next section, we shall encounter contributing structures that are very 
strange looking indeed. 

Problem 6.15 The icni^ation potential of the all} I radical is 188 kail/mole. 
(a) Hou does the allvl cation compare in stability \\ith the simple alk\l cations of Sec. 
5.18? (b) Is the cation adequately represented by the structure CH : 
Describe its structure in both valence-bond and orbital terms. (Check your 
in Sec. 8.21.) 

Problem 6.16 Benzene, C 6 H , is a flat molecule \\ith all bond angles 120 ami 
all carbon carbon bonds 1.39 A long. Its heat of h>drogenation (absorption of three 
moles of hydrogen) is 49.8 kcal mole, as compared with values of 28.6 foi c>clohexene 
(one mole of hydrogen) and 55.4 for l,3-c>clohe\udicnc Uv\o moles of hvdrogen). 
(a) Is ben/ene adequately represented by the Kekule formula sho\\n? (b) Suggest a 



CH 



H 

CH 2 ^' 

-^" 



^C" HC^ /CH H- Ccx. ^C-H 

CH 2 CM C 

i 
H 

C\clohcxene l,3-C>cIohc\adicne Benzene 

(Kckulc formula) 

better structure for benzene in both valence-bond and orbital terms. (Check your 
answer in Sees. 10.7-10.8.) 



6.28 Resonance stabilization of alky] radicals. Hyperconjugation 

The relative stabilities of tertiary, secondary, and primary alkyl radicals are 
accounted for on exactly the same basis as the stability of the allyl radical: 



SEC. 6.28 RESONANCE STABILIZATION OF ALKYL RADICALS 217 

delocalization of electrons, this time through overlap between the p orbital occupied 
by the odd electron and a a orbital of the alkyl group (Fig. 6.6). Through tbis 





Figure 6.6. Hyperconjugation in an alkyl free ratlical. (a) Separate a and 
p orbitals. (b) Overlapping orbitals. 



overlap, individual electrons can, to an extent, help bind together three nuclei, 
two carbons and one hydrogen. This kind of delocalization, involving a bond 
orbitals, is called hypcrconjugation. 

In resonance language, we would say that the ethyl radical, for example, is a 
Inbnd of not only the usual structure, I, but also three additional structures, II, 

HH HH H H HH 

! i i 

H-C C H- C--C H-C-C H-C-C 

i . i ' I : I 

H H H H H H -H H 

I II III IV 

ill, and IV, in which a double bond joins the two carbons, and the odd electron 
is held by a hydrogen atom. 

Individually, each of these "no-bond" resonance structures appears strange 
Inil, l.ikcn together, ihev mean that the carbon hydrogen bond is something less 
th.ui a single bond, thai the carbon carbon bond has some double bond charac- 
u-i, and thai the odd electron is parll\ accommodated by hydrogen atoms. Contri- 
bution from these unstable structures is not nearly so important as from, say, 
the equivalent structures for the allyl radical, and the resulting stabilization is 
not nearly so large. It is believed, however, to stabilize the ethyl radical to the 
extent of 6 kcal relative to the methyl radical (104 - 98, Sec. 3.24), for which such 
resonance is not possible. 

If we extend this idea to the isopropyl radical, we find that instead of three 
hypcrconjugation structures we now have six. (Draw them.) The larger number of 
contributing structures means more extensive delocalization of the odd electron, 
and hence greater stabilization of the radical. In agreement with this expectation, 
\se find that the bond dissociation energy of the isopropyl-hydrogen bond is only 
95 kcal, indicating a resonance energy of 9 kcal/mole (104 95). 

For the tert-buiyl radical there should be nine such hyperconjugation struc- 
tures. (Draw them.) Here we find a bond dissociation energy of 91 kcal, indicating 
a resonance stabilization of 13 kcal/mole (104 -, 91). 

In summary, the relative stabilities of the free radicals we have studied are 
determined by delocalization of electrons. Delocalization takes place through 



218 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

overlap of the p orbital occupied by the odd electron: overlap with the n cloud of a 
double bond in the allyl radical, or overlap with a bonds in alkyl radicals. 

Problem 6.17 It has been postulated that the relative stabilities of alkyl cat- 
ions are determined not only by inductive effects but also by resonance stabilization. 
How might you account for the following order of stability of cations? 

tert-buty] > isoprop>l > ethyl > methyl 



6.29 Ozonolysis. Determination of structure by degradation 

Along with addition and substitution we may consider a third general kind of 
alkene reaction, cleavage: a reaction in which the double bond is completely 
broken and the alkene molecule converted into two smaller molecules. 

The classical reagent for cleaving the carbon-carbon double bond is ozone. 
Ozonolysis (cleavage "by ozone) is carried out in two stages: first, addition of ozone 
to the double bond to form an ozonide ; and second, hydrolysis of the ozonide to 
yield the cleavage products. 

Ozone gas is passed into a solution of the alkene in some inert solvent like 
carbon tetrachloride; evaporation of the solvent leaves the ozonide as a viscous oil. 
This unstable, explosive compound is not purified, but is treated directly with 
water, generally in the presence of a reducing agent. 

In the cleavage products a doubly-bonded oxygen is found attached to each 
of the originally doubly-bonded carbons: 



Ozonolysis 



-U- 




o^c- 



Alkene Molozonide Ozonide Cleavage products 

(Aldehydes and ketones) 

These compounds containing the C~O group are called aldehydes and ketones; 
at this point we need only know that they are compounds that can readily be 
identified (Sec. 19.17). The function of the reducing agent, which is frequently zinc 
dust, is to prevent formation of hydrogen peroxide, which would otherwise react 
with the aldehydes and ketones. (Aldehydes, RCHO, are often converted into 
acids, RCOOH, for ease of isolation.) 

Knowing the number and arrangement of carbon atoms in these aldehydes 
and ketones, we can work back to the structure of the original alkene. For example, 
for three of the isomeric hexylenes: 

H H 

CH 3 CH 2 CH 2 G=0 + 0=CCH 3 < H2 /Zn ^- CH 3 CH 2 CH 2 CH==CHCH 3 
Aldehydes 2-Hexene 



H 



0=CCH 2 CH 3 < H2 /Zn <^- CH 3 CH 2 CH^CHCH 2 CH 3 
Aldehydes 3-Hexene 



SEC. 6.30 ANALYSIS OF ALKENES 219 

H CH 3 CH 3 

CH 3 CH 2 (Uo + 0==i--CH 3 < H ' /Zn <&- CH 3 CH 2 CH==C--CH 3 
Aldehyde Ketone 2-Methyl-2-pentenc 

One general approach to the determination of the structure of an unknown 
compound is degradation, the breaking down of the unknown compound int9 a 
number of smaller, more easily identifiable fragments. Ozonolysis is a typical 
means of degradation. 

Another method of degradation that gives essentially the same information 
although somewhat less reliable is vigorous oxidation by permanganate, which 
is believed to involve formation and cleavage of intermediate glycols (Sec. 6.20). 



x^ 

i 



OH I 



Carboxylic acids, RCOOH, are obtained instead of aldehydes, RCHO. A terminal 
=CH 2 group is oxidized to CO 2 . For example: 

CH 3 CH 3 

CH 3 COOH + O=C-CH 3 < KM " 4 CH 3 CH-=C-CH 3 

Carboxylic Ketone 2-Methyl-2-butene 

acid 

CH 3 CH 2 CH 2 COOH + CO 2 < KMn * CH 3 CH 2 CH 2 CH-CH 2 
Carboxylic Carbon 1-Pentene 

acid dioxide 

Problem 6.18 What products would you expect from each of the dimers of 
isobutylene (Sec. 6.15) upon cleavage by: (a) ozonolysis, (b) KMnO 4 ? 



6.30 Analysis of alkencs 

The functional group of an alkene is the carbon-carbon double bond. To 
characterize an unknown compound as an alkene, therefore, we must show that 
it undergoes the reactions typical of the carbon-carbon double bond. Since there 
are so many of these reactions, we might at first assume that this is an easy job. 
But let us look at the problem more closely. 

First of all, which of the many reactions of alkenes do we select? Addition 
of hydrogen bromide, for example 9 Hydrogenation ? Let us imagine ourselves 
in the laboratory, working with gases and liquids and solids, with flasks and test 
tubes and bottles. 

We could pass dry hydrogen bromide from a tank through a test tube of an 
unknown liquid. But what would we see? How could we tell whether or not a 
reaction takes place? A colorless gas bubbles through a colorless liquid; a di fie rent 
colorless liquid may or may not be formed. 

We could attempt to hydrogenate the unknown compound. Here, we might 
say, we could certainly tell whether or not reaction takes place: a drop in the 
hydrogen pressure would show us that addition had occurred. This is true, and 
hydrogenalion can be a useful analytical tool. But a catalyst must be prepared, 



220 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

and a fairly elaborate piece of apparatus must be used ; the whole operation might 
take hours. 

Whenever possible, we select for a characteritation test a reaction that is 
rapidly and conveniently carried out, and that gives rise to an easily observed change. 
We select a test that requires a few minutes and a few test tubes, a test in which 
a color appears or disappears, or bubbles of gas are evolved, or a precipitate forms 
or dissolves. 

Experience has shown that an alkene is best characterized, then, by its property 
of decolorizing both a solution of bromine in carbon tetrachloride (Sec. 6.5) and 
a cold, dilute, neutral permanganate solution (the Baeyer test. Sec. 6.20). Both 
tests are easily carried out; in one, a red color disappears, and in the other, a purple 
color disappears and is replaced by brown manganese dioxide. 

X O=C X + Br 2 /CCl 4 > -C-C- 

' x U 

Alkenc Red Colorless 

C=C + MnO 4 " v MnO 2 4- C C or other products 

OH d>H 

Alkene Purple Brown ppt. Colorless 

Granting that we have selected the best tests for the characterization of alkenes, 
let us go on to another question. We add bromine in carbon tetrachloride to an 
unknown organic compound, let us say, and the red color disappears. What 
does this tell us? Only that our unknown is a compound that reacts with bromine. 
It may be an alkene. But it is not enough merely to know that a particular kind 
of compound reacts with a given reagent; we must also know what other kinds of 
compounds also react with the reagent. In this case, the unknown may equally 
well be an alkyne. (It may also be any of a number of compounds that undergo 
rapid substitution by bromine; in that case, however, hydrogen bromide would be 
evolved and could be detected by the cloud it forms when we blow our breath over 
the test tube.) 

In the same way, decolorization of permanganate does not prove that a com- 
pound is an alkene, but only that it contains some functional group that can be 
oxidized by permanganate. The Compound may be an atttene; but it may instead 
be a& atkyne, an aldehyde, or any of a number of easily oxidized compounds. 
It may even be a compound that is contaminated with an Impurity that fa oxidised; 
alcohols, for example, are not oxidized under tlrtse conditions, but often contain 
impurities that ore. We can usually rule out this by making sure that more than a 
drop or two of the reagent is decolorized* 

By itself, a single characterization test seldom proves that an unknown is 
one particular kind of compound. It may limit tte number of possibilities, so 
that a final decision can then be made on tbe ba*m of ad4itloaal tests. Of, con- 
versely, if certain possibilities have already been eHmto*trf, a tingle tort may ponnk 
a final choice to be made. Thus, the bromine or pwmaganate test would to 
sufficient to differentiate an alkene from an alkane, ot an aJfcene from an alky! 
halide, or an alkene from an alcohol. 



PROBLEMS 221 

The tests most used in characterizing alkenes, then, are the following: (a) rapid 
decolorization of bromine in carbon tetrachloride without evolution of HBr, 
a test also given by alkynes; (b) decolorization of cold, dilute, neutral, aqueous 
permanganate solution (the Baeyer test), a test also given by alkynes and aldehydes. 
Also helpful is the solubility of alkenes in cold concentrated sulfuric acid, a test 
also given by a great many other compounds, including all those containing oxygen 
(they form soluble oxonium salts) and compounds that are readily sulfonated 
(Sees. 12.11 and 17.8). Alkanes or alkyl halides are not soluble in cold concen- 
trated sulfuric acid. 

Of the compounds we have dealt with so far, alcohols also dissolve in sulfuric 
acid. Alcohols can be distinguished from alkenes, however, by the fact that 
alcohols give a negative test with bromine in carbon tetrachloride and a negative 
Baeyer test so long as we are not misled by impurities. Primary and secondary 
alcohols ore oxidized by chromic anhydride, CrO 3 , in aqueous sulfuric acid: 
within two seconds, the clear orange solution turns blue-green and becomes 
opaque. 

ROH -I- HCrO 4 - > Opaque, blue-green 
1 or 2 Clear, 
orange 

Tertiary alcohols do not give this test; nor do alkenes. 

Problem 6.19 Describe simple chemical tests (if any) that would distinguish j 
between: (a) an alkene and an alkane; (b) an alkene and an alkyl halide; (c) an alkenc " 
and a secondary alcohol; (d) an alkene, an alkane, an alkyl halide, and a secondary 
alcohol. Tell exactly what you would do and see. 

Problem 6.20 Assuming the choice to be limited to alkane, alkene, alkyl halide, 
secondary alcohol, and tertiary alcohol, characterize compounds A, B, C, D, and E 
on the basis of the following information: 



KMn0 4 CrO 3 





Qual. 




Compound 


elem. anal. 


H 2 S0 4 


A 





Insoluble 


B 




Soluble 


C 


Cl 


Insoluble 


D 





Soluble 


E 





Soluble 



Once characterized as an alkene, an unknown may then be identified as a 
previously reported alkene on the basts of its physical properties, including its 
infrared spectrum and molecular weight. Proof of structure of a new compound 
is best accomplished by degradation: cleavage by ozone or permanganate, followed 
by identification of the fragments formed (Sec. 6.29). 

(Spectroscopic analysis of alkenes will be discussed in Sees. 13.15-13.16.) 

PROBLEMS 

1. Draw a structural formula and give (when you can) an alternative name for: 

(a) ethylene bromide (e) propylene glycol 

(b) ethyl bromide (f ) propylene bromohydrin 

(c) bromoe^hylene (g) vinyl bromide 

(d) ethylene glycol (h) ally! chloride 



222 AUKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

2. Give structures and names of the products (if any) expected from reaction of 
isobutytene with: 

(a) H 2 , Ni (g) HI (m) H 2 SO 4 ( C 8 H 16 ) 

(b) C1 2 (h) HI (peroxides) (n) isobutane + HP 

(c) Br 2 (i) H 2 SO 4 (o) cold alkaline KMnO 4 

(d) I 2 (j) H 2 0, H+ (p) hot KMn0 4 

(e) HBr (k) Br 2 , H 2 O (q) HCO 2 OH 

(0 HBr (peroxides) (I) Br 2 + NaCI(aq) (r) O 3 ; then Zn, H 2 O 

3. Which alkene of each pair would you expect to be more reactive toward addition 
ofH 2 SO 4 ? 

(a) ethylene or propylene (e) vinyl chloride or 1,2-dichloroethene 

(b) ethylene or vinyl bromide (f) 1-pentene or 2-methyM-butene 

(c) propylene or 2-butene (g) ethylene or CH 2 =CHCOOH 

(d) 2-butene or isobutylene (h) propylene or 3,3,3-trifluoropropene 

4. Give structures and names of the principal products expected from addition of 
HI to: 

(a) 2-butene (e) 3-methyl-l-butene (2 products) 

(b) 2-pentene (f) vinyl bromide 

(c) 2-methyl-l-butene (g) 2,3-dimethyI-l-butene 

(d) 2-methyl-2-butene . (h) 2,4,4-trimethyl-2-pentene 

5. Draw the structure of 6-methyl-2-heptene. Label each set of hydrogen atoms to 
show their relative reactivities toward chlorine atoms, using (1) for the most reactive, 
(2) for the next, etc. 

6. Account for the fact that addition of CBrCl 3 in the presence of peroxides takes 
place faster to 2-ethyl-l-hexene than to 1-octene. 

7. In methyl alcohol solution (CH 3 OH), bromine adds to ethylene to yield not only 
ethylene bromide but also Br CH 2 CH 2 OCH 3 . How can you account for this? Write 
equations for all steps. 

8. As an alternative to the one-step 1,2-hydride shift described in Sec. 5.22, one might 
instead propose in view of the reactions we have studied in this chapter that carbonium 
ions rearrange by a two-step mechanism, involving the intermediate formation of an 
alkene: 




=c- + H+ > -c-c- 



When (by a reaction we have not yet taken up) the isobutyl cation was generated 
in D 2 O containing D 3 O+, there was obtained /er/-butyl alcohol containing no deuterium 
attached to carbon. How does this experiment permit one to rule out the two-step 
mechanism? 

9. In Sec. 6.17 a mechanism was presented for free-radical addition of hydrogen 
bromide. Equally consistent with the evidence given there is the following alternative 
mechanism: 

(2a) Rad- + HBr > Rad-Br + H- 

1 . i I 

(3a) H- + -0=C > -C-C- 

H 



PROBLEMS 223 

i I 
(4a) C-C- + HBr 



II i I 

~ ~ ~ CC- -L- H 



H H ir 

then (3a), (4a), (3a), (4a), etc. 

(a) In steps (2a) and (4a) an alkyl radical abstracts bromine instead of hydrogen 
from hydrogen bromide. On the basis of bond dissociation energies (Table 1.2, p. 21), 
is this mechanism more or less likely than (2) -(4) on p. 203? Explain. 

(b) The esr study (p. 204) showed that the intermediate free radical from a given 
alkene is the same whether HBr or DBr (deuterium bromide) is being added to the dodble 
bond. Explain how this evidence permits a definite choice between mechanism (2a)-(4a) 
and mechanism (2) -(4). 

10. (a) Write all steps in the free-radical addition of HBr to propylene. (b) Write 
all steps that would be involved in the free-radical addition of HC1 to propylene. 

(c) List A// for each reaction in (a) and (b). Assume the following bond dissociation 
energies: * bond, 68 kcal; 1 R Br, 69 kcal; 1 R Cl, 82 kcal; 2 R-H, 95 kcal. 

(d) Suggest a possible reason why the peroxide effect is observed for Hflr but not for 
HC1. 

11. When isobutylene and chlorine are allowed to react in the dark at in the ab- 
sence of peroxides, the principal product is not the addition product but methallyl 
chloride (3-chloro-2-methyl-l-propene). Bubbling oxygen through the reaction mixture 
produces no change. 

This reaction was carried out with labeled isobutylene (l- 14 C-2-methyl-l-propene, 
(CH 3 ) 2 G-= 14 CH 2 ), and the methallyl chloride contained was collected, purified, and sub- 
jected to ozonolysis. Formaldehyde (H 2 C=O) and chloroacetone (C1CH 2 COCH 3 ) were 
obtained; all (97% or more) of the radioactivity was present in the chloroacetone. 

(a) Give the structure, including the position of the isotopic label, of the methallyl 
chloride obtained, (b) Judging from the evidence, is the reaction ionic or free radical? 
(c) Using only steps with which you are already familiar, outline a mechanism that 
accounts for the formation of this product, (d) Can you suggest one reason why isobutyl- 
ene is more prone than 1- or 2-butene to undergo this particular reaction? (e) Under 
similar conditions, and in the presence of oxygen, 3,3-dimethyl-l-butene yields mostly 
the addition product, but also a small yield of 4-chloro-2,3-dimethyl-l-bufene. In light 
of your answer to (c) how do you account for the formation of this minor product? 

12. How do you account for the following facts: formic acid, HCOOH, contains 
one carbon -oxygen bond of 1.36 A and another of 1.23 A, yet sodium formate, HCOO~ 
Na + , contains two equal carbon-oxygen bonds, each of 1.27 A. (Check your answer 
in Sec. 18.13.) 

H-/" 

\>H 

Formic acid 

13. (a) When 1-octene is allowed to react with N-bromosuccinimide, there is 
obtained not only 3-bromo- 1-octene but also l-bromo-2-octene. How can you account 
for this? (b) Propylene, CH 3 CH- 14 CH 2 , labeled with carbon-14 (a radioactive isotope) 
is converted into allyl bromide by free-radical bromination. What would you predict 
about the position of the tagged atom ( 14 C) in the product? 

14. Give the structure of the alkene that yields on ozonolysis: 

(a) CH 3 CH 2 CH 2 CHO and HCHO 

(b) CH 3 -CH-~CHO and CH 3 CHO* 

CH 3 

(c) OnlyCH, CO-CH 3 

(d) CH 3 CHO and HCHO and OHC CH 2 -CHO 

(e) What would each of these alkenes yield upon cleavage by KMnO 4 ? 



224 ALKENES II. REACTIONS OF CARBON-CARBON DOUBLE BOND CHAP. 6 

15. Describe simple chemical tests that would distinguish between: 

(a) 2-chloropentane and //-heptane (d) allyl bromide and 1-hexene 

(b) 2-hexene and tert-butyl bromide (e) sec-butyl alcohol and //-heptane 

(c) isobutane and isobutylene (f ) 1-octene and //-pentyl alcohol 

(g) /m-pentyl alcohol and 2,2-dimethylhexane 

(h) //-propyl alcohol and allyl alcohol (CH 2 ^CHCH 2 OH) 

Tell exactly what you would do and see. (Qualitative elemental analysis is a simple 
chemical test; degradation is not.) 

16. A hydrocarbon, A, adds one mole of hydrogen in the presence of a platinum 
catalyst to form //-hexane. When A is oxidi/ed vigorously with KMnO 4 , e single car- 
boxylic acid, containing three carbon atoms, is isolated. Give the structure and name of A. 
Show your reasoning, including equations for all reactions. 

17. Outline all steps in a possible laboratory synthesis of each of the following 
compounds, using only the organic source given, plus any necessary solvents and inorganic 
reagents. (See general instructions about synthesis below.) 

(a) ethylene from ethane 

(b) propylene from propane 

(c) ethyl iodide from ethane 

(d) 2-bromopropane from propane (Note: simple monobromination of propane yields, 
of course, a mixture of 1 -bromopropane and 2-bromopropane, and is therefore not 
satisfactory for this synthesis. The mixture might, however, be used as an inter- 
mediate.) 

(e) 1,2-dibromopropane from propane 

(f) 1,2-dibromobutane from 1 -bromobutane 

(g) 2-iodobutane from 1-chlorobutane 
(h) 2-methylpentane from propylene 

(i) 3-methyl heptane from //-butyl bromide 

(j) l,2-dibromo~2-methylpropane from isobutane 

(k) 2-iodobutane from //-butyl alcohol 

(1) //-propyl bromide from isopropyl bromide 

(m) propylene chlorohydrin from //-propyl iodide 

(n) isohexane from (CH 3 ) 2 C(OH)CH 2 CH 2 GH 3 

(o) 2,2-dimethylbutane from 3-chloro-2,2-dimethylbutane 



About Synthesis 

Each synthesis should be the one that gives a reasonably pure product in 
reasonably good yield. 

It is not necessary to complete and balance each equation. Simply draw 
the structure of the organic compounds, and write on the arrow the necessary 
reagents and any critical conditions. For example: 

CH 3 CH 2 OH Jill!!!!!* CH 2 -CH 2 " 2>N S CH 3 CH 3 

At this stage you may be asked to make a particular compound by a method 
that would never actually be used for that compound: for example, the synthesis 
of ethane just above. But if you can work out a way to make ethane from ethyl 
alcohol, then, when the need anses, you will also know how to make a compli- 
cated alkane from a complicated alcohol, and, in fact, how to replace an OH 
group by H in just about any compound you encounter. Furthermore, you 
will have gained practice in putting together what you have learned about several 
different kinds of compounds. 



Chapter 



i 



Stereochemistry II. 
Preparation and Reactions 
of Stereoisomers 



7.1 Stereoisomerism 

Stereoisomers, we have learned, are isomers that differ only in the way their 
atoms are oriented in space, ^o far, our study has been limited to finding out what 
the various kinds of stereoibomers are, how to predict their existence, how to name 
them, and, in a general way, how their properties compare. 

In Chap. 4, we learned th.it Stereoisomers exist of the kind called enantio- 
mers (mirror-image isomer.^, that they can be optically active, and that both their 
existence and their optical activity are the result of the chirality of certain mole- 
cules, that is, of the non supeninposability of such molecules on their mirror 
images. We learned how to predict, from a simple examination of molecular struc- 
ture, whether or not a particular compound can display this kind of isomerism. 
We learned how to specify the configuration of a particular enantiomer by use of 
the letters R and S. 

~ We learned about diaster earners*. Stereoisomers that are'waf mirror images. 
Some of these (Sees. 4.17 and 4.18) were of the kind that contained more than one 
chiral center. Others (Sec. 5.6) were the kind, geometric isomers, that owe their 
existence to hindered rotation about double bonds. 

In Sees. 4.20 and 5.6, we learned that Stereoisomers can be classified not only 
as to whether or not they are mirror images, but also and quite independently 
of the other classification as to how they are interconverted. Altogether, we have: 
(a) configuration^ isomers, interconverted by inversion (turning-inside-out) at a 
chiral center; (b) geometric isomers, interconverted in principle by rotation 
about a double bond ; and (c) conformational isomers, interconverted by rotations 
about single bonds. 

The operation required rotation is the same for interconversion of geo- 
metric and conformational isomers, and it has been suggested that they be called 
collectively rotational (or torsional) isomers. Geometric isomers are thus double-bond 
rotational isomers, and conformational isomers are single-bond rotational isomers. 



215 



226 STEREOCHEMISTRY II , CHAP. 7 

On the other hand, from the very practical standpoint of isolability, geometric 
isomers are more akin to configurational isomers: interconversion requires bond 
breaking a IT bond in the case of geometric isomers and hence is always a diffi- 
cult process. Conformational isomers are interconverted by the (usually) easy 
process of rotation about single bonds. 

For convenience, we laid down (Sec. 4.20) the following "ground rule" for 
discussions and problems in this book: unless specifically indicated otherwise, 
the terms " stereoisomcrs" " enantiomers" and " diastereomers" will refer only to 
configurational isomers, including geometric isomers, and will exclude conforma- 
tional isomers. The latter will be referred to as "conformational isomers," "'con- 
formers," " conformational enantiomers," and "conformational diastereomers." 



7.2 Reactions involving stereoisomers 

Now let us go on from the existence of stereoisomers, and look at their ///- 
voli-ement in chemical reactions: reactions in which stereoisomers are formed, 
and reactions in which stereoisomers are consumed', reactions in which the reagent 
is of the ordinary (i.e., optically inactive) kind. and those in which the reagent is 
optically active. 

We shall take up: 

(a) the conversion of an achiral molecule into a chiral molecule, with the 
generation of a chiral center; 

(b) reactions of chiral molecules in which bonds to the chiral center are not 
broken, and see how such reactions can be used to relate the configuration of one 
compound to that of another; 

(c) reactions of the kind in (b) in which a second chiral center is generated; 

(d) reactions of chiral compounds with optically active reagents. 

Then we shall examine the stereochemistry of several reactions we have al- 
ready studied free-radical halogenation of alkanes, and electrophilic addition of 
halogens to alkencs -and see how stereochemistry can be used to get information 
about reaction mechanisms. In doing this, we shall take up: 

(e) a reaction of a chiral compound in which a bond to a chiral center is 
broken ; 

(f) a reaction of an achiral compound in which two chiral centers are gener- 
ated at the same time. 

7.3 Generation of a chiral center. Synthesis and optical activity 

One of the products of chlorination of /i-butane is the chiral compound, 
sec-butyl chloride. It can exist as two enantiomers, I and II, which are specified 

CH 3 CH 2 CH 2 CH3 c^ .heat or light > CH3CH2 _cH-CHj + w -Butyl chloride 
/i-Butane 1. 

Achiral 5^Butyl chloride 

Chiral 

(Sec. 4.16) as S and R, respectively. 



SEC. 7.3 GENERATION OF A CHIRAL CENTER 227 

CH, CH 3 

Cl Cl- 





C 2 H 5 



1 II 

S-( f )-jec-butyl chloride R-( - )-sec -butyl chloride 

Each enantiomer should, of course, be optically active. Now, if we were to 
put the sec-butyl chloride actually prepared by the chlorination of w-butane into 
a polarimeter, would it rotate the plane of polarized light? The answer is no, 
because prepared as described it would consist of the racemic modification. The 
next question is: why is the racemic modification formed ? 

In the first step of the reaction, a chlorine atom abstracts hydrogen to yield 
hydrogen chloride and a sec-butyl free radical. The carbon that carries the odd 
electron in the free radical is s/? 2 -hybridized (trigonal, Sec. 2.21), and hence a part 
of the molecule \sflat, the trigonal carbon and the three atoms attached to it lying 
in the same plane. In the second step, the free radical abstracts chlorine from a 
chlorine molecule to yield sec-butyl chloride. But chlorine may become attached 
to either face of the flat radical, and, depending upon which face, yield either of two 
products: R or S (see Fig. 7.1). Since the chance of attachment to one face is 
exactly the same as for attachment to the other face, the enantiomers are obtained 
in exactly equal amounts. The product is the racemic modification. 



Figure 7.1. Generation of a chiral 
center. Chlorine becomes attached 
to either face of flat free radical, via 
(a) or (6), to give enantiomers, and 
in equal amounts. 




Enantiomers 
Formed in equal amounts 



If we were to apply the approach just illustrated to the synthesis of any 
compound whatsoever and on the basis of any mechanism, correct or incorrect 
we would arrive at the same conclusion: as long as neither the starting material nor 
the reagent (nor the environment) is optically active, we should obtain an optically 



228 STEREOCHEMISTRY II CHAP. 7 

inactive product. At some stage of the reaction sequence, there will be two alter- 
native paths, one of which yields one enantiomer and the other the opposite 
enantiomer. The two paths will always be equivalent, and selection between them 
random. The facts agree with these predictions. Synthesis of chiral compounds from 
achiral reactants always yields the racemic modification. This is simply one aspect 
of the more general rule: optically inactive reactants yield optically inactive pro- 



Problem 7.1 Show in detail why racemic Mr-butyl chloride would be obtained 
if: (a) the sec-butyl radical were not flat, but pyramidal; (b) chlorination did not in* 
volve a free rar-butyl radical at all, but proceeded by a mechanism in which a chlorine 
atom displaced a hydrogen atom, taking the position on the carbon atom formerly 
occupied by that hydrogen. 

To purify the sec-butyl chloride obtained by chlorination of w-butane, we 
would carry out a fractional distillation. But since the enantiomeric sec-butyl 
chlorides have exactly the same boiling point, they cannot be separated, and are 
collected in the same distillation fraction. If recrystallization is attempted, there 
can again be no separation since their solubilities in every (optically inactive) 
solvent are identical. It is easy to see, then, that whenever a racemic modification 
is formed in a reaction, we will isolate (by ordinary methods) a racemic modification. 

If an ordinary chemical synthesis yields a racemic modification, and if this 
cannot be separated by our usual methods of distillation, crystallization, etc., how 
do we know that the product obtained is a racemic modification? It is optically 
inactive; how do we know that it is actually made up of a mixture of two optically 
active substances? The separation of enantiomers (called resolution) can be 
accomplished by special methods; these involve the use of optically active reagents, 
and will be discussed later (Sec. 7.9). 

P*oMem 7.2 Isopentane is allowed to undergo free-radical chlorination, and the 
reaction mixture is separated by careful fractional distillation, (a) How many frac- 
tions of formula C 5 H n Cl would you expect to collect? (b) Draw structural formulas, 
stereochemical where pertinent, for the compounds making up each fraction. Specify 
each enantiomer as R or S. (c) Which if any, of the fractions, as collected, would 
show optical activity ? (d) Account in detailjust as was done in the preceding section 
for the optical activity or inactivity of each fraction. 



7.4 Reactions of chiral molecules. Bond breaking 

Having made a chiral compound, sec-butyl chloride, let us see what happens 
when it, in turn, undergoes free-radical chlorination. A number of isomeric di- 
chlorobutanes are formed, corresponding to attack at various positions in the 
molecule. (Problem: What are these isomers?) 

CH 3 CHr-CH-CH 3 cb ' hcat or light > CH 3 CH2-CH-CH 2 C1 f other products 

CI Cl 

jrc-Butyl chloride 1 ,2-DichIorobutane 

Let us take, say, (S^sec-butyl chloride (which, we saw in Sec. 7.3, happens to 
rotate light to the right), and consider only the part of the reaction that yields 
1,2-dichlorobutane. Let us make a model (I) of the starting molecule, using a 



SEC. 7,5 RELATING CONFIGURATIONS 229 

single ball for C2H 5 but a separate ball for each atom in -CH 3 . Following the 
familiar steps of the mechanism, we remove an H from CH3 and replace it 
with a Cl. Since we break no bond to the chiral center in either step, the model 
we arrive at necessarily has configuration II, in which the spatial arrangement 



CH 2 CI 




(R)-l,2-Dichlorobutane 



about the chiral center is unchanged or, as we say, configuration is retained 
with CH 2 C1 now occupying the same relative position that was previously occu- 
pied by CH 3 . It is an axiom of stereochemistry that molecules, too, behave in 
just this way, and that a reaction that does not involve the breaking of a bond /< a 
chiral center proceeds with retention of configuration about that chiral center. 

(If a bond to a chiral center is broken in a reaction, we can make no geneul 
statement about stereochemistry, except that configuration can be and more 
than likely will be changed. As discussed in Sec. 7.10, just what happ. r.> 
depends on the mechanism of the particular reaction.) 

Problem 7.3 We carry out free-radical chlorination of (S)-jrer-butyl chloride, 
and by fractional distillation isolate the various isomeric products, (a) Draw stereo- 
chemical formulas of the 1,2-, 2,2-, and 1,3-dichlorobutanes obtained in this way. 
Give each enantiomer its proper R or S specification, (b) Which of these fractions, 
as isolated, \\ill be optically active, and which will be optically inactive? 

Now, let us see how the axiom about bond breaking is applied in relating i ne 
configuration of one chiral compound to that of another. 

7.5 Reactions of chiral molecules. Relating configurations 

We learned (Sec. 4.14) that the configuration of a particular enantiomer 
can be determined directly by a special kind of x-ray diffraction, which was first 
applied in 1949 by Bijvoet to ( -I )-tartaric acid. But the procedure is difficult and 
time-consuming, and can be applied only to certain compounds. In spite of this 
limitation, however, the configurations of hundreds of other compounds are 
now known, since they had already been related by chemical methods to (+)- 
tartaric acid. Most of these relationships were established by application of the 
axiom given above; that is, the configurational relationship between two optically 
active compounds can be determined by converting one into the other by reactions 
that do not involve breaking oj a bond to a chiral center, 

Let us take as an example (-)-2.-methyl-l-butanol (the enantiomer found in 
fusel oil) and accept, for the moment, that it has configuration III, which we would 
specify S. We treat this alcohol with hydrogen chloride >and obtain the alky I 
chloride, l-chloro-2-methylbutane. Without knowing the mechanism of this 
reaction, we can see that the carbon-oxygen bond is the one that is broken. No 
bond to the chiral center is broken, and therefore configuration is retained, with 



230 STEREOCHEMISTRY II CHAP. 7 



occupying the same relative position in the product that was occupied 
by CH 2 OH in the reactant. We put the chloride into a tube, place this tube in 
a polarimeter, and find that the plane of polarized light is rotated to the right; 





HCI > 

I i 
C 2 H S 

III lv 

S-(-)-2-Methyl-l-butanol S-(-h)-l-Chloro-2-methylbutanc 

that is, the product is ( + )-l-chloro-2-methylbutane. Since ( )-2-methyl-l-butanol 
has configuration III, (+ )-l-chloro-2-methylbutane must have configuration IV. 

Or, we oxidize ( )-2-methyl-l-butanol with potassium permanganate, obtain 
the acid 2-methylbutanoic acid, and find that this rotates light to the right. Again, 
no bond to the chiral center is broken, and we assign configuration V to (4- )-2- 
methylbutanoic acid. 







II! y 

(S)-( - )-2-Methyl-l-butanoI (S)-( + )-2-Methylbutanoic acid 

We can nearly always tell whether or not a bond to a chiral center is broken by 
simple inspection of the formulas of the reactant and product, as we have done in these 
cases, and without a knowledge of the reaction mechanism. We must be aware of the 
possibility, however, that a bond may break and re-form during the course of a reaction 
without this being evident on the surface. This kind of thing does not happen at random, 
but in certain specific situations which an organic chemist learns to recognize. Indeed, 
stereochemistry plays a leading role in this learning process: one of the best ways to 
detect hidden bond-breaking is so to design the experiment that if such breaking occurs, 
it must involve a chiral center. 

But how do we know in the first place that (-)-2-methyl-l-butanol has 
configuration III? Its configuration was related in this same manner to that of 
another compound, and that one to the configuration of still another, and so on, 
going back ultimately to ( 4- )-tartaric acid and Bijvoet's x-ray analysis. 

We say that the (-)-2-methyl-l-butanol, the (+)-chloride, and the (+)-acid 
have similar (or the same) configurations. The enantiomers of these compounds, 
the (+)-alcohol, (-)-chloride, and (-)-acid, form another set of compounds with 
similar configurations. The (-)-alcohol and, for example, the (-)-chloride are 
said to have opposite configurations. As we shall find, we are usually more inter- 
ested in knowing whether two compounds have similar or opposite configurations 
than in knowing what the actual configuration of either compound actually is. 
That is to say, we are more interested in relative configurations than in absolute 
configurations. 



SEC. 7.6 OPTICAL PURITY 231 

In this set of compounds with similar configurations, we notice that two are 
dextrorotatory and the third is levorotatory. The sign of rotation is important 
as a means of keeping track of a particular isomer just as we might use boiling 
point or refractive index to tell us whether we have cis- or /ra/w-2-butene, now 
that their configurations have been assigned but the fact that two compounds 
happen to have the same sign or opposite sign of rotation means little; they may or 
may not have similar configurations. 

The three compounds all happen to be specified as S, but this is simply because 
CH 2 C1 and COOH happen to have the same relative priority as CH 2 OH. 
If we were to replace the chlorine with deuterium (Problem: How could this be 
done?), the product would be specified R, yet obviously it would have the same 
configuration as the alcohol, halide, and acid. Indeed, looking back to sec-butyl 
chloride and 1 ,2-dichlorobutane, we see that the similar configurations I and 
II are specified differently, one S and the other R; here, a group ( CH 3 ) that 
has a lower priority than ~C 2 H 5 is converted into a group ( CH 2 C1) that has a 
higher priority. We cannot tell whether two compounds have the same or opposite 
configuration by simply looking at the letters used to specify their configurations; 
we must work out and compare the absolute configurations indicated by those 
letters. 

Problem 7.4 Which of the following reactions could safely be used to relate con- 
figurations? 

(a) ( + )-C 6 H 5 CH(OH)CH 3 + PBr 3 > C 6 H 5 CHBrCH 3 

(b) ( + )-CH,CH 2 CHClCH 3 + C 6 H 6 + A1C1 3 > C 6 H 5 CH(CH 3 )CH 2 CH 3 

(c) (-)-C 6 H 5 CH(OC 2 H 5 )CH 2 OH + HBr > C 6 H 5 CH(OC 2 H 5 )CH 2 Br 

(d) (+>CH 3 CH(OH)CH 2 Br + NaCN > CH 3 CH(OH)CH 2 CN 

(e) (40-CH 3 CH,C-K)CH(CH 3 )C 2 H 5 + OH~ > CHjCH.COCT 

.^ x + CH 3 CH 2 CH'OHCH 3 

(f) (-)-CH 3 CH 2 CHBrCH, + C 2 H 5 O~Na h ^ C 2 H 5 O CH(CH 3 )CH 2 CH 3 

(g) ( + )-CH 3 CH,CHOHCH 3 -^> CH 3 CH 2 CH(ONa)CH 3 - c ^^-> 

C 2 H 5 -0 CH(CH 3 )CH 2 CH 3 

Problem 7.5 What general conclusion must you draw from each of the follow- 
ing observations? (a) After standing in an aqueous acidic solution, optically active 
CH 3 CH 2 CHOHCH 3 is found to have lost its optical activity, (b) After standing in 
solution with potassium iodide, optically active -C 6 HnCHlCH 3 is found to have 
lost its optical activity, (c) Can you suggest experiments to test >our conclusions? 
(See Sec. 3.29.) 



7.6 Optical purity 

Reactions in which bonds to chiral centers are not broken can be used to get 
one more highly important kind of information: the specific rotations of optically 
pure compounds. For example, the 2-methyl-l-butanol obtained from fusel oil 
(which happens to have specific rotation -5.756) is optically pure like most 
chiral compounds from biological sources that is, it consists entirely of the one 
enantiomer, and contains none of its mirror image. When this material is treated 
with hydrogen chloride, the l-chloro-2-methylbutane obtained is found to have 
specific rotation of -f 1.64. Since no bond to the chiral center is broken, every 



2*2 STEREOCHEMISTRY II CHAP, 7 

molecule of alcohol with configuration III is converted into a molecule of chloride 
with configuration IV; since the alcohol was optically pure, the chloride of specific 
rotation -f 1.64 is also optically pure. Once this maximum rotation has been estab- 
lished, anyone can determine the optical purity of a sample of l-chloro-2-methyl- 
hutane in a few moments by simply measuring its specific rotation. 

If a sample of the chloride has a rotation of +0.82, that is, 50% of the maxi- 
mum, we say that it is 50 % optically pure. We consider the components of the mix- 
ture to be (+ )-isomer and ( )-isomer (not ( + )-isomer and (-)-isomer). (Problem: 
What are the percentages of (-f)-isomer and (-)-isomer in this sample?) 

Problem 7.6 Predict the specific rotation of the chloride obtained by treatment 
with hydrogen chloride of 2-methyl-l-butanol of specific rotation -f 3.12. 



7.7 Reactions of chiral molecules. Generation of a second chiral center 

Let us return to the reaction we used as our example in Sec. 7.4, free -radical 
chlorination of sec-butyl chloride, but this time focus our attention on one of the 
other products, one in which a second chiral center is generated: 2,3-dichloro- 
butane. This compound, we have seen (Sec. 4.18), exists as three slereoisomers, 
tncA'o and a pair of enantiomers. 

<:H 3 CH 2 ~~CH-CH 3 H2JJL5LM ^ CH 3 -CHCH-CH 3 -h other products 

I i i 

CI O Cl 

c-But> 1 chloi I Jc 2,3-Dichlorobutane 

Let us suppose that we take optically active .\vc-butyl chloride (the (S)-isomer, 
s i^ , carry out the chlorination, and by fractional distillation scpuiate the 2,3- 
dichlorobutanes from all the other products (the 1,2-isomer, 2,2-is>omer, etc.). 
Which stereoisomers can we expect to have? 

Figure 7.2 shows the course of reaction. Three important points are illustrated 
Ahich apply in all cases where a second chiral center is generated. first 51 nee no 
bonds to the original chiral center, C-2, are broken, its configuration u retained 
,n all the products. Second, there are two possible configurations aooul U>< new 
chirai center, C-3, and both of these appear; in this particular case, 'hey r esilt 
from attacks (a) and (b) on opposite sides of the flat portion of the free radical, 
giving the diastereomeric S,S and R,S (or meso) products. Third, the diastw-reo^eric 
products will be formed in unequal amounts; in this case because attack (a\ and 
attack (b) are not equally likely. 

In Sec. 7.3 we saw that generation of the first chiral center in a compound 
>ields equal amounts of enantiomers, that is, yields an optically inactive raccmic 
modification. Now we see that generation of a new chiral center in a compound 
'hit is already optically active yields an optically active product containing unequal 
amounts of diastereomers. 

Suppose (as is actually the case) that the products from (S)-.yeobui\ i chloride 
>how an S:meso ratio of 29:71. What would we get from chlorination of (RJ-vtr- 
'Mityl chloride? We would get (R,R-) and /wew-products, and the R,R:/m'st> ratio 
w^uld be exactly 29:71. Whatever factor favors mew-product over (S,S)-produ:t 
vill favor meyo-product over (R,R)~product, and to exactly the same extent. 



SEC. 7.8 FORMATION OF ENANTIOMERS AND DIASTEREOMERS 233 




H 

C H, 

Figure 7.2. Generation of a second 
chiral center. Configuration at ori- 
ginal chiral center unchanged. . 
Chlorine becomes attached via (a) or 7 \ 
(b) to give diastereomers, and in un- / \ 
equal amounts. 





(S r S> 

Diastereomers 
Formed in unequal amounts 



Finally, what can we expect to get from optically inactive, racertiic see-butyl 
chloride? The (S)-isomer that is present would yield (S,S> and /new-products in 
the ratio of 29:71 ; the (R>isomer would yield (R,R)- and weso-products, and in 
the ratio of 29:71. Since there are exactly equal quantities of (S> and (R)-reactants, 
the two sets of products would exactly balance each other, and we would obtain 
products rn the ratio of 29: 7K Optically inactive reactants yield 



opticaUy inactive products. 

One point i^c^rilrei further discussion. Why an? the diastereomerte products 
formed in unequal amounts? It is because the itttertnediate 3-chloro~2-butyl 
radical in Fig, 7.2 already contains a chiral center. The free radical is chiral. and 
lacks the symmetry that is necessary for atUttfc at the two faces to be equally 
likely. (Make a model of the radical and assure yourself that this is so.) 

Ill the following section, tWs point is discussed in more detail 

To understand better how formation of diastcreomers differs from forma- 
tion of enaatioiners, let us contrast the reaction of the chiral 3-chlorc-2-buty! 
radical shown in Fig, 7.2 with the reaction of the achiral sec-butyl radical. 



234 



STEREOCHEMISTRY II 



CHAP. 7 



In Sec. 7.3, we said that attachment of chlorine to either face of the sec- 
butyl radical is equally likely. This is in effect true, but deserves closer examina- 
tion. Consider any conformation of the free radical: I, for example. It is clear 
that attack by chlorine from the top of I and attack from the bottom are not 





sec-Butyl radical 
Achiral 

equally likely. But a rotation of 180 about the single bond converts 1 into II; 
these are two conformations of the same free radical, and are, of course, in equili- 
brium with each other. They are mirror images, and hence of equal energy and 
equal abundance; any preferred attack from, say, the bottom of I to give the 
(R)-product will be exactly counterbalanced by attack from the bottom of 11 to 
give the (S)-product. 

The "randomness of attack" that yields the racemic modification from 
achiral reactants is not necessarily due to the symmetry of any individual icactant 
molecule, but rather to the random distribution of such molecules between mirror- 
image conformations (or to random selection between mirror-image transition 
states). 

Now, let us turn to reaction of the chiral 3-chloro-2-butyl radical (Fig. 7.2). 
Here, the free radical we are concerned with already contains a chiral center, 
about which it has the (S)-configuration ; attack is not random on such a radical 
because mirror-image conformations are not present they could only come from 
(R) free radicals, and there are none of those radicals present. 

Preferred attack from, say, the bottom of conformation III a likely prefer- 
ence since this would keep the two chlorine atoms as far apart as possible in the 
transition state would yield /w^o-2,3-dichlorobutane. A rotation of 180 about 
the single bond would convert III into IV. Attack from the bottom of IV would 



H <> 





HI 



IV 



3-Chloro2-butyl radical 
Chiral 

yield the (S,S)-isomer. But III and IV are not mirror images, are not of equal 
energy, and are not of equal abundance. In particular, because of lesser crowding 
between the methyl groups, we would expect HI to be more stable and hence 
more abundant than IV, and the meso product to predominate over the (S,S)- 
isomer (as it actually does). 



SEC. 7.9 RESOLUTION 235 

We might have made a different guess about the preferred direction of attack, 
and even a different estimate about relative stabilities of conformations, but we 
would still arrive at the same basic conclusion: extept by sheer coincidence, the 
two diastereomers would not be formed in equal amounts. 

In this discussion, we have assumed that the relative rates of competing reactions 
depend on relative populations of the conformations of the reactants. This assumption 
is correct here, if, as seems likely, reaction of the free radicals with chlorine is easier and 
faster than the rotation that intc .'converts conformations. 

If, on the other hand, reaction with chlorine were a relatively difficult reaction and 
much slower than interconversion of conformations, then relative rates would be deter- 
mined by relative stabilities of the transition states. We would still draw the same general 
conclusions. In the reaction of the achiral sec-butyl radical, the transition states are 
mirror images and therefore of the same stability, and the rates of formation of the two 
products would be exactly the same. In the reaction of the chiral 3-chloro-2-butyl radical, 
the transition states are not mirror-images and therefore not of the same stability, and 
rates of formation of the two products would be different. (n the latter case, we would 
even make the same prediction, that the meso product would predominate, since the same 
relationship between methyl groups that would make conformation III more stable would 
also make the transition state resembling conformation III more stable.) 

Problem 7.7 Answer the following questions about the formation of 2,3-dichloro- 
butane from (R)-.w-butyl chloride, (a) Draw conformations (V and VI) of the inter- 
mediate radicals that correspond to III and IV above, (b) What is the relationship be- 
tween V and VI? (c) How will the V:V1 ratio compare with the III: IV ratio? 
(d) Assuming the same preferred direction of attack by chlorine as on III and IV, 
which stereoisomeric product would be formed from V? From VI? (e) Which 
product would you expect to predominate? (f) In view of the ratio of products 
actually obtained from (S)-s?c-butyl chloride, what ratio of products must be obtained 
from (R)-^c-butyl chloride? 

Problem 7.8 Each of the following reactions is carried out, and the products 
are separated by careful fractional distillation or recrystallization. For each reaction 
tell how many fractions will be collected. Draw stereochemical formulas of the com- 
pound or compounds making up each fraction, and give each its R/S specification. 
Tell whether each fraction, as collected, will show optical activity or optical inactivity. 

(a) monochlorination of (R)-sec-buty\ chloride at 300; 

(b) monochlorination of racemic sec-butyl chloride at 300; 

(c) monochlorination of racemic l-chloro-2-methyIbutane at 300; 

(d) addition of bromine to (S)-3-bromo-l-butene. 



7.9 Reactions of chiral molecules with optically active reagents. Resolution 

So far in this chapter we have discussed the reactions of chiral compounds 
only with optically inactive reagents. Now let us turn to reactions with optically 
active reagents, and examine one of their most useful applications: resolution of a 
racemic modification, that is, the separation of a racemic modification into enantio- 
mers. 

We know (Sec. 7.3) that when optically inactive reactants form a chiral 
compound, the product is the racemic modification. We know that the enantio- 
mers making up a racemic modification have identical physical properties (except 
for direction of rotation of polarized light), and hence cannot be separated by 
the usual methods of fractional distillation or fractional crystallization. Yet 
throughout this book are frequent references to experiments carried out using 



236 STEREOCHEMISTRY II CHAP. 7 

optically active compounds like (+)-jeobutyl alcohol, ( )-2-bromooctane, 
(-)-a-phenylethyl chloride, (H-)-a-phenylpropionamide. How are such optically 
active compounds obtained? 

Some optically active compounds are obtained from natural sources, since 
living organisms usually produce only one enantiomer of a pair. Thus only 
(-)-2-methyl-l-butanol is formed in the yeast fermentation of starches, and only 
(+)-lactic acid, CHsCHOHCOOH, in the contraction of muscles; only (-)- 
malic acid, HOOCCH 2 CHOHCOOH, is obtained from fruit juices, only (-)- 
quinine from the bark of the cinchona tree. Indeed, we deal with optically active 
substances to an extent that we may not realize. We eat optically active bread 
and optically active meat, live in houses, wear clothes, and read books made of 
optically active cellulose. The proteins that make up our muscles and other tissues, 
the glycogen in our liver and in our blood, the enzymes and hormones that enable 
us to grow, and that regulate our bodily processes all these are optically active. 
Naturally occurring compounds are optically active because the enzymes that 
bring about their formation and often the raw materials from which they are 
made are themselves optically active. As to the origin of the optically active 
enzymes, we can only speculate. 

Amino acids, the units from which proteins are made, have been reported present 
in meteorites, but in such tiny amounts that the speculation has been made that "what 
appears to be the pitter-patter of heavenly feet is probably instead the print of an earthly 
thumb." Part of the evidence that the ammo acids found in a meteorite by Cyril Pon- 
namperuma (of NASA) are really extraterrestrial in origin is that they are optically in- 
active not optically active as earthly contaminants from biological sources would be. 

From these naturally occurring compounds, other optically active compounds 
can be made. We have already seen, for example, how (~)-2-methyl-l-butanol 
can be converted without loss of configuration into the corresponding chloride 
or acid (Sec! 7.5); these optically active compounds can, in turn, be converted into 
many others. 

Most optically active compounds are obtained by the resolution of a racemic 
modification, that is, by a separation of a racemic modification into enantiomers. 
Most such resolutions are accomplished through the use of reagents that are them- 
selves optically active; these reagents are generally obtained from natural sources. 

The majority of resolutions that have been carried out depend upon the reaction 
of organic bases with organic acids to yield salts. Let us suppose, for example, 
that we have prepared the racemic acid, ()-HA. Now, there are isolated from 
various plants very complicated bases called alkaloids (that is, alkali-like), among 
which are cocaine, morphine, strychnine, and quinine. Most alkaloids are 
produced by plants in only one of two possible enantiomeric forms, and hence 
they are optically active. Let us take one of these optically active bases, say a 
levorotatory one, ( )-B, and mix it with our racemic acid ()-HA. The acid is 
present in two configurations, but the base is present in only one configuration; 
there will result, therefore, crystals of two different salts, [(-)-BH + (H-)-A~] 



What is the relationship between these two salts? They are not superimpos- 
able, since the acid portions are not superimposable. They are not mirror images, 
since the base portions are not mirror images. The salts are stereoisomers that are 
not enantiomers, and therefore are diaster corners. 



SEC. 7.10 MECHANISM OF FREE-RADICAL CHLORINATION 237 

H + 



(4-)-HA 



(-i-)-HA K->'BH' 

-! (-)-B > 

[( )-BH 



> (-)-HA : ( )-BH+ 

Enantiomers: A\lkaloid Diasiereomeis. Resolved Alkaloid 

/// a tat cnuc bast 1 \tyw/ <//>/<* enantiomerb as a \alt 

modification 

These diastereomeric salts have, of course, different physical properties, 
including solubility in a given solvent. They can therefore be separated by frac- 
tional crystallization. Once the two salts are separated, optically active acid can 
be recovered from each salt by addition of strong mineral acid, which displaces 
the weaker organic acid. If the salt has been carefully purified by repeated crystal- 
lizations to remove all traces of its diastereomer, then the acid obtained from it is 
optically pure. Among the alkaloids commonly used for this purpose are ( )- 
brucine, ( )-quinine, ( )-strychnine, and ( + )-cinchonine. 

Resolution of organic bases is carried out by reversing the process just de- 
scribed : using naturally occurring optically active acids, ( )-malic acid, for example. 
Resolution of alcohols, which we shall find to be of special importance in synthesis, 
poses a special problem: since alcohols are neither appreciably basic nor acidic, 
they cannot be resolved by direct formation of salts. Yet they can be resolved by 
a rather ingenious adaptation of the method we have just described: one attaches 
to them an acidic "handle," which permits the formation of salts, and then when it 
is no longer needed can be removed. 

Compounds other than organic bases, acids, or alcohols can also be resolved. 
Although the particular chemistry may differ from the salt formation just described, 
the principle remains the same: a racemic modification is converted by an optically 
active reagent into a mixture of diastereomers which can then be separated. 

7.10 Reactions of chiral molecules. Mechanism of free-radical chlorination 

So far, we have discussed only reactions of chiral molecules in which bonds 
to the chiral center are not broken. What is the stereochemistry of reactions in 
which the bonds to the chiral center are broken? The answer is: it depends. 
It depends on the mechanism of the reaction that is taking place; because of this, 
stereochemistry can qften give us information about a reaction that we cannot 
get in any other way. 

For example, stereochemistry played an important part in establishing the 
mechanism that was the basis of our entire discussion of the halogenation of 
alkanes (Chap. 3). The chain-propagating steps of this mechanism are: 

(2a) X- + RH > HX + R- 

(3a) R- + X 2 > RX + X- 



238 



STEREOCHEMISTRY II 



CHAP. 7 



Until 1940 the existing evidence was just as consistent with the following alter- 
native steps: 

(2b) X- + RH > RX + H- 

(3b) H- + X 2 > HX + X 

To differentiate between these alternative mechanisms, H. C. Brown, M. S. 
Kharasch, and T. H. Chao, working at the University of Chicago, carried out the 
photochemical halogenation of optically active S-(H-)-1-chloro-2-methylbutanc. 
A number of isomeric products were, of course, formed, corresponding to attack 
at various positions in the molecule. (Problem: What were these .products?) 
They focused their attention on just one of these products: j,2-dichloro-2-mcthyl- 
butane, resulting from substitution at the chiral center (C-2). 



CH 3 
CH 3 CH2CHCH 2 C1 



h , light 



CH 3 
CH 3 CH 2 CCH 2 C1 



(S)-(4-)-l-Chloro-2-methylbutane ()-l,2-Dich1oro-2-mcthylbutane 
Optically active Optically inactive 

They had planned the experiment on the following basis. The two mechanisms 
differed as to whether or not a free alkyl radical is an intermediate. The most 
likely structure for such a radical, they thought, was /to as, it turns out, it very 
probably isand the radical would lose the original chirality. Attachment of 
chlorine to either face would be equally likely, so that an optically inactive, racemic 
product would be formed. That is to say, the reaction would take place wiffi 
racemization (see Fig. 7.3). 



/- ii \. Intermediate 
3 C. 2 H 5 )fc f ree radical 

Chiralilv 
lost 



CHjd 



S-( + )- 1 -Chloro-2-methylbutane 
Optically actlre 




-Cl 



CH 2 C1 CH 2 CI 

Racemic product 
; Optically inactke 

Figure 7.3. Racemization through free-radical formation. Chlorine be- 
comes attached to either face of free radical, via (a) or (6), to give enantio- 
mers, and in equal amounts. 



SEC. 7.11 SYN- AND /i/VJ/-ADDITION 239 

For the alternative mechanism, in which chlorine would become attached to 
the molecule while the hydrogen was being displaced, they could make no pre- 
diction, except that formation of an optically inactive product would be highly 
unlikely: there was certainly no reason to expect that back-side attack (on the face 
opposite the hydrogen) would take place to exactly the same extent as front-side 
attack. (In ionic displacements, attack is generally back-side.) 

By careful fractional distillation they separated the 1 ,2-dichloro-2-methyl- 
butane from the reaction mixture, and found it to be optically inactive. From this 
they concluded that the mechanism involving free alkyl radicals, (2a), (3a), is the 
correct one. This mechanism is accepted without question today, and the work 
of Brown, Kharasch, and Chao is frequently referred to as evidence of the stereo* 
chemical behavior of free radicals, with the original significance of the work 
exactly reversed. 

We can begin to see how stereochemistry provides the organic chemist with 
one of his most powerful tools for finding out what is going on in a chemical reac- 
tion. 

Problem 7.9 This work does not prove that free radicals are flat. Racemization 
is consistent with what Bother structure for free radicals? Explain. (Hint: See Sec. 
2.21.) 

Problem 7.10 Altogether, the free-radical chlorination of (S)-(+M-chloro- 
2-methylbutane gave six fractions of formula C 5 HioCl 2 . Four fractions were found to 
be optically active, and two fractions optically inactive. Draw structural formulas 
for the compounds making up each fraction. Account in detail for optical activity 
or inactivity in each case. 



7.11 Stereoselective and stereospecific reactions* syn- and anti-Addition 

As our second example of the application of stereochemistry to the study of 
reaction mechanisms, let us take another familiar reaction: addition of halogens 
to alkenes. In this section we shall look at the stereochemical facts and, in the next, 
see how these facts can be interpreted. 

Addition of bromine to 2-butene yields 2,3-dibromobutane. Two chiral cen- 
ters are generated in the reaction, and the product, we know, can exist as a meso 
compound and a pair of enantiomers. 

CH 3 CH=CHCH 3 + Br 2 > CH 3 -CH CH-CH 3 

2-Butene ^ ^ 

2,3-Dibromobutane 

The reactant, too, exists as diastereomers: a pair of geometric isomers. If 
we start with, say, cw r -2-butene, which of the stereoisomeric products do we 
get? A mixture of all of them? No. cw-2-Butene yields only racemic 2,3-dibromo- 
butane; none of the meso compound is obtained. A reaction that yields predomi- 
nantly one stereoisomer (or one pair of enantiomers) of several diastereomeric 
possibilities is called a Stereoselective reaction. 

Now, suppose we start with f/my-2-butene. Does this, too, yield the racemic 
dibromide? No. mm.?-2-Butene yields only /ner0-2,3-dibromobutane. A reaction in 
which stereochemically different reactants give stereochemically different products 
is called a stereospecific reaction. 



STEREOCHEMISTRY II 



CHAP. 7 



' Addition of bromine to alkenes is both stereoselective and stereospecific. We 
say it is completely stereoselective since, from a given alkene, we obtain only one 
diastereomer (or one pair of enantiomers). We say it is stereospecific, since just 
which stereoisomer we obtain depends upon which stereoisomeric alkene we start 
with. 

In the above definition, stereochemically different means, in practice, diastereo- 
merically different. The term stereospecific is not applied to reactions, like those in Sees. 
7.4 and 7.5, in which enantiomerically different reactants give enantiomerically different 
products. 

All stereospecific reactions are necessarily stereoselective, but the reverse is not 
true. There are reactions from which one particular stereoisomer is the predominant 
product regardless of the stereochemistry of the reactant; there are reactions in which 
the reactant cannot exist as stereoisomers, but from which one particular stereoisomer is 
the predominant product. Such reactions are stereoselective but not stereospecific. 

To describe stereospecificity in addition reactions, the concepts of jyw-addition 
and anti-addition are used. These terms are not the names of specific mechanisms. 
They simply indicate the stereochemical facts: that the product obtained is the 
one to be expected if the two portions of the reagent were to add to the same face 
of the alkene (syri) or to opposite faces (anti). 




Y-Z 




5>7i-Addition 




Addition of bromine to the 2-butenes involves a/M-addition. If we start 
(Fig. 7.4) with cw-2-butene, we can attach the bromine atoms to opposite faces 
of the alkene either as in (a) or in (b) and thus obtain the enantiomers. Since, 
whatever the mechanism, (a) and (b) should be equally likely, we obtain the racemic 
modification. 

Starting with rraws-2-butene (Fig. 7.5), we can again attach the bromine atoms 
to opposite faces of the alkene in two ways but, whichever way we choose, we ob- 
tain the w<?S(?-dibrornide. 

0////-Addition is the general rule for the reaction of bromine or chlorine with 
simple alkenes. We i>hall encounter other examples of stereospecific additions, 
both anti and \vn. We shall find that other reactions besides addition can be 



SEC. 7.11 



51W- AND ^//-ADDITION 



241 



anfi-Addition 




CH 3 

m-2-Butene 



"7^"* . H 

H^/^-CH 3 Br 



CH 3 



-Br 
-H 



Br 



CH 3 

I 



CH 3 



H 




CH 3 



Br- 
H- 



-H 
-Br 



-CH 3 

CH 3 

II II 

I and II are enantiomers 
Racemic 2,3-dibromobutane 

Figure 7.4. aw/i-Addition to cw-2-butene. Attachment as in (a) or (b) 
equally likely: gives racemic modification. 



anrAAddition 

Br 




H 

r/.t-2-Butcne 




CHj 



Br 



CH, 



H- 



H- 



-Br 



Br 



CH 3 



CH 




Br 



CH 3 



HI 



Figure 7.5. anti- Addition to //<wy-2-butene. Attachment as in (c) or (d) 
gives meso piuiiuct. 



242 STEREOCHEMISTRY II ' CHAP. 7 

stereospecific and also that some can be non-stereospecific. Whatever the stereo- 
chemistry of a reaction, it must, of course, be accounted for by a satisfactory 
mechanism. 

Problem 7.11 On treatment with permanganate, c/s-2-butene yields a glycol of 
m.p. 34, and /ra/w-2-butene yields a glycol of m.p. 19. Both glycols are optically 
inactive. Handling as described in Sec. 7.9 converts the glycol of m.p. 19 (but not 
the one of m.p. 32) into two optically active fractions of equal but opposite rota- 
tion. 

(a) What is the configuration of the glycol of m.p. 19? Of m.p. 32? 

(b) Assuming these results are typical (they are), what is the stereochemistry of 
hydroxylation with permanganate ? 

(c) Treatment of the same alkenes with peroxy acids gives the opposite results: 
the glycol of m.p. 19 from c/s-2-butene, and the glycol of m.p. 32 from trans-2- 
butene. What is the stereochemistry of hydroxylation with peroxy acids? 



7.12 Mechanism of halogen addition 

We saw earlier (Sec. 6.13) that addition of halogens to alkenes is believed to 
proceed by two steps : first, addition of a positive halogen ion to form an organic 

-fcU 

X 

(2) -C-C -I- X- > -C C 

| CD I I 

X XX 

cation; then combination of this cation with a negative halide ion. We saw some 
of the facts that provide evidence for this mechanism. 

In the last section, we learned another fact: halogens add to simple alkenes 
with complete stereospecificity, and in the anti sense. Let us reexamine the mechan- 
ism in the light of this stereochemistry, and focus our attention on the nature of 
the intermediate cation. This intermediate we represented simply as the carbonium 
ion. A part of a carbonium ion, we remember (Sec. 5.16), is flat: the carbon that 
carries the positive charge is s/? 2 -hybridized, and this trigonal carbon and the 
three atoms attached to it lie in the same plane. 

Now, is the observed stereochemistry consistent with a mechanism involving 
such an intermediate? Let us use addition of bromine to c/j-2-butene as an exam- 
ple. A positive bromine ion is transferred to, say, the top face of the alkene to 



(1) C=C + X-X > C C + X- 

/ \ 



~~A "X 




CH 3 \ CH 3 

ci$-2-Butene Cation I (S,S)-2,3-Dibromobutane 



SEC. 7.12 



MECHANISM OF HALOGEN ADDITION 



243 



form the carbonium ion I. Then, a bromide ion attacks the bottom face of the 
positively charged carbon to complete the anti addition; attack at this face is 
preferred, we might say, because it permits the two bromines to be as far apart 
as possible in the transition state. (We obtain the racemic product: the S,S-di~ 
bromide as shown, the R,R-dibromide through attachment of positive bromine 
to the near end of the alkene molecule.) 

But this picture of the reaction is not satisfactory, and for two reasons. First, 
to account for the complete stereospecificity of addition, we must assume that 
attack at the bottom face of the cation is not just preferred, but is the only line of 
attack: conceivable, but especially in view of other reactions of carbonium ions 
(Sec. 14.13) not likely. Then, even if we accept this exclusively bottom-side at- 
tack, we are faced with a second problem. Rotation about the carbon-carbon 
bond would convert cation I into cation II; bottom-side attack on cation II would 



Br 




Br 



Br 

wev0-2,3-Dibromobutane 

yield not the racemic dibromide but the meso dibromide in effect ^-addition, 
and contrary to fact. 

To accommodate the stereochemical facts, then, we would have to make two 
assumptions about halogen addition: after the carbonium ion is formed, it is 
attacked by bromide ion (a) before rotation about the single bond can occur, and 
(b) exclusively from the side away from the halogen already in the cation. Neither 
of these assumptions is very likely; together, they make the idea of a simple car- 
bonium ion intermediate hard to accept. 

In 1937, to account better for the observed stereochemistry, I. Roberts and 
G. E. Kimball at Columbia University proposed the following mechanism. In 
step (1) of the addition of bromine, for example, positive bromine attaches itself 



(1) 



(2) 




Br-Br 



A bromonium ion 




not to just one of the doubly-bonded carbon atoms, but to both, forming a cyclig 
bromonium ion. In step (2), bromide ion attacks this bromonium ion to yield the 
dibromide. 



244 STEREOCHEMISTRY II CHAP. 7 

Now, how does the bromonium ion mechanism account for an //-addition? 
Using models, let us first consider addition of bromine to m-2-butene (Fig, 7.6). 




CH 3 

ro-2-Butene 




IV and V are enantiomers 
Racemic 2.3-dibromobutane 

Figure 7.6. Addition of bromine to c/s-2-butene via cyclic bromonium 
ion. Opposite-side attacks (a) and (b) equally likely, give enantiomers in 
equal amounts. 



In the first step, positive bromine becomes attached to either the top or bottom 
face of the alkene. Let us see what we would get if bromine becomes attached to 
the top face. When this happens, the carbon atoms of the double bond tend to 
become tetrahedral, and the hydrogens and methyls are displaced downward. 
The methyl groups are, however, still located across from each other, as they were 
in the alkene. In this way, bromonium ion III is formed. 

Now bromonium ion HI is attacked by bromide ion. A new carbon-bromine 
bond is formed, and an old carbon-bromine bond is broken. This attack occurs on 
the bottom face of III, so that the bond being formed is on the opposite of carbon 
from the bond being broken. Attack can occur by path (a) to yield structure IV 
or by path (b) to yield structure V. We recognize IV and V as enantiomers. Since 
attack by either (a) or (b) is equally likely, the enantiomers are formed in equal 
amounts, and thus we obtain the racemic modification. The same results are ob- 
tained if positive bromine initially becomes attached to the bottom face of cfr-2- 
butene. (Show with models that this is so.) 

Next, let us carry through the same operation on trans-2-buiene (Fig. 7.7). 
This time, bromonium ion VI is formed. Attack on it by path (c) yields VII, 
attack by (d) yields VIII. If we simply rotate either VII or VIII about the carbon- 
carbon bond, we readily recognize the symmetry of the compound. It is meso- 
2,3-dibromo-butane; VII and VIII are identical. The same results are obtained if 



SEC. 7.12 MECHANISM OP HALOGEN ADDITION 245 

positive bromine is initially attached to the bottom face of //ms^-butene. (Show 
with models that this is so.) 



H 
Br-Br 




CH, 




frwtf-2-Butene Vf 




H 

Butene 

trans-Bromorimm ion 



. VII and VIII are the same 
/Mtw-2,3-Dibromobutane 

Figure 7.7. Addition of bromine to /ra/tt-2-butene via cyclic bromonium 
ion. Opposite-side attacks (c) and (<0 give same product. 

Problem 7.12 (a) What is the relationship between the bromonium ions formed 
by attachment of positive bromine to the top and bottom faces of /r<nw2-butene? 
In what proportions are they formed? (b) Answer the same questions for cfc-2-butene. 
(c) For mww-2-pentene. (d) For m-2-pentene. 

Problem 7.13 (a) Predict the products of addition of bromine to fra/u-2-pentene. 
Is attack by bromide ion by the two paths equally likely? Account for the fact that 
inactive material is actually obtained, (b) Do the same for c/>2-pentcne. 

The concept of a halonium ion solves both of the problems associated with 
an open carbonium ion: a halogen bridge prevents rotation about the carbon- 
carbon bond, and at the same time restricts bromide ion attack exclusively to the 
opposite face of the cation. This opposite-side approach, we shall find (Sec. 14.10), 
is typical of attack by bases (nucleophiles) on tetrahedral carbon. 

That such cyclic intermediates can give rise to a/m-addition is demonstrated 
by hydroxylation with peroxy acids (Problem 7.11, p. 242): there, analogous 
intermediate* perfectly respectable compounds called epoxides (Chap. 17) 
can actually be isolated and studied. 

-v- 

An cpoxide 

Cyclic halonium ions were first proposed, then, simply as the most reason- 
able explanation for the observed stereochemistry. Since that time, however, 



246 STEREOCHEMISTRY II CHAP, 7 

more positive evidence has been discovered. In 1967, Olah (p. 160) prepared cations 
whose nmr spectra indicate that they are indeed cyclic halonium ions. For example: 

liquid SO 2 

(CH 3 ) 2 C~CHCH 3 + SbF 5 (CH 3 ) 2 CCHCH 3 SbF 6 ~ 

i Br V 

The idea of a bromonium or chloronium ion may appear strange to us, in 
contrast to the already familiar oxonium and ammonium ions. The tendency for 
halogen to share two pairs of electrons and acquire a positive charge, we might 
say, should be weak because of the high electronegativity of halogens. But the 
evidence here, and in other connections (Sec. 11.21 and Sec. 25.6) shows that 
this tendency is appreciable. In halogen addition we are concerned with this ques- 
tion: which is more stable, an open carbonium ion in which carbon has only a 
sextet of electrons, or a halonium ion in which each atom (except hydrogen, of 
course) has a complete octet ? It is not a matter of which atom, halogen or carbon, 
can better accommodate a positive charge; it is a matter of completeness or in- 
completeness of octets. 

In halonium ion formation we see one more example of what underlies all 
carbonium ion behavior: the need to get a pair of electrons to complete the octet 
of the positively charged carbon. 

There are exceptions to the rule of or/-addition of halogens, but exceptions that arc 
quite understandable. If the alkene contains substituents that can strongly stabilize the 
open carbonium ion as, for example, in a benzyl cation (Sec. 12.19) then addition 
proceeds with little or no stereospecificity. Carbon is getting the electrons it needs, but 
in a different way. 

Problem 7.14 Olah treated compounds of the formula (CH 3 ) 2 CXCF(CH 3 )2 
with SbF 5 . He observed the formation of halonium ions when X=C1, Br, or I, but an 
open carbonium ion when X=F. How do you account for the difference in behavior 
of the difluoro compound? (Hint: See Sec. 1.15.) 



PROBLEMS 

1. Each of the following reactions is carried out* and the products are separated by 
careful fractional distillation or recrystal fixation. For each reaction tell how many frac- 
tions will be collected. Draw stereochemical formulas of the compound or compounds 
making up each fraction, and give each its R/S specification. Tell whether each fraction, 
as collected, will show optical activity or optical inactivity. 

(a) /f-pentane + C1 2 (300) >C 5 H U C1; 

(b) 1-chloropentane + C1 2 (300) > C 5 Hi Cl 2 ; 

(c) (S)-2-chloropentane + C1 2 (300) >C 5 H 10 CI 2 ; 

(d) (R)-2-chloro-2,3-dimethylpentane + C1 2 (300) ^C 7 H, 4 C1 2 ; 

(e) m*^HOCH 2 CHOHCHOHCH 2 OH + HNO 3 HOCH 2 CHOHCHOHCOQH; 

(f) (R)-jec-butyl chloride + KOH (ale); 

(g) (S)-3-chloro-l-butene -r HC1; 

(h) racemic QHsCOCHOHQJHj + H 2 , catalyst > C 6 H 5 CHOHCHOHC 6 H 5 . 

2. In Problem 7.1 1 we saw that hydroxylation with permanganate is syn, and hydroxy- 
lation with peroxy acids is anti. Keeping in mind that reaction ofepoxides(Sec. 17. 12) is 
acid-catalyzed, give a detailed mechanism for hydroxylation with peroxy acids. (Check 
your answer in Sec. 1 7. 1 2.) 



PROBLEMS 

3. Give the absolute configuration and R/S specification of compounds /v 

(a) (R)-HOCH 2 CHOHCH~-- CH ? -f cold alkaline KMnO 4 > A.(optically acti, 
B (optically inactive); 

(b) (S)-l-chloro-2-methylbutane 4- Li, then + Cul > C; 

(c) C 4- (S)-l-chloro~2-methylbutane *D; 

(d) (R,R)-HOCH 2 CHOHCHOHCH 2 OH -f HBr > E(HOCH : CHOHCHOHCH 2 Br); 

(e) (R)-3-methyl-2-ethyI-l-pentene -f H 2 /Ni ^ F (optically active) 4- G (optically 
inactive). 

4. An excess of the racemic acid CH 3 CHC1COOH is allowed to react with (S)-2- 
methyM-butanol to form the ester, CH 3 CHaC~OCH,CH(CH 3 )CH 2 CH 3 , and the rc- 

II 

O 

action mixture is carefully distilled. Three fractions are obtained, each of which is opti- 
cally active. Draw stcreochemical formulas of the compound or compounds, making 
up each fraction. 

5. Addition oi chlorine water to 2-butenc yields not only 2,3-dichlorobutane but 
the chlorohydnn, 3-chloro-2-butanol. r/s-2*Butene gives only the threo chlorohydrin, 
and rra/tf-2-bulenc gives only the erythro chlorohydrin. What is the stereochemistry of 
chlorohydnn formation, and how do you account for it? 



Cl- 
H~ 



-H H 

-OH H- 



CH 3 

-Cl 



-OH 



(.HI CH 3 

and enanucmer and enantiomer 

Thrto Ervthro 

3-Chloro-2-butanol 



6. (a) How do you account for the fact that when allyl bromide is treated with dilute 
H : SO 4 , there is obtained not only l-bromo-2-propanol, but also 2-bromo-l-propanol? 
(b) In contrast, allyl chloride yields only one product, l-chloro-2-propanol. How do you 
account for this difference between the chloride and the bromide? 

7. (a) Alfred Hassncr (University of Colorado) has found iodine azide, IN 3 , to 
add to terminal alkenes with the orientation shown, and with complete stereospecificity 

RCH=CH 2 -f IN 3 > RCHCH.l 

N 3 
(anti) to the 2-butenes. Suggest a mechanism for this reaction. 

(b) In polar solvents like nitromethane, BrN 3 adds with the same orientation and 
stereospecificity as JNV In non-polar solvents like u-pentcne, however, orientation is 
reversed, and addition is non-stercospecific. In solvents of intermediate polarity like 
methylene chloride, mixtures of products are obtained; light or peroxides favor forma- 
tion of RCHBrCH 2 N 3 ; oxygen favors' formation of RCH(N 3 )CH 2 Br. Account in detail 
for these observations. 



250 ALKYNES AND DIENES CHAP. 8 

carbon-carbon double bond of ethylene (163 kcal) or the carbon-carbon single 
bond of ethane (88 kcal), and therefore is shorter than either. 

Again; 1 the quantum mechanical structure is verified by direct evidence. 
Electron diffraction, \-ray diffraction, and spectroscopy show acetylene (Fig. 8.4) 



* 




Figure 8.4. Acetylene molecule: 
shape and size. 

180 



to be a linear molecule. The CC distance is 1.21 A, as compared with 1.34 A 
in ethylene and 1.53 A in ethane. As in the case of the double bond, the structure 
of the triple bond is verifiedal though this time in a negative way by the evidence 
of isomer number. As we can readily see from models, the linearity of the bonding 
should not permit geometric isomerisni; no such isomers have ever been found. 

The CH distance in acetylene is 1.08 A, even shorter than in ethylene 
(1.103 A); because of their greater s character, sp orbitals are smaller than sp 2 
orbitals.and sp-hybridized carbon forms shorter bonds than A/? 2 -hybridized carbon. 
The C H bond dissociation energy in acetylene is not known, but we would ex- 
pect it to be even greater than in ethylene. Oddly enough, the same sp hybridization 
that almost certainly makes cleavage of the CH bond to form free radicals 
(homolysis) more difficult, makes cleavage to form ions (hetcrolysis) easier, as we 
shall see (Sec. 8.10). 

Homely sis: 

HCr- C : H > HC- C- 4 H- one electron to 

each fragment 

Heterolysis: 

HC- C:H * HC-i:C: -f H + both electrons to 

one fragment 

Problem 8.1 Compare the electronic configurations of CO : , \\hich is a Iinc'ir 
molecule (check your ansvscr to Problem 1.6, p. 25), and HiO, \\hich has a bond angle 
of 105 . 



8.3 Higher alkyncs. Nomenclature 

Like the alkanes and alkenes, the alkynes form a homologous series, the 
increment again being CH 2 . 

The alkynes are named according to two systems. In one, they are considered 
to be derived from acetylene by replacement of one or both hydrogen atoms by 
alkyl groups. 



H-C-C-C 2 H 5 CH 3 GsC-CH 3 CH 3 -O-C CH(CH 3 ) 2 

Ethylacetylene Dimethylacetylene Methylisopropylacetylene 

1-Butyne 2-Butyne 4-Methyl-2-pentyne 

For more complicated alkynes the IUPAC names are used. The rules are 
exactly the same as for the naming of alkenes, except that the ending -yne replaces 



SEC. 8.5 



INDUSTRIAL SOURCE OF ACETYLENE 



251 



-ene. The parent structure is the longest continuous chain that contains the triple 
bond, and the positions both of substituents and of the triple bond are indicated 
by numbers. The triple bond is given the number of Ihe first triply-bonded carbon 
encountered, starting from the end of the chain nearest the triple bond. 



8.4 Physical properties of alkynes 

Being compounds of low polarity, the alkynes have physical properties that 
are essentially the same as those of the alkanes and alkenes. They are insoluble 
in water but quite soluble in the usual organic solvents of low polarity: ligroin, 
ether, benzene, carbon tetrachloride. They are less dense than water. Their boil- 
ing points (Table 8.1) show the usual increase with increasing carbon number, 



Name 


Table 8.1 ALKYNES 
Formula 


M.p., 
C 


B.p., 
C 


Density 
(at 20) 


Acetylene 


HC==CH 


- 82 


- 75 




Propyne 


HfeCCHa 


V101.5 


- 23 




-Butyne 


HfeCCH 2 CH 3 


r!22 


9 




-Pentyne 


HC=C(CH 2 ) 2 CH 3 


8 


40 


0.695 


-Hexyne 


HC-EC(CH 2 ) 3 CH 3 


414 


72 


.719 


-Heptyne 


HC=C(CH 2 ) 4 CH 3 


* 


100 


.733 


-Octyne 


HGEEC(CH 2 ) 5 CH 3 


J^P 


126 


.747 


Nonyne 


HC^C(CH 2 ) 6 CH 3 


rTjW 


151 


.763 


-Decyne 


HCz=C(CH 2 ) 7 CH 3 


IT 36 


182 


.770 


2-Butyne 


CH 3 C~CCH 3 


- 24 


27 


.694 


2-Pentyne 


CH 3 C=EECCH 2 CH 3 


-101 


55 


.714 


3-Methyl-l -butyric 


HC^CCH(CH 3 ) 2 




29 


.665 


2-Hexyne 


CH 3 C=C(CH 2 ) 2 CH 3 


- 92 


84 


.730 


3-Hexyne 


CH 3 CH 2 GFECCH 2 CH 3 


- 51 


81 


.725 


3,3-Dimethyl-l-butyne 


HC=CC(CH 3 ) 3 


- 81 


38 


.669 


4-Octyne 


CH 3 (CH 2 ) 2 CEEC(CH 2 ) 2 CH 3 




131 


.748 


5-Decyne 


CH 3 (CH 2 ) 3 C=C(CH 2 ) 3 CH 3 




175 


.769 



and the usual effects of chain-branching; they are very nearly the same as the 
boiling points of alkanes or alkenes with the same carbon skeletons. 



8.5 Industrial source of acetylene 

The alkyne of chief industrial importance is the simplest member of the 
family, acetylene. It can be prepared by the action of water on calcium carbide, 
CaC 2 , which itself is prepared by the reaction between calcium oxide and coke at 
the very high temperatures of the electric furnace. The calcium oxide and coke 
are in turn obtained from limestone and coal, respectively. Acetylene is thus 
obtained by a few steps from three abundant, cheap raw materials: water, coal, 
limestone. 



252 ALKYNES AND DIENES CHAP. 8 

Coal - > coke 




CaCz _^ H-C-C-H 

Limestone - > CaO 

An alternative synthesis, based on petroleum, is displacing the carbide process. 
This involves the controlled, high-temperature partial oxidation of methane. 

6CH 4 + O 2 1SOO > 2HC-CH H- 2CO + 10H 2 

(The economic feasibility of this process is partly due to use of side-products: 
carbon monoxide and hydrogen for production of alcohols, and some hydrogen 
as fuel to maintain the oxidation temperature.) 

Enormous quantities of acetylene are consumed each year. Dissolved under 
pressure in acetone contained in tanks,- it is sold to be used as fuel for the oxyacet- 
ylene torch. It is the organic starting material for the large-scale synthesis of 
important organic compounds, including acetic acid and a number of unsaturated 
compounds that are used to make plastics and synthetic rubber. Many of the 
synthetic uses of acetylene have grown out of work done in Germany before and 
during World War II by W. Reppe (at the I, G. Farbenindustrie). Aimed at 
replacing petroleum (scarce in Germany) by the more abundant coal as the prim- 
ary organic source, this work has revolutionized the industrial chemistry of acet- 
ylene. 

8.6 Preparation of alkynes 

A carbon-carbon triple bond is formed in the same way as a double bond: 
elimination of atoms or groups from two adjacent carbons. The groups eliminated 

W X W X 

-C-C- > -0=C- > -CsC- 

U 

and the reagents used are essentially the same as in the preparations of alkenes. 



PREPARATION OF ALKYNES 



1. Dehydrohalogenation of alkyl dihalides. Discussed in Sec. 8.6. 



r H H 



H H H 

NaNH 2 % 



-~C_C_- ^H(alO > ^Q^^ 

Jci I 

.Example: 

CH 3 CH=CH 2 22+ CH 3 CH-CH 2 KOH(alc) > CH 3 CH=CHBr NaNH *> CH 3 (feCH 

1 1 1-Bromo-l-propene Propyne 

Jtsr oT 

1 ,2-Dibromopropane 
(Propylene bromide) 



SEC. 8.6 PREPARATION OF ALKYNES 253 

2. Reaction of sodium acetylides with primary alkyl halides. Discussed in Sec. 8.12. 
==C:-Na+ + RX > -C-C-R + NaX 

metal (R mus t 

be T) 
Examples: 

HC==C:-Na+ + CH 3 CH2CH 2 CH 2 Br > HC=CCH 2 CH 2 CH 2 CH 3 
Sodium acetylide /f-Butyl bromide 1-Hexyne 

(w-Butylacetylene) 



CH 3 (CH 2 ) 4 OEEC:-Na+ + CH 3 (CH 2 ) 3 CH 2 C1 

Sodium n-pentylacetylide /i-Pentyl chloride 6-Dodecyne 

3. Dehalogenation of tetrahalides. Discussed in Sec. 8.6. 
X X 

-C-C- + 



2Zn > feC- + 2ZnX 2 

Example: 

Br Br 

CHy-C-CH -^> CH 3 C==CH 



ir ir Propyne 



Dehydrohalogenation of vicinal dihalides is particularly useful since the di- 
halides themselves are readily obtained from the corresponding alkenes by addition 
of halogen. This amounts to conversion by several steps of a double bond into 
a triple bond. 

Dehydrohalogenation can generally be carried out in two stages as shown. 




A vinyl halide 
Very unreactive 

Carried through only the first stage, it is a valuable method for preparing un- 
saturated halides. The halides thus obtained, with halogen attached directly to 
doubly-bonded carbon, are called vinyl halides, and are very unreactive (Sec. 25.5). 
Under mild conditions, therefore, dehydrohalogenation stops at the vinyl halide 
stage; more vigorous conditions use of a stronger base are required for alkyne 
formation. 

Reaction of sodium acetylides with alkyl halides permits conversion of smaller 
alkynes into larger ones. Practically, the reaction is limited to the use of primary 
halides because of the great tendency for secondary and tertiary halides to undergo 
a side reaction, elimination; this point will be discussed further (Sec. 8.12) after 
we have learned something about the nature of acetylides. 

Dehalogenation of tetrahalides is severely limited by the fact that these halides 
are themselves generally prepared from the alkynes. As is the case with the double 



254 ALKYNES AND DIENES CHAP. 8 

bond and a dihalide, the triple bond may be protected by conversion into a tetra- 
halide with subsequent regeneration of the triple bond by treatment with zinc. 



8.7 Reactions of alkynes 

Just as alkene chemistry is the chemistry of the carbon-carbon double bond, 
so alkyne chemistry is the chemistry of the carbon-carbon triple bond. Like 
aikenes, alkynes undergo electrophilic addition, and for the same reason: avail- 
ability of the loosely held n electrons. For reasons that are not understood, the 
carbon-carbon triple bond is less reactive than the carbon-carbon double bond 
toward electrophilic reagents. 

Reasonably enough, the triple bond is more reactive than the double bond 
toward reagents that are themselves electron-rich. Thus alkynes undergo a set 
of reactions, nucleophilic addition, that are virtually unknown for simple aikenes. 
Although time does not permit us to go into these particular reactions here, we 
shall take up nucleophilic addition later in connection with other kinds of com- 
pounds (Chaps. 19 and 27). 

Besides addition, alkynes undergo certain reactions that are due to the acidity 
of a hydrogen atom held by triply-bonded carbon. 



REACTIONS OF ALKYNES 
Addition Reactions 



Y Z 

- 



~C=C- + YZ * -C=C 



1. Addition of hydrogen. Discussed in Sec. 8.9. 

H H 

-G==C- 252 -C-C- 

Alkyne ^ ^ 

Alkane 

H 
NaorU ' NH > X C=C Trans 



HI ^ ^-rs Cjs 



Pd or Ni-B (P-2) / 

H 

Examples: 

,, ^l-NL 

2-Butyne /i-Butane 



SEC. 8.7 REACTIONS OF ALKYNES 255 



3-Hexyne 

H 2 . Ni-B (P-2) ^ 



C 2 H 5 X H 
fjww-3-Hexene 

Chief product 



H H 

r/5-3-Hexene 
9*-99% pure 

2. Addition of halogens. Discussed in Sec. 8.8. 

X X 
_C~C- -^ -C=C ^> -C-O- X 2 = C1 2 , Br 2 

XX XX 

Example: 

Br Br 

Br 2 Br 2 I I 

/"if /~* /"'LJ n '2 ^ /^Uf /" /^LI Pr 2 /"'U /^ /^Uf 

1^05^=1^11 > \sFls L^=Crl > 1^03 C Uri 

Br Br Br Br 

3. Addition of hydrogen halides. Discussed in Sec. 8.8. 

H X 

-U- 



_C=E=C > -C=C _c-C- HX = HC1, HBr, HI 

H X H(X)X 



1 H(X) J - 

I 

I 

CH3C==CH > CH3C=CH 2 > CH 3 C CH3 

Cl Ci 

4. Addition of water. Hydration. Discussed in Sec. 8.13. 

H 

H 
H2SO " H ' SO S H-i-C-H 



Acetaldehyde 
H H 

CH 3 -c==c--H 4- H 2 o H2S 4>H8S0 s H 



Acetone 



256 ALKYNES AND DIENES CHAP. 8 

Reactions as Acids 
-OEC-H + base > -C=C.- 

5. Formation of heavy metal acetylides. Discussed in Sec. 8.11. 

-C==C-H + M+ > -C==C-M + H + 
Examples: 

H-C-C-H + 2Ag + ^^ Ag-G=C--Ag + 2H+ 

Silver acetylide Identification 

of terminal 
CH 3 feC-H + Cu(NH 3 ) 2 + - > CH 3 C==C-Cu + NH 4 + + NH 3 

Cuprous 
methylacetylide 

6. Formation of alkali metal acetylides. Discussed in Sec. 8.10. 
Examples: 



H-feC-H + Na HqNH3 



Sodium acetylide 



CH 3 -CHC=C H + NaNH 2 ether > CH 3 -CH-C=CrNa + + NH 3 

CH 3 CH 3 

Sodium isopropylacetylide 



8.8 Addition reactions of alkynes 

Addition of hydrogen, halogens, and hydrogen halides to alkynes is very 
much like addition' to alkenes, except that here two molecules of reagent can be 
consumed for each triple bond. As shown, it is generally possible, by proper 
selection of conditions, to limit reaction to the first stage of addition, formation of 
alkenes. In some cases at least, this is made simpler because of the way that the 
atoms introduced in the first stage affect the second stage. 

Problem 8.2 (a) Write the equation for the two-stage addition of bromine to 
2-butyne. (b) How will the first two bromine atoms affect the reactivity of the double 
bond? (c) How will this influence the competition for halogen between 2-butyne and 
2,3-dibromo-2-butene? (d) In what proportions would you mix the reagents to help 
limit reaction to the first stage? (e) Would you bubble 2-butyne into a solution of Br 2 
in CC1 4 , or drip the bromine solution into a solution of 2-butyne? 



8.9 Reduction to alkenes 

Reduction of an alkyne to the double-bond stage can unless the triple bond 
is* at the end of a chain yield either a os-alkene or a f/ms-alkene. Just which 
isomer predominates depends upon the choice of reducing agent. 

Predominantly fraws-alkene is obtained by reduction of alkynes with sodium 
or lithium in liquid ammonia. Almost entirely os-alkene (as high as 98%) is 
obtained by hydrogenation of alkynes with several different catalysts : a specially 



SEC. 8.10 ACIDITY OF ALKYNES 257 

prepared palladium called Lindlar's catalyst \ or a nickel boride called P-2 catalyst 
reported by H. C. Brown (see p. 507) and his son, C. A. Brown. 



>< 

v/ 



Pd/C or Ni-B 

H H 



Cis 



Each of these reactions is, then, highly stereoselective. The stereoselectivity 
in the r/s-reduction of alkynes is attributed, in a general way, to the attachment of 
two hydrogens to the same side of an alkyne sitting on the catalyst surface; pre- 
sumably this same stereochemistry holds for the hydrogenation of terminal 
alkynes, RCCH, which cannot yield cis- and /raws-alkenes. 

The mechanism that gives rise to /raws-reduction is not understood. 

Problem 8.3 Most methods of making alkenes (Sees. 5.14 and 5.23) yield pre- 
dominantly the more stable isomer, usually the trans. Outline all steps in the conver- 
sion of a mixture of 75% /ra/w-2-pentene and 25% f/>-2-pentene into essentially pure 
c/.s-2-pentene. 



8.10 Acidity of alkynes. Very weak acids 

In our earlier consideration of acids (in the Lowry-Bronsted sense, Sec. 1.22), 
we took acidity to be a measure of the tendency of a compound to lose a hydrogen 
ion. Appreciable acidity is generally shown by compounds in which hydrogen is 
attached to a rather electronegative atom (e.g., N, O, S, X). The bond holding the 
hydrogen is polar, and the relatively positive hydrogen can separate as the positive 
ion; considered from another viewpoint, an electronegative element can better 
accommodate the pair of electrons left behind. In view of the electronegativity 
series, F > O > N > C, it is not surprising to find that HF is a fairly strong acid, 
H 2 O a comparatively weak one, NH 3 still weaker, and CH 4 so weak that we would 
not ordinarily consider it an acid at all. 

In organic chemistry we are frequently concerned with the acidities of com- 
pounds that do not turn litmus red or taste sour, yet have a tendency even though 
small to lose a hydrogen ion. 

A triply-bonded carbon acts as though it were an entirely different element 
a more electronegative one from a carbon having only single or double bonds. 
As a result, hydrogen attached to triply-bonded carbon, as in acetylene or any 
alkyne with the triple bond at the end of the chain (RC=C H), shows appreciable 
acidity. For example, sodium reacts with acetylene to liberate hydrogen 
form the compound sodium acetylide. 

HCE-C-H + Na > HfeCrNa* + H 2 
Sodium acetylide 



258 ALKYNES AND DIENES CHAP. 8 

Just how strong an acid is acetylene? Let us compare it with two familiar 
compounds, ammonia and water. 

Sodium metal reacts with ammonia to form sodamide, NaNH 2 , which is the 
salt of the weak acid, H NH 2 . 

NH 3 + Na > Na + NH 2 - + iH 2 
Sodamide 

Addition of acetylene to sodamide dissolved in ether yields ammonia and sodium 
acetylide. 



HCC-H + Na+NH 2 - ~^ H-NH 2 

Stronger Stronger Weaker Weaker 

acid base acid base 

The weaker acid, H NH 2 , is displaced from its salt by the stronger acid, 
HCC H. In other language, the stronger base, NH 2 ~, pulls the hydrogen ion 
away from the weaker base, HC~C~; if NH 2 ~ holds the hydrogen ion more 
tightly than HC^C", then H NH 2 must necessarily be a weaker acid than 
HC-C-H. 

Addition of water to sodium acetylide forms sodium hydroxide and regenerates 

H-OH + HC==C-Na+ -51* HC^C-H + Na + OH~ 

Stronger Stronger Weaker Weaker 

acid base acid base 

acetylene. The weaker acid, HCC H, is displaced from its salt by the stronger 
acid, H-OH. 

Thus we see that acetylene is a stronger acid than ammonia, but a weaker acid 
than water. 

Acidity H 2 O > HC^CH > NH 3 

Other alkynes that have a hydrogen attached to triply-bonded carbon show com- 
parable acidity. 

The method we have just described for comparing acidities of acetylene, 
ammonia, and water is a general one, and has been used to determine relative 
acidities of a number of extremely weak acids. One compound is shown to be a 
stronger acid than another by its ability to displace the second compound from salts. 

A-H + B-M+ * B-H + A-M+ 

Stronger Weaker 

acid acid 

How can we account for the fact that hydrogen attached to triply-bonded 
carfeon is especially acidic? How can we account for the fact that acetylene is a 
stronger acid than, say, ethane? A possible explanation can be found in the 
electronic configurations of the anions. 

If acetylene is a stronger acid than ethane, then the acetylide ion must be a 
weaker base than the ethide ion, C 2 H 5 . In the acetylide anion the unshared 



SEC. 8.11 FORMATION OF HEAVY METAL ACETYLIDES 259 

pair of electrons occupies an sp orbital; in the ethide anion the unshared pair of 
electrons occupies an sp 3 orbital. The availability of this pair for sharing with 
acids determines the basicity of the anion. Now, compared with an sp 3 orbital, 



HC==C:H <- H+ + 

Acetylene Acetylide ion 

Stronger Weaker 

acid base 

CH 3 CH 2 :H - H+ + CH 3 CH 2 :- 

Ethane Ethide ion 

Weaker Stronger 

acid base 

an sp orbital has less p character and more s character (Sec. 5.4). An electron in 
a p orbital is at some distance from the nucleus and is held relatively loosely; an 
electron in an s orbital, on the other hand, is close to the nucleus and is held more 
tightly. The acetylide ion is the weaker base since its pair of electrons is held more 
tightly, in an sp orbital. 

Problem 8.4 When 1-hexyne was added to a solution of //-propylmagnesium 
bromide, a gas was evolved. The density of the gas showed that it had a molecular 
weight of 44. When it was bubbled through aqueous KMnO 4 or Br 2 in CC1 4 , there 
was no visible change, (a) What was the gas? (b) Write an equation to account for its 
formation, (c) How could you have predicted such a reaction? 

Problem 8.5 What do you suppose the structure of calcium carbide is? Can 
you suggest another name for it? What is the nature of its reaction with water? 



8.11 Formation of heavy metal acetylides 

The acidic acetylenes react with certain heavy metal ions, chiefly Ag* and 
Cu + , to form insoluble acetylides. Formation of a precipitate upon addition of 
an alkyne to a solution of AgNO 3 in alcohol, for example, is an indication of hydro- 
gen attached to triply-bonded carbon. This reaction can be used to differentiate 
terminal alkynes (those with the triple bond at the end of the chain) from non- 
terminal alkynes. 

CH 3 CH 2 feC-H -^-* CH 3 CH 2 C=C-Ag [ HN 3 > CH 3 CH 2 C~C-H + Ag+] 

1-Butyne Precipitate 1-Butync 

A terminal alkyne 

CH 3 -feC-CH 3 -^U no reaction 

2-Butyne 
A non-terminal alkyne 

If allowed to dry, these heavy metal acetylides are likely to explode. They 
should be destroyed while still wet by warming with nitric acid; the strong mineral 
acid regenerates the weak acid, acetylene. 



260 ALKYNES AND DIENES CHAP. 8 



8.12 Reaction of sodium acetylides with alkyl halides. Substitution vs. 
elimination 

Sodium acetylides are used in the synthesis of higher alkynes. For example: 

Na+ + C 2 H 5 :X: * HC==CC 2 H 5 + Na+:Xr 
1-Butyne 



C 2 H 5 C=C : - Na + + CH 3 : X : > C Z H 5 C^CCH 3 + Na + : X : - 

2-Pentyne 

This reaction involves substitution of acetylide ion for halide ion. It results 
from attack by the acetylide ion on carbon.. 



Attack on C: 

substitution 




Alky ae 



Since sodium acetylide is the salt of the extremely weak acid, acetylene, the 
acetylide ion is an extremely strong base, stronger in fact than hydroxide ion. 
In our discussion of the synthesis of alkenes from alkyl halides (Sec. 5.13), we saw 
that the basic hydroxide ion causes elimination by abstracting a hydrogen ion. 
It is not surprising that the even more basic acetylide ion can also cause elimination. 



x __, / _ 

> X- V C ^ C \ f HCS5 CH 

Alkene 
-C-CH 

The acetylide ion, then, can react with an alkyl halide in two ways: by attack 
at carbon to give substitution, or by attack at hydrogen to give elimination. We have 
seen that the order of reactivity of alkyl halides toward elimination (Sec. 5.14) 
is 3 > 2 > 1. In substitution (of the present kind), we shall find (Sec. 14.11) 
the order of reactivity is just the opposite: 1 > 2 > 3. It is to be expected, then, 
that : where substitution and elimination are competing reactions, the proportion of 
elimination increases as the structure of an alkyl halide is changed from primary to 

Elimination increases 

* Elimination (E2) 

RX= 1 2 3 vs. 

< Substitution (S N 2) 

Substitution increases 

secondary to tertiary. Many tertiary halides fastest at elimination and slowest 
at substitution yield exclusively alkenes under these conditions. 

When the attacking reagent is a strong base like hydroxide or acetylide, that 
is, when the reagent has a strong affinity for hydrogen ion, elimination is partic- 
ularly important. Practically speaking, only primary halides give good yields of 



SEC. 8.13 HYDRATION OF ALKYNES. TAUTOMERISM 261 

the substitution product, the alkyne. With secondary and tertiary halides, elimina- 
tion predominates to such an extent that the method is essentially useless. We shall 
encounter this competition between substitution and elimination again and again 
in our study of organic chemistry. 

Reflecting their ability to form carbon-carbon bonds (Sec. 3. 17),. copper 
acetylides, too, are used to couple with organic halides, particularly with the 
ordinarily unreactive vinyl and aryl halides (Chap. 25). 

8.13 Hydration of alkynes. Tautomerism 

Addition of water to acetylene to form acetaldehydc, which can then be 
oxidized to acetic acid, is an extremely important industrial process. 

From the structure of acetaldehyde, it at first appears that this reaction 
follows a different pattern from the others, in which two groups attach themselves 
to the two triply-bonded carbons. Actually, however, the product can be accounted 
for in a rather simple way. 

H H 

H "-P*"* ""* 80 ', H-C-e?-H ^r> H-C-C 



Acetylene H (pl^ - > HO 

Vinyl alcohol Acetaldehyde 

If hydration of acetylene followed the same pattern as hydration of alkenes, 
we would expect addition of H and OH to the triple bond to yield the structure 
that we would call vinyl alcohol. But all attempts to prepare vinyl alcohol result 
like hydration of acetylene in the formation of acetaldehyde. 

A structure with OH attached to doubly-bonded carbon is called an cnol 
(ene for the carbon carbon double bond, -ol for alcohol). It is almost always 
true that when we try to make a compound with the enol structure, we obtain 
instead a compound with the keto structure (one that contains a C O group). 

II II 

C=C OH -^r* C C^O Keto-enol tautomerism 

I 
H 

Enol structure Keto structure 

There is an equilibrium between the two structures, but it generally lies very much 
in favor of the keto form. Thus, vinyl alcohol is formed initially by hydration of 
acetylene, but it is rapidly converted into an equilibrium mixture that is almost 
all acetaldehyde. 

Rearrangements of this enol-keto kind take place particularly easily because 
of the polarity of the O H bond. A hydrogen ion separates readily from oxygen 
to form a hybrid anion; but when a hydrogen ion (most likely a different one) 
returns, it may attach itself either to oxygen or to carbon of the anion. When it 
returns to oxygen, it may readily come off again; but when it attaches itself to 



-C=C O H ^Z [ OO-OJ + H+ __> C-C=O Keto-enol 

H 

Stronger acid Weaker acid 



I tautomerism 

H 



262 ALKYNES AND DIENES CHAP. 8 

carbon, it tends to stay there. We recognize this reaction as another example of 
the conversion of a stronger acid into a weaker acid (Sec. 8.10). 

Compounds whose structures differ markedly in arrangement of atoms, but which 
exist in equilibrium, are called tautomers. The most common kind of tautomerism 
involves structures that differ in the point of attachment of hydrogen. In these 
cases, as in keto-enol tautomerism, the tautomeric equilibrium generally favors 
the structure in which hydrogen is bonded to carbon rather than to a more electro- 
negative atom; that is, equilibrium favors the weaker acid. 

Problem 8.6 Hydration of propyne yields the ketone acetone, CHjCOCH 3 , 
rather than the aldehyde CH 3 CH 2 CHO. What does this suggest about the orientation 
of the initial addition ? 



DIENES 

8.14 Structure and nomenclature of dienes 

Dienes are simply alkenes that contain two carbon-carbon double bonds. 
They therefore have essentially the same properties as the alkenes we have already 
studied. For certain of the dienes, these alkene properties are modified in important 
ways; we shall focus our attention on these modifications. Although we shall 
consider chiefly dienes in this section, what we shall say applies equally well to 
compounds with more than two double bonds. 

Dienes are named by the ILJPAC system in the same way as alkenes, except 
that the ending -diene is used, with two numbers to indicate the positions of the 
two double bonds. This system is easily extended to compounds containing any 
number of double bonds. 

CH 2 =CH CH=CH 2 CH 2 =CH CH 2 CH=CH 2 CH 2 =CH CH=CH CH=CH 2 

1,3-Butadiene 1,4-Pentadiene 1,3,5-Hexatriene 

Dienes are divided into two important classes according to the arrangement 
of the double bonds, Double bonds that alternate with single bonds are said to 
be conjugated; double bonds that are separated by more than one single bond are 
said to be isolated. 



Conjugated I 

double bonds Isolated 

double bonds 

A third class of dienes, of increasing interest to organic chemists, contain 
cumulated double bonds; these compounds are known as allenes: 



-JwO- 



Cumulated double bonds: allenes 



8.15 Preparation and properties of dienes 

Dienes are usually prepared by adaptations of the methods used to make 
simple alkenes. For example, the most important diene, 1,3-butadiene (used to 



SEC. 8.16 STABILITY OF CONJUGATED DIENES 263 

make synthetic rubber, Sec. 8.25), has been made in this country by a cracking 
process, and in Germany by dehydration of an alcohol containing two OH 
groups: 



CH 3 CH 2 CH 2 CH 3 
/i-Butane 



heat 



CH 3 CH 2 CH=CH 2 

1-Butene 



CH 2 =CH-CH=CH 2 

1,3-Butadiene 



CH 3 CH-CHCH 3 

2-Butene 

CH,CH 2 CH 2 CH 2 hea '- ac " L 

OH OH 1,3-Butadiene 

The chemical properties of a diene depend upon the arrangement of its double 
bonds. Isolated double bonds exert little effect on each other, and hence each 
reacts as though it were the only double bond in the molecule. Except for the 
consumption of larger amounts of reagents, then, the chemical properties of the 
non-conjugated dienes are identical with those of the simple alkenes. 

Conjugated dienes differ from simple alkenes in three ways: (a) they are more 
stable, (b) they undergo 1,4-addition, and (c) toward free radical addition, they are 
more reactive. 



8.16 Stability of conjugated dienes 

If we look closely at Table 6.1 (p. 183), we find that the heats of hydrogenation 
of alkenes having similar structures are remarkably constant. For monosubsti- 
tuted alkenes (RCH CH 2 ) the values are very close to 30 kcal/mole; for disubsti- 
tuted alkenes (R 2 C CH 2 or RCH-CHR), 28 kcal/mole; and for trisubstituted 
alkenes (R 2 C ^CHR), 27 kcal/mole. For a compound containing more than one 
double bond we might expect a heat of hydrogenation that is the sum of the 
heats of hydrogenation of the individual double bonds. 

For non-conjugated dienes this additive relationship is found to hold. As 
shown in Table 8.2, 1,4-pentadiene and 1,5-hexadiene, for example, have heats of 
hydrogenation very close to 2 x 30 kcal, or 60 kcal/mole. 

Table 8.2 HEATS OF HYDROGENATION OF DIENES 

A// of Hydrogenation, 
Diene kcal/mole 



1,4-Pentadiene 60.8 

1,5-Hexadiene 60.5 

1,3-Butadiene 57.1 

1,3-Pentadiene 54.1 

2rMethyl- 1 ,3-butadiene (Isoprene) 53.4 

2,3-Dimethyl-l,3-butadiene 53.9 

1 ,2-Propadiene (Allene) 71.3 



264 ALKYNES AND DIENES CHAP. 8 

For conjugated dienes, however, the measured values are slightly lower than 
expected. For 1,3-butadiene we might expect 2 x 30, or 60 kcal: the actual value, 
57 kcal, is 3 kcal lower. In the same way the values for 1 ,3-pentadiene and 2,3- 
dimethyl- 1,3-butadiene are also below the expected values by 2-4 kcal. 

Heats of Hydrogenation 

CH 2 ==CH-CH==CH 2 CH 3 CH=CH CH=CH 2 

Expected: 30 + 30 = 60 kcal Expected: 28 4- 30 = 58 kcal 

Observed: 57 Observed: 54 

CH 3 CH 3 

CH 2 =C C=CH 2 

Expected: 28 + 28 = 56 kcal 
Observed: 54 

What do these heats of hydrogenation tell as about the conjugated dienes? 
Using the approach of Sec. 6.4, let us compare, for example, 1 ,3-pentadiene (heat 
of hydrogenation, 54 kcal) and 1,4-pentadiene (heat of hydrogenation, 61 kcal). 
They both consume two moles of hydrogen and yield the same product, w-pentane. 
If 1,3-pentadiene evolves less energy than 1,4-pentadiene, it can only mean that it 
contains less energy; that is to say, the conjugated 1,3-pentadiene is more stable 
than the non-conjugated 1,4-pentadiene. 

In the next three sections we shall see how two different factors have been 
invoked to account for the relative stabilities of conjugated dienes, and of simple 
alkenes as well : (a) delocalization of TT electrons, and (b) strengthening of v bonds 
through changes in hybridization of carbon, 

Unusual stability of conjugated dienes is also strongly indicated by the fact 
that, where possible, they are the preferred diene products of elimination reactions 
(Sec. 5.14). 

Problem 8.7 Predict the major product of dehydrohalogenation of 4-bromo-l- 
hexene. 

Problem 8.8 (a) Predict the heat of hydrogenation of allene, CHf==C==CH 2 . 
(b) The actual value is 71 kcal. What can you say about the stability of a cumulated 
diene? 



8.17 Resonance in conjugated dienes 

Let us focus our attention on the four key carbon atoms of any conjugated 
diene system. We ordinarily write the^C?! C 2 and C 3 C 4 bonds as double, and 
the C 2 C 3 bond as single: 

1234 

~M i r 

This would correspond to an orbital picture of the molecule (see Fig. 8.50), in 
which IT bonds are formed by overlap of the p orbitals of C\ and C 2 , and 
overlap of the p orbitals of C 3 and C 4 . 

In the allyl radical we saw that resonance resulted from the overlap of the 
p orbital of a carbon atom with p orbitals on both sides. We might expect that, 



SEC. 8.17 



RESONANCE IN CONJUGATED DIENES 



265 



in the same way, there could be a certain amount of overlap between the 
p orbitals of C 2 and C 3 , as shown in Fig. 8.56. The resulting delocalization of the 
TT electrons makes the molecule more stable: each pair of electrons attracts and 
is attracted by not just two carbon nuclei, but four. 





(a) 



Figure 8.5. Conjugated diene. (a) Overlap of p orbitals to form two 
double bonds. (/>) Overlap of p orbitals to form conjugated system: de- 
localization of TT electrons. 



Using the language of conventional valence-bond structures, we say that a 
conjugated diene is a resonance hybrid of I and II. The dotted line in II represents 



1234 

c=c c=c- 

III! 

I 



c-c=c c- 

I I I I 



a formal bond, and simply means that an electron on C } and an electron on C 4 have 
opposite spins, that is to say, are paired. 

To the extent that II contributes to the structure, it gives a certain double-bond 
character to the C 2 C 3 bond and a certain single-bond character to the Cj C 2 
and 3 C 4 bonds; most important, it makes the molecule more stable than we 
would expect I (the most stable contributing structure) to be. 

Formation of a bond releases energy and stabilizes a system; all other things 
being equal, the more bonds, the more stable a structure. Consideration of number 
of bonds is one of the criteria (Sec. 6.27) that can be used to estimate relative 
stability and hence relative importance ofjjcontributing structure. On this basis 
we would expect II with 10 bonds (the formaTbond does not count) to be less stable 
than I with 1 1 bonds. The resonance energy for such a hybrid of non-equivalent 
structures should be less than for a hybrid made up of equivalent structures. The 
structure of a conjugated diene should resemble 1 more than II, since the more 
stable structure I makes the larger contribution to the hybrid. 

Consistent with partial double-bond character, the C 2 C 3 bond in 1,3- 
butadiene is 1.48 A long, as compared with 1.53 A for a pure single bond. The 
resonance energy of a conjugated diene is only 2-4 kcal/mole, compared with 
10 kcal/mole for the allyl radical. (However, for an alternative interpretation, 
see s Sec. 8.19.) 




266 ALKYNES AND DIENES CHAP. 8 



8.18 Resonance in alkenes. Hyperconjugation 

Heats of hydrogenation showed us (Sec. 6.4) that alkenes are stabilized not 
only by conjugation but also by the presence of alkyl groups : the greater the number 
of alkyl groups attached to the doubly-bonded carbon atoms, the more stable the 
alkene. To take the simplest example, the heat of hydrogenation of propylene is 
2.7 kcal lower than that of ethylene, indicating that (relative to the corresponding 
alkane) propylene is 2.7 kcal more stable than ethylene. 

Stabilization by alkyl groups has been attributed to the same fundamental 
factor as stabilization by a second double bond : delocalization of electrons, this 
time through overlap between a p orbital and a a orbital of the alkyl group. 



Figure 8.6. Hyperconjugation in an 
alkene: overlap between p and TT 
orbitals. 



Through this overtop, individual electrons can, to an extent, help bind together 
four nuclei. Delocalization of this kind, involving a bond orbitals, we recognize 
as hyperconjugation (Sec. 6.28). 

Translated into resonance terminology, such hyperconjugation is represented 
by contribution from structures like II. (As before, the dotted line in II represents 

H H H H H H 

III III 

H-C-C=C-H H C=C- C H 

I I 

H " 

321 321 

I II 

and two more 
equivalent 
structures 

a formal bond, indicating that electrons on the two atoms are paired.) Considered 
by itself, a structure like II is indeed strange, since there is no real bond joining the 
hydrogen- to carbon. This is, however, simply a rough way of indicating that the 
carbon-hydrogen bond is something less than a single bond, that the C 2 C 3 
bond has some double-bond character, and that the C\ C 2 bond has some 
single-bond character. 

Consistent with partial double-bond character, the carbon-carbon "single" 
bond in propylene is 1.50 A long, as compared to 1.53 A for a pure single bond. 

The greater the number of alkyl groups attached to the doubly-bonded car- 
bons, the greater the number of contributing structures like II, the greater the 
delocalization of electrons, and the more stable the alkene. 

Hyperconjugation of the kind described above is called sacrificial hyperconjugation, 
since there is one less real bond in structures like II than in I. In contrast, the kind of 



SEC. 8.19 AN ALTERNATIVE INTERPRETATION 267 

hyperconjugation we encountered in connection with free radicals and carbon ium ions 
involves no *' sacrifice" of a bond and is called isovalent hyperconjugation. 

8.19 Stability of dienes and alkenes: an alternative interpretation 

We have seen that the carbon-hydrogen bond length decreases as we proceed 
along the series ethane, ethylene, acetylene, and we attributed this to changes in 
hybridization of carbon (see Table 8.3). As the/? character of the bonding orbital 

Table 8.3 CARBON -HYDROGEN SINGLE BOND LENGTHS AND HYBRIDIZATION 
Compound Length, A Hybridization 



CH 3 -CH 3 


1.112 


spi-s 


CH 2 -=CH 2 


1.103 


.V/>2-jj 


HC-CH 


1.079 


sp-s 



decreases, the orbital size decreases, and the bond becomes shorter (Sec. 5.4). 

The carbon-carbon single-bond length also decreases along an analogous 
series, ethane, propylene, propyne (Table 8.4). We notice that these differences 

Table 8.4 CARBON CARBON SINGLE BOND LENGTHS AND HYBRIDIZATION 
Compound Length, A Hybridization 



CHj CH 3 


1.53 


.s/i 3 sp* 


CH 2 --CH-CH 3 


1.50 


sp 2 -sp* 


HC.-==C-CHi 


1.46 


sp-sp* 



arc bigger than for carbon-hydrogen bonds. Here, the bond-shortening has been 
attributed to hyperconjugation, as discussed in Sec. 8.18. 

U has been argued, most notably by M. J. S. Dewar of the University of Texas, 
that there is no need to invoke hyperconjugation in molecules like these, and that 
the changes in C C bond length - like the. changes in C H bond length are 
due simply to changes in hybridization of carbon. 

Furthermore, Devvar has proposed that such shortening of bonds is accom- 
panied by a proportional increase in bond energies (); that is, shortening a bond 
makes the molecule more stable. Change in hybridization affects bond lengths 
more and hence affects molecular stability more when carbon-carbon bonds are 
involved than when carbon hydrogen bonds are involved. An alkyl substituent 
stabilizes an alkene, relative to the corresponding alkane. because sp 2 hybridization 
strengthens a carbon-carbon bond more than a carbon- hydrogen bond. 

In a similar way, the unusual stability of conjugated dienes is attributed, not 
to delocalization of the -n electrons, but to the fact that &p 2 -sp 2 hybridization makes 
the C 2 C 3 bond short (1.48 A) and strong. 

There is little doubt that both factors, delocalization of n electrons and change 
in a bonds, are at work. The question is: what is the relative importance of each? 
The answer may well turn out to be: both are important. 

In the case of molecules like the allyl radical, where clearly no single structure is 
acceptable, Dewar has not questioned the importance of ^-electron delocalization, 



268 



ALKYNES AND DIENES 



CHAP. 8 



although he considers o-bond stability to play a larger part than has been recognized. 
He also accepts a more important role for isovalent hyperconjugation in free radicals 
and carbonium ions than for the sacrificial hyperconjugation we have so far discussed. 



8.20 Electrophilic add|bn to conjugated dienes. 1,4-Addition 

When 1 ,4-pentadiene is treated with bromine under conditions (what are they?) 
that favor formation of the dihatide, there is obtained the expected product, 4,5- 
dibromo-1-pentene. Addition of more bromine yields the 1,2,4,5-tetrabromo- 



CH 2 =CH-CH 2 -CH-CH 2 -^ 

CH 2 -CH-CH 2 -CH=CH 2 
Br Br 



CH 2 CH-CH 2 -CH CH 2 

Ir ir ir ir 



pentane. This is typical of the behavior of dienes containing isolated double bonds: 
the double bonds react independently, as though they were in different molecules. ' 

When 1,3-butadiene is treated with bromine under similar conditions, there 
is obtained not only the expected 3,4-dibromo-l-butene, but also l,4-dibromo-2- 
butene. Treatment with HC1 yields not only 3-chloro-l-butene, but also l-chloro* 
2-butene. Hydrogenation yields not only 1-butene but also 2-butene, 

~V CH 2 -CH-CH=CH 2 and CH 2 -CH=CH~CH 2 

Br Br Br Br 

1,2-addition 1,4-addition 



CH 2 =CH-CH-=CH 2 

1,3-Butadiene 



HC1 



CH 2 -CH-CH=CH 2 and CH 2 -CH=CH-CH 2 

H Cl H Cl 

1,2-addition 1,4-addition 



H 2 

cat. x 



CH 2 -CH-CH^CH 2 and 



H 



H 

1,2-addition 



H 



1,4-addition 



Study of many conjugated dienes and many reagents shows that such behavior 
is typical: in additions to conjugated dienes, a reagent may attach itself not only to a 
pair of adjacent carbons (1,2-addition), but also to the carbons at the two ends of the 
conjugated system (1,4-addition). Very often the 1,4-addition product is the major 
one. 



YZ 



I I I I 

c c oc- 

Y Z 

1,2-addition 



and 



Y 2 

1,4-addition 



8.21 Allyl cations. Delocalization in carbonium ions 

How can we account for the products obtained? We have seen (Sees. 6.10 
and 6.1 1) that electrophilic addition is a two-step process, and that the first step 



SEC. 8.21 ALLYL CATIONS. DELOCALIZATION IN CARBONIUM IONS 269 

takes place in the way that yields the more stable carbonium ion. Let us apply this 
principle to the addition, for example, of HC1 to 2,4-hexadiene, which yields 
4-chloro-2-hexene and 2-chloro-3-hexene : 

CH 3 -CH-CH-CH-CH- CH 3 -^U CH 3 - CH~CH-CH=CH-CH 3 

2,4-Hexadiene H Q 

4-Chloro-2-hexene 

CH 3 -CH-CH==CH--CH-CH 3 

H Cl 

2-Chloro-3-hexene 

These products show that hydrogen adds to C-2 to yield carbonium ion I, 
rather than to C-3 to yield carbonium ion II: 

[> CH 3 -CH-CH~CH=CH-CH 3 
CH 3 ~CH=CH-CH=CH-CH 3 I H+ 1 

X> CH 3 -CH-CH-CH=CH-CH 3 

H 

U 

Since both I and II are secondary cations, how can we account for the preference? 
I is not simply a secondary cation, but is an allyl cation as well, since the carbon 
bearing the positive charge is attached to a doubly-bonded carbon. 

Let us look more closely at such cations, using the parent allyl cation, 
CH 2 CH CH 2 +, as our example. Bond dissociation energies showed us that 
allyl radicals are unusually stable, and we attributed this stability to resonance 
between equivalent structures (Sees. 6.24-6.25). The ionization potential (188 
kcal) of the allyl radical enables us to calculate that the allyl cation, too, is un- 
usually stable. Even though we have just drawn its structure as that of a primary 
cation, it is 24 kcal more stable than the ethyl cation, and just about as stable as 
the isopropyl cation. We can now expand the sequence of Sec. 5.18. 

Stability of allyl Cf + 

carbonium ions 3 > 2 o > 1 > CH 3 



Like the allyl radical, the allyl cation is a resonance hybrid of two exactly 
equivalent structures: 

[CH 2 -=CH-CH 2 + + CH 2 -CH=CH 2 ] equivalent to CH 2 ~CH CHj 

in iv e 

In either of the contributing structures, there is an empty p orbital on the electron- 
deficient carbon. Overlap of this empty p orbital with the ft cloud of the double 
bond results in delocalization of the * electrons: each of them helps to hold to- 
gether all three carbon nuclei (Fig. $.7). We see how fatness is required to permit 



270 



ALKYNES AND DIENES 



CHAP, 



the overlap that provides electrons to the electron-deficient carbon and stabilize* 
the cation. 




equivalent to 




[CH S ^-CH CH 2 4 'CH>-CH CH-] cqunalent to CH 2 -CH CH- 



Figure 8.7 Allyl cation. The p orbital of the middle carbon overlaps p 
orbitals on both sides to permit delocahzation of electron^. 



The relative stabilities of simple alky! cations has also been attributed to 
dclocalization, this time by overlap of the/? orbital with IT bonds, that is, through 
hyper conjugation (Sec. 6.28). 

Problem 8.9 Draw resonance structures to show how the order of stability of 
alkyl cations could be accounted for on the basis of hypcrconjugutien. 

The products obtained from addition to conjugated dicnes arc always con- 
sistent with the formation of the most stable intermediate carbonium ion: an allyl 
cation. This requires the first step to be addition to one of the ends of the conjugated 
system, 

> _i__ i^_ !^!__ 



i I i i 

-c~c c-< 



Y+ 



Aih f \ to end of l 



u.s/tv;/ 



Y 



An allyl idibomiim ion 



The first step of addition to 2,4-hcxadiene yields, then, not cation I, but the 
hybrid cation VI in which the charge is spread over two carbons: 

rCH 3 CH CH-CH ~CH-CH 3 CH 3 CH CH- CH- CH -CH 3 1 



I 
H 



H 



equivalent to 

/~*L4 'I.T f'UI - f~*IJT /^*LI /"'T T 

v-rT} - v~n v_xrT v-jci^v.n v^rij 

H & 

VI 

In the second step, the negative chloride ion can attach itself to either of these 
carbons and thus yield the 1,2- or 1,4-product. 



SEC. 8.22 



1,2- YS. 1,4-ADD1TION 



271 



CH r -CH - 
H 



-Cl "- 



VI 



CH 3 -CH~-CH-CH==CH -CH 3 

H Cl 

1,2-Addition product 



CH 3 -CH CHCH CH-CH 3 

H Cl 

1,4-Addition product 

We have not shown u/iv 1,4-addition occurs; we have simply shown that it 
is not unreasonable that it t/<*es happen. In summary: 



1! 






Addition to end of 
conjugated s vsiem 



I 

Y 

Allyl carbonium ion 



Y Z 

1,2-Addition 



Y Z 

1,4-Addition 



Thus the hybrid nature of the allyl cation governs both steps of electrophiiic 
addition to conjugated dienes: the first, through stabilization; the second, by 
permitting attachment to either of two carbon atoms. 

Problem 8.10 Account for tho fact that 2-methyI-l,3-butadiene reacts (a) with 
HC1 to yield only 3-chloro-3-methyi-l-butene and l-chloro-3-methyl-2-butene; 
(b) with bromine to yield only 3,4-dibromo-3-methyl-l-butene and l,4-dibromo2- 
methyl-2-butene. 



8.22 1,2- vs. 1,4-Addition. Rate vs. equilibrium 

A very important principle emerges when we look at the relative amounts of 
1,2- and 1,4-addition products obtained. 

Addition of HBr to 1,3-butadiene yields both the 1,2- and the 1,4-products; 
the proportions in which they are obtained are markedly affected by the temperature 

HBr 

+ 
CH 2 =CH-CH=CH 2 



80% 





. 

H 

20% CH 2 -CH-CH-CH 2 
H Br 



40 



20% CH 2 -CH-CH=CH 2 

H Br 
80% CH 2 ~CH-CH-~CH 2 



272 ALKYNES AND EHENES CHAP. 8 

at which the reaction is carried out. Reaction at a low temperature (80) yields 
a mixture containing 20% of the 1,4-product and 80% of the 1,2-product. Reaction 
at a higher temperature (40) yields a mixture of quite different composition, 
80% 1,4- and 20% 1,2-product. At intermediate temperatures, mixtures of inter- 
mediate compositions are obtained. Although each isomer is quite stable at low 
temperatures, prolonged heating of either the 1,4- or the 1,2-compound yields the 
same mixture. How are these observations to be interpreted ? 

The fact that either compound is converted into the same mixture by heating 
indicates that this mixture is the result of equilibrium between the two compounds. 
The fact that the 1,4-compound predominates in the equilibrium mixture- indicates 
that it is the more stable of the two. 

The fact that more 1,2- than 1,4-product is obtained at 80 indicates that 
the 1,2-product is formed faster than the 1,4-product; since each compound re- 
mains unchanged at 80, the proportions in which they are isolated show the 
proportions in which they were initially formed. As the reaction temperature is 
raised, the proportions in which the products are initially formed may remain the 
same, but there is faster conversion of the initially formed products into the 
equilibrium mixture. 

The proportions of products actually isolated from the low-temperature 
addition are determined by the rates of addition, whereas for the high-temperature 
addition they are determined by the equilibrium between the two isomers. 

Let us examine the matter of 1,2- and 1,4-addition more closely by drawing 
a potential energy curve for the reactions involved (Fig. 8.8). The carbonium 
ion initially formed reacts to yield the 1,2-product faster than the 1,4-product; 
consequently, the energy of activation leading to the 1 ,2-product must be less than 




CH 2 -CH---CH==CH2 *- CH2-CH-CH---CH2 - CH 2 -CH 2 ==CH CH 2 



H Br 



I 



H e H 

1 ,2-addition + Br~ 1 ,4-addition 

product | product 



+ HBr 



< Progress of 1 ,2-addition Progress of 1,4-addition * 

Figure 8.8. Potential energy changes during progress of reaction : 1 ,2- vs. 
1,4-addition. 




SEC. 8,22 1,2- VS. 1,4-ADDITION 273 

that leading to the 1,4- product. We represent this by the lower hill leading from 
the ion to the 1,2-product. More collisions have enough energy to climb the low 
hill than the high hill, so that the 1,2-compound is formed faster than the 1,4- 
compound. The 1,4-product, however, is more stable than the 1,2-product, and 
hence we must place its valley at a lower level than that of the 1,2-product. 

We shall see later (Sec. 14.12) that alkyl halidcs, and particularly allyl halides, 
can undergo ionization. Now ionization of either bromo compound yields the 
same carbonium ion; the most likely- and simplest way in which the 1,2- and 
1,4-products reach equilibrium is through this ion. 



CH 2 -CH-CH-CH 2 

H Br 

1,2- 



CH 2 ==CH-CH==CH 2 
+ HBr 

Ionization of the bromides involves climbing the potential hills back toward 
this carbonium ion. But there is a higher hill separating the ion from the 1,4- 
product than from the 1,2-product; consequently, the 1,4-product will ionize more 
slowly than the 1,2-product. Equilibrium is reached when the rates of the opposing 
reactions are equal. The 1,2-product is formed rapidly, but ionizes rapidly. The 
1,4-product is formed slowly, but ionizes even more slowly; once formed, the 1,4- 
product tends to persist. At temperatures high enough for equilibrium to be reached 
that is, high enough for significantly fast ionization the more stable 1,4- 
product predominates. 

We have not tried to account for the fact that the 1,2-product is formed faster 
than the 1,4-product, or for the fact that the 1,4-product is more stable than the 
1,2-product (although we notice that this is consistent with our generalization that 
disubstituted alkenes are more stable than monosubstituted alkenes). We have 
accepted these facts and have simply tried to show what they mean in terms of 
energy considerations. Similar relationships have been observed for other dienes 
and reagents. 

These facts illustrate two important points. First, we must be cautious wh'en 
we interpret product composition in terms of rates of reaction; we must be sure 
that one product is not converted into the other after its formation. Second, the 
more stable product is by no means always formed faster. On the basis of much 
evidence, we have concluded that generally the more stable a carbonium ion or 
free radical, the faster it is formed; a consideration of the transition states for the 
various reactions has shown (Sees. 3.26, 5.21, and 6.11) that this is reasonable. 
We must not, however, extend this principle to other reactions unless the evidence 
warrants it. 

Problem 8.11 Addition of one mole of bromine to 1,3,5-hexatricne yields only 
5,6-dibromo-l > 3-hcxadiene and l,6-dibromo2,4-hexadiene. (a) Are these products 



274 ALKYNES AND D1ENES CHAP. 8 

consistent %vith the formation of the most stable intermediate carbonium ion? 
(b) What other product or products would also be consistent? (c) Actually, which 
factor appears to be in control, rate or position of equilibrium? 



8,23 Free-radical addition to conjugated dienes: orientation 

Like other alkenes, conjugated dienes undergo addition not only by electro- 
philic reagents but also by free radicals. In free-radical addition, conjugated 
dienes show two special features: they undergo 1,4-addition as weJl as 1,2-addition, 
and they are much more reactive than ordinary alkenes. We can account for both 
features orientation and reactivity by examining the structure of the intermediate 
free radical. 

Let us take, as an example, addition of!BrCCl 3 to 1,3-butadiene in the presence 
of a peroxide. As we have seen (Sec. 6.18), the peroxide decomposes (step 1) 
to yield a free radical, which abstracts bromine from BrCCl} (step 2) to generate a 
CC1 3 radical. 

(1) Peroxide > Rad- 

(2) Rad- + BrCClj - > RaU-Br + -CCI 3 



The 'CCli radical thus formed adds to the butadiene (step 3). Addition to 
one of the ends of the conjugated system is the preferred reaction, since this yields 
a resonance-stabilized allyl free radical. 

/CC1 3 
i, 
(3) C 




Allylic free radical 
The allyl free radical then abstracts bromine from a molecule of BrCCl 3 



(step 4) to complete the addition, and in doing so forms a new -CCls radical which 
can carry on the chain. In step (4) bromine can become attached to either C-2 
or C-4 to yield either the 1,2- or 1,4-product. 

(4) CJ 3 C~CH;r-CH^CH^CH 2 BrCC>i > CI 3 C-CH 2 -~CH-CH=CH 2 

Br 

Ally lie free radical 1 ,2-Addition product 

and C! 3 C-CH 2 --CH=CH-CH 2 ~-Br 

1,4-Addition product 

8.24 Free-radical addition to conjugated dienes: reactivity 

if BrCCl 3 is allowed to react with a 50:50 mixture of 1,3-butadiene and a 
simple alkene like 1-octene, addition occurs almost exclusively to the 1,3-butadiene. 
Evidently the CC1 3 radical adds much more rapidly to the conjugated diene than 



SEC. 8.25 



FREE-RADICAL POLYMERIZATION OF DIENES 



275 



to the simple alkene. Similar results have been observed in a great many radical 
additions. 

How can we account for the unusual reactivity of conjugated dienes? In our 
discussion of halogenation of the simple alkenes (Sec. 3.27), we found that not only 
orientation but also relative reactivity was related to the stability of the free radical 
formed in the first step. On this basis alone, we might expect addition to a con- 
jugated diene, which yields a stable allyl free radical, to occur faster than addition 
to a simple alkene. 

On the other hand, we have just seen (Sec. 8.16) that conjugated dienes are 
more stable than simple alkenes. On this basis alone, we might expect addition to 
conjugated dienes to occur more slowly than to simple alkenes. 

The relative rates of the two reactions depend chiefly upon the ac t's- Stabili- 
zation of the incipient allyl free radical lowers the energy level of the transition 
state; stabilization of the diene lowers the energy of the reactants. Whether the 
net aot is larger or smaller than for addition to a simple alkene depends upon 
which is stabilized more (see Fig. 8.9). 



ac t (diene) < ac t (alkene) 




Stabilization of 
- transition state 
formed front diene 



Alkene 



Stabilization 
of diene 




Allyl 

free 

radical 



Conjugated 
diene 



Progress of reaction > 

Figure 8.9. Molecular structure and rate of reaction. Transition state 
from diene stabilized more than diene itself: ac t is lowered. (Plots aligned 
with each other for easy comparison.) 

The fact is that conjugated dienes are more reactive than simple alkenes. In 
the present case, then and in most cases involving alkenes and free radicals, or 
alkenes and carbonium ions the factors stabilizing the transition state are more 
important than the factors stabilizing the reactant. However, this .is not always 
true. (It does not seem to be true, for example, in electrophilic addition to con- 
jugated dienes.) 



8.25 Free-radical polymerization of dienes. Rubber and rubber substitutes 

Like substituted ethylenes, conjugated dienes, too, undergo free-radical 
polymerization. From 1,3-butadiene, for example, there is obtained a polymer 



276 ALKYNES AND DIENES CHAP. 8 



CH 2 =CH--CH=CH 2 

1 ,3-Butadiene Polybutadiene 

whose structure indicates that 1,4-addition occurs predominantly: 



Rad* CH 2 ==CH CH=CH 2 CH 2 =CH-CH=CH 2 CH 2 ==CHCH==CH 2 

1 ,3-Butadiene 1 



Such a polymer differs from the polymers of simple alkenes in one very important 
way: each unit still contains one double bond. 

Natural rubber has a structure that strongly resembles these synthetic poly- 
dienes. We could consider it to be a polymer of the conjugated diene 2-methyl- 
1,3-butadiene, isoprene. 

r^u r r*u 

v-ri 3 1 *-"Ti 3 

CH 2 =O~CH CH 2 L-CH 2 C=CH-CH 2 

Isoprene c/5-Polyisoprene 

Natural rubber 

The double bonds in the rubber molecule are highly important, since appar- 
ently by providing reactive allylic hydrogens they permit vulcanization, the forma- 
tion of sulfur bridges between different chains. These cross-links make the rubber 
harder and stronger, and do away with the tackiness of the untreated rubber. 

CH, CH, 



CH 3 CH 3 

Natural rubber 

CH 3 CH 3 

~CH-C-CH-CH 2 -CH-C- 

s I s 1 

-CHC=CH CH 2 CH 2 C==CH~CH- 

CH 3 CH 3 

Vulcanized rubber 

Polymerization of dienes to form substitutes for rubber was the forerunner of 
the enormous present-day plastics industry. Polychloroprene (Neoprene, Duprene) 
was the first commercially successful rubber substitute in the United States. 

Cl I" Cl I 

CH 2 =C-CH=CH 2 LCH2 C=CH-CH 2 J n 

Chloroprene Polychloroprene 

The properties of rubber substitutes like those of other polymers are determined, 
in part, by the nature of the substituent groups. Polychloroprene, for example, is 



SEC. 8.26 ISOPRENE AND THE ISOPRENE RULE 277 

inferior to natural rubber in some {properties, but superior in its resistance to oil, 
gasoline, and other organic solvents. 

Polymers of isoprene, too, can be made artificially: they contain the same 
unsaturated chain and the same substituent (the CH 3 group) as natural rubber. 
But poly isoprene made by the free-radical process we have been talking about 
was in the properties that really matter a far cry from natural rubber. It 
differed in stereochemistry, natural rubber has the eft-configuration at (nearly) 
every double bond; the artificial material was a mixture of cis and trans. Not 
until 1955 could a true synthetic rubber be made; what was needed was an entirely 
new kind of catalyst and an entirely new mechanism of polymerization (Sec. 32.6). 
With these, it became possible to carry out a stereoselective polymerization of 
isoprene to a material virtually identical with natural rubber: cw-l,4-polyiso- 
prene. 

CH 3 H CH 3 ^ H 

^ ^CX/ CH 2 CH 2 

/"""U f*U f* f^ /"*T 

\~,IM.2, V^*l2 Vx \~t v^X"! 

CH 3 H 

Natural rubber 
All els-configurations 



8.26 Isoprene and the isoprene rule 

The isoprene unit is one of nature's favorite building blocks. It occurs not 
only in rubber, but in a wide variety of compounds isolated from plant and animal 
sources. For example, nearly all the terpenes (found in the essential oils of many 
plants) have carbon skeletons made up of isoprene units joined in a regular, head- 
to-tail way. Recognition of this fact the so-called isoprene rule has been of 
great help in working out structures of terpenes. 

H 3 C CH 3 CH 3 CH 

I 



H 2 C / 

I 



Vitamin A 



CH 3 ( CH 3 

CH 2 =C~CH 2 ~CH 2 -hCH 2 ~CH-CH 2 -CH 2 OH 

Citroncllol : a terpene A*j 

(found in oil of geranium) / V 

H 3 C CH 3 

X-Terpincne: a terpene 
(found in coriander oil) 



278 



ALKYNES AND DIENES 



CHAP. 8 



A fascinating area of research linking organic chemistry and biology is the 
study of the biogenesis of natural products: the detailed sequence of reactions by 
which a compound is formed in living systems, plant or animal. All the isoprene 
units in nature, it appears, originate from the same compound, " isopentenyl" 
pyrophosphate. 

CH 3 O O 

CH 2 =C-CH 2 -CH 2 -~<>-P-0-P-OH 

A 4 

Isopentyl pyrophosphate 

Work done since about 1950 has shown how compounds as seemingly different 
from rubber as cholesterol (p. 507) are built up, step by step, from isoprene units. 



O 

Isopentenyl 
pyrophosphate 




Squalene 



Lanosterol 



Cholesterol 



Problem 8.12 (a) Mark off the isoprene units making up the squalene mole- 
cule. (b) There is one deviation from the head-to-tail sequence. Where is it? Does 
its particular location suggest anything to you in general terms about the bio- 
genesis of this molecule? (c) What skeletal changes, if any, accompany the conversion 
of squalene into lanosterol ? Of lanosterol into cholesterol ? 

8.27 Analysis of alkynes and dienes 

Alkynes and dienes respond to characterization tests in the same way as 
alkenes: they decolorize bromine in carbon tetrachloride without evolution of 
hydrogen bromide, and they decolorize cold, neutral, dilute permanganate; they 
are not oxidized by chromic anhydride. They are, however, more unsaturated 
than alkenes. This property can be detected by determination of their molecular 
formulas (C n H 2n -2) an ^ by a quantitative hydrogenation (two moles of hydrogen 
are taken up per mole of hydrocarbon). 

Proof of structure is best accomplished by the same degradative methods that 
are used in studying alkenes. Upon ozonolysis alkynes yield carboxylic acids, 
whereas alkenes yield aldehydes and ketones. For example: 



CH 3 CH 2 C==CCHj 

2-Pentyne . 



CH 3 CH 2 COOH + HOOCCHj 
Carboxylic acids 



PROBLEMS 279 

Ozonolysis of dienes yields aldehydes and ketones, including double-ended ones 
containing two C=O groups per molecule. For example: 

CH 3 
CH 2 =C-~CH=CH 2 -- 




A terminal alkyne (RG=CH) is characterized, and differentiated from isomers, 
by its conversion into insoluble silver and cuprous acetylides (Sec. 8.1 1). 

(Spectroscopic analysis of alkynes and dienes is discussed in Sees. 13.15-13.16.) 

Problem 8.13 Contrast the ozonolysis products of the following isomers: 
(a) 1-pentyne, (b) 2-pentyne, (c) 3-methyM butyne, (d) 1,3-pentadiene, (e) 1,4-penta- 
diene, (f) isoprene (2-methyl-l,3-butadiene). 

Problem 8.14 Predict the ozonolysis products from polybutadiene, (C 4 H<,) B : 
(a) if 1,2-addition is involved in the polymerization; (b) if 1,4-addition is involved. 

Problem 8.15 Ozonolysis of natural rubber yields chiefly (90%) the compound 

H CH 3 

O=C-CH 2 CH 2 C=O 

What does this teJl us about the structure of rubber? 



PROBLEMS 

1. (a) Draw structures of the seven isomeric alkynes of formula C 6 Hiq. (b) Give 
the IUPAC and derived name of each, (c) Indicate which ones will react with Ag+ or 
Cu(NH 3 ) 2 *. (d) Draw structures of the ozonolysis products expected from each. 

2. (a) Draw structures of all isomeric dienes of formula C 6 Hio, omitting cumulated 
dienes. (b) Name each one. (c) Indicate which ones are conjugated, (d) Indicate which 
ones can show geometric isomerism, and draw the isomeric structures, (e) Draw structures 
of the ozonolysis products expected from each, (f) Which isomers (other than cis-trans 
pairs) could not be distinguished on the basis of (e)? 

3. Write equations for all steps in the manufacture of acetylene starting from lime- 
stone and coal. 

4. Outline all steps in the synthesis of propyne from each of the following com- 
pounds, using any needed organic or inorganic reagents. Follow the other directions 
given on page 224. 

(a) 1,2-dibromopropane (e) /i-propyl alcohol 

(b) propylene (f) U-dichloropropane 

(c) isopropyl bromide (g) acetylene 

(d) propane (h) 1,1,2.2-tetrabromopropane 

5. Outline all steps in the synthesis from acetylene of each of the following com- 
pounds, using any needed organic or inorganic reagents. 

(a) ethylene (e) 1,2-dichloroethane (j) cw-2-butene 

(b) ethane (f) acetaldehyde (k) rtwM-2-butene 

(c) ethylidene bromide (g) propyne (1) 1-pentyne 
(1,1-dibromoethane) (h) 1-butyne (m) 2-pentyne 

(d) vinyl chloride (0 2-butyne (n) 3-hexyne 



280 ALKYNES AND DIENES CHAP. 8 

6. Give structures and names of the organic products expected from the reaction 
(if any) of 1 -butyne with : 

(a) 1 mole H 2 , Ni (i) product (h) + HNO 3 

(b) 2 moles H 2 , Ni (j) NaNH 2 

(c) 1 mole Br 2 (k) product (j) + C 2 H 5 Br 

(d) 2 moles Br 2 (0 product (j) -f /er/-butyi chloride 

(e) 1 mole HC1 (m) C 2 H 5 MgBr 

(f) 2 moles HC1 (n) product (m) + H 2 O 

(g) H 2 0, H+,Hg + + (o) 3 ,thenH 2 
(h) Ag+ (p) hot KMnO 4 

7. Answer Problem 6 for 1,3-butadiene instead of 1 -butyne. 

8. Answer Problem 6 for 1,4-pentadiene instead of 1 -butyne. 

9. Give structures and names of the products from dehydrohalogenation of each 
of the following halides. Where more than one product is expected, indicate which will 
be the major product. 

(a) 1-chlorobutane; 2-chlorobutane 

(b) 1-chlorobutane; 4-chloro-l-butene 

(c) 2-bromo-2-methylbutane; 3-bromo-2-methylbutane 

(d) l-bromo-2-methylbutane; 4-bromo-2-methyl butane 

(e) l-ch!oro-2,3-dimethyl butane; 2-chIoro-2,3-dimethyIbutane 

(f) 4-chloro-l-butene; 5-chloro-l-pentene 

10. Which alkyl halide of each pair in Problem 9 would you expect to undergo 
dehydrohalogenation faster? 

11. Give structures of the chief product or products expected from addition of one 
mole of HC1 to each of the following compounds: 

(a) 1,3-butadiene; 1-butene (c) 1,3-butadiene; 2-methy 1-1 ,3-butadiene 

(b) 1,3-butadiene; 1,4-pentadiene (d) 1,3-butadiene; 1,3-pentadiene 

12. Answer Problem 11 for the addition of BrCCl 3 in the presence of peroxides 
(Sec. 6.18) instead of addition of HCL 

13. Which compound of each pair in Problem 12 would you expect to be more reac- 
tive toward addition of BrCCl 3 ? 

14. (a) The heat of hydrogenation of acetylene (converted into ethane) is 75.0 
kcal/mole. Calculate A// for hydrogenation of acetylene to ethylene. (b) How does the 
stability of an alkyne relative to an alkene compare with the stability of an alkene relative 
to an alkane? (c) Solely on the basis of your answer to (b), would you expect acetylene 
to be more or less reactive than ethylene toward addition of a free methyl radical, CH 3 ? 
(d) Draw the structure of the free radical expected from addition of CH 3 - to acetylene: 
from addition of CH 3 - to ethylene. Judging only from the relative stabilities of the 
radicals being formed, would you expect CH 3 - to add to acetylene faster or slower than 
to ethylene? (e) CH 3 - has been found to add more slowly to acetylene than to ethylene. 
Which factor reactant stability or radical stability is more important here? 

15. (a) Make a model of allene, CH 2 =^C~CH 2 , a cumulated diene. What is the 
spatial relationship between the pair of hydrogens at one end of the molecule and the 
pair of hydrogens at the other end? (b) Substituted allenes of the type RCH==C=X:HR 
have been obtained in optically active form. Is this consistent with the shape of the 
molecule in (a)? Where are the chiral centers in the substituted allene? (c) Work out 
the electronic configuration of allene. (Hint: How many atoms are attached to the middle 
carbon? To each of the end carbons?) Does this lead to the same shape of molecule 
that you worked out in (a) and (b)? 



PROBLEMS 281 

16. A useful method of preparing 1-alkenes involves reaction of Grignard reagents 
with the unusually reactive halide, allyl bromide: 

RMgX + BrCH 2 CH==CH 2 > R-CH 2 CH=CH 2 

When 1-hexene (b.p. 63.5) is prepared in this way, it is contaminated with /i-hexane 
(b.p. 69) and 1,5-hexadiene (b.p. 60); these are difficult to remove because of the close- 
ness of boiling points. The mixture is treated with bromine and the product distilled. 
There are obtained three fractions: b.p. 68-69; b.p. 77-78 at 15 mm pressure; and a 
high-boiling residue. 

(a) What does each of these fractions contain? (b) What would you do next to get 
pure 1-hexene? (c) Show how this procedure could be applied to the separation of 
/f-pentane{b.p. 36) and l-pentene(b.p. 30); l-decene(b.p. 171) and 5-decyne (b.p. 175). 

17. Outline all steps in a possible laboratory synthesis of each of the following, 
using alcohols of four carbons or fewer as your only organic source, and any necessary 
inorganic reagents. (Remember: Work backwards.) 

(a) m?.w?-3,4-dibromohexane; 

(b) (2R,3R;2S,3S)-2,3-heptanediol, a racemic modification. 

18. Treatment with phosphoric acid converts 2,7-dimethyl-2,6-octadiene into I. 

H 3 C CH 3 



H 2 C CH 2 

I 
1 , 1 -Dimethyl-2-isopropenylcyclopentane 

Using reaction steps already familiar to you, suggest a mechanism for this reaction. 

19. Gutta percha is a non-elastic naturally-occurring polymer used in covering golf 
balls and underwater cables. It has the same formula, (C 5 H 8 ) n , and yields the same 
hydrogenation product and the same ozonolysis product (Problem 8.15, page 279) as 
natural rubber. Using structural formulas, show the most likely structural difference 
between gutta percha and rubber. 

20. Describe simple chemical tests that would distinguish between : 

(a) 2-pentyne and w-pentane (e) 1,3-pentadiene and 1-pentyne 

(b) 1-pentyne and 1-pentene (f) 2-hexyne and isopropyl alcohol 

(c) 1-pentyne and 2-pentyne (g) allyl bromide and 2,3-dimethyl- 

(d) 1,3-pentadiene and K-pentane 1,3-butadiene 

Tell exactly what you would do and see. 

21. Describe chemical methods (not necessarily simple tests) that would distinguish 
between: 

(a) 2-pentyne and 2rpentene (c) 1,4-pentadiene and 2-pentyne 

(b) 1 ,4-pentadiene and 2-pentene (d) 1,4-pentadiene and 1,3-pentadiene 

22. On the basis of physical properties, an unknown compound is believed to be 
one of the following: 

/i-pentane (b.p. 36) 1-pentyne (b.p. 40) 

2-pentene (b.p. 36") methylene chloride (b.p, 40) 

1-chloropropene (b.p. 37) 3,3-dimethyl-l-butene (b.p. 41) 

trimethylethylene (b.p. 39) 1,3-pentadiene (b.p. 42) 



282 ALKYNES AND DIENES CHAP. 8 

Describe how you would go about finding out which of the possibilities the unknown 
actually is. Where possible, use simple chemical tests; where necessary, use more elabor- 
ate chemical methods like quantitative hydrogenation and cleavage. Tell exactly what 
you would do and see. 

23. A hydrocarbon of formula C 6 Hi absorbs only one mole of H 2 upon catalytic 
hydrogenation. Upon ozonolysis the hydrocarbon yields 

H H 

I I 

O=C-CH 2 ~-CH 2 CH 2 -CH 2 O=O 



What is the structure of the hydrocarbon ? (Check your answer in Sec. 9.17.) 

24. A hydrocarbon was found to have a molecular weight of 80-85. A 10.02-mg 
sample took up 8.40 cc of H 2 gas measured at 0^ and 760 mm pressure. Ozonolysis 
yielded only 

H-C-H and H-C-C-H 



U 



What was the hydrocarbon ? 

25. Myrcene, C 10 H 16 , a terpene isolated from oil of bay, absorbs three moles of 
hydrogen to form C 10 H 2 2. Upon ozonolysis myrcene yields: 

CH 3 C-CH 3 H-C H H C CH 2 CH 2 -C C H 

A A i U 

(a) What structures are consistent with these facts ? 

(b) On the basis of the isoprene rule (Sec. 8.26), what is the most likely structure for 
myrcene? 

26. Dihydromyrcene, C 10 H 18 , formed from myrcene (Problem 25), absorbs two 
moles of hydrogen to form Ci H 2 2. Upon cleavage by KMnO 4 , dihydromyrcene yields: 

CH 3 -C-CH 3 CH 3 -C-OH CH 3 -C CH 2 CH 2 -C OH 

(a) Keeping in mind the isoprene rule, what is the most likely structure for dihydro- 
myrcene? (b) Is it surprising that a compound of this structure is formed by reduction 
of myrcene? 

27. At the beginning of the biogenesis of squalene (Sec. 8.26) isopentenyl pyrophos- 
phate, CH 2 =C(CH 3 )CH 2 CH 2 OPP, is enzymatically isomerized to dimethylallyl pyro- 
phosphate, (CH 3 ) 2 C=CHCH 2 OPP. These two compounds then react together to yield 
geranyl pyrophosphate, (CH 3 ) 2 C^CHCH 2 CH 2 C(CH 3 )=^CHCH 2 OPP. (a) Assuming that 
the weakly basic pyrophosphate anion is, like the protonated hydroxyl group, a good 
leaving group, 

R-OPP > Re + OPP- 

can you suggest a series of familiar steps by which geranyl pyrophosphate might be formed ? 

(b) Geranyl pyrophosphate then reacts with another molecule of isopentenyl pyro- 
phosphate to form farnesyl pyrophosphate. What is the structure of farnesyl pyro- 
phosphate? (c) What is the relationship between farnesyl pyrophosphate and squalene? 
(d) An enzyme system from the rubber plant catalyzes the conversion of isopentenyl 
pyrophosphate into rubber; dimethylallyl pyrophosphate appears to act as an initiator 
for the process. Can you suggest a "mechanism** for the formation of natural rubber? 



Chapter 

Alicyclid Hydrocarbons 



9.1 Open-cbflin and cyclic compounds 

In the compounds that we have studied in previous chapters, the carbon 
atoms are attached to one another to form chains; these are called open-chain 
compounds, In many compounds, however, the carbon atoms are arranged to 
form rings; these are called cyclic compounds. 

In this chapter we shall take up the alicyclic hydrocarbons (aliphatic cyclic 
hydrocarbons). Much of the chemistry of cycloalkanes and cycloalkenes we al- 
ready know, since it is essentially the chemistry of open-chain alkanes and alkenes. 
But the cyclic nature of some of these compounds confers very special properties 
on them. It is because of these special properties that, during the past fifteen 
years, alicyclic chemistry has become what Professor Lloyd Ferguson, of the 
California State College at Los Angeles, has called "the playground for organic 
chemists." It is on some of these special properties that we shall focus our attention. 

9.2 Nomenclature 

Cyclic aliphatic hydrocarbons are named by prefixing cycle- to the name of 
the corresponding open-chain hydrocarbon having the same number of carbon 
atoms as the ring. For example: 

HjC CH 2 




I ^ / 

/*! u /"*.____/" VI 

i^tij 112^ x^fi2 




Cydobutane Cyclopcnlcnc 

Substituents on the ring are named, and their positions are indicated by numbers, 

203 



282 



ALICYCLIC 



Table 9.1 CYCLIC ALIPHATIC HVbftoCARBONR 



Name 


M.p., 
C 


B.p., 
C 


penaity 
(at 20C) 


Cyclopropane 


-127 


- 33 




Cyclotnitane 


- 80 


13 




Cyclopentane 


_ 94 


49 


0.746 


Cyclohexane 


6.5 


81 


.778 


Cycloheptane 


- 12 


118 


.810 


Cyclooctane 


14 


149 


,830 


Methylcyclopentane 


-142 


72 


.749 


m~l ,2-Dimethyfcyclopentane 


- 62 


99 


.772. 


transA ,2-Dimethylcyclopentane 


-120 


92 


.750 


Methylcyclohexane 


-126 


100 


.769 


Cyclopentene 


- 93 


46 


.774 


1 ,3-Cyclopentadiene 


- 85 


42 


.798 


Cyclohexene 


-104 


83 


.810 


1,3-Cyclohexadiene 


- 98 


80.5 


.840 


1,4-Cyelohexadiene 


- 49 


7 


.847 



the lowest combination of numbers being used, fti simple cycloalkenes and 
cycloalkynes the doubly- and triply-bonded carbons are considered to occupy 
positions 1 and 2. For example: 



H 




ChlorocycJopropane 

r 

,CH 2 

.^ 
CH 2 

,3-Dimcthylcyclohexane 



Eihykryctopemcne 




1,3-Cyclohcxadicne 



For convenience, aliphatic rings are often represented by simple geometric 
figures: a triangle for cyclopropane, a square for cyclobutane, a pentagon for 
cyclopentane, a hexagon for cyclohexane, and so on. ft is understood that two 
hydrogens are located at each corner of the figure unless some other group is 
indicated. For example: 



o 



Cyclopentane 







Cyclohexane 




3-Ethykyclopentenc 
CHj 



1,3-Cyclopcntadicne 




-CH 3 

1 ,3-Dimcthylcyclohexane 



1,3-Cyclohcxadiene 



SEC. 9.2 



NOMENCLATURE 



285 



Polycyclic compounds contain two or more rings that share two or more 
carbon atoms. We can illustrate the naming system with norbornane, whose sys- 
tematic name is bicyclo[2.2.1]heptane: (a) heptane, since it contains a total of seven 
carbon atoms; (b) bicydo, since it contains two rings, that is, breaking two carbon- 






Bicyclo[2.2.1]heptane 
Norbornane 



Bicyclo[2.2.2]octa-2-ene Tricycto[2.2.1.0 2 ' 6 ]heptane 

Nortricyclene ' 



carbon bonds converts it into an open-chain compound; (c) [2.2.1], since the 
number of carbons between bridgeheads (shared carbons) is two (C-2 and C-3), 
two (C-5 and C-6), and one (C-7). 

Polycyclic compounds in a variety of strange and wonderful shapes have 
been made, and their properties have revealed unexpected facets of organic chemis- 
try. Underlying much of this research there has always been the challenge; can 
such a compound be madel 






Cubanc 



Basketane 



Adamantane 



The ultimate polycyclk; aliphatic system is diamond which is, of course, not a hydro- 
carbon at all, but one of the allotropic forms of elemental carbon. In diamond each 




Diamond 

carbon atom is attached to four others by tetrahedral bonds of the usual single bond length, 
1.54 A. (Note the cyclohcxane chairs, Sec. 9.11.) 



286 ALICYCLIC HYDROCARBONS CHAP. 9 



9.3 Industrial source 

We have already mentioned (Sec. 3.13) that petroleum from certain areas, 
(in particular California) is rich in cycloalkanes, known to the petroleum industry 
as naphthenes. Among these are cyclohexane, methylcyclohexane, methylcyclo- 
pentane, and 1,2-dimethylcyclopentane. 

These cycloalkanes are converted by catalytic reforming into aromatic hydro- 
carbons, and thus provide one of the major sources of these important compounds 
(Sec. 12.4). For example: 



Mo,O y Al,0 >t W 

300ib/in.' * ^65^H3 + JH 2 Dehydrogenatioo 

CH 2 Toluene 

Methykyclohexane Aromatic 

Aliphatic 

Just as elimination of hydrogen from cyclic aliphatic compounds yields aro- 
matic compounds, so addition of hydrogen to aromatic compounds yields cyclic 
aliphatic compounds, specifically cyclohexane derivatives. An important example 
of this is the hydrogenation of benzene to yield pure cyclohexane. 



HjC^ 

f^ t i in Ni 1 50 25u 7 i ~ 

C 6 H 6 -r 3H 2 2Satm . > i I Hydrogenation 

Benzene H 2 C\ /^-^2 

Aromatic CH 2 

Cyclohexane 
Aliphatic 

As we might expect, hydrogenation of substituted benzenes yields substituted 
cyclohexanes. For example: 



C 6 H 5 OH 4- 3H 2 

Phenol 
Aromatic 

Cyclohcxanol 
Aliphatic 

From cyclohcxanol many other cyclic compounds containing a six-membered 

ring can be made. 



9.4 Preparation 

Preparation of alicyclic hydrocarbons from other aliphatic compounds 
generally involves two stages: (a) conversion of some open-chain compound or 



SEC. 9.5 REACTIONS 287 

compounds into a compound that contains a ring, a process called cyclizatipn; 
(b) conversion of the cyclic compound thus obtained into the kind of compound 
that we want: for example, conversion of a cyclic alcohol into a cyclic alkene* or 
of a cyclic alkene into a cyclic alkane^ 

Very often, cyclic compounds are made by the adapting of a standard method 
of preparation to the job of closing a ring. For example, we have seen (Sec. 3.17) 
that the alkyl groups of two alkyl halides can be coupled together through con- 
version of one halide into an organometallic compound (a lithium dialkylcopper) : 



CH 3 CH 2 -C1 - CH 3 CH 2 -M - , CH 3 CH 2 

CH 3 CH 2 -C1 - ' CH 3 CH 2 
Ethyl chloride w-Butane 

2 moles 

The same method applied to a tf/halide can bring about coupling between two 
alkyl groups that are part of the same molecule: 



a CH 2 " ZnX X 

H 2 -&JSU H 2 < H 2 < 

CH 2 -C1 CH 2 -X CH 2 

1 ,3-Dichloropropane Cyclopropane 

In this case zinc happens to do a good job. Although this particular method works 
well only for the preparation of cyclopropane, it illustrates an important prin- 
ciple : the carrying out of what is normally an intermolecular (between-molecules) 
reaction under such circumstances that it becomes an intramolecular (within-a* 
molecule) reaction. As we can see, it involves tying together the ends of a di- 
functional molecule. 

Alicyclic hydrocarbons are prepared from other cyclic compounds (e.g., 
halides or alcohols) by exactly the same methods that are used for preparing 
open-chain hydrocarbons from other open-chain compounds. 

Problem 9.1 Starting with cyclohexanol (Sec. 9.3), how would you prepare: 
(a) cyclohexene, (b) 3-bromocyclohexene, (c) 1,3-cyclohexadiene? 

Problem 9.2 Bromocyclobutane can be obtained from open-chain compounds. 
How would you prepare cyclobutane from it? 

The most important route to rings of many different sizes is through the im- 
portant class of reactions called cycloadditions: reactions in which molecules are 
added together to form rings. We shall see one example of cycloaddition in Sees. 
9.15-9.16, and others later on. 



9.5 Reactions 

With certain very important and interesting exceptions, alicyclic hydrocarbons 
undergo the same reactions as their open-chain analogs. 



2*8 



ALICYCLIC HYDROCARBONS 



CHAP. 9 



Cycloalkanes undergo chiefly free-radical substitution (compare Sec. 3.19). 
For example: 



:HCI + HCI 




HjC 

Cyclopropane 



H 2 



Chlorocyclopropanc 




Br 2 



300V 



H 2 C CH 2 

Cyclopcntane 



+ 

- CH 2 

Rromocyclopentane 



HB, 



Cycloalkenes undergo chiefly addition reactions, both electrophilic and free 
radical (compare Sec. 6.2); like other alkenes, they can also undergo cleavage and 
allylic substitution. For example: 



Br 2 



HjC 
HzC 



CHBr 



Cyclohexcnc 

r, ^/ C< ^ 



C 2 

1 ,2-Dibromocyclohexane 



CH3 



+ HI 
H^ CH 2 

1 -Methylcyclopentene 




HCH 3 

3,5-Dimethylcyclopentene 



H 2 

1 -lodo- 1 -methylcyclopcntane 



? C3 
O==CCH CH 2 CH C=O 



A dialdehyde 



The two smallest cycloalkanes, cyclopropane and cyclobutane, show certain 
chemical properties that are entirely different from those of the other members of 
their family. Some of these exceptional properties fit into a pattern and, as we 
shall see, can be understood in a general way. 

The chemistry of bicyclic compounds is even more remarkable, and is right now 
one of the most intensively studied areas of organic chemistry (Sec. 28.13). 



9.6 Reactions of small-ring compounds. Cyclopropane and cyclobutane 

Besides the free-radical substitution reactions that are characteristic of cyclo- 
alkanes and of alkanes in general, cyclopropane and cyclobutane undergo certain 



Sfifc *.? flAEYER STRAIN THEORY 1W 

addition reactions. These addition reactions destroy the cycldprofttrie and ctyplp* 
butane HHg systems, and yield open-chain products. For example: 

Ni.H 2 .My^ 




H 

Cycjppropanc 



H H 

Propane 

.* **!'. CH 2 CH 2 CH a 

Cl Cl 

] ,3-DichloropropQfte 

^ onc ' HiS S CHaCH^CHz 



li 6H 

n-rtopyl ateohoi 

th each of these reactions a carbon-carbon bond is broken, and the two atoms of 
the reagent appear at the ends of the propane chain: 




In general, cyclopropane undergoes addition less readily than propylene: 
chlorination, for example, requires a Lewis acid catalyst to polarize the chlorine 
molecule (compare Sec. 1 1.1 1). Yet the reaction with sulfuric acid and other aqueous 
protic acids takes place considerably faster for cyclopropane than for propylene. 
(Odder still, treatment with bromine and FeBr 3 yields a grand mixture of bromo- 
propanes.) 

Cyclobutane does not undergo most of the ring-opening reactions of cycle- 
propane; it is hydrogenated, but only under mote vigorous conditions than those 
required for cyclopropane, thus cyclobutape undergoes addition less readily than 
cyclopropane and, with some exceptions, cyclopropane less readily than an 
alkene. The remarkable thing is tfiat these pycloalkanes undergo addition at all. 

0i7 Baeyer strain theory 

In 1885 Adolf von Baeyer (of the University of Munich) proposed a theory to 
account for certain aspects of the chemistry of cyclic compounds. The part of 
his theory dealing with the ring-opening tendencies of cyclopropane and cyclo- 
butane is generally accepted today, although it is dressed in more modern language. 
Other parts of his theory have been shown to be based on false assumptions, 
and have been discarded. 

Baeyer's argument was essentially the following. In general, when carbon 
is bonded to four other atoms, the angle between any pair of bonds is the tetra- 
hedral angle 109.5. But the ring of cyclopropane is a triangle with three angles 
of 60, and the ring of cyclobutane is a square with four angles of 90. In cyclo- 
propane or cyclobutane, therefore, one pair of bonds to each carbon cannot 
assume the tetrahedral angle, but must be compressed to 60 or 90 to fit the 
geometry of the ring. 



290 ALICYCLIC HYDROCARBONS CHAP. 9 

These deviations of bond angles from the "normal" tetrahedral value cause 
the molecules to be strained, and hence to be unstable compared with molecules 
in which the bond angles are tetrahedral. Cyclopropane and cyclobutane undergo 
ring-opening reactions since these relieve the strain and yield the more stable open- 
chain compounds. Because the deviation of the bond angles in cyclopropane 
(109.5 - 60 = 49.5) is greater than in cyclobutane (109.5 - 90 19.5), 
cyclopropane is more highly strained, more unstable, and more prone to undergo 
ring-opening reactions than is cyclobutane. 

The angles of a regular pentagon (108) are very close to the tetrahedral angle 
(109.5), and hence cyclopentane should be virtually free of angle strain. The 
angles of a regular hexagon (120) are somewhat larger than the tetrahedrai angle, 
aftd hence, Baeyer proposed (incorrectly), there should be a certain amaunt of 
strain in cyclohexane. Further, he suggested (incorrectly) that as one proceed 4 
to cycloheptane, cyclooctane, etc., the deviation of the bond angles from 109.5 
would become progressively larger, and the molecules would become progressively 
more strained. 

Thus Baeyer considered that tings smaller or larger than cyclopentane or 
cyclohexane were unstable; it was because of this instability that the three- and 
fouNmembered rings underwent ring-opening reactions; it was because of this 
instability that great difficulty had been encountered in the synthesis of the larger 
How does Baeyer's strain theory agree with the facts? 



9$ Heals of combustion and relative stabilities of the cycloalkaites 

We recall (Sec. 2.6) that the heat of combustion is the quantity of heat evolved 
when one mole of a compound is burned to carbon dioxide and water. Lik$ 
heats of hydrogenation (Sees. 6.4 and 8.16), heats of combustion can often fur- 
nish valuable information about the relative stabilities of organic compounds. 
Let us see if the heats of combustion of the various cycloalkanes support Baeyer's 
proposal that rings smaller or larger than cyclopentane and cyplohexane are un- 
stable. 

Examination of the data for a great many compounds has shown that the heat 
of combustion of an aliphatic hydrocarbon agrees rather closely with that calcu- 
lated by assuming a certain characteristic contribution from each structural unit. 
For open-chain alkanes each methylene group, CH 2 , contributes very close to 
157.4 kcal/mole to the heat of combustion. Table 9.2 lists the heats of combus- 
tion tjiat have been measured for some of the cycloalkanes. 

Table 9.2 HEATS dt COMBUSTION OF CYCLOALKANES 



Ring Heat of combustion 
size per CH 2 , kcal/mole 


Ring Heat of combustion 
size per CH2, kcal/mole 


3 


166.6 


10 


158.6 


4 


164.0 


11 


158.4 


5 


158.7 


12 


157.6 


6 


157.4 


13 


157.8 


1 


158.3 


14 


157;4 


8 


158.6 


15 


157*5 


9 


158.8 


17 * 


157.2 




Open-chain 


157.4 





SEC 9.8 HEATS OF COMBUSTION 291 

We notice that for cyclopropane the heat of combustion per CH 2 group 
is 9 kcal higher than the open-chain value of 157.4; for cyclobutane it is 7 kcal 
higher than the open-chain value. Whatever the compound in which it occurs, a 
CH 2 group yields the same products on combustion: carbon dioxide and water. 

~CH 2 - + |O 2 * CO 2 + H 2 O -I- heat 

If cyclopropane and cyclobutarie evolve more energy per CH 2 group than an 
open-chain compound, it can mean only that they contain more energy per ~rCH 2 ~ 
group. In agreement with the Baeyer angle-strain theory, then, cyclopropane and 
cyclobutane are less stable than open-chain compounds; it is reasonable to suppose 
that their tendency to undergo ring-opening reactions is related to this instability. 

According to Baeyer, rings larger than cyclopentane and cyclohexane also 
should be unstable, and hence also should have high heats of combustion; further- 
more relative instability and, with it, heat of combustion should increase steadily 
with ring size. However, we see from Table 9.2 that almost exactly the opposite 
is true. For none of the rings larger than four carbons does the heat of combustion 
per CH 2 ~ deviate much from the open-chain value of 157.4. Indeed, one of the 
biggest deviations is for Baeyer's "most stable" compound, cyclopentane: 1.3 
kcal per CH 2 , or 6.5 kcal for the molecule. Rings containing seven to eleven 
carbons have about the same value as cyclopentane, and when we reach rings of 
twelve carbons or more, heats of combustion are indistinguishable from the open- 
chain Values. Contrary to Baeyer's theory, then, none of these rings is appreciably 
less stable than open-chain compounds, and the larger ones are completely free of 
strain. Furthermore, once they have been synthesized, these large-ring cyclo- 
alkanes show little tendency to undergo the ring-opening reactions characteristic 
of cyclopropane and cyclobutane. 

What is wrong with Baeyer's theory that it does not apply to rings larger than 
four members? Simply this: the angles that Baeyer used for each ring were based 
on the assumption that the rings were ./to/. For example, the angles of a regular 
(flat) hexagon are 120, the angles for a regular decagpn are 144. But the cyclo- 
hexane ring is not a regular hexagon, and the cyclodecane ring is not a regular 
decagon. These rings are not Hat, but are puckered (see Fig. 9.1) so that each bond 
angle of carbon can be 109.5. 





(a) 
Figure 9.1. Puckeitd rings* (a) Qttlohexane* (b) Cyclodecane. 

* A three-membered ring niust be j>l*ttat, since three points (the three carbon 
nuclei) define a plane. A fQwrmcmfreied ring need not be planar, but puckeriqg 



192 AMCYCUC imWOCAFBONS CHAP* 9 

here would increase (angle) strain- A five-membered ring need not be planar, but 
in this case a planar arrangement would permit the bond angles to have nearly the 
tetrahedral value, All rings larger than this are puckered. (Actually, as we shall 
see, cyclobutane and cyclopentane are puckered, too, but this is in spite ^/increased 
angle strain.) 

Iflarge rings are stable, why are they difficult to synthesize? Here we encoun- 
ter Baeyer's second false assumption. The fact that a compound is difficult to 
synthesize does not necessarily mean that it is unstable. The closing of a ring 
requires that two ends of a chain be brought close enough to each other for a bond 
to form. The larger the ring one wishes to synthesize, the longer must be the chain 
from which it is made, and the less is the likelihood of the two ends of the chain 
approaching each other. Ufider these conditions the end of one chain is more 
likely to encounter the ewj pf Q different chain, add thus yield an entirely different 
product (see Fig, 9*2). 




CH 2 Y CH 2 Y CH 2 Y CH 2 ~CH 2 

Figure 94. Ring closure (upper) vs. chain lengthening (lower). 



the methods that are used successfully to make large rings take this fact into 
consideration. Reactions are carried out in highly dilute solutions where collisions 
between two different chains are unlikely; under these conditions the ring-closing 
reaction, although slow, is the principal one. Five- and six-membered rings are 
the kind most commonly encountered in organic chemistry because they are large 
enough to be free of angle strain, Qnd small enough that ring closure is likely. 



9.9 Orbital picture of angle strain 

What is the meaning of Baeyer's angle strain in terms of the modern picture 
of tha coVdlent bond ? 

We have seen (Sec. 1.8) that, for a bond to form* two atoms must be located 
so that an orbital of one overlaps an orbital of the other. For a given pair of 
atoms, the greater the overlap of atomic orbitals, the stronger the bond. When 
carbon is bonded to four other atoms, its bonding orbitals (sp* orbitals) are 
directed to the corners of a tetrahedron; the angle between any pair of orbitals 
is thus 109.5. Formation of a bond with another carbon atom involves over- 
lap of one of these sp* orbitals with a similar sp* orbital of the other carbon atom. 



SEC 9.9 ORBITAL PICTURE OF ANGLE STRAIN 293 

% 

This overlap is most effective, and hence the bond is strongest, when the two 
atoms are located so that an sp* orbital of each atom points toward the other 
atom. This means that when carbon is bonded to two other carbon atoms the 
C-C-C bond angle should be 109.5. 

In cyclopropane, however, the CC C bond angle cannot be 109.5, but 
instead must be 60. As a result, the carbon atoms cannot be located to permit 





( 



Figure 9.3. Angle strain, (a) Maximum overlap permitted for open- 
chain or large-ring compounds. (6) Poor overlap for cyclopropane ring. 
Bent bonds have much p character. 



their sp 3 orbitals to point toward each other (see Fig. 9.3). There is less over- 
lap and the bond is weaker than the usual carbon-carbon bond. 

The decrease in stability of a cyclic compound attributed to angle strain 
is due to poor overlap of atomic orbitals in the formation of the carbon-carbon 
bonds. 

On the basis of quantum mechanical calculations, C. A. Coulson and W, A. 
Moffitt (of Oxford University) proposed bent bonds between carbon atoms of cyclo- 
propane rings; this idea is supported by electron density maps based on X-ray 
studies. Carbon uses sp 2 orbitals for carbon-hydrogen bonds (which are short 
and strong), and orbitals with much p character (sp 4 to sp 5 ) for the carbon-carbon 
bonds. The high p character of these carbon-carbon bonds, and their location 
largely outside the ring seems to underlie much of the unusual chemistry of these 
rings. The carbon-carbon bond orbitals can overlap orbitals on adjacent atoms; 
the resulting delocalization is responsible for the effects of cyclopropyl as a substi- 
tuent. The carbon-carbon bond orbitals provide a site for the attack by acids 
that is the first step of ring-opening. (Indeed, "edge-protonated" cyclopropanes 
seem to be key intermediates in many reactions that do not, on the surface, seem 
to involve cyclopropane rings.) 

Ring-opening is due to the weakness of the carbon-carbon bonds, but the 
way in which it happens reflects the unusual nature of the bonds; all this steins 
utimately from the geometry of the rings and angle strain. 



294 , ALICYCUC HYDROCARBONS CHAP. 9 

9 JO Factors affecting stability of conformations 

To go more deeply into the chemistry of cyclic compounds, we must use 
conformational analysis (Sec. 4.20). As preparation for that, let us review the 
factors that determine the stability of a conformation. 

Any atom tends to have bond angles that match those of its bonding orbitals: 
tetrahedral (109.5) for j/7 3 -hybridized carbon, for example. Any deviations from 
the "normal" bond angles are accompanied by angle strain (Sees. 9.8-9.9). 

Any pair of tetrahedral carbons attached to each other tend to have their bonds 
staggered. That is to say, any ethane-like portion of a molecule tends, like* ethane, 
to take up a staggered conformation. Any deviations from the staggered arrange- 
ment are accompanied by torsional strain (Sec. 3.3). 

Any two atoms (or groups) that are not bonded to each other can interact in 
several ways, depending on their size and polarity, and how closely they-are brought 
together. These non-bonded interactions can be either repulsive or attractive, 
and the result can be either destabilization or stabilization of the conformation. 

Non-bonded atoms (or groups) that just touch each other that is, that are 
about as far apart as the sum of their van der Waals radii attract each other. If 
brought any closer together, they repel each other: such crowding together is 
accompanied by van der Waals strain (steric strain) (Sees. 1.19, 3.5). 

Non-bonded atoms (or groups) tend to take, positions that result in the most 
favorable dipole-dipole interactions: that is, positions that minimize dipole-dipole 
repulsions or maximize dipole-dipole attractions. (A particularly powerful attrac- 
tion results from the special kind of dipole-dipole interaction called the hydrogen 
bond (Sec. 1.19). 

All these factors, working together or opposing each other, determine the 
net stability of a conformation. To figure out what the most stable conformation 
of a particular molecule should be, one ideally should consider all possible com- 
binations of bond angles, angles of rotation, and even bond lengths, and see which 
combination results in the lowest energy content. A start in this direction 
feasible only by use of computers has been made, most notably by Professor 
James F. Hendrickson (of Brandeis University). 

Both calculations and experimental measurements show that the final result is 
a compromise, and that few molecules have the idealized conformations that we 
assign them and, for convenience, usually work with. For example, probably no 
tetravalent carbon compound except one with four identical substituents has 
exactly tetrahedral bond angles: a molecule accepts a certain amount of angle 
strain to relieve van der Waals strain or dipole-dipole interaction. In the gauche 
conformer of w-butane (Sec. 3.5), the dihedral angle between the methyl groups is 
not 60, but almost certainly larger: the molecule accepts some torsional strain 
to ease van der Waals strain between the methyl groups. 



9.11 Conformations of cycloalkanes 

Let us look more closely at the matter of puckered rings, starting with cyclo- 
hexane, the most important of the cycloalkanes. Let us make a model of the mcijp. 
cule, and examine the conformations that are free of angle strain. 



SEC. 9.11 



CONFORMATIONS OF CYCLOALKANES 



295 










Chair conformation 



Twist-boat conformation 



Boat conformation 
An energy maximum 

Figure 9.4. Conformations of cyclohexane that are free of angle strain. 

First, there is the chair form (Fig. 9.4). If we sight along each of the carbon- 
carbon bonds in turn, we see in every case perfectly staggered bonds: 





Chair 
cyclohexane 



Staggered 
ethane 



The conformation is thus not only free of angle strain but free of torsional strain as 
well. It lies at an energy minimum, and is therefore a conformational isomer. 
The chair form is the most stable conformation of cyclohexane, and, indeed, of 
nearly every derivative of ryclohexane. 

Next, let us flip the "left" end of the molecule up (Fig. 9.4) to make the boat 
conformation. (Like all the transformations we shall carry out in this section, 
this involves only rotations about single bonds; what we are making are indeed 
conformations.) This is not a very happy arrangement. Sighting along either of 
two carbon-carbon bonds, we see sets of exactly eclipsed bonds, 

Flagpole 
bonds 





296 



ALICYCLIC HYDROCARBONS 



CHAP. 9 



and hence we expect considerable torsional strain: as much as in two ethane mole- 
cules. In addition, there is van der Waals strain due to crowding between the 
"flagpole" hydrogens, which lie only 1.83 A apart, considerably closer than the 
sum of their van der Waals radii (2.5 A). The boat conformation is a good deal 
less stable (6.9 kcal/mole, it has been calculated) than the chair conformation. It 
is believed to lie, not at an energy minimum, but at an energy maximum; it is thus 
not a con former, but a transition state between two conformers. 

Now, what are these two conformers that lie energetically speaking on 
either side of the boat conformation ? To see what they are, let us hold a model of 
the boat conformation with the flagpole hydrogens (// and H b ) pointing up, and 
look down through the ring. We grasp C-2 and C-3 in the right hand and C-5 





Move together 
Boat 



Twist-boat 



Cyclohexane 



did O6 in the left hand, and twist the molecule so that, say, C-3 and C-6 go 
down, and C-2 and C-5 come up. As we do this, H a and H 6 move diagonally 
apart, and we see (below the ring) a pair of hydrogens, H c and H d (on C-3 and 
C-6, respectively), begin to approach each other. (If this motion is continued, we 
make a new boat conformation with H c and H d becoming the flagpole hydrogens.) 
When the H a H b distance is equal to the H c H d distance, we stop and examine 
the molecule. We have minimized the flagpole-flagpole interactions, and at the 
same time have partly relieved the torsi onal strain at the C 2 C 3 and C 5 C 6 
bonds. 

Flagpole 
bonds 





Boat 
cyclohexanc 



Twist-boat 
cyclohcxane 



This new configuration is the twist-boat form. It is a conformer, lying at 
energy minimum 5.5 kcal above the chair conformation. The twist-boat conformed* 
is separated from another, enantiomeric twist-boat conformer by an energy barrier 
1.6 kcal high, at the top of which is the boat conformation. 



SEC. 9.11 



UNFUKMATIOKIS J1F CYCLOALKXNIH 



297 



Between the chair form and the twi*t*boat forth llfes the highest barrier of all; 
a transition state conformation (the half-chair) which, with angle strain and tor* 
siopal strain, lies about 1 1 kcal above the chair form. 

The overall relationships are summarized In Fig. 9.5. Equilibrium exists 
between the chair and twist-boat forms, with the more stable chair form being 
favored 10,000 to 1 at room temperature. 




Conform* 



Conformer 



Conformer 



Figure 9.5. Potential energy relationships among conformations of 
cyclohexane. 

If chair cyclohexane is, conformationally speaking, the perfect specimen of a 
cycloalkane, planar cyclopcntane (Fig. 9.6) must certainly be the poorest: there is 



298 ALICYCLIC HYDROCARBONS CHAP. 9 

exact bond eclipsing between every pair of carbons. To (partially) relieve this 
torsional strain, cyclopentane takes on a slightly puckered conformation, even at 
the cost of a little angle strain. (See also Problem 10; p. 316.) 





Figure 9.6. Planar cyclopentane: much torsional strain. Molecule actu- 
ally puckered. 

Evidence of many kinds strongly indicates that cyclobutane is not planar, 
but rapidly changes between equivalent, slightly folded conformations (Fig. 9.7). 
Here, too, torsional strain is partially relieved at the cost of a little angle strain. 





Figure 9.7. Cyclobutane: rapid transformation between equivalent non- 
planar "folded" conformations. 

Rings containing seven to twelve carbon atoms are also subject to torsional 
strain, and hence these compounds, too, are less stable than cyclohexane; scale 
models also reveal serious crowding of hydrogens inside these rings. Only quite 
large ring systems seem to be as stable as cyclohexane. 

9.12 Equatorial and axial bonds in cyclohexane 

Let us return to the model of the chair conformation of cyclohexane (see Fig. 
9.8). Although the cyclohexane ring is not flat, we can consider that the carbon 
atoms lie roughly in a plane. If we look at the molecule in this way, we see that 
the hydrogen atoms occupy two kinds of position: six hydrogens lie in the plane, 





Equatorial bonds Axial bonds 

Figure 9.8. Equatorial and axia! bonds in cyclohexane. 



SEC. 9.12 



EQUATORIAL AND AXIAL BONDS IN CYCLOHEXANE 



299 



while six hydrogens lie above or below the plane. The bonds holding the hydro- 
gens that are in the plane of the ring lie in a belt about the "equator" of the ring, 
and are called equatorial bonds. The bonds holding the hydrogen atoms that are 
above and below the plane are pointed along an axis perpendicular to the plane 
and are called axial bonds. In the chair conformation each carbon atom has one 
equatorial bond and one axial bond. 

Cyclohexane itself, in which only hydrogens are attached to the carbon atoms, 
is not only free of angle strain and torsional strain, but free of van der Waals 
strain as well. Hydrogens on adjacent carbons are the same distance apart (2.3 A) 
as in (staggered) ethane and, if anything, feel mild van der Waals attraction for 
each other. We notice that the three axial hydrogens on the same side of the mole- 
cule are thrown rather closely together, despite the fact that they are attached to 
alternate carbon atoms; as it happens, however, they are the same favorable 
distance apart (2.3 A) as the other hydrogens are. 

If, now, a hydrogen is replaced by a larger atom or group, crowding occurs. 
The most severe crowding is among atoms held by the three axial bonds on the 
same side of the molecule; the resulting interaction is called 1, 3-d i axial inter- 
action. Except for hydrogen, a given atom or group has more room in an equatorial 
position than in an axial position. 

As a simple example of the importance of 1,3-diaxial interactions, let us con- 
sider methylcyclohexane. In estimating relative stabilities of various conformations 
of this compound, we must focus our attention on methyl, since it is the largest 
substituent on the ring and hence the one most subject to crowding. There are two 





Equatorial CH 3 Axial CH 3 

Figure 9.9. Chair conformations of methylcyclohexane. 

possible chair conformations (see Fig. 9.9), one with CH 3 in an equatorial posi- 
tion, the other with CH 3 in an axial position. As shown in Fig. 9.10, the two 
axial hydrogens (on C-3 and C-5) approach the axial CH 3 (on C-l) more 
closely than any hydrogens approach the equatorial CH 3 . We would expect 




CH 3 




Axial CH 3 



Figure 9.10. 1,3-Diaxial interaction in methylcyclohexane. Axial CHj 
more crowded than equatorial CH 3 . 



300 



ALICYCLIC HYDROCARBONS 



CHAP. 9 



the equatorial conformation to be the more stable, and it is, by about 1.8 kcal. 
Most molecules (about 95/ at room temperature) exist in the conformation with 
methyl in the uncrowded equatorial position. 

In an equatorial position, we see, CH 3 points away from its nearest neighbors: 
the two hydrogens one axial, and one equatorial on the adjacent carbons. This is 
not true of CH 3 in an axial position, since it is held by a bond that is parallel to the 
bonds holding its nearest neighbors: the two axial hydrogens. 

Conformational analysis can account not only for the fact that one confor- 
mation is more stable than another, but often with a fair degree of accuracy 
for just how much more stable it is. We have attributed the 1.8-kcal energy dif- 
ference between the two conformations of methylcyclohexane to 1,3-diaxial inter- 
actions between a methyl group and two hydrogens. If, on that basis, we assign 
a value of 0.9 kcal/mole to each 1,3-diaxial methyl-hydrogen interaction, we shall 
find that we can account amazingly well for the energy differences between con- 
formations of a variety of cyclohexanes containing more than one methyl group. 

We notice that 0.9 kcal is nearly the same value that we earlier (Sec. 3.5) assigned to a 
gauche interaction in /z-butane; examination of models shows that this is not just acci- 
dental. 

Let us make a model of the conformation of methylcyclohexane with axial methyl. 
If we hold it so that we can sight along the Q 2 bond, we see something like this, 
represented by a Newman projection: 






Axial CH 3 



CH 3 

Gauche 

//-Butane 



The methyl group and C-3 of the ring have the same relative locations as the two methyl 
groups in the gauche conformation of /;-butane (Sec. 3.5). If we now sight along the 
(?! C 6 bond, v\e see a similar arrangement but with C-5 taking the place of C-3. 

Next, let us make a model of the conformation with equatorial methyl. This time, 
if we sight along the Cj C2 bond, we see this: 





Tquatonal ("H, 



Ann 
n- Butane 



Here, methyl and C-3 of the ring have the same relative locations as the two methyl 
groups in the anti conformation of //-butane. And if we sight along the C r C 6 bond, we 
see methyl and C-5 in the anti relationship. 



SEC. 9.13 STEREOISOMERISM OF CYCLIC COMPOUNDS 301 



Thus, for each 1,3-diaxial methyl- hydrogen interaction there is a " 
intei action between the methyl group and a carbon atom 'of the ring. Of the two ap- 
proaches, however, looking for 1,3-diaxial interactions is much the easier and has the 
advantage, when we study substituents other than methyl, of focusing our attention on the 
sizes of the groups being crowded together. 

In general, then, it has been found that (a) chair conformations are more 
stable than twist conformations, and (b) the most stable chair conformations 
are those in which the largest groups are in equatorial positions. There are excep- 
tions to both these generalizations (which we shall encounter later in problems), 
but the exceptions are understandable ones. 

Problem 9.3 For other alkylcyclohexane^ the difference in energy between 
equatorial and axial conformations has been found to be: ethyl, 1.9 kcal/mole; iso- 
propyl, 2.1 kcal/mole; and /m-butyl, more than 5 kcal/mole. Using models, can you 
account for the big increase at /m-butyl? (Hint: Don't forget freedom of rotation 
about all the single bonds.) 



9.13 Stereoisomerism of cyclic compounds: cis- and 

Let us turn for the moment from conformational analysis, and look at con- 
figurational isomerism in cyclic compounds. 

We shall begin with the glycol of cyclopentene, 1,2-cyclopentanediol. Using 
models, we find that we can anange the atoms of this molecule as in I, in which 
both hydroxyls lie below (or above) the plane of the ring, and as in II, in which 
one hydroxyl lies above and the other lies belpw the plane of the ring. 





OH OH H OH 

l II 

m-1 ,2-Cyclopcntanediol trans- \ ,2-Cvclopentanediol 

I and II cannot be superimposed, and hence are isomers. They differ only in 
the way their atoms are oriented in space, and hence are stereoisomers. No amount 
of rotation about bonds can interconvert 1 and II, and hence they are not conforma- 
tional isomers. They are configurational isomers; they are interconverted only by 
breaking of bonds, and hence arc isolable. They are not mirror images, and hence 
are diastcreomers; they should, therefore, have different physical properties, as the 
two glycols actually have. Configuration I is designated the ra-configuration, and 
H is designated the /raw-configuration. (Compare cis- and /ra/75-alkenes. Sec. 5.6.) 

Problem 9.4 You have two bottles labeled " 1.2-Cyclopentanediol," one con- 
taining a compound of m.p. 30 , the other a compound of m.p. 55 ; both compounds 
are optically inactive. Ho\\ could you decide, beyond any doubt, which bottle should 
he labeled "w" and \\hich * 



Problem 9.5 (a) Starting from cyclopcntanol, outline a synthesis of stcreo- 
chemically pure r/.v-!,2-eyclopentancdioi. (b) Of stcreochemicully pure mww-1.2-cyclo- 
nentunediol. 



302 



ALICYCL1C HYDROCARBONS 



CHAP. 9 



Stereoisomerism of this same sort should be possible for compounds other 
than glycols, and for rings other than cyclopentane. Some examples of isomers 
that have been isolated are: 





COOH 

m-l,3-( \clopentdnedicarbox} lie 
acid 



COOH 




H Br 

/tt//M-I,2-Dibromoc\clopcntanc* 



COOH 



//<///s-lJ-( \clopentunedicarbo\y lie 
acid 



HOOC 




COOH 




r/s-l,3-C\clobutaned]carbox\lie 
acid 



//.//;\-l,3-C>clobulanedicarbo\>lic 
acid 




H 



CH 3 

c i\-\ ,2-I)imctln lc\clopropanc 




CH 3 



CH 3 



H 



//</m-I,2-I)imetlulc\cIopmpdne 



If we examine models of C/A- and /r0/7s-l,2-cyclopentanediol more closely, 
we find that each compound contains two chiral centers. We know (Sec. 4.18) 
that compounds containing more than one chiral center are often -but not always 
chiral. Are these glycols chiral? As always, to test for possible chirality, we con- 
struct a model of the molecule and a model of its mirror image, and see if the two 
are superimposable. When we do this for the //my-glycol, we find that the models 

m irror 




OH 




OH 



Not superimpo\ahle 

Enantiomers : resolvable 

tran?-\ ,2-Cyclopentancdiol 



SEC. 9.14 



CONFORMATIONAL ANALYSIS 



303 



are not superimposable. The trans glycol is chiral, and the two models we have 
constructed therefore correspond to enarttiomers. Next, we find that the models 
are not interconvertible by rotation about single bonds. They therefore represent, 
not conformational isomers, but configurational isomers; they should be capable 
of isolation resolution and, when isolated, each should be optically active. 

Next let us look at c/s-l,2-cyclopentanediol. This, too, contains two chiral 
centers; is it also chiral? This time we find that a model of the molecule and a 
model of its mirror image are superimposable. In spite of its chiral centers, c/s-1,2- 

mirroi 





OH OH OH 



Superimposable 

A meso compound 

cis-] ,2-Cyclopentanediol 

cyclopentanediol is not chiral; it cannot exist in two enantiomeric forms, and can- 
not be optically active. It is a meso compound. 

We might have recognized c/>l,2-cyclopentanediol as a meso structure on sight 
from the fact that one half of the molecule is the mirror image of the other half (Sec. 4.18): 




OH 



A meso compound 
m-l,2-Cvclopentanedioi 

Thus, of the two 1,2-cyclopentanediols obtainable from cyclopentene, only 
one is separable into enantiomers, that is, is resolvable', this must necessarily be 
the /ra/u-glycol. The other glycol is a single, inactive, nonresolvable compound, 
and it must have the cis configuration. 

What is the relationship between the meso ra-glycol and either of the enan- 
tiomeric /ra/w-glycols? They are diaster -earners, since they are stereoisomers that 
are not enantiomers. 

Problem 9.6 Five of the eight structures sho\\n at the top of p. 302 arc achiral. 
Which are these? 



9.14 Stereoisomcrism of cyclic compounds. Conformational analysis 

So far, we have described the relative positions of groups in cis- and trans- 
isomers in terms of flat rings: both groups are below (or above) the plane of the 



304 



ALICYCLIC HYDROCARBONS 



CHAP. 9 



ring, or one group is above and the other is below the plane of the ring. In view 
of what we have said about puckering, however, we realize that this is a highly 
simplified picture even for four- and five-membered rings, and for six-membered 
rings is quite inaccurate. 

Let us apply the methods of conformational analysis to the stereochemistry 
of cyclohexane derivatives; and, since we are already somewhat familiar with inter- 
actions of the methyl group let us use the dimethylcyclohexanes as our examples. 

If we consider only tuc 1 ore stable, chair conformations, we find that a 
particular molecule of rra/is-l,2-dimethylcyclohexane, to take our first example, 
can exist in two conformations (see Fig. 9.11). In one, both CH 3 groups are in 





H 

Diequatorial Diaxial 

Figure 9.11. Chair conformations of /raw5-l,2-dimethylcyclohexane. 

equatorial positions, and in the other, both -CH 3 groups are in axial positions. 
Thus, we see, the two CH 3 groups of the trans-isomer are not necessarily on 
opposite sides of the ring; in fact, because of lesser crowding between CH 3 
groups and axial hydrogens of the ring (less 1,3-diaxial interaction), the more 
stable conformation is the diequatorial one. 

A molecule of cw-l,2-dimethylcyclohexane can also exist in two conformations 
(see Fig. 9.12). In this case, the two are of equal stability (they are mirror images) 
since in each there is one equatorial and one axial CH 3 group. 




CH 3 

Equatorial-axial 




CH 3 

Axial-equatorial 
Figure 9. 12. Chair conformations of cis- 1 ,2-dimethyicyclohexane. 



In the most stable conformation of /ra/w-l,2-dimethylcyclohexane, both 
CH 3 groups occupy uncrowded equatorial positions. In either conformation 
of the aj-l,2-dimethylcyclohexane, only one CH 3 group can occupy an equatorial 
position. It is not surprising to find that /r/75-l,2-dimethylcyclohexane is more 
stable than c/.s-l,2-dimethylcyclohexane. 

It is interesting to note that in the most stable conformation (diequatorial) 
of the frarts-isomer, the CH 3 groups are exactly the same distance apart as they 
are in either conformation of the c/5-rsomer. Clearly, it is not repulsion between 



SEC. 9.14 CONFORMATIONS AN \LYSIS 305 

the CHj groups as one might incorrectly infer from planar representations 
that causes the difference in stability between the trans- and m-isomers: the cause 
is 1,3-diaxial interactions '(Sec. 9.12). 

Now, just how much more stable is the /ra/w-isomer? In the ns-l,2-dimethyl- 
cyclohexane there is one axial methyl group, which means two 1,3-diaxial 
methyl-hydrogen interactions: one with each of two hydrogen atoms. (Or, what is 
equivalent (Sec. 9.12), there are two butone-gauche interactions between the methyl 
groups and carbon atoms of the ring.) In addition, there is one butane-gauche. 
interaction between the two methyl groups. On the basis of 0.9 kcal for each 
1,3-diaxial methyl-hydrogen interaction or butane-gfli/c7/e interaction, we calculate 
a total of 2.7 kcal of van der Waals strain for the m-l,2-dimethylcyclohexane. 
In the (diequatorial) /ro/w-isomer there are no 1,3-diaxial methyl-hydrogen 
interactions, but there is one butane-gauche interaction between the methyl 
groups; this confers 0.9 kcal of van der Waals strain on the molecule. We sub- 
tract 0.9 kcal from 27 kcal and conclude that the /ra//s-isomer should be more 
stable than the r/5-isomer by 1.8 kcal/mole, in excellent agreement with the meas- 
ured value of 1.87 kcal. 

Problem 9.7 Compare stabilities, of .the possible chair conformations of: (a) cis- 
1,2-dimethylcyclohexane; (b) //ww- 1,2-dimethylcyclohexane; (c) m-l,3-dimethyl- 
cyclohexane; (d) /ra//s-l,3-dimethylcyclohexane; (e) m-l,4-dimethylcyclohexanc; 
(f) /r0Hs-l,4-dirnethylcyclohexane. (g) On the basis of 0.9 kcal/mole per 1,3-diaxial 
methyl- hydrogen interaction, predict (where you can) the potential energy difference 
between the members of each pair of conformations. 

Problem 9.8 On theoretical grounds, K. S. Pitzer (then at the University of 
California) calculated that the energy difference between the conformations of cis- 
1,3-dimethylcyclohexane should be about 5.4 kcal, much larger than that between the 
chair conformations of //W7.y-l,2-dimethylcyclohexane or of //Y///j-l,4-dimethylcycJo- 
hexane. (a) What special factor must Pitzer have recognized in the r/5-l,3-isomer? 
(b) Using the 0.9 kcal value where it applies, what value must you assign to the factor 
you invoked in (a), if you are to arrive at the energy difference of 5.4 kcal for the cis- 
1,3-conformations? (c) The potential energy difference between cis- and fra/is-lj,3,5- 
tetramethyicyclohexane was then measured by Norman L. A/linger (at Wayne State 
University) as 3.7 kcal/mole. This measurement was carried out because of its direct 
bearing on the matter of m-l,3-dimethylcyciohexane. What is the connection between 
this measurement and parts (a) and (b)? Does Allinger's measurement support Pitzer's 
calculation? 

Problem 9.9 Predict the relative stabilities of the cis- and /rww-isomers of: 
(a) 1,3-dimethylcyclohexane; (b) 1,4-dimethylcyclohexane. (c) On the basis of 0.9 
kcal/mole per 1,3-diaxial methyl-hydrogen interaction or butane-gauche interaction, 
and assuming that each stereoisomer exists exclusively in its more stable conforma- 
tion, predict the potential energy difference between members of each pair of stereo- 
isomers. 

Conformational analysis of cyclohexane derivatives containing several different 
substituents follows along the same lines as that of the dimethylcyclohexanes. 
We need to keep in mind that, of two groups, the larger one will tend to call the 
tune. Because of its very large 1,3-diaxial interactions (Problem 9.3, p. 301), the 
bulky /T/-butyl group is particularly prone to occupy an equatorial position. 
If as is usually the case other substituents are considerably smaller than /erf- 
butyl, the molecule is virtually locked in a single conformation: the one with an 



306 



ALICYCLIC HYDROCARBONS 



CHAP. 9 



equatorial tert-buiyl group. Consider cyclohexanes I and II containing a 4-tert- 
butyl group cis or trans to another substituent G. In each diastereomer, ten- 
butyl holds G exclusively in the axial or in the equatorial position, yet, because 



(CH 3 ) 3 C 




(CH 3 ) 3 C 




A cis-4-tert -butyl 
substituted cyclohexane 



A trans-4-tert-buty\ 
substituted cyclohexane 



of its distance, exerts little electronic effect on G. Following a suggestion by 
Professor Saul Winstein (of the University of California, Los Angeles), te/Y-butyl 
has been widely used as a holding group, to permit the study of physical and 
chemical properties associated with a purely axial or purely equatorial substituent. 

Problem 9.10 Use the energy differences given in Problem 9.3 (p. 301) to calcu- 
late values for the various alkyl-hydrogen 1,3-diaxial interactions, and from these 
calculate the difference in energy between the two conformations of: 

(a) r/5-4-/*r/-butyImethylcyclohexane; 

(b) 

(c) 



Now, what can we say about the possible chirality of the 1,2-dimethylcyclo- 
hexanes? Let us make a model of /raw5-l,2-dimethylcyclohexane in the more 
stable diequatorial conformation, say and a model of its mirror image. We find 



mirror 



H 




CH 3 



CH 3 



H 




Not superimposable; not interconvertible 

trans- \ ,2-Dimethylcyclohexane 
A resolvable racemic modification 

they are not superimposable, and therefore are enantiomers. We find that they 
are not interconvertible, and hence are configurational isomers. (When we flip 
one of these into the opposite chair conformation, it is converted, not into its 
mirror image, but into a diaxial conformation.) Thus, fr0Aw-l,2-dimethylcyclo- 
hexane should, in principle, be resolvable into (configurational) enantiomers, 
each of which should be optically active. 

Next, let us make a model of cw-l,2-dimethylcyclohexane and a model of its 
mirror image. We find they are not superimposable, and hence are enantiomers. 



SEC. 944 



CONFORMATIONAL ANALYSIS 

mirror 



307 




CH 3 




CH 3 



CH 3 | 



Not superimposahle; bin intercom ertihle 

t'/A-l,2-DimethyIc)cIohcxanc 
A non-resoli ahfc racemic modification 

In contrast to what we have said for the trans-compound, however, we find that 
these models are interconvertible by flipping one chair conformation into the other. 
These are conformational enantiomers and hence, except possibly at low tempera- 
tures, should interconvert too rapidly for resolution and measurement of optical 
activity. 

Thus, just as with the cis- and rra/j.s-l,2-cyclopentanediols (Sec. 9.13), we 
could assign configurations to the cis- and fra?,s-l,2-dimethylcyclohexanes by 
finding out which of the two is resolvable. The c/s-l,2-dimethylcyclohexane is 
not literally a meso compound, but it is a non-resolvable racemic modification, 
which for most practical purposes amounts to the same thing. 

To summarize, then, 1,2-dimethylcyclohexane exists as a pair of (configura- 
tional) diastereomers: the cis- and /ra/w-isomers. The c/.v-isomer exists as a pair 
of conformational enantiomers. The trans-homer exists as a pair of configura- 
tional enantiomers, each of which in turn exists as two conformational diastereo- 
mers (axial -axial and equatorial-equatorial). 

Because- of the ready interconvertibility of chair conformations, it is possible 
to use planar drawings to predict the configurational stereoisomerism of cyclo- 

mirror 




OH 



Superimposablif 
m- 1 ,2-ryclohexanediol 






OH H 



\r>t super imposahle 
trans- 1 ,2-Cyclohexancdiol 



308 ALICYCLIC HYDROCARBONS CHAP. 9 

hexane derivatives. To understand the true geometry of such molecules, however, 
and with ii the matter of stability, one must use models and formulas like those in 
Figs. 9.11 and 9. 12. 



Problem 9.11 Which of the following compounds are resolvable, and which 
are non-resolvable? Which are truly mcso compounds? Use models as well as draw- 
ings. 

(a) m-l,2-cyclohexanediol (d) m7w$-l,3-cyclohexanediol 

(b) m7//y-l,2-cyclohexanediol (e) m-1,4-cyciohexanediol 

(c) c/>l,3-cyclohexancdiol (f) //ww-1,4-cyclohexanediol 

Problem 9.12 Tell which, if any, of the compounds of Problem 9.1 1 exist as: 

(a) a single conformation ; 

(b) a pair of confonnationa! enantiomers; 

(c) a pair of ccnformational diastercomers; 

(d) a pair of (configurational) enantiomers, each of which exists as a single conforma- 
tion; 

(e) a pair of (configurational) enantiomers, each of which exists as a pair of c on form a- 
tional diastereomers; 

(f) none of the above answers. (Give the correct answer.) 

Problem 9.13 Draw structural formulas for all stereoisomers of the following. 
Label any meso compounds and indicate pairs of enantiomers. Do any (like ro-1,2- 
dimethylcyclohexane) exist as a non-resolvable racemic modification? 

(a) ci>2-chlorocyclohexanol (d) rrawj-3-chlorocyclopentanol 

(b) /rfl/w-2-chlorocyclohexanol (e) m-4-chlorocyclohexanol 

(c) c/>3-chlorocyclopentanol (f) /rc//tf-4-chlorocycIohexanol 



9.15 Carbcncs. Methylene 

The difference between successive members of a homologous series, we have 
seen, is the CH 2 unit, or methylene. But methylene is more than just a building block 
for the mental construction of compounds; it is an actual molecule, and its chem- 
istry and the chemistry of its derivatives, the carbenes, has become one of the 
most exciting and productive fields of organic research, 

Methylene is formed by the photolysis of either diazomethane, CH 2 N 2 , or 
ketene, CH 2 - O O. (Notice that the two starting materials and the two other 



Diazomethane Methylene 

uuravtolctllght 



CH 2 + CO 

Kctene Methylene 

products, nitrogen and carbon monoxide, are pairs ofisoelectronic molecules, that 
is, molecules containing the same number of valence electrons.) 

Methylene as a highly reactive molecule was first proposed in the 1930s to 
account for the fact that something formed by the above reactions was capable of 
removing certain metal mirrors (compare Problem 16, p. 72). Its existence was 
definitel) established in 1959 by spcctroscopic studies. 



SEC. 9.15 



CARBENES. METHYLENE 



309 



4 IF I Ml* CMz WITH 
NW* AND BOIL TWE 
ATOMS IN OSMOTIC 
FO&, J SHOULD GET 
SPECKLED 




(c> Wart Disney Productions 

Figure 9.13. Evidence of early (1944) research on methylene. CH 2 by D. 
Duck. (As unearthed by Professors P. P. Caspar and G. S. Hammond of 
the California Institute of Technology.) 

These studies revealed that methylene not only exists but exists in two different 
forms (different spin states), generally referred to by their spectroscopic designa- 
tions: singlet methylene, in which the unshared electrons are paired: 

H 

HI 03 "~*y, . - 
/ /1.12A 

CH 2 : H:C: H-'-C: 

Singlet methylene 
Unshared electrons paired 

and triplet methylene, in which the unshared electrons are not paired. 



cn r H:C:H H-H-^H 

Triplet methylene 

Unshared electrons not paired' 

a diradical 

Triplet methylene is thus a free radical: in fact, it is a {//radical. As a result of 
the difference in electronic configuration, the two kinds of molecules differ in 
shape and in chemical properties. Singlet methylene is the less stable form, and 
is often the form Jirst generated, in the initial photolysis. 

The exact chemical properties observed depend upon which form of methylene 




310 ALICYCLIC HYDROCARBONS CHAP. 9 

is reacting, and this in turn depends upon the experimental conditions. In the 
liquid phase, the first-formed singlet methylene reacts rapidly with the abundant 
solvent molecules before it loses energy. In the gas phase especially in the presence 
of an inert gas like nitrogen or argon singlet methylene loses energy through 
collisions and is converted into triplet methylene, which then reacts. 

When methylene is generated in the presence of alkenes, there are obtained 
cyclopropanes. For example: 

CH 3 CH=CHCH 3 + CH 2 N 2 -^-* CH 3 CH CHCH 3 -r N 2 
2-Butene Diazomethane CH 

1 ,2-Dimethylcyclopropane 

This is an example of the most important reaction of methylene and other carbenes: 
addition to the carbon-carbon double bond. Its most striking feature is that it can 
occur with two different kinds of stereochemistry. 

X O=C / + CH 2 v C C Addition 

/ -\ * ' 

For example, photolysis of diazomethane in liquid cw-2-butene gives only 
c/$-l,2-dimethylcyclopropane, and in liquid fra/w-2-butene gives only trans-1,2- 
dimethylcyclopropane. Addition here is stereospecific and syn. Photolysis of 
diazomethane in gaseous 2-butene either cis or trans gives both cis- and trans- 
1,2-dimethylcyclopropanes. Addition here is non-stereospecific. 

There seems to be little doubt that the following interpretation, due to P. S. 
Skell of Pennsylvania State University, is the correct one. 

It is singlet methylene that undergoes the stereospecific addition. Although 
neutral, singlet methylene is electron-deficient and hence electrophilic; like other 

J Singlet methylene 
> ~\ / C ~ Stereospecific 
CH 2 syn-addition 

electrophiles, it can find electrons at the carbon-carbon double bond. The stereo- 
chemistry strongly indicates simultaneous attachment to both doubly-bonded 
carbon atoms. (However, on both theoretical and experimental grounds, the 
transition state is believed to be unsymmetrical : attachment to one carbon has 
proceeded further than attachment to the other, with the development of con- 
siderable positive charge on the second carbon.) 

It is triplet methylene that undergoes the non-stereospecific addition. Triplet 

CH, ' V-C 7 C-C -C^-^C- Triplet methylene 

Non-stereospecific 
addition 



methylene is a diradical, and it adds by a free-radical-like two-step mechanism: 
actually, addition followed by combination. The intermediate diradical 1 lasts 
long enough for rotation to occur about the central carbon-carbon bond, and 





SEC. 9.16 SUBSTITUTED CARBENES. a-ELIMINATION 311 

both cis and trans products are formed. (Problem: Using the approach of Sec. 
7.12, assure yourself that this is so.) 

Besides addition, methylene undergoes another reaction which, quite literally, be- 
longs in a class by itself: insertion. 

-C-H+-CH, - -C-CH 2 -H Insertion 

i " I 

Methylene can insert itself mio every carbon-hydrogen bond of most kinds of molecules. 
We cannot take time to say more here about this remarkable reaction, except that when 
addition is the desired reaction, insertion becomes an annoying side-reaction. 

Problem 9.14 In the gas phase, with low alkene concentration and in the pres- 
ence of an inert gas, addition of methylene to the 2-butenes is, we have seen, non- 
stereospecific. If, however, there is present in this system a little oxygen, addition 
becomes completely stereospecific (syn). Account in detail for the effect of oxygen. 
{Hint: See Sec. 2.14.) 

9.16 Substituted carbenes. a-Elimination 

A more generally useful way of making cyclopropanes is illustrated by the 
reaction of 2-butene with chloroform in the presence of potassium terf-butoxide 
(f-Bu = ter/-butyl): 

CH 3 CH-CHCH 3 t CHC1 3 a+ CH 3 CH~CHCH 3 /-BuOH T KC1 
2-Butene Chloroform C 

Cl Cl 
3,3-Dichloro- 1 ,2-dimethylcyclopropane 

The dichlorocyclopropanes obtained can be reduced to hydrocarbons or hydro- 
lyzed to ketones, the starting point for many syntheses (Chap. 19). 

Here, too, reaction involves a divalent carbon compound, a derivative of 
methylene: dichlorocarbene, :CC1 2 . It is generated in two steps, initiated by 
attack on chloroform by the very strong base, /erf-butoxide ion, and then adds 
to the alkene. 



(1) r-BuO:- + H:CC1 3 < - :CC1 3 - + /-BuO:H 

(2) :CC1 3 - - > :CC1 2 + CT 

Dichlorocarbene 

(3) CH 3 CH=CHCH 3 + :CC1 2 - > CH 3 CH~CHCH 3 

c. A c. 

It is believed that, because of the presence of the halogen atoms, the singlet 
form, with the electrons paired, is the. more stable form of dichlorocarbene, and 
is the one adding to the double bond. (Stabilization by the halogen atoms is 
presumably one reason wh> dihalocarbenes do not generally undergo the inser- 
tion reaction that is so characteristic of unsubstituted singlet methylene.) 

The addition of dihalocarbenes, like that of singlet methylene, is stereospecific 
and syn. 



312 ALICYCLIC HYDROCARBONS CHAP. 9 

Problem 9.15 (a) Addition of :CCI ? to cyclopentene yields a single compound. 
What is it? (b) Addition of :CBrCl to cvclopentene yields a mixture of stereoiso- 
mers. In light of (a), how do >ou account-for this? What are the isorners likely to be? 
(Hint: Use models.) 

In dehydrohalogenation of alkyl halides (Sec. 5.13), we have already encoun- 
tered a reaction in which hydrogen ion and halide ion are eliminated from a mole- 
cule by the action of base; there H and X were lost from adjacent carbons, 
and so the process is called fi-eUniinaiion. In the generation of the methylene 
shown here, both H and X are eliminated from the same carbon, and the process 
is called a-eliminaiion. (Later on, in Sec. 24.12, we shall see some of the evidence 
for the mechanism of ^-elimination shown above.) 

X 



5S~* C-C Beta-elimination 

I i ' ^ 

-C-H ^ -c: Alpha-elimination 

JC 

Problem 9.16 (a) Why does CHC1 3 not undergo ^-elimination through the action 
of base? (b) What factor would you expect to make a-elimination from CHC1 3 easier 
than from, say, CH 3 CI? 

There are many ways of generating what appear to be carbenes. But in some 
cases at least, it seems clear that no free carbene is actually an intermediate; 
instead, a carbenoid (carbene-like) reagent transfers a carbene unit directly to a 
double bond. For example, in the extremely useful Simmons-Smith reaction 

CH 2 I 2 + Zn(Cu) > ICH 2 ZnI 



ICH 2 ZnI 



I I 



-c 



H 

I Zn----I 



2 



C- + ZnI 2 



\ 



CP 



(H. . Simmons and R. D. Smith of the du Pont Company) the carbenoid is an 
organozinc compound which delivers methylene stereospecifically (and without 
competing insertion) to the double bond. 



9.17 Analysis of alicyclic hydrocarbons 

A cyclopropane readily dissolves in concentrated sulfuric acid, and in this 
resembles an alkene or alkyne. It can be differentiated from these unsaturated 
hydrocarbons, however, by the fact that it is not oxidized by cold, dilute, neutral 
permanganate. 

Other alicyclic hydrocarbons have the same kind of properties as their open- 
chain counterparts, and they are characterized in the same way: cycloalkanes by 
their general inertness, and cycloalkenes and cycloalkynes by their response to 



SEC. 9.17 ANALYSIS OF ALICYCLIC HYDROCARBONS 313 

tests for unsaturation (bromine in carbon tetrachloride, and aqueous permanga- 
nate). That one is dealing with cyclic hydrocarbons is shown by molecular formu- 
las and by degradation products. 

The properties of cyclohexane, for example, show clearly that it is an alkane. 
However, combustion analysis and molecular weight determination show its 
molecular formula to be C 6 H 12 . Only a cyclic structure (although not necessarily 
a six-membered ring) is consistent with both sets of data. 

Similarly, the absorption of only one mole of hydrogen shows that cyclohexane 
contains only one carbon-carbon double bond; yet its molecular formula is 
C 6 Hi , which in an open-chain compound would correspond to two carbon- 
carbon double bonds or one triple bond. Again, only a cyclic structure fits the facts. 

Problem 9.17 Compare the molecular formulas of: (a) /r-hcxane and c>clo- 
hexare; (b) //-pentanc and c>clopentane; (c) I-hcxene and cyclohexene; (d)dodecane, 
-hexylcyclohexane, and c>clohcx>lcyclohcAanc. (e) In general, how can you deduce 
the number of rings in a compound from its molecular formula and degree of un- 
saturation? 

Problem 9.18 What is the molecular formula of: (a) cyclohexane; (b) methyl- 
cyclopentane; (c) 1,2-dimethylcyclobutane? (d) Docs the molecular formula give any 
information about the size of ring in a compound? 

Problem 9.19 The ycilovv plant pigments -, /?-, and y-caionnc, and the red pig- 
ment of tomatoes, lycop&w, are converted into Vitamin A in the liver. All four have 
the molecular formula C 40 Hs ft . Upon catalytic hydrogenation, - and /^-carotene 
yield C 40 H78, y-carotene welds C 40 H S o, and Ijcopene yields C 4() H S 2- How many rings, 
if any, are there in each compound? 

Cleavage products of cycloalkenes and cydoalkynes also reveal the cyclic 
structure. Ozonolysis of cyclohexene, for example, does not break the molecule 
into two aldehydes of lower carbon number, but simply into a single six-carbon 
compound containing two aldehyde groups. 



H 2 cr 



H 2 0, Zn 



A di~aldehyde 
,CH 



CH 2 

Cyclohexene 



KMnO 4 



COOH 



-* 1 



.COOH 
CH 2 

A di-acid 

Problem 9.20 Predict the ozonolysis products of: (a) cyclohexene; (b) 1-methyl- 
cyclopentene; (c) 3-meth>lcyclopentene; (d) 1,3-cyclohexadiene; (e) 1 ,4-cyclohexa- 
diene. 

Problem 9.21 Both cyclohexene and 1,7-octadiene yield the di-aldehyde 
OHC(CH 2 ) 4 CHO upon ozonolysis. What other facts would enable you to distinguish 
between the two compounds? 

(Analysis of cyclic aliphatic hydrocarbons by spectroscopy will be discussed 
in Sees. 13.15-13.16.) 



314 ALICYCLIC HYDROCARBONS CHAP. 9 

PROBLEMS 

1. Draw structural formulas of: 

(a) methylcyclopentane (f) cyclohexylcyclohexane 

(b) 1-methylcyclohexene (g) cyclopentylacetylene 

(c) 3-methylcyclopentene (h) IJ-dimethyM-chlorocycloheptane 

(d) m?/?.y-l,3-dichlorocyclobutane (i) bicyclo[2.2.1]hepta-2,5-diene 

(e) cAs-2-bromo-l -methylcyclopentane (j) l-chlorobicyclo[2.2.21octane 

2. Give structures and names of the principal organic products expected from each 
of the following reactions: 

(a) cyclopropane + C1 2 , FeCl 3 (j) 1-methylcyclohexene + Br 2 (aq) 

(b) cyclopropane -f C1 2 (300 J ) (k) 1-methylcyclohexene -f HBr 

(c) cyclopropane + cone. H 2 SO 4 (peroxides) 

(d) cyclopentane + C1 2 , FeCl 3 (I) 1,3-cyclohexadiene + HC1 

(e) cyclopentane + C1 2 (300) (m) cyclopentanol + H 2 SO 4 (heat) 

(f) cyclopentane + cone. H 2 SO 4 (n) bromocyclohexane + KOH(alc) 

(g) cyclopentene + Br 2 /CCl 4 (o) cyclopentene + cold KMnO 4 
(h) cyclopentene -f- Br 2 (300) (p) cyclopentene + HCO 2 OH 

(i) 1-methylcyclohexene -f HCI (q) cyclopentene + hot KMnO 4 

(r) chlorocyclopentane + (C 2 H 5 ) 2 CuLi 
(s) 1-methylcyclopentene + cold cone. H 2 SO 4 
(0 3-methylcyclopentene + O 3 , then H 2 O/Zn 
(u) cyclohexene + H 2 SO 4 - > C, 2 H 20 
(v) cyclopentene + CHG1 3 -f /-BuOK 
(w) cyclopentene -f- CH 2 I 2 4- Zn(Cu) 

3. Outline all steps in the laboratory synthesis of each of the following from cyclo- 
hexanol. 

(a) cyclohexene (g) adipic acid, HOOC(CH?) 4 COOH 

(b) cyclohexane (h) bromocyclohexane 

(c) mj/w-l,2-di bromocyclohexane (i) 2-chlorocyclohexanol 

(d) m-l,2-cyclohexanediol (j) 3-bromocyclohexene 

(e) /w/w-l,2-cyclohexanediol (k) 1,3-cyclohexadiene 

(f) OHC(CH 2 ) 4 CHO (1) cyclohexylcyclohexane 

(m) norcarane, bicyclo[4. 1 .Ojheptane 

4. Give structure of all isomers of the following. For cyclohexane derivatives, 
planar formulas (p. 307) will be sufficient here. Label pairs of enantiomers, and meso 
compounds. 

(a) dichlorocyclopropanes (d) dichlorocyclohexanes 

(b) dichlorocyclobutanes (e) chloro-l,l-dimethylcyclohexanes 

(c) dichlorocyclopentanes (f) 1 ,3,5-trichlorocyclohexanes 

(g) There are a number of stereoisomeric 1,2,3,4,5,6-hexachlorocyclohexanes. 
Without attempting to draw all of them, give the structure of the most stable isomer, and 
show its preferred conformation. 

5. (a) 2,5-Dimethyl-l,l-cyclopentanedicar boxy lie acid (I) can be prepared as two 
optically inactive substances (A and B) of different m.p. Draw their structures, (b) Upon 
heating, A yields two 2,5-dimeth>lcyclopentanecarboxylic acids (II), and B yields only 
one. Assign structures to 4 and B. 



HOOC COOH H COOH 





PROBLEMS 315 

6. (a) The following compounds can be resolved into optically active enantiomers. 




3,3'-Diaminospiro[3.3]heptane 4-Methylcyclohexylideneaceticacid 

Using models and then drawing three-dimensional formulas, account for this. Label 
the chiral center in each compound. 

(b) Addition of bromine to optically active 4-methylcyclohexylideneacetic acid yields 
two optically active dibromides. Assuming a particular configuration for the starting 
material, draw stereochemical formulas for the products. 

7. (a) m7//$-l,2-Dimethylcyclohexane exists about 99% in the diequatorial con- 
formation. /rfl/w-l,2-Dibromocyclohexane (or /r0/w-l,2-dichlorocyclohexane), on the 
other hand, exists about equally in the diequatorial and diaxial conformations; further- 
more, the fraction of the diaxial conformation decreases with increasing polarity of the 
solvent. How do you account for the contrast between the dimethyl and dibromo (or 
dirhloro) compounds? (Hint: See Problem 11, p. 141.) 

(b) If //YW$-3-a>4-dibromo-te/7-butylcyclohexane is subjected to prolonged heating, 
t is converted into an equilibrium mixture (about 50:50) of itself and a diastereomer. 
What is the diastereomer likely to be? How do you account for the approximately equal 
stability of these two diastereomers? (Here, and in (c), consider the more stable confor- 
mation of each diastereomer to be the one with an equatorial /erf-butyl group.) 

(c) There are two more diastereomeric 3,4-dibromo-/?/7-butylcyclohexanes. What 
are they? How do you account for the fact that neither is present to an appreciable 
extent in the equilibrium mixture? 

8. The compound decalin, C 10 H 18 , consists of two fused cyclohexanc rings: 



00 



Deculm 



(a) Using models, show how there can be two isomeric decalins, cis and trans, (b) How 
many different conformations free of angle strain are possible for m-dccalin ? For trans- 
decalin ? (c) Which is the most stable conformation of c w-decalin ? Of /raw-decalin ? (Hint 
Consider each ring in turn. What are the largest substituents on each ring?) (d) Account for 
the fact that fra/K-decalin is more stable than r/s-decalin. (e) The difference in stability 
between cis- and /ra/w-decalin is about 2 kcal/mole; conversion of one into the other 
takes place only under very vigorous conditions. The chair and twist-boat forms of cyclo- 
hexane, on the other hand, differ in stability by about 6 kcal/mole, yet are readily inter- 
converted at room temperature. How do you account for the contrast? Draw energy 
curves to illustrate your answer. 

9. Allinger (p. 305) found the energy difference between cis- and trans-} ,3-d\-tert~ 
butylcyclohexane to be 5.9 kcal/mole, and considers that this value represents the energy 
difference between the chair and twist-boat forms of cyclohexane. Defend Allinger's 
position. 

10. It has been suggested that in certain substituted cyclopentanes the ring exists 
preferentially in the "envelope" form: 




316 ALICYCLIC HYDROCARCONS CHAP. 9 

Using models, suggest a possible explanation for each of the following facts: 

(a) The attachment of a methyl group to the badly strained cyclopentane ring raises 
the heat of combustion very little more than attachment of a methyl group to the un- 
strained cyclohexane ring. (Hint: Where is the methyl group located in the "envelope" 
form?) 

(b) Of the 1 ,2-dimethyIcyclopentanes, the /ra//s-isomer is more stable than the cis. 
Of the 1,3-dimethylcyclopentanes, on the other hand, the rfa-isomer is more stable than 
the trans. 

(c) The c /j-isomer of methyl 3-methylcyclobutanecarboxylate 



COOCH 3 




CH 3 

is more stable than the /ra/w-isomer 

11. Each of the following reactions is carried out, and the products are separated by 
careful distillation, recrystallization, or chromatography. For each reaction tell how 
many fractions will be collected. Draw a stereochemical formula of the compound or 
compounds making up each fraction. Tell whether each fraction, as collected, will be 
optically active or optically inactive. 

(a) (R)-3-hydroxycyclohexene + KMnO 4 * C 6 H, 2 O 3 ; 

(b) (R)-3-hydroxycyclohexene + HCO 2 OH > C 6 H, 2 O 3 ; 

(c) (S,S)-l,2-dichlorocyclopropane 4- C1 2 (300) > C 3 H 3 Cl3; 

(d) racemic 4-methylcyclohexene + Br 2 /CCl 4 . 

12. Outline all steps in a possible laboratory synthesis of each of the following, 
using alcohols of four carbons or fewer as your only organic source, and any necessary 
inorganic reagents. (Remember: Work backwards.) 

(a) cis-l ,2-di(w-propyl)cyclopropane ; 

(b) racemic m?/7$-l-methyl-2-ethyl-3,3-dichIorocycIopropane* 

13. Describe simple chemical tests that would distinguish between : 

(a) cyclopropane and propane 

(b) cyclopropane and propylene 

(c) 1 ,2-dimethylcyclopropane and cyclopentane 

(d) cyclobutane and 1-butene 

(e) cyclopentane and 1-pentene 

(f) cyclopentane and cyclopentene 

(g) cyclohexanol and w-butylcyclohexane 

(h) 1,2-dimethylcyclopentene and cyclopentanol 

(i) cyclohexane, cyclohexene, cyclohexanol, and bromocyclohexane 

14. How many rings does each of the following contain? 

(a) Camphane, Ci Hi 8 , a terpene related to camphor, takes up no hydrogen, (b) Chole- 
stane, C 27 H 48 , a steroid of the same ring structure as cholesterol, cortisone, and the sex 
hormones, takes up no hydrogen, (c) fi-Phellandrene, Ci Hi 6 , a terpene, reacts with 
bromine to form Ci H 16 Br 4 . (d) Ergocalciferol (so-called "Vitamin D 2 "), C 28 H 44 O, 
an alcohol, gives C 28 H 52 O upon catalytic hydrogenation. (e) How many double bonds 
does ergocalciferol contain ? 

15. On the basis of the results of catalytic hydrogenation, how many rings does each 
of the following aromatic hydrocarbons contain ? 

(a) benzene (C 6 H 6 ) > C 6 H, 2 (e) phenanthrene (C, 4 H, ) > C, 4 H 24 

(b) naphthalene (C, H 8 ) > Ci H, 8 (f) 3,4-benzpyrene (C 20 H 12 ) > C 20 H 32 

(c) toluene (C 7 H 8 ) C 7 H, 4 (g) chrysene (C 18 H 12 ) > Ci 8 H 30 

(d) anthracene (Ci 4 H 10 ^ > C H H 24 

(Check your answers by use of the index.) 



PROBLEMS 317 

16. (a) A hydrocarbon of formula Ci H l6 absorbs only one mole of H 2 upon hydro- 
genation. How many rings does it contain? (b) Upon ozonolysis it yields 1,6-cyclo- 
decanedione (III). What is the hydrocarbon? 



O CH 3 C- -CH 2 CH 2 CH 

L (C H 2)4 J A WOOF 

Hi IV 

17. Limonene, Ci Hi 6 , a terpene found in orange, lemon, and grapefruit peel, 
absorbs only two moles of hydrogen, forming p-menthane, CioH 20 . Oxidation by per- 
mangdnate converts limonene into IV. (a) How many rings, if any, are there in limonene? 

(b) What structures are consistent with the oxidation product ?'(c) On the basis of the 
isoprene rule (Sec. 8.26), which structure is most likely for limonene? For /?-menthane? 
(d) Addition of one mole of H 2 O convcits limonene into a-terpineol. What are the most 
likely structures for a-terpineol? (e) Addition of two moles of H 2 O to limonene yields 
terpin hydrate. What is the most likely structure for terpin hydrate? 

18. a-Terpinene, C JO Hi 6 , a terpene found in coriander oil, absorbs only two moles 
of hydrogen, forming p-menthane, C 10 H 2 o- Ozonolysis of a-terpinene yields V; per- 
manganate cleavage yields VI. 

CH 3 CH(CH 3 ) 2 

CH 3 -C-CH 2 CH 2 -C-CH(CH 3 ) 2 HOOC C CH 2 CH 2 C COOH 

00 OH OH 

V VI 

(a) How many rings, if any, are there in a-terpinene? (b) On the basis of the cleavage 
products, V and VI, and the isoprene rule, what is the most likely structure for a-terpinene? 

(c) How do you account for the presence of the OH groups in VI? 

19. Using only chemistry that you have already encountered, can you suggest a 
mechanism for the conversion of nerol (Ci Hi 8 O) into a-terpineol (Ci Hi 8 O) in the 
presence of dilute H 2 SO 4 ? 



CH 3 C=CH-CH 2 -CH 2 -C=CH-CH 2 OH 

Nerol (found in bcrgamot) 




H 3 C CH 3 

-Terpineol 



Chapter 

10 



Benzene 

Arorriatic Character 



10.1 Aliphatic and aromatic compounds 

Chemists have found it useful to divide all organic compounds into two broad 
classes: aliphatic compounds and aromatic compounds. The original meanings of 
the words "aliphatic" (fatty) and "aromatic" (fragrant) no longer have any 
significance. 

Aliphatic compounds are open-chain compounds and those cyclic compounds 
that resemble the open-chain compounds. The families we have studied so far 
alkanes, alkenes, alkynes, and their cyclic analogs are all members of the aliphatic 
class. 

(Aromatic compounds are benzene and compounds that resemble benzene in 
chemical behavior. Aromatic properties are those properties of benzene that 
distinguish it from aliphatic hydrocarbons. Some compounds that possess 
aromatic properties have structures that seem to differ considerably from the 
structure of benzene: actually, however, there is a basic similarity in electronic 
configuration (Sec. 10.10).^ 

'^Aliphatic hydrocarbons, as we have seen, undergo chiefly addition and free- 
radical substitution; addition occurs at multiple bonds, and free-radical substitu- 
tion occurs at other points along the aliphatic chain. In contrast, we shall find that 
aromatic hydrocarbons are characterized by a tendency to undergo ionic substitution. 
We shall find this contrast maintained in other families of compounds (i.e., acids, 
amines, aldehydes, etc.); the hydrocarbon parts of their molecules undergo reactions 
characteristic of either aliphatic or aromatic hydrocarbons. ) 

It is important not to attach undue weight to the division between aliphatic 
and aromatic compounds. Although extremely useful, it is often less important 
than some other classification. For example, the similarities between aliphatic 
and aromatic acids, or between aliphatic and aromatic amines, are more important 
than the differences. 

318 



10.2 Structure of benzene 

It is obvious from our definition of aromatic Compounds that any study of 
their chemistry must begin with a study of benzene. Benzene has been known since 
1825; its chemical and physical properties are perhaps better known than those of 
any other single organic compound. In spite of this, no satisfactory structure for 
benzene had been advanced until about 1931, and it was ten to fifteen years before 
this structure was generally used by organic chemists. 

The difficulty was not the complexity of the benzene molecule, but rather the 
limitations of the structural theory as it had so far developed. Since an under- 
standing of the structure of benzene is important both in our study of aromatic 
compounds and in extending our knowledge of the structural theory, we shall 
examine in some detail the facts upon which this structure of benzene is built. 

10.3 Molecular formula. Isomer number. Kekule structure 

(a) Benzene has the molecular formula C^H^. From its elemental composition 
and molecular weight, benzene was known to contain six carbon atoms and six 
hydrogen atoms. The question was: how are these atoms arranged? 

In 1858, August Kekule (of the University of Bonn) had proposed that 
carbon atoms can join to one another to form chains. Then, in 1865, he offered 
an answer to the question of benzene: these carbon chains can. sometimes be closed, 
to form rings. 

"I was sitting writing at my textbook, but the work did not progress; my 
thoughts were elsewhere. I turned my chair to the fire, and dozed. Again the atoms 
were gamboling before my eyes. This time the smaller groups kept modestly in the 
background. My mental eye, rendered more acute by repeated visions of this kind, 
could now distinguish larger structures of manifold conformations; long rows, 
sometimes more closely fitted together; all twisting and turning in snake-like motion. 
But look ! What was that ? One of the snakes had seized hold of its own tail, and 
the form whirled mockingly before my eyes. As if by a flash of lightning I 
woke ; . . . I spent the rest of the night working out the consequences of the hypothesis. 
Let us learn to dream, gentlemen, and then perhaps we shall learn the truth." 
August Kekule, 1865. 

Kekule's structure of benzene was one that we would represent today as I. 



U ' 
^x^^O H 

c \ 




C 



H-C - x 

H H 

H 

I II IN 

Kekule formula '" Dewar" formula 



IV V 

Other structures are, of course, consistent with the formula C 6 H 6 : for example, 
II- V. Of all these, KekuU's structure was accepted as the most nearly satisfactory; 



320 BENZENE CHAP. 10 

the evidence was of a kind with which we are already familiar: isomer number 
(Sec. 4.2). 

^j$ Benzene yields only one monosubstitution product, QH 5 Y. Only one 
bromobenzene, C$H 5 Br, is obtained when one hydrogen atom is replaced by 
bromine; similarly, only one chlorobenzene, C 6 H 5 C1, or one nitrobenzene, 
C 6 H 5 NO 2 , etc., has ever been made. This fact places a severe limitation on the 
structure of benzene: each hydrogen must be exactly equivalent to every other 
hydrogen, since the replacement of any one of them yields the same product. 

Structure V, for example, must now be rejected, since it would yield two 
isomeric monobromo derivatives, the 1-bromo and the 2-bromo compounds; all 
hydrogens are not equivalent in V. Similar reasoning shows us that II and III 
are likewise unsatisfactory. (How many monosubstitution products would each 
of these yield?) I and IV, among others, are still possibilities, however. 

(c) Benzene yields three isomeric disubstitution products, C 6 H 4 Y 2 or C 6 H 4 YZ. 
Three and only three isomeric dibromobenzenes, C 6 H 4 Br 2 , three chloronitroben- 
zenes, C$H4C1NO2, etc., have ever been made. This fact further limits our choice 
of a structure; for example, IV must now be rejected. (How many disubstitution 
products would IV yield ?) 

At first glance, structure I seems to be consistent with this new fact; that is, 
we can expect three isomeric dibromo derivatives, the 1,2- the 1,3-, and the 1,4- 
dibromo compounds shown: 



Br Br 

C-H H-Ci 



Br Br 

II 
--- -* 

\>*:-H 

I 



H H Br 

1 ,2-Dibromobenzene 1,3-Di bromobenzene 1,4-Di bromobenzene 

Closer examination of structure I shows, however, that two 1,2-dibromo isomers 
(VI and VII), 'differing in the positions of bromine relative to the double bonds, 
should be possible: 



f 






"V:-H 






H H 

VI VII 

But Kekul visualized the benzene molecule as a dynamic thing: ". . . the form 
whirled mockingly before my eyes . . V He described it in terms of two structures, 
VIII and IX, between which the benzene molecule alternates. As a consequence, 
the two 1,2-dibromobenzenes (VI and VID would be in rapid equilibrium and hence 
could not be separated. 



SEC. 10.4 REACTIONS OF BENZENE 321 



C^ 



I I. 

H H 

VIII IX 



I II 

.C-H 




VI VII 

Later, when the idea of tautomerism (Sec. 8.13) became defined, it was 
assumed that Kekule's "alternation" essentially amounted to tautomerism. 

On the other hand, it is believed by some that Kekule had intuitively anticipated by 
some 75 years our present concept of delocalized electrons, and drew two pictures (VIlI 
and IX) as we shall do, too- as a crude representation of something that neither picture 
alone satisfactorily represents. Rightly or wrongly, the term "Kekule structure" has 
come to mean a (hypothetical) molecule with alternating single and double bonds just 
as the term "D^war benzene" has come to mean a structure (II) that James Dewar 
devised in 1867 as an example of what benzene was not. 

0.4 Stability of the benzene ring. Reactions of benzene 

Kekule's structure, then, accounts satisfactorily for facts (a), (b), and (c) in 
Sec. 10.3. But there are a number of facts that are still not accounted for by this 
structure; most of these unexplained facts seem related to unusual stability of the 
benzene ring. The most striking evidence of this stability is found in the chemical 
reactions of benzene. 

(d) Benzene undergoes substitution rather than addition. Kekule's structure of 
benzene is one that we would call "cyclohexatriene." We would expect this 
cyclohexatriene, like the very similar compounds, cyclohexadiene and cyclohexene, 
to undergo readily the addition reactions characteristic of the alkene structure. 
As the examples in Table 10.1 show, this is not the case; under conditions that 
cause an alkene to undergo rapid addition, benzene reacts either not at all or very 

slowly. 

Table 10.1 CYCLOHEXENE vs. BENZENE 



Reagent 


Cyclohexene gives 


Benzene gives 


KMnO 4 
(cold, dilute, 
aqueous) 
Br 2 /CCl 4 
(in the dark) 
HI 


Rapid oxidation 

Rapid addition 
Rapid addition 


No reaction 

No reaction 
No reaction 



Rapid hydrogenation Slow hydrogenation 
at 25, 20 lb/in.2 at 100-200, 1500 lb/in.2 



322 BENZENE CHAP. 10 

In place of addition reactions, benzene readily undergoes a new set of reactions, 
all involving substitution. The most important are shown below. 



REACTIONS OF BENZENE 
1. Nitration. Discussed in Sec. 11.8. 



HON0 2 C 6 H 5 NO 2 + H 2 O 

Nitrobenzene 



2. Sulfonation. Discussed in Sec. 11.11. 



C 6 H 6 + HOSO 3 H -L> C 6 H 5 SO 3 H + H 2 O 
Benzenesulfonic acid 



3. Halogenation. Discussed in Sec. 11.10. 




C 6 H 5 C1 + HC1 
Chlorobenzene 



HBr 

Bromobenzene 

4. Friedel-Crafts alkylation. Discussed in Sees. 11.9 and 12.6. 

C 6 H 6 + RCI ^Q-+ C 6 H 5 R + HCI 
An alkylbenzene 

5. Friedel-Crafts acylation. Discussed in Sec. 19.6. 



C 6 H 6 + RCOC1 > CoHsCOR + HCI 

An acyl chloride A ketone 



In each of these reactions an atom or group has been substituted for one of the 
hydrogen atoms of benzene. The product can itself undergo further substitution 
of the same kind ; the fact that it has retained the characteristic properties of benzene 
indicates that it has retained the characteristic structure of benzene. 

It would appear that benzene resists addition, in which the benzene ring sys- 
tem would be destroyed, whereas it readily undergoes substitution, in which the 
ring system is preserved. 

10.5 Stability of the benzene ring. Heats of hydrogenation and combustion 

Besides the above qualitative indications that the benzene ring is more stable 
than we would expect cyclohexatriene to be, there exist quantitative data which 
show how much more stable. 

(e) Heats of hydrogenation and combustion of benzene are lower than expected, 
We recall (Sec. 6.3) that heat of hydrogenation is the quantity of heat evolved when 
one mole of an unsaturated compound is hydrogenated. In most cases the value is 
about 28-30 kcal for each double bond the compound contains. It is not surpris- 
ing, then, that cyclohexene has a heat of hydrogenation of 28.6 kcal and cyclo- 
hexadiene has one about twice that (55.4 kcal.) 



SEC. 10.6 



CARBON-CARBON BOND LENGTHS IN BENZENE 



323 



We might reasonably expect cyclohexatriene to have a heat of hydrogenation 
about three times as large as cyclohexene, that is, about 85.8 kcal. Actually, 
the value for benzene (49.8 kcal) is 36 kcal less than this expected amount. 

This can be more easily visualized, perhaps, by means of an energy diagram 
(Fig. 10.1), in which the height of a horizontal line represents the potential energy 
content of a molecule. The broken lines represent the expected values, based upon 
three equal steps of 28.6 kcal. The final product, cyclohexane, is the same in all 
three cases. 



Potential energy ** 


Cyclohexatriene -f 3H* 




Resonance 
energy < 
36 kcal Benzene 


~r 

Cyclohexadiene -f 2H> 
-f3H 2 




__ _ _ i 
Cyclohexene 4- Ha 


49.8 
(Obs) 


85.8 55.4 

(Calc)i (Obs) 

* 


57.2 28.6 
(Calc) (Obs) 

* 


Cyclohexane 



Figure 10.1. Heats of hydrogenation and stability: benzene, cyclohexa- 
diene, and cyclohexene. 

The fact that benzene evolves 36 kcal less energy than predicted can only mean 
that benzene contains 36 kcal less energy than predicted; in other words, benzene is 
more stable by 36 kcal than we would have expected cyclohexatriene to be. The 
heat of combustion of benzene is also lower than that expected, and by about the 
same amount. 



Problem 10.1 

(a) benzene + H 2 
hexene. 



From Fig. 10. 1 determine the A H of the following reactions: 
> 1,3-cyclohexadiene; (b) 1,3-cyclohexadiene + H 2 > cyclo- 



Problem 10.2 For a large number of organic compounds, the heat of combustion 
actually measured agrees rather closely with that calculated by assuming a certain 
characteristic contribution from each kind of bond, e.g., 54.0 kcal for each C H bond, 
49.3 kcal for each C C bond, and 117.4 kcal for each O=C bond (c/.y-l,2-disubsti- 
tuted). (a) On this basis, what is the calculated heat of combustion for cyclohexatriene? 
(b) How does this compare with the measured value of 789.1 kcal for benzene? 



Carbon-carbon bond lengths in benzene 

(/) All carbon- carbon bonds in benzene are equal and are intermediate Mength 
between single and double bonds. Carbon-carbon double bonds in a wide variety 
of compounds are found tcrbe about 1.34 A iong. Carbon-carbon single bonds, 
in which the nuclei are held together by only one pair of electrons, are considerably 
longer: 1.53 A in ethane, for example, 1.50 A in propylene, i.48 A in 1,3-butadiene. 



324 BENZENE CHAP. 10 

If benzene actually possessed three single and three double bonds, as in a 
Kekule structure, we would expect to find three short bonds (1.34 A) and three 
long bonds (1.48 A, probably, as in 1,3-butadiene). Actually, x-ray diffraction 
studies show that the six carbon-carbon bonds in benzene are equal and have a 
length of 1.39 A, and are thus intermediate between single and double bonds. 



10.7 ^ Resonance structure of benzene 

The Kekute structure of benzene, while admittedly unsatisfactory, was gener- 
ally used by chemists as late as 1945. The currently accepted structure did not 
arise from the discovery of new facts about benzene, but is the result of an exten- 
sion or modification of the structural theory; this extension is the concept of 
resonance (Sec. 6.23). 

H H 



C H H C^ X C H 

II I I II 

HCX 4CH 



The Kekule structures I and II, we now immediately recognize, meet the 
cgfldiiions* for -resonance: structures that differ only in the arrangement of elec- 
trons Benzene is a hybrid oFTand II. Since ; T aricTII a? ^HacTlT^quTvalent, and 
hence of exactly the same stability, they make equal contributions to the hybrid. 
And, also since I and II are exactly equivalent, stabilization due to resonance 
should be large. 

The puzzling aspects of benzene's properties now fall into place. The six 
bond lengths are identical because the six bonds are identical: they are one-and-a- 
half bonds and their length, 1.39 A, is intermediate between the lengths of single 
and double bonds. 

When it is realized that all carbon-carbon bonds in benzene are equivalent, 
there is no longer any difficulty in accounting for the number of isomeric disub- 
stitution products. It is clear that there should be just three, in agreement with 
experiment: 





3r 
1, 2- Dibromo benzene 1,3-Dibromobenzene 1 ,4-Dibromobenzene 

Finally, the "unusual" stability of benzene is not unusual at all: it is what 
one would expect of a hybrid of equivalent structures. The 36 kcal of energy that 
Benzene does not contain compared with cyclohexatriene is resonance energy. 
It is the 36 kcal of resonance energy that is responsible for the new set of proper- 
ties we <$11 aromatic properties^ 



SEC. 10.8 ORBITAL PICTURE OF BENZENE 325 

Addition reactions convert an alkene into a more stable saturated compound. 
Hydrogenation of cyclohexene, for example, is accompanied by the evolution of 
28.6 kcal; the product lies 28.6 kcal lower than the reactants on the energy scale 
(Fig. 10.1). 

But addition would convert benzene into a less stable product by destroying 
the resonance-stabilized benzene ring system; for example, according to Fig. 10.1 
the first stage of hydrogenation of benzene requires 5.6 kcal to convert benzene into 
the less stable cyclohexadiene. As a consequence, it is easier for reactions of 
benzene to take an entirely different course, one in which the ring system is retained : 
substitution. 

(This is not quite all of the story in so far as stability goes. As we shall see in 
Sec. 10.10, an additional factor besides resonance is necessary to make benzene 
what it is.) 

10.8 Orbital picture of benzene 

A more detailed picture of the benzene molecule is obtained from a considera- 
tion of the bond orbitals in this molecule. 

Since each carbon is bonded to three other atoms, it uses sp 2 orbitals (as in 
ethylcne, Sec. 5.2). These lie in the same plane, that of the carbon nucleus, and 
arc directed toward the corners of an equilateral triangle. If we arrange the six 
carbons and six hydrogens of benzene to permit maximum overlap of these 
orbitals, we obtain the structure shown in Fig. 10.2a. 



a 
120' 




120 



(fl) W 

Figure 10.2. Benzene molecule, (a) Only a bonds shown, (b) p orbitals 
overlap to form TT bonds. 

Benzene is aflat molecule, with every carbon and every hydrogen lying in the 
same plane. It is a very symmetrical molecule, too, with each carbon atom lying 
at the angle of a regular hexagon; every bond angle is 120. Each bond orbital is 
cylindrically symmetrical about the line joining the atomic nuclei and hence, as 
before, these bonds are designated as a bonds. 

The molecule is not yet complete, however. There are still six electrons to be 
accounted for. In addition to the three orbitals already used, each carbon atom 
has a fourth orbital, a p orbital. As we know, this p orbital consists of two equal 
lobes, one lying above and the other lying below the plane of the other three 
orbitals, that is, above and below the plane of the ring; it is occupied by a single 
electron. 

As in the case of ethylene, the/? orbital of one carbon can overlap thep orbital 
of an adjacent carbon atom, permitting the electrons to pair and an additional n 



326 BENZENE CHAP. 10 

bond to be formed (see Fig. 10.26). But the overlap here is not limited to a 
pair ofp orbitals as it was in ethylene; the/> orbital of any one carbon atom over- 
laps equally well the p orbitals of both carbon atoms to which it is bonded. The 
result (see Fig. 10.3)' i$ two continuous doughnut-shaped electron clouds, one. 
lying above and the other below the plane of the atoms. 



Figure 10.3. Benzene molecule. IT 
-H clouds above and below plane of 




As with the allyl radical, it is the overlap of the p orbitals in both directions, 
and the resulting participation of each electron in several bonds that corresponds 
to our description of the molecule as a resonance hybrid of two structures. Again 
it is the delocalization of the n electrons their participation in several bonds 
that makes the molecule more stable. 

To accommodate six IT electrons, there must be three orbitals (Sec. 2^.5). Their 
sum is, however, the symmetrical TT clouds we have described. 

The orbital approach reveals the importance of the planarity of the benzene 
ring. The ring is flat because the trigonal (sp 2 ) bond angles of carbon just fit the 
120 angles of a regular hexagon; it is this flatness that permits ihe overlap of the 
p orbitals in both directions, with the resulting delocalization and stabilization. 

The facts are consistent with the orbital picture of the benzene molecule. 
X-ray and electron diffraction show benzene (Fig. 10.4) to be a completely flat, 



Figure 10.4. Benzene molecule: 
shape and size. 
H 

symmetrical molecule with all carbon-carbon bonds equal, and all bond angles 
120. 

As we shall see, the chemical properties of benzene are just what we would 
expect of this structure. Despite delocalization, the TT electrons are nevertheless 
more loosely held than the a electrons. The ir electrons are thus particularly 
available to a reagent that is seeking electrons: the typical reactions of the benzene 
ring are those in which it serves as a source of electrons for electrophiiic (acidic) 
reagents. Because of the resonance stabilization of the benzene Ving, these reactions 
lead to substitution, in which the aromatic character of the benzene ring is preserved. 

Problem 10.3 The carbon-h>drogen bond dissociation energy for benzene (112 
kcal) is considerably larger than for cyclohexane. On the basis of the orbital picture 
of benzene, what is one factor that may be responsible for this? What piece of physical 
evidence tends to support your answer? (Hint: Look at Fig. 10.4 and tee See. 5.4.) 




SEC. 10.10 AROMATIC CHARACTER, THE HOCKEL 4n + 2 RULE 327 

Problem 10.4 The molecules of pyridine, C 5 H 5 N, are flat, with all bond angles 
about 120. All carbon-carbon bonds are 1.39 A long and the two carbon-nitrogen 
bonds are 1.36 A long. The measured heat of combustion is 23 kcal lower than that 
calculated by the method of Problem 10.2 on page 323. Pyridine undergoes such 
substitution reactions as nitration and sulfonation (Sec. 10.4). (a) Is pyridine ade- 
quately represented by formula I? (b) Account for the properties of pyridine by both 
valence-bond and orbital structures. (Check your answer in Sec. 31.6.) 







l 



Problem 10.5 The compound borazole, B 3 N 3 H 6 , is shown by electron diffrac- 
tion to have a flat cyclic structure with alternating boron and nitrogen atoms, and all 
boron-nitrogen bond lengths the same, (a) How would you represent borazole by 
valence-bond structures? (b) In terms of orbitals? (c) Ho\v many jr electrons are there, 
and which atoms have they "come from" ? 



10.9 Representation of the benzene ring 

For convenience we shall represent the benzene ring by a regular hexagon 
containing a circle (I); it is understood that a hydrogen atom is attached to each 
angle of the hexagon unless another atom or group is indicated. 

means a hybrid of || ] and 



I represents a resonance hybrid of the Kekule structures II and III. The straight 
lines stand for the a bonds joining carbon atoms. The circle stands for the cloud 
of six delocalized n electrons. (From another viewpoint, the straight lines stand 
for single bonds, and the circle stands for the extra half-bonds.) 

I is a particularly useful representation of the benzene ring, since it emphasizes 
the equivalence of the various carbon-carbon bonds. The presence of the circle 
distinguishes the benzene ring from the cyclohexane ring, which is often represented 
today by a plain hexagon. 

There is no complete agreement among chemists about how to represent the benzene 
ring. The student should expect to encounter it most often as one of the Kekule formulas. 
The representation adopted in this book has certain advantages, and its use seems to be 
gaining ground. It is interesting that very much the same representation was advanced 
as long ago as 1899 by Johannes Thiele (of the University of Munich), who used a broken 
circle to stand for partial bonds ("partial valences"). 

10.10 Aromatic character. The Httckel 4n + 2 rule 

We have defined aromatic compounds as those that resemble benzene. But 
just which properties of benzene must a compound possess before we speak of it 
as being aromatic? Besides the compounds that contain benzene rings, there are 
many other substances that are called aromatic; yet some of these superficially 
bear little resemblance to benzene. 

What properties do all aromatic compounds have in common? 



328 



BENZENE 



CHAP. 10 



From the experimental standpoint, aromatic compounds are compounds 
whose molecular formulas would lead us to expect a high degree of unsaturation, 
and yet which are resistant to the addition reactions generally characteristic of 
unsaturated compounds. Instead of addition reactions, we often find that these 
aromatic compounds undergo electrophilic substitution reactions like those of 
benzene. Along with this resistance toward addition and presumably the cause 
of it we find evidence of unusual stability: low heats of hydrogenation and low 
heats of combustion. Aromatic compounds are cyclic generally containing 
five-, six-, or seven-membered rings and when examined by physical methods, 
they are found to have flat (or nearly fiat) molecules. Their protons* show the 
same sort of chemical shift in nmr spectra (Sec. 13.8) as the protons of benzene 
and its derivatives. 

From a theoretical standpoint, to be aromatic a compound must have a 
molecule that contains|y>c7/c clouds of delocalized n electrons abgi^jmalbelow the 
plane of the w0/ecM/e])fuftrTeTmr^ 

nde^on^That is to say, for the particular degree of stability that characterizes 
an aromatic compound, delocalization alone is not enough. There must be a 
particular number of TT electrons: 2, or 6, or 10, etc. This requirement, called the 
4/1 + 2 rule or HUckel rule (after Erich Hiickel, of the Institut fiir theoretische 
Physik, Stuttgart), is based on quantum mechanics, and has to do with the filling 
up of the various orbitals that make up the TT cloud (Sec. 29.6). The HUckel rule 
is strongly supported by the facts. 

Let us look at some of the evidence supporting the Hiickel rule. Benzene has 
six TT electrons, the aromatic sextet \ six is, of course, a HUckel number, corre- 
sponding to n = 1. Besides benzene and its relatives (naphthalene, anthracene, 
phenanthrene, Chap. 30), we shall encounter a number of heterocyclic compounds 
(Chap. 31) that are clearly aromatic; these aromatic heterocycles, we shall see, are 
just the ones that can provide an aromatic sextet. 

Or, as further examples, consider these six compounds, for each of which just 
one contributing structure is shown: 

H H H 






Cyclopentadienyl 
cation 

Four n electrons 




Cyclopentadienyl 
radical 

Five * electrons 



H 



Cyclopentadienyl 
anion 

Six IT electrons 
Aromatic 



H 



Cycloheptatrienyl 

cation 

(Tropylium ion) 
Six n electrons 
Aromatic 





Cycloheptatrienyl 
radical 

Seven n electrons 



Cycloheptatrienyl 
anion 

Eight IT electrons 



SEC. 10.10 



AROMATIC CHARACTER. THE HCCKEL 4/1 + 2 RULE 



329 



Each molecule is a hybrid of either five or seven equivalent structures, with the 
charge or odd electron on each carbon. Yet, of the six compounds, only two give 
evidence of unusually high stability: the cyclopentadienyl anion and the cyclo- 
heptatrienyl cation (tropylium ion). 

For a hydrocarbon, cyclopentadiene is an unusually strong acid (K a = 10~ 15 ), 
indicating that loss of a hydrogen ion gives a particularly stable anion. (It is, 
for example, a much stronger acid than cycloheptatriene, K a = 10~ 45 , despite the 
fact that the latter gives an anion that is stabilized by seven contributing structures.) 
Dicyclopentadienyliron (ferrocene), [(C 5 H 5 )~] 2 Fe + + , is a stable molecule that has 
been shown to be a "sandwich" of an iron atom between two flat five-membered 
rings. All carbon-carbon bonds are^lA A long. The rings of ferrocene undergo 
two typically aromatic substitution reactions: sulfonation and the Friedel-Crafts 
reaction. 





Ferrocene 

Of the cycloheptatrienyl derivatives, on the other hand, it is the cation that is 
unusual. Tropylium bromide, C 7 H 7 Br, melts above 200, is soluble in water but 
insoluble in non-polar solvents, and gives an immediate precipitate of AgBr when 
treated with- silver nitrate. This is strange behavior for an organic bromide, and 
strongly suggests that, even in the solid, we are dealing with an ionic compound, 
R+Br~, the cation of which is actually a stable carbonium ion. 

Consider the electronic configuration of the cyclopentadienyl anion (Fig. 
10.5). Each carbon, trigonally hybridized, is held by a or bond to two other carbons 






(W 



(c) 



Figure 10.5. Cyclopentadienyl anion. (a) Two electrons in p orbital of 
one carbon ; one electron in p orbital of each of the other carbons, (b) Over- 
lap ofp orbitals to form -n bonds, (c) ir clouds above and below plane of 
ring; total of six IT electrons, the aromatic sextet. 

"and one hydrogen. The ring is a regular pentagon, whose angles (108) are not a 
bad fit for the 120 trigonal angle; any instability due to imperfect overlap (angle 
strain) is more than made up for by the delocalization that is to follow. Four 
carbons have ope electron each in p orbitals; the fifth carbon (the "one" that lost 
the proton, but actually, of course, indistinguishable from the others) has two 



286 ALICYCUC HYDROCARBONS CHAP. 9 

9.3 Industrial source 

We have already mentioned (Sec. 3.13) that petroleum from certain areas, 
(in particular California) is rich in cycloalkanes, known to the petroleum industry 
as naphthenes. Among these are cyclohexane, methylcyclohexane, methylcyclo- 
pentane, and 1,2-dimethylcyclopentane. 

These cycloalkanes are converted by catalytic reforming into aromatic hydro- 
carbons, and thus provide one of the major sources of these important compounds 
(Sec. 12.4). For example: 




[ 2 
HCH 3 



w 

3H 2 DehydrogewtioB 



CH 2 Toluene 

Methykyclohexane Aromatic 

Aliphatic 



Just as elimination of hydrogen from cyclic aliphatic compounds yields aro- 
matic compounds, so addition of hydrogen to aromatic compounds yields cyclic 
aliphatic compounds, specifically cyclohexane derivatives. An important example 
of this is the hydrogenation of benzene to yield pure cyclohexane. 



ru 4- 1W Ni. 150-250 -, , - 

C 6 H 6 + 3H 2 25a tm. ' ..I L* Hydrogenatioii 

Benzene 
Aromatic 



As we might expect, hydrogenation of substituted benzenes yields substituted 
cyclohexanes. For example: 




C 6 H 5 OH 4 

Phenol _ . 

Aromatic CH 2 

Cyclohexanol 
Aliphatic 

From cyclohcxanol many other cyclic compounds containing a six-membered 

ring can be made. 



9.4 Preparation 

Preparation of alicyclic hydrocarbons from other aliphatic compounds 
generally involves two stages: (a) conversion of some open-chain compound or 



SEC. 9.16 SUBSTITUTED CARBENES. a-ELIMINATlON 311 

both cis and trans products are formed. (Problem: Using the approach of Sec. 
7.12, assure yourself that this is so.) 

Besides addition, methylene undergoes another reaction which, quite literally, be- 
longs in a class by itself: insertion. 

I I 

-C-H + CH 2 - * -_C-CH 2 -H Insertion 

Methylene can insert itself into every carbon-hydrogen bond of most kinds of molecules. 
We cannot take time to say more here about this remarkable reaction, except that when 
addition is the desired reaction, insertion becomes an annoying side-reaction. 

Problem 9.14 In the gas phase, with low alkene concentration and in the pres- 
ence of an inert gas, addition of methylene to the 2-butenes is, we have seen, non- 
stereospecific. If, however, there is present in this system a little oxygen, addition 
becomes completely stereospecific (syn). Account in detail for the effect of oxygen. 
(Hint: See Sec. 2.14.) 

9.16 Substituted carbenes. a-Elimination 

A more generally useful way of making cyclopropanes is illustrated by the 
reaction of 2-butene with chloroform in the presence of potassium /ir/-butoxide 
(f-Bu ter/-butyl): 



CH 3 CH^CHCH 3 -T CHCI 3 "" CH 3 CH-CHCH 3 r-BuOH -r KC1 
2-Butene Chloroform C 

Cl Cl 
3,3-DichIoro- 1 ,2-dimcthylcyclopropanc 

The dichlorocyclopropanes obtained can be reduced to hydrocarbons or hydro- 
lyzed to ketones, the starting point for many syntheses (Chap. 19). 

Here, too, reaction involves a divalent carbon compound, a derivative of 
methylene: dichlorocarbene, :CCI 2 . It is generated in two steps, initiated by 
attack on chloroform by the very strong base, /er/-butoxide ion, and then adds 
to the alkene. 

(1) /-BuOr + H:CC1 3 =. :CC1 3 - -I- f-BuO:H 

(2) :CC1 3 - - > :CC1 2 + Cr 

Dichlorocarbene 

(3) CH 3 CH=CHCH 3 + :CC1 2 - > CH 3 CH--CHCH 3 

C 

a x ci 

It is believed that, because of the presence of the halogen atoms, the singlet 
form, with the electrons paired, is the. more stable form of dichlorocarbene, and 
is the one adding to the double bond. (Stabilization by the halogen atoms is 
presumably one reason wh> dihalocarbenes do not generally undergo the inser- 
tion reaction that is so characteristic of unsubstituted singlet methylene.) 

The addition of dihalocarbenes, like that of singlet methylene, is stereospecific 
and syn. 



312 ALICYCLIC HVDROCARBONS CHAP. 9 

Problem 9.15 (a) Addition of :CC1 2 to cyclopentene yields a single compound. 
What is it? (b) Addition of :CBrO to c>clopentene yields a mixture of stcreoiso- 
mers. In light of (a), how do you account* for this? What are the isorners likely to be? 
(Hint: Use models.) 

In dehydrohalogenation of alkyl halides (Sec. 5.13), we have already encoun- 
tered a reaction in which hydrogen ion and halide ion are eliminated from a mole- 
cule by the action of base; there H and X were lost from adjacent carbons, 
and so the process is called ^-elimination. In the generation of the methylene 
shown here, both H and --X are eliminated from the same carbon, and the process 
is called a-eliminarion. (Later on, in Sec. 24.12, we shall see some of the evidence 
for the mechanism of a-elimi nation shown above.) 



Beta-elimination 

/ \ 



-C-H 5f^> -C: 
>C 



Alpha-elimination 



Problem 9.16 (a) Why does CHCI 3 not undergo 0-elimination through the action 
of base? (b) What factor would you expect to make a-elimination from CHC1 3 easier 
than from, say, CH 3 Cl? 

There are many ways of generating what appear to be carbenes. But in some 
cases at least, it seems clear that no free carbene is actually an intermediate; 
instead, a carbenoid (carbene-like) reagent transfers a carbene unit directly to a 
double bond. For example, in the extremely useful Simmons-Smith reaction 

CH 2 l2 + Zn(Cu) - > ICH 2 ZnI 



ICH 2 Znl 



I 

c- 



I-Zn -I 




ZnI 2 



(H. E. Simmons and R. D. Smith of the du Pont Company) the carbenoid is an 
organozinc compound which delivers methylene stereospecifically (and without 
competing insertion) to the double bond. 



9.17 Analysis of alicyclic hydrocarbons 

A cyclopropane readily dissolves in concentrated sulfuric acid, and in this 
resembles an alkene or alkyne. It can be differentiated from these unsaturated 
hydrocarbons, however, by the fact that it is not oxidized by cold, dilute, neutral 
permanganate. 

Other alicyclic hydrocarbons have the same kind of properties as their open- 
chain counterparts, and they are characterized in the same way: cycloalkanes by 
their general inertness, and cycloalkenes and cycloalkynes by their response to 



SEC. 9.17 ANALYSIS OF ALICYCLIC HYDROCARBONS 313 

tests for unsaturation (bromine in carbon tetrachloride, and aqueous permanga- 
nate). That one is dealing with cyclic hydrocarbons is shown by molecular formu- 
las and by degradation products. 

The properties of cyclohexane, for example, show clearly that it is an alkane. 
However, combustion analysis and molecular weight determination show its 
molecular formula to be C 6 H 12 . Only a cyclic structure (although not necessarily 
a six-membered ring) is consistent with both sets of data. 

Similarly, the absorption of only one mole of hydrogen shows that cyclohexane 
contains only one carbon-carbon double bond; yet its molecular formula is 
Q>Hio which in an open-chain compound would correspond to two carbon- 
carbon double bonds or one triple bond. Again, only a cyclic structure fits the facts. 

Problem 9.17 Compare the molecular formulas of: (a) //-hexane and cyclo- 
hexane; (b) //-pentanc and c>clopentane; (c) I-hexene and cyclohexene; (d) dodecane, 
w-hexy Icyclohexane, and c>clohcx>lcyclohe\anc. (e) In general, how can you deduce 
the number of rings in a compound from its molecular formula and degree of un- 
saturation? 

Problem 9.18 What is the molecular formula of: (a) cyclohexane; (b) methyl- 
cyclopentane; (c) 1,2-dimethylcyclobutane? (d) Does the molecular formula give any 
information about the size of ring in a compound ? 

Problem 9.19 The yellow plant pigments -, /?-, and y-carohnc, and the reel pig- 
ment of tomatoes, lycoptnc, aie converted into Vitamin A in the iiver. All four have 
the molecular formula C m H 5ft . Upon catalytic hydrogenation, - and ^-carotene 
yield C 40 H 7 s, y-carotene >ields C 40 H HO and lycopene yields C 40 H s: - How many rings, 
if any, are there in each compound ? 

Cleavage products of cycloalkenes and cycloalkynes also reveal the cyclic 
structure. Ozonolysis of cyclohexene, for example, does not break the molecule 
into two aldehydes of lower carbon number, but simply into a single six-carbon 
compound containing two aldehyde groups. 

H 2 



CH 2 

Cyclohexene 



A di-aldehyde 
,CH 



KMn0 4 



. .COOH 
CH 2 

A di-acid 

Problem 9.20 Predict the ozonolysis products of: (a) cyclohexene; (b) 1-methyl- 
cyclopentene; (c) 3-methylc>clopentene; (d) 1,3-cyclohexadiene; (e) 1 ,4-cyclohexa- 
diene. 

Problem 9.21 Both cyclohexene and 1,7-octadiene yield the di-aldehyde 
OHC(CH 2 ) 4 CHO upon ozonolysis. What other facts would enable you to distinguish 
between the two compounds? 

(Analysis of cyclic aliphatic hydrocarbons by spectroscopy will be discussed 
in Sees. 13.15-13.16.) 



314 ALICYCLIC HYDROCARBONS CHAP. 9 

PROBLEMS 

1. Draw structural formulas of: 

(a) methylcyclopentane (f) cyclohexyteyclohexane 

(b) 1 -methylcyclohexene (g) cyclopentylacetylene 

(c) 3-methylcyclopentene (h) l,l-dimethyl-4-chlorocycloheptane 

(d) /rfl/7j-l,3-dichlorocyclobutane (i) bicyclo[2.2.1]hepta-2,5-diene 

(e) r/5-2-bromo-l -methylcyclopentane (j) l-chlorobicyclo[2.2.21octane 

2. Give structures and names of the principal organic products expected from each 
of the following reactions: 

(a) cyclopropane + C1 2 , FeCI 3 (j) 1 -methylcyclohexene + Br 2 (aq) 

(b) cyclopropane + C1 2 (300 J ) (k) 1 -methylcyclohexene 4 HBr 

(c) cyclopropane + cone. H 2 SO 4 (peroxides) 

(d) cyclopentane + C1 2 , FeCl 3 (I) 1,3-cyclohexadiene + HC1 

(e) cyclopentane + C1 2 (300) (m) cyclopentanol + H 2 SO 4 (heat) 

(f) cyclopentane + cone. H 2 SO 4 (n) bromocyclohexane + KOH(alc) 

(g) cyclopentene + Br 2 /CCl 4 (o) cyclopentene + cold KMnO 4 
(h) cyclopentene + Br 2 (300) (p) cyclopentene + HCO 2 OH 

(i) 1 -methylcyclohexene + HC1 (q) cyclopentene + hot KMnO 4 

(r) chlorocyclopentane + (C 2 H 5 ) 2 CuLi 
(s) 1-methylcyclopentene + cold cone. H 2 SO 4 
(t) 3-methylcyclopentene -f O 3 , then H 2 O/Zn 
(u) cyclohexene + H 2 SO 4 - > C i2 H 20 
(v) cyclopentene + CHG1 3 -f /-BuOKf 
(w) cyclopentene + CH 2 l2 + Zn(Cu) 

3. Outline all steps in the laboratory synthesis of each of the following from cyclo- 
hexanol. 

(a) cyclohexene (g) adipic acid, HOOC(CH ? ) 4 COOH 

(b) cyclohexane (h) bromocyclohexane 

(c) trans-l,2-d\ bromocyclohexane (i) 2-chlorocyclohexanol 

(d) cw-l,2-cyclohexanediol (j) 3-bromocyclohexene 

(e) /rn/;5-l,2-cyclohexanediol (k) 1,3-cyclohexadiene 

(f) OHC(CH 2 ) 4 CHO (1) cyclohexylcyclohexane 

(m) norcarane, bicyclo[4.1.0]heptane 

4. Give structure of all isomers of the following. For cyclohexane derivatives, 
planar formulas (p. 307) will be sufficient here. Label pairs of enantiomers, and meso 
compounds. 

(a) dichlorocyclopropanes (d) dichlorocyclohexanes 

(b) dichlorocyclobutanes (e) chloro-lj-dimethylcyclohexanes 

(c) dichlorocyclopentanes (f) 1,3,5-trichlorocyclohexanes 

(g) There are a number of stereoisomeric 1,2,3,4,5,6-hexachlorocyclohexanes. 
Without attempting to draw all of them, give the structure of the most stable isomer, and 
show its preferred conformation. 

5. (a) 2,5-Dimethyl-l,l-cyclopentanedicarboxylic acid (I) can be prepared as two 
optically inactive substances (A and B) of different m.p. Draw their structures, (b) Upon 
heating, A yields two 2,5-dimethylc>clopentanecarboxylic acids (II), and B yields only 
one. Assign structures to 4 and B. 



HOOC COOH H COOH 




PROBLEMS 315 

6. (a) The following compounds can be resolved into optically active enantiomers. 




3,3'-Diaminospiro[3.3]heptane 4-Methylcyclohexylideneacetic acid 

Using models and then drawing three-dimensional formulas, account for this. Label 
the chiral center in each compound. 

(b) Addition of bromine to optically active 4-methylcyclohexylideneacetic acid yields 
two optically active dibromides. Assuming a particular configuration for the starting 
material, draw stereochemical formulas for the products. 

7. (a) fraws-l,2-Dimethylcyclohexane exists about 99% in the diequatorial con- 
formation. /raAi,y-l,2-Dibromocyclohexane (or mwj-l,2-dichlorocyclohexane), on the 
other hand, exists about equally in the diequatorial and diaxial conformations; further- 
more, the fraction of the diaxial conformation decreases with increasing polarity of the 
solvent. How do you account for the contrast between the dimethyl and dibromo (or 
dirhloro) compounds? (Hint: See Problem 11, p. 141.) 

(b) If frafw-3-c/>4-dibromo-fc77-butylcyclohexane is subjected to prolonged heating, 
t is converted into an equilibrium mixture (about 50:50) of itself and a diastereomer. 
What is the diastereomer likely to be? How do you account for the approximately equal 
stability of these two diastereomers ? (Here, and in (c), consider the more stable confor- 
mation of each diastereomer to be the one with an equatorial /er/-butyl group.) 

(c) There are two more diastereomeric 3,4-dibromo-/er/-butylcyclohexanes. What 
are they? How do you account for the fact that neither is present to an appreciable 
extent in the equilibrium mixture? 

8. The compound decalin, Ci H 18 , consists of two fused cyclohexane rings: 



CO 



Decalm 



(a) Using models, show how there can be two isomeric decalins, cis and tram, (b) How 
many different conformations free of angle strain are possible for m-decalin? For trans- 
decalin ? (c) Which is the most stable conformation of c w-decalin ? Of /ra/?.s-decalin ? (Hint 
Consider each ring in turn. What are the largest substituents on each ring?) (d) Account for 
the fact that mws-decalin is more stable than m-decalin. (e) The difference in stability 
between cis- and /ra/w-decalin is about 2 kcal/mole; conversion of one into the other 
takes place only under very vigorous conditions. The chair and twist-boat forms of cyclo- 
hexane, on the other hand, differ in stability by about 6 kcal/mole, yet are readily inter- 
converted at room temperature. How do you account for the contrast? Draw energy 
curves to illustrate your answer. 

9. Allinger (p. 305) found the energy difference between cis- and /n?/f5-l,3-di-/*r/- 
butylcyclohexane to be 5.9 kcal/mole, and considers that this value represents the energy 
difference between the chair and twist-boat forms of cyclohexane. Defend Allinger's 
position. 

10. It has been suggested that in certain substituted cyclopentanes the ring exists 
preferentially in the "envelope" form: 




316 ALICYCLIC HYDROCARCONS CHAP. 9 

Using models, suggest a possible explanation for each of the following facts: 

(a) The attachment of a methyl group to the badly strained cyclopentane ring raises 
the heat of combustion very little more than attachment of a methyl group to the un- 
strained cyclohexane ring. (Hint: Where is the methyl group located in the "envelope" 
form?) 

(b) Of the 1 ,2-dimethylcyclopentanes, the /ra//s-isomer is more stable than the cis. 
Of the 1,3-dimethyIcyclopentanes, on the other hand, the m-isomer is more stable than 
the trans. 

(c) The c fr-isomer of methyl 3-methylcyclobutanecarboxylate 



-,COOCH 3 



CH 3 
is more stable than the trans-isomer 

11. Each of the following reactions is carried out, and the products are separated by 
careful distillation, recrystallization, or chromatography. For each reaction tell how 
many fractions will be collected. Draw a stereochemical formula of the compound or 
compounds making up each fraction. Tell whether each fraction, as collected, will be 
optically active or optically inactive. 

(a) (R)-S-hydroxycyclohexene + KMnO 4 > C 6 H 12 O 3 ; 

(b) (R)-3-hydroxycyclohexene + HCO 2 OH > C 6 H 12 O 3 ; 

(c) (S,S)-l,2-dichlorocyclopropane + C1 2 (300) > C 3 H3C1 3 ; 

(d) racemic 4-methylcyclohexene 4- Br 2 /CCJ 4 . 

12. Outline all steps in a possible laboratory synthesis of each of the following, 
using alcohols of four carbons or fewer as your only organic source, and any necessary 
inorganic reagents. (Remember: Work backwards.) 

(a) cis-l ,2-di(//-propyl)cyclopropane ; 

(b) racemic fra/?s-l-methyl-2-ethyl-3,3-dichlorocyclopropane* 

13. Describe simple chemical tests that would distinguish between : 

(a) cyclopropane and propane 

(b) cyclopropane and propylene 

(c) 1,2-dimethylcyclopropane and cyclopentane 

(d) cyclobutane and 1-butene 

(e) cyclopentane and 1-pentene 

(f) cyclopentane and cyclopentene 

(g) cyclohexanol and //-butylcyclohexane 

(h) 1,2-dimethylcyclopentene and cyclopentanol 

(i) cyclohexane, cyclohexene, cyclohexanol, and bromocyclohexane 

14. How many rings does each of the following contain? 

(a) Camphane, C 10 H J8 , a terpene related to camphor, takes up no hydrogen, (b) Chole- 
stane, C 27 H 48 , a steroid of the same ring structure as cholesterol, cortisone, and the sex 
hormones, takes up no hydrogen, (c) fi-Phellandrene, Ci Hi 6 , a terpene, reacts with 
bromine to form C, Hi 6 Br 4 . (d) Ergocakiferol (so-called "Vitamin D 2 "), C 2 8H 44 O, 
an alcohol, gives C 2 8H 52 O upon catalytic hydrogenation. (e) How many double bonds 
does ergocalciferol contain? 

15. On the basis of the results of catalytic hydrogenation, how many rings does each 
of the following aromatic hydrocarbons contain? 

(a) benzene (C 6 H 6 ) > C 6 Hi 2 (e) phenanthrene (C H H 10 ) > C J4 H 24 

(b) naphthalene (C 10 H 8 ) > C, H, 8 (f) H-benzpyrene (C 20 H 12 ) > C 20 H 32 

(c) toluene (dHd > C 7 H M (g) chrysene (C 18 H, 2 ) > C I8 H 30 

(d) anthracene (C 14 H 10 > * Q 4 H 24 

(Check your answers by use of the index.) 



SEC. 10.10 



AROMATIC CHARACTER. THE HCCKEL 4 + 2 RULE 



329 



Each molecule is a hybrid of either five or seven equivalent structures, with the 
charge or odd electron on each carbon. Yet, of the six compounds, only two give 
evidence of unusually high stability: the cyclopentadienyl anion and the cyclo- 
heptatrienyl cation (tropylium ion). 

For a hydrocarbon, cyclopentadiene is an unusually strong acid (K a = 10" ls ), 
indicating that loss of a hydrogen ion gives a particularly stable anion. (It is, 
for example, a much stronger acid than cycloheptatriene, K a = 10 ~ 45 , despite the 
fact that the latter gives an anion that is stabilized by seven contributing structures.) 
Dicyclopentadienyliron (ferrocene), [(C 5 H 5 )~] 2 Fe + + , is a stable molecule that has 
been shown to be a "sandwich" of an iron atom between two flat five-membered 
rings. All carbon-carbon bonds are* 1.4 A long. The rings of ferrocene undergo 
two typically aromatic substitution reactions: sulfonation and the Friedel-Crafts 
reaction, \ 





Ferrocene 

Of the cycloheptatrienyl derivatives, on the other hand, it is the cation that is 
unusual. Tropylium bromide, C 7 H 7 Br, melts above 200, is soluble in water but 
insoluble in non-polar solvents, and gives an immediate precipitate of AgBr when 
treated with- silver nitrate. This is strange behavior for an organic bromide, and 
strongly suggests that, even in the solid, we are dealing with an ionic compound, 
R + Br~, the cation of which is actually a stable carbonium ion. 

Consider the electronic configuration of the cyclopentadienyl anion (Fig. 
10.5). Each carbon, trigonally hybridized, is held by a a bond to two other carbons 





(a) (b) (c) 

Figure 10.5. Cyclopentadienyl anion. (a) Two electrons in p orbital of 
one carbon ; one electron in p orbital of each of the other carbons, (b) Over- 
lap of p orbitals to form TT bonds, (c) IT clouds above and below plane of 
ring; total of six TT electrons, the aromatic sextet. 

and one hydrogen. The ring is a regular pentagon, whose angles (108) are not a 
bad fit for the 120 trigonal angle; any instability due to imperfect overlap (angle 
strain) is more than made up for by the delocalization that is to follow. Four 
carbons have one electron each in p orbitals; the fifth carbon (the "one" that lost 
the proton, but actually, of course, indistinguishable from the others) has two 



330 BENZENE CHAP. 10 

electrons. Overlap of the p ofbitals gives rise to n clouds containing a total of 
six electrons, the aromatic sextet. 

In a similar way, we arrive at the configuration of the t ropy li urn ion. It is a 
regular heptagon (angles 128.5). Six carbons contribute one p electron each, and 
the seventh contributes only an empty p orbital. Result: the aromatic sextet. 

The ions are conveniently represented as: 




Cyclopentadienyl Cycloheptatrienyl 

ani6n* cation 

(Tropylium ion) 

Six is the Hiickel number most often encountered, and for good reason. To 
provide p orbitals, the atoms of the aromatic ring must be trigonally (sp 2 ) hybrid- 
ized, which means, ideally, bond angles of 120. To permit the overlap of the 
p orbitals that gives rise to the ir cloud, the aromatic compound must be flat, or 
nearly so. The number of trigonally hybridized atoms that will fit a flat ring 
without undue angle strain (i.e., with reasonably good overlap for n bond 
formation) is five, six, or seven. Six is the Hiickel number of tr electrons that can 
be provided as we have just seen by these numbers of atoms. (It is surely no 
coincidence that benzene, our model for aromatic character, is the "perfect" 
specimen: six carbons to provide six ?r electrons and to make a hexagon whose 
angles exactly match the trigonal angle.) 

Now, what evidence is there that other Hiickel numbers 2, 10, 14, etc. are 
also "magic" numbers? We cannot expect aromatic character necessarily to 
appear here in the form of highly stable compounds comparable to benzene and its 
derivatives. The rings will be too small or too large to accommodate trigonally 
hybridized atoms very well, so that any stabilization due to aromaticity may be 
largely offset by angle strain or poor overlap of p orbitals, or both. 

We must look for stability on a comparative basis as was done above with 
the cyclopentadienyl and cycloheptatrienyl derivatives and may find evidence of 
aromaticity only in the fact that one molecular species is less unstable than its 
relatives. The net effect of a great deal of elegant work is strongly to support the 
4/i -f 2 rule. The question now seems rather to be: over how unfavorable a 
combination of angle strain and multiple charge can aromaticity manifest itself? 

Problem 10.6 Ronald Breslow (of Columbia University) found that treatment 
of 3-chlorocyclopropene with SbCI 5 yields a stable crystalline solid, I, of formula 



H C! 

3-Chloioc>clopropene 

C 3 H 3 SbCl , insoluble in non-polar solvents but soluble in polar solvents like nitro- 
methane, acetonitrile, or sulfur dioxide. The nmr spectrum of I shows three exactly 
equivalent protons. 3-Chlorocyclopropene reacts with AgBF 4 to give AgC! and a 
solution \vtih .in nmr spectrum identical to that of I. Treatment of I with chloride 
ion regenerates 3-chlorocyi'iopropcnc. 



SEC. 10.11 



NOMENCLATURE OF BENZENE DERIVATIVES 



331 



Conversion of I into C 3 H 3 + by electron impact (Sec. 5.16) requires 235 kcal/ 
mole, as compared with 255 kcal/mole for conversion of allyl chloride into CjH 5 +. 

(a) Give in detail the most likely structure of I, and show how this structure ac- 
counts for the various observations, (b) Of what theoretical significance are these 
findings? 

Problem 10.7 \ 9 3 t 5J-Cyclooctatetraene, C 8 H 8 , has a heat of combustion (com- 
pare Problem 10.2, p. 322) of 1095 kcal; it rapidly decolorizes cold aqueous KMnO 4 
and reacts with Br 2 /CCl 4 to yield C 8 H 8 Br 8 . (a) How should its structure be repre- 
sented? (b) Upon what theoretical grounds might one have predicted its structure and 
properties? (c) Treatment of cyclooctatetraene with potassium metal has been found 
to yield a stable compound 2K+C 8 H 8 ~ ~ Of what significance is the formation of 
this salt? (d) Using models, suggest a possible shape (or shapes) for cyclooctatetraene. 
What shape would you predict for the C 8 H 8 ~ ~ anion? 

10.11 Nomenclature of benzene derivatives 

In later chapters we shall consider in detail the chemistry of many of the 
derivatives of benzene. Nevertheless, for our present discussion of the reactions 
of the benzene ring it will be helpful for us to learn to name some of the more 
important of these derivatives. 

For many of these derivatives we simply prefix the name of the substituent 
group to the word -benzene, as, for example, in chlorobenzene, bromobenzene, 
iodobenzene, or nitrobenzene. Other derivatives have special names which may 

Cl Br J NO 2 



Chlorobenzene Bromobenzene Iodobenzene Nitrobenzene 

show no resemblance to the name of the attached substituent group. For example, 
methylbenzene is always known as toluene, aminobenzene as aniline* hydroxy- 
benzene as phenol, and so on. The most important of these special compounds 
are: 

CH 3 NH 2 QH COOH SO 3 H 



Toluene 



Aniline 



Phenol 



Ben/oic acid 



Ben/cncsulfortic 
acid 



If several groups are attached to the benzene ring, we must not only tell what 
they are, but also indicate their relative positions. The three possible isomers of a 
disubstituted benzene are differentiated by the use of the names ortho, meta, and 
para. For example: 





o-Di bromobenzene 

0/7/70 



m-Dibromoben/ene 
meta 



/j-Dibromobenzene 
pare. 



332 



BENZENE 



CHAP. 10 



If the two groups are different, and neither is a group that gives a special narne to 
the molecule, we simply name the two groups successively and end the word with 
-benzene, as, for example, chloronitrobenzene, bromoiodobenzene, etc. If one of 
the two groups is the kind that gives a special name to the molecule, then the 
compound is named as a derivative of that special compound, as, for example, 
nitrotoluene, bromophenol, etc. 
I 



NO 2 




S0 3 H 



/>-Bromoiodobenzene 



w-Chloronitrobcn/cne 



CH 3 




N0 2 



o-Nitrotoluene 




/^Chlorobenzenesulfonic 
acid 



NH 2 




Br 

/7-Broniophenol 



w-Nitrobenzoic 
acid 



r>-lodoamlme 



If more than two groups are attached to the benzene ring, numbers are used 
to indicate their relative positions. For example : 




)H 







1 ,2,4-Tribromobenzene 



NO 2 

2-Chloro-4-nitrophenol 



2,6~Dinitrotoluene 



N0 2 





3-Bromo-5-chIoronitrobenzene 2,4,6- Tnbromoaniline 

If all the groups are the same, each is given a number, the sequence being the one 
that gives the lowest combination of numbers; if the groups are different, then the 
last-named group is understood to be in position 1 and the other numbers conform 
to that, as, for example, in 3-bromo-5-chloronilrobenzene. If one of the groups 
that gives a special name is present, then the compound is named as having the 
special group in position 1 ; thus in 2,6-dinitrotoluene the methyl group is considered 
to be at the 1 -position. 



Problem 10.8 You have three bottles containing the three isomeric dibromo- 
benzenes; they have the melting points + 87, + 6, and - T. By a great deal of work, 



SEC. 10.12 QUANTITATIVE ELEMENTAL ANALYSIS 333 



you prepare six dibromonitroben7enes (CaHjB^NC^) and find that, of the six, one 
is related to (derived from or convertible into) the dibromobenzene of m.p. +87, 
two to the isomer of m.p. +6, arid three to the isomer of m.p. -7. 

Label each bottle with the correct name of ortho, meta, or para. 

(This work was actually carried out by Wilhelm Korner, of the University of 
Milan, and was the first example of the Kdrner method of absolute orientation.) 



10.12 Quantitative elemental analysis: nitrogen and sulfur 

This chapter has dealt with the structure of benzene and with some of its 
reactions. It is well to remind ourselves again that all this discussion has meaning 
only because it is based upon solid facts. As we saw earlier (Sec. 2.24), we can 
discuss the structure and reactions of a compound only when we know its molec- 
ular formula and the molecular formulas of its products. 

To know a molecular formula we must know what elements are present in 
the compound, and in what proportions. In Sec. 2.25 we saw how various elements 
can be detected in an organic compound, and in Sec. 2.26 how the percentage of 
carbon, hydrogen, and halogen can be measured. 

^Quantitative analysis for nitrogen is carried out either (a) by the Dumas 
method or (b) by the Kjeldahl method. The Kjeldahl method is somewhat more 
convenient, particularly if many analyses must be carried out; however, it cannot be 
used for all kinds of nitrogen compounds. 

In the Dumas method, the organic compound is passed through a tube con- 
taining, first, hot copper oxide and, next, hot copper metal gauze. The copper 
oxide oxidizes the compound (as in the carbon-hydrogen combustion, Sec. 2.26), 
converting combined nitrogen into molecular nitrogen. The copper gauze reduces 
any nitrogen oxides that may be formed, also to molecular nitrogen. The nitrogen 
gas is collected and its volume is measured. For example, an 8.32-mg sample of 
aniline yields 1.11 cc of nitrogen at 21 and 743 mm pressure (corrected for "the 
vapor pressure of water). We calculate the volume at standard temperature and 
pressure, 

vol. N 2 at S.T.P. = 1.1 1 x 273 27 + 3 21 x ^ = 1.01 cc 
and, from it, the weight of nitrogen, 

wt. N = ~~ x (2 x 14.01) = 0.00126 g or 1.26 mg 



and, finally, the percentage of nitrogen in the sample 

100-15.2% 



Problem 10.9 Why is the nitrogen in the Dumas analysis collected over 50% 
aqueous KOH rather than, say, pure water, aqueous NaCl, or mercury? 

In the Kjeldahl method, the organic compound is digested with concentrated 
sulfuric acid, which converts combined nitrogen into ammonium sulfate. The 
solution is then made alkaline. The ammonia thus liberated is distilled, and its 
amount is determined by titration with standard acid. For example, the ammonia 



334 BENZENE CHAP. 10 

formed from a 3.51-mg sample of aniline neutralizes 3.69 ml of 0.0103 N acid. 
For every milliequivalent of acid there is a milliequivalent of ammonia, and a 

milligram-atoms N = mjlliequivalents NH 3 = milliequivalents acid 

- 3.69 x 0.0103 - 0.0380 

milligram-atom of nitrogen. From this, the weight and, finally, the percentage of 
nitrogen in the compound can be calculated. 

wt. N milligram-atoms N x 14.01 = 0.0380 x 14.01 = 0.53 mg 

%N " 531 X 10 " 15 ' 1% 

Sulfur in an organic compound is converted into sulfate ion by the methods 
used in halogen analysis (Sec. 2.26) : treatment with sodium peroxide or with nitric 
acid (Carius method). This is then converted into barium sulfate, which is weighed. 

Problem 10.10 A Dumas nitrogen analysis of a 5.72-mg sample of p-phenylene- 
diamine gave 1.31 cc of nitrogen at 20 and 746 mm. The gas 'was collected over satu- 
rated aqueous KOH solution (the vapor pressure of water, 6 mm). Calculate the 
percentage of nitrogen in the compound. 



1 10.11 A Kjeldahl nitrogen analysis of a 3.88-mg sample of ethanolamine 
required 5.73 ml of 0.0110 N hydrochloric acid for titration of the ammonia produced. 
Calculate the percentage of nitrogen in the compound. 

Problem 10.12 A Carius sulfur analysis of a 4.81 -mg sample of p-toluenesulfonic 
acid gave 6.48 mg of BaSO 4 . Calculate the percentage of sulfur in the compound. 

Problem 10.13 How does each of the above answers compare with the theoret- 
ical value calculated from the formula of the compound? (Each compound is listed 
in the index.) 



PROBLEMS 



1. Draw structures of: 



(a) p-dinitrobenzene (g) rnesitylene ( 1, 3 >5-tri methyl benzene) 

(b) m-bromonitrobenzene (h) 3,5-dinitrobenzenesulfonic acid 

(c) 0-chIorobenzoic acid (i) 4-chloro-2,3-dinitrotoluene 

(d) m-nitrotoluene (j) 2-amino-5-bromo-3-nitrobenzoic acid 

(e) p-bromoaniline (k) p-hydroxybenzoic acid 

(f) m-iodophenol (0 2,4,6-trinitrophenol (picric acid) 

2. Give structures and names of all the possible isomeric: 

(a) xylenes (dimethylbenzenes) (d) dibromonitrobenzenes 

(b) aminobenzoic acids (H 2 NC 6 H 4 COOH) (e) bromochlorotoluenes 

(c) trimethylbenzenes (f) trinitrotoluenes 

3. (a) How many isomeric monosubstitution products are theoretically possible 
from each of the following structures of formula C 6 H 6 ? (b) How many disubstitution 
products? (c) Which structures, if any, would be acceptable for benzene on the basis of 
isomer number? 



PROBLEMS 335 



CH 2 

A 



v 
CH 2 H 2 C 



4. Give structures and names of all theoretically possible products of the ring 
mononitration of: 

(a) o-dichlorobenzene (g) 0-chloronitrobenzene 

(b) w-dichlorobenzene (h) w-chloronitrobenzene 

(c) p-dichlorobenzene (i) p-chloronitrobenzene 

(d) 0-bromochlorobenzene (j) 1,3,5-trimethylbenzene 

(e) m-bromochlorobenzene (k) 4-bromo-l,2-dimethylbenzene 

(f) p-bromochlorobenzene (1) p-ethyltoluene 

5. Give structures and names of all benzene derivatives that theoretically can have 
the indicated number of isomeric ring-substituted derivatives. 

(a) C 8 H 10 : one monobromo derivative (e) C 9 Hi 2 : two mononitro derivatives 

(b) C 8 H 10 : two monobromo derivatives (f) C 9 H J2 : three mononitro derivatives 

(c) C 8 H 10 : three monobromo derivatives (g) C 9 H J2 : four mononitro derivatives 

(d) C 9 Hi 2 : one mononitro derivative 

6. There are three known tribromobenzenes, of m.p. 44, 87, and 120. Could 
these isomers be assigned structures by use of the Korner method (Problem 10.8, p. 332)? 
Justify your answer. 

7. For a time the prism formula VI, proposed in 1869 by Albert Ladenburg of 
Germany, was considered as a possible structure for benzene, on the grounds that it 
would yield one monosubstitution product and three isomeric disubstitution products. 

CH CH 



CH- 



CH 



CH 

VI 

(a) Draw Ladenburg structures of three possible isomeric dibromobenzenes. 

(b) On the basis of the Korner method of absolute orientation, label each Ladenburg 
structure in (a) as ortho, meta, or para. 

(c) In light of Chap. 4, can the Ladenburg formula actually pass the test of isomer 
number? 

(Derivatives of Ladenburg " benzene," called prismanes, have actually been made.) 

8. In 1874Griess (p. 1077) reported that he had decarboxylated the six known di- 
aminoben/oic acids, C 6 Hi(NH 2 )>COOH, to the diaminobenzenes. Three acids gave a 
diamine of m.p. 63, two acids gave a diamine of m.p. 104, and one acid gave a diamine 
of m.p. 142. Draw the structural formulas for the three isomeric diaminobenzenes and 
label each with its melting point. 



336 BENZENE: CHAP. 10 

9. For which of the following might you expect aromaticity (geometry permitting) ? 

(a) The annulenes containing up to 20 carbons. (Annulenes are monocyclic compounds 
of the general formula [ CH=CH ].) 

(b) The monocyclic polyenes C 9 H 10 , C 9 H 9 +, C 9 H 9 ~. 

10. The properties of pyrrole, commonly represented by VII, 



H 

vn 

show that it is aromatic. Account for its aromaticily on the basis of orbital theory. (Hint: 
See Sec. 10.10. Check your answer in Sec. 31.2.) 

11. When benzene is treated with chlorine under the influence of ultraviolet light, 
a solid material of m.wt. 291 is formed. Quantitative analysis gives an empirical forrmila 
of CHC1. (a) What is the molecular formula of the product? (b) What is a possible struc- 
tural formula? (c) What kind of reaction has taken place? (d) Is the product aromatic? 
(e) Actually, the product can be separated into six isomeric compounds, one of whicn is 
used as an insecticide (Gammexane or Lindane). How do these isomers differ fzoin each 
other? (f ) Are more than six isomers possible? 

12. Can you account for the following order of acidity. (Hint: See Sec. 8.10.) 

acetylene > benzene > /7-pentane 



Chapter 

n 



Electrophilic 
Aromatic Substitution 



11.1 Introduction 

We have already seen that the characteristic reactions of benzene involve 
substitution, in which the resonance-stabilized ring system is preserved. What kind 
of reagents bring about this substitution? What is the mechanism by which these 
reactions take place? 

Above and below the plane of the benzene ring there is a cloud of TT electrons. 
Because of resonance, these TT electrons are more involved in holding together 
carbon nuclei than are the TT electrons of a carbon-carbon double bond. Still, in 
comparison with a electrons, these n electrons are loosely held and are available 
to a reagent that is seeking electrons. 



Figure 11.1. Benzene ring: TT cloud is source of 
electrons. 




H 



It is not surprising that in its typical reactions the benzene ring serves as a 
source of electrons, that is, as a base. The compounds with which it reacts are 
deficient in elections, that is, are electrophilic reagents or acids. Just as the typical 
reactions of the alkenes are electrophilic addition reactions, so the typical reactions 
of the benzene ring are electrophilic substitution reactions. 

These reactions are characteristic not only of benzene itself, but of the benzene 
ring wherever it is foundand, indeed, of many aromatic rings, benzenoid and 
non-benzenoid. 

337 



338 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

Electrophilic aromatic substitution includes a wide variety of reactions: 
nitration, halogenation, sulfonation, and Friedel-Crafts reactions, undergone by 
nearly all aromatic rings; reactions like nitrosation and diazo coupling, undergone 
only by rings of high reactivity; and reactions like desulfonation, isotopic exchange, 
and many ring closures which, although apparently unrelated, are found on closer 
examination to be properly and profitably viewed as reactions of this kind. In 
synthetic importance electrophilic aromatic substitution is probably unequaled by 
any other class of organic reactions. It is the initial route of access to nearly all 
aromatic compounds: it permits the direct introduction of certain substituent 
groups which can then be converted, by replacement or by transformation, into 
other substituents, including even additional aromatic rings. 



ELECTROPHILIC AROMATIC SUBSTITUTION 

Ar s aryl, any aromatic group with attachment directly to ring carbon 
1. Nitration. Discussed in Sec. 11.8. 



ArH + HONO 2 *+ ArNO 2 + H 2 O 
A nitro compound 



2. Sulfonation. Discussed in Sec. 11.9. 



ArH + HOSOjH s 3 , ArSO 3 H + H 2 O 
A sulfonic acid 

3. Halogenation. Discussed in Sec. 11.11. 

ArH + C1 2 ^- ArCl + HC1 
An aryl chloride 

ArH + Br 2 -^-* ArBr + HBr 
An aryl bromide 

4. Friedel-Craftealkylation. Discussed in Sec. 11.10. 

ArH + RC1 AICI J> ArR + HC1 
An a! kyl benzene 

5. Friedel-Crafts acyiation. Discussed in Sec. 19.6. 

ArH + RCOCI AIC1 3 > ArCOR + HCI 
An acyl chloride A ketone 

6. Protonation. Discussed in Sec. 11.12. 

ArSO 3 H + H + H 2 o > ArH + H 2 SO 4 Desulfonation 

ArH + D* - > ArD + H + Hydrogen exchange 



SEC. 11.2 EFFECT OF SUBSTITUENT GROUPS 

7. Thallation. Discussed in Sec. 11.13. 



339 



ArH + T1(OOCCF 3 ) 3 

Thallium 
trifluoroacetate 



ArTl(OOCCF 3 ) 2 + CF 3 COOH 

An arylthallium 
ditrifluoroacetate 



8. Nitrosation. Discussed in Sees. 23.11 and 24.10. 

ArH -f HONO > ArN=O + H 2 O Only for highly 

A nitroso compound reactive ArH 

9. Diazo coupling. Discussed in Sec. 23.17. 

ArH + Ar'N 2 + X~ > ArN=NAr' + HX Only for highly 

A diazonium salt An azo compound reactive ArH 

10. Kolbe reaction. Discussed in Sec. 24.11. Only for phenols. 

11. Reimer-Tiemann reaction. Discussed in Sec. 24.12. Only for phenols. 



11.2 Effect of substituent groups 

Like benzene, toluene undergoes electrophilic aromatic substitution : sulfona- 
tion, for example. Although there are three possible monosulfonation pro- 
ducts, this reaction actually yields appreciable amounts of only two of them : the 
o- and /Msomers. 




and 



Toluene 



SO 3 H 

/>-Toluene- 

sulfonic 

acid 

62% 




and 6% m-isomcr 



0-Toluene- 

sulfonic 

acid 

32% 



Benzene and toluene are insoluble in sulfuric acid, whereas the sulfonic acids 
are readily soluble; completion of reaction is indicated simply by disappearance of 
the hydrocarbon layer. When shaken with fuming sulfuric acid at room tempera- 
ture, benzene reacts completely within 20 to 30 minutes, whereas toluene is found 
to react within only a minute or two. 

Studies of nitration, halogenation, and Friedel-Crafls alkylation of toluene 
give analogous results. In some way the methyl group makes the ring more 
reactive than unsubstituted benzene, and directs the attacking reagent to the 
ortho and para positions of the ring. 

On the other hand, nitrobenzene, to take a different example, has been found 
to undergo substitution more slowly than benzene, and to yield chiefly the meta 
isomer. 



340 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

Like methyl or nitro, any group attached to a benzene ring affects the reactivity 
of the ring and determines the orientation of substitution. When an electrophilic 
reagent attacks an aromatic ring, it is the group already attached to the ring that 
determines how readily the attack occurs and where it occurs. 

A group that makes the ring more reactive than benzene is called an activating 
group. A group that makes the ring less reactive than benzene is called a 
deactivating group. 

A group that causes attack to occur chiefly at positions ortho and para to it is 
called an ortho,para director. A group that causes attack to occur chiefly at 
positions meta to it is called a meta director. 

In this chapter we shall examine the methods that are used to measure these 
effects On reactivity and orientation, the results of these measurements, and a theory 
that accounts for these results. The theory is, of course, based on the most likely 
mechanism for electrophilic aromatic substitution ; we shall see what this mechanism 
is, and some of the evidence supporting it. First let us look at the facts. 



11.3' Determination of orientation 

To determine the effect of a group on orientation is, in principle, quite simple: 
the compound containing this group attached to benzene is allowed to undergo 
substitution and the product is analyzed for the proportions of the three isomers. 
Identification of each isomer as ortho, meta, or para generally involves comparison 
with an authentic sample of that isomer prepared by some other method from a 
compound whose structure is known. In the last analysis, of course, all these 
identifications go back to absolute determinations of the Kdrner type (Problem 
10.8, p. 332). 

In this way it has been found that every group can be put into one of two 
classes: ortho,para directors or meta directors. Table 11.1 summarizes the orienta- 
tion of nitration in a number of substituted benzenes. Of the five positions open 
to attack, three (60%) are ortho and para to the substituent group, and two (40%) 
are meta to the group; if there were no selectivity in the substitution reaction, we 

Table 11.1 ORIENTATION OF NITRA'IION OF C 6 H 5 Y 
Y Ortho Para Ortho plus para Meta 



OH 


50-55 


45-50 


100 


trace 


NHCOCH 3 


19 


79 


98 


2 


-CH 3 


58 


38 


96 


4 


-F 


12 


88 


100 


trace 


Cl 


30 


70 


100 


trace 


Br 


37 


62 


99 


1 


I 


38 


60 


98 


2 


NO 2 


6.4 


0.3 


6.7 


93.3 


-N(CH 3 ) 3 + 





11 


11 


89 


CN 








19 


81 


COOH 


19 


1 


20 


80 


SO 3 H 


21 


7 


28 


72 


CHO 








28 


72 



SEC. 11.4 DETERMINATION OF RELATIVE REACTIVITY 341 

would expect the ortho and para isomers to make up 60% of the product, and the 
meta isomer to make up 40%. We see that seven of the groups direct 96-100% 
of nitration to the ortho and para positions; the other six direct 72-94% to the 
meta positions. 

A given group causes the same general kind of orientation predominantly 
ortho.para or predominantly meta whatever the electrophilic reagent involved. 
The actual distribution of isomers may vary, however, from reaction to reaction. 
In Table 11.2, for example, compare the distribution of isomers obtained from 
toluene by sulfonation or bromination with that obtained by nitration. 

Table 11.2 ORIENTATION OF SUBSTTTUTION IN TOLUENE 
Ortho Meta Para 



Nitration 


58 


4 


38 


Sulfonation 


32 


6 


62 


Bromination 


33 





67 



11.4 Determination of relative reactivity 

A group is classified as activating if the ring it is attached to is more reactive 
than benzene, and is classified as deactivating if the ring it is attached to is less 
reactive than benzene. The reactivities of benzene and a substituted benzene are 
compared in one of the following ways. 

The time required for reactions to occur under identical conditions can be 
measured. Thus, as we just saw, toluene is found to react with fuming sulfuric acid 
in about one-tenth to one-twentieth the time required by benzene. Toluene is 
more reactive than benzene, and -zCH^ js therefore an activating group. 

The severity of conditions required for comparable reaction to occur within the 
same period of time can be observed. For example, benzene is nitrated in less 
than an hour at 60 by a mixture of concentrated sulfuric acid and concentrated 
nitric acid; comparable nitration of nitrobenzene requires treatment at 90 with 
fuming nitric acid and concentrated sulfuric acid. Nitrobenzene is evidently less 
reactive than benzene, and the nitro group, NO 2 , is a deactivating group. 

For an exact, quantitative comparison under identical reaction conditions, 
competitive reactions can be carried out, in which the compounds to be compared 
are allowed to compete for a limited amount of a reagent (Sec. 3.22). For example, 
if equimolar amounts of benzene and toluene are treated with a small amount of 
nitric acid (in a solvent like nitromethane or acetic acid, which will dissolve both 



HN0 3 



HN0 3 




TH-N0 2 o,m,and/> 




342 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

organic and inorganic reactants), about 25 times as much nitrotoluene as nitro- 
benzene is obtained, showing that toluene is 25 times as reactive as benzene. On 
the other hand, a mixture of benzene and chlorobenzene yields a product in which 
nitrobenzene exceeds the nitrochlorobenzenes by 30: 1, showing that chlorobenzene 
is only one-thirtieth as reactive as benzene. The chloro group is therefore classified 
as deactivating, the methyl group as activating. The activation or deactivation 
caused by some groups is extremely powerful: aniline, C 6 H 5 NH 2 , is roughly one 
million times as reactive as benzene, and nitrobenzene, C 6 H 5 NO 2 , is roughly 
one-millionth as reactive as benzene. 



11.5 ' Classification of substituent groups 

The methods described in the last two sections have been used to determine 
the effects of a great number of groups on electrophilic substitution. As shown in 
Table 1 1.3, nearly all groups fall into one of two glasses: activating and ortho.para- 
directing, or deactivating and wete-directing. The halogens are in a class by them- 
selves, being deactivating but 0r//70,/?<7ra-directing. 

Table 11.3 EFFECT OF GROUPS ON ELECTROPHILIC AROMATIC SUBSTITUTION 

Activating: Ortho.para Directors Deactivating: Meta Directors 

Strongly activating NO 2 

NH 2 ( NHR, NR 2 ) N(CH 3 ) 3 + 

OH CN 

COOH ( COOR) 
Moderately activating SO 3 H 

OCH 3 ( OC 2 H 5 , etc.) CHO, COR 

NHCOCH 3 

Deactivating: Ortho,para Directors 

Weakly activating F, Cl, Br, I 

-C 6 H 5 
CH 3 ( C 2 H 5 ,etc.) 

Just by knowing the effects summarized in these short lists, we can now predict 
fairly accurately the course of hundreds of aromatic substitution reactions. We 
now know, for example, that bromination of nitrobenzene will yield chiefly the 
Aw-isomer and that the reaction will go more slowly than the bromination of ben- 
zene itself; indeed, it will probably require severe conditions to go at all. We 
now know that nitration of C 6 H 5 NHCOCHj, (acetanilide) will yield chiefly the 0- 
and /Msomers and will take place more rapidly than nitration of benzene. 

Although, as we shall see, it is possible to account for these effects in a reason- 
able way, it is necessary for the student to memorize the classifications in Table 
11.3 so that he m&y deal rapidly with synthetic problems involving aromatic 
compounds. 



11.6 Orientation in disubstituted benzenes 

The presence of two substituents on a ring makes the problem of orientation 
more complicated, but even here we can frequently make very definite predictions. 
First of all, the two substituents may be located so that the directive influence of 



C. 11.6 



ORIENTATION IN DISUBSTITUTED BENZENES 



343 



e reinforces that of the other; for example, in I, II, and III the orientation clearly 
ist be that indicated by the arrows. 



N0 2 






in 



On the other hand, when the directive effect of one group opposes that of the 
icr, it may be difficult to predict the major product; in such cases complicated 
xtures of several products are often obtained. 

Even where there are opposing effects, however, it is still possible in certain 
les to make predictions in accordance with the following generalizations. 

(a) Strongly activating groups generally win out over deactivating or weakly 
tivating groups. The differences in directive power in the sequence 

-NH 2 , -OH > -OCH 3 , -NHCOCH 3 > -C 6 H 5 , -CH 3 > meta directors 
j great enough to be used in planning feasible syntheses. For example: 



OH 



HNO >f H,SO 4 



OH 

(of* 



CH 3 


CH 3 

Sole product 


NHCOCH 3 

r\] ** FeBr y 


NHCOCH 3 

f~*\ T 

Chief product 


CH 3 




Br 2 . FcBr, 



OH 



CHO 



Chief product 



ere must be, however, a fairly large difference in the effects of the two groups for 
ar-cut results; otherwise one gets results like these: 




HNOj, HjSO^ 




A 

and 




(b) There is often little substitution between two groups that are meta to each 
er. In many cases it seems as though there just is not enough room between 



344 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

two groups located meta to each other for appreciable substitution to occur there, 
as illustrated by IV and V: 




62% 

Nitration 
IV 



11.7 Orientation and synthesis 

As we discussed earlier (Sec. 3.14), a laboratory synthesis is generally aimed 
at obtaining a single, pure compound. Whenever possible we should avoid use of 
a reaction that produces a mixture, since this lowers the yield of the compound we 
want and causes difficult problems of purification. With this in mind, let us see 
some of the ways in which we can apply our knowledge of orientation to the 
synthesis of pure aromatic compounds. 

First of all, we must consider the order in which we introduce these various 
substituents into the ring. In the preparation of the brom on itro benzenes, for 
example, it is obvious that if we nitrate first and then brominate, we will obtain the 
w-isomer; whereas if we brominate first and then nitrate, we will obtain a mixture 
of the o- and p-isomers* The order in which we decide to carry out the two steps, 
then, depends upon which isomer we want. 



N 2 




m-Bromonitrobenzene 

Br Br 

- Br " Fc ^ HN0 3 .H,so 4> [Q] N0 2 and (O) 

N0 2 

Bromonitrobenzcne 
ortho- para- 

38% 62% 

Next, if our synthesis involves conversion of one group into another, we must 
consider the proper time for this conversion. For example, oxidation of a methyl 
group yields a carboxyl group (Sec. 12.10). In the preparation of nitrobenzoic 
acids from toluene, the particular product obtained depends upon whether 
oxidation or nitration is carried out first. 

Substitution controlled by an activating group yields a mixture of ortho and 
para isomers; nevertheless, we must often make use of such reactions, as in the 
examples just shown. It is usually possible to obtain the pure para isomer from 
Jthe mixture by fractional crystallization. As the more symmetrical isomer, it is 
tKe less soluble (Sec. 12.3), and crystallizes while the solvent still retains the soluble 



SEC. 11.8 MECHANISM OF NITRATION 345 

ortho isomer. Some para isomer, of course, remains in solution to contaminate 
the ortho isomer, which is therefore difficult to purify. As we shall see, special 
approaches are often used to prepare ortho isomers. 

COOH COOH 

KMnO 4 r/^l HNO,, 



m-Nitrobenzoic acid 
r-u- 
T IUenC HN " H * SO ? mi 2 and 



K 2 Cr 2 7 





NO 2 

o-Nitrobenzoic acid p-Nitrobenzoic acid 

In the special case of nitro compounds, the difference in boiling points is often 
large enough that both ortho and para isomers can be obtained, pure by fractional 
distillation. As a result, many aromatic compounds are best prepared not by 
direct substitution bu.t by conversion of one group into another, in the last analysis 
starting from an original nitro compound; we shall take up these methods of 
conversion later. 

A goal of aromatic synthesis is control of orientation: the preparation, at 
will and from the same substrate, of a pure ortho, a pure meta, or a pure para 
isomer. Steps toward this goal have been taken very recently by Edward C. Taylor 
(Princeton University) and Alexander McKillop (University of East Anglia), 
chiefly through the chemistry of thallium: thallium as the cation in organic salts; 
thallium salts as Lewis acids; arylthallium compounds (Sec. 11.13) as reactive 
organometallic intermediates. One approach to regicspecific substitution involves 
complexing attachment through a Lewis acid-base reaction of the attacking 
reagent by some other molecule. Complexing of the reagent by the substituent 
group prior to reaction tends to favor attack at the nearest position : ortho. Com- 
plexing of the reagent by a bulky molecule tends to favor attack at the least 
crowded position : para. If reaction can be carried out so that orientation is governed, 
not by relative rates of reaction as it usually is but by position of equilibrium, 
then the most stable isomer is favored: often the meta isomer. We shall see examples 
of all these ways of controlling orientation. 

fl.8 Mechanism of nitration 

Now that we have seen the effects that substituent groups exert on (^dentation 
and reactivity in electrophilic aromatic substitution, let us see how we can account 



346 



ELECTROPHILIC AROMATIC SUBSTITUTION * 



CHAP. 11 



for these effects. The first step in doing this is to examine the mechanism for the 
reaction. Let us begin with nitration, using benzene as the aromatic substrate. 

The commonly accepted mechanism for nitration with a mixture of nitric and 
sulfuric acids (the widely used "mixed acid" of the organic chemist) involves the 
following sequence of reactions: 



(1) HONO 2 + 2H 2 SO 4 



2HSO 4 - + ONO 2 

Nitronium ion 



(2) 



(3) 



0N0 2 + C 6 H 6 



OV 
C 6 H< 



HS<V 



H 

f 

5 

X NO 2 
C 6 H 5 NO 2 -f H 2 SO 4 



Slow 



Fast 



NO 2 



Step (1) generates the nitronium ion, cNO 2 , which is the electrophilic particle 
that actually attacks the benzene ring. This reaction is simply an acid-base 
equilibrium in which sulfuric acid serves as the acid and the much weaker nitric 
acid serves as a base. We may consider that the very strong acid, sulfuric acid, 
causes nitric acid to ionize in the sense, HO % ~ . . . + NO 2 , rather than in the usual 
way, H+ . . . ~ONO 2 . The nitronium ion is well known, existing in salts such as 
nitronium perchlorate, NO 2 + C1O 4 ~, and nitronium fluoborate, NO 2 +BF 4 ~. 
Indeed, solutions of these stable nitronium salts in solvents like nitromethane or 
acetic acid have been found by George Olah (of Case Western Reserve University) 
to nitrate aromatic compounds smoothly and in high yield at room temperature. 

Needing electrons, the nitronium ion finds them particularly available in the TT 
cloud of the benzene ring, and so in step (2) attaches itself to one of the carbon 
atoms by a covalent bond. This forms the carbonium ion, 

H 



N0 2 

often called a benzenonium ion. 

Just what is the structure of this carbonium ion ? We find that we can repre- 
sent it by three structures (I, II, and III) that differ from each other only in position 
of double bonds and positive charge. The actual ion must then be a reson;mcqf 
hybrid of these three structures. 




H N0 2 



H N0 2 




represented as 



lir 



H N0 2 



IV 



This means, of course, that the positive charge is not localized on one carbon 
atom, but is distributed over the molecule, being particularly strong on the carbon 



SEC. 11.9 MECHANISM OF SULFONA1ION 347 

atoms ortho and para to the carbon bearing the NO 2 group. (As we shall see 
later, this ortho.para distribution is significant.) The dispersal of the positive 
charge over the molecule by resonance makes this ion more stable than an ion with 
a localized positive charge. It is probably because of this stabilization that the 
carbonium ion forms at all, in view of the stability of the original benzene itself. 
Sometimes the hybrid carbonium ion is represented as IV, where the broken line 
stands for the fractional bonds due to the delocalized TT electrons. 

Thus far the reaction is like addition to alkenes: an electrophilic particle, 
attracted by the n electrons, attaches itself to the molecule to form a carbonium 
ion. But the fate of this carbonium ion is different from the fate of the ion formed 
from an alkene. Attachment of a basic group to the benzenonium ion to yield the 
addition product would destroy the aromatic character of the ring. Instead, the 
basic ion, HSO 4 ", abstracts a hydrogen ion (step 3) to yield the substitution pro- 
duct, which retains the resonance-stabilized ring. Loss of a hydrogen ion, as we have 
seen, is one of the reactions typical of a carbonium ion (Sec. 5.20); it is the preferred 
reaction in this case. 

As with other carbonium ion reactions we have studied, it is the formation 
of the carbonium ion (step 2) that is the more difficult step; once formed, the 
carbonium ion rapidly loses a hydrogen ion (step 3) to form the products. (We 
shall see proof of this in Sec. 11.16.) 

Electrophilic substitution, then, like electrophilic addition, is a stepwise 
process involving an intermediate carbonium ion. The two reactions differ, how- 
-ever, in the fate of the carbonium ion. While the mechanism of nitration is, per- 
haps, better established than the mechanisms for other aromatic substitution 
reactions, it seems clear that all these reactions follow the same course. 

Problem 11.1 Nitration by nitric acid alone is believed to proceed by essentially 
the same mechanism as nitration in the presence of sulfuric acid. Write an equation for 
the generation of NO 2 * from nitric acid alone. 



Mechanism of sulfonation 

Sulfonation of many aromatic compounds involves the following steps: 

(1) 2H 2 S0 4 ;r H 3 + + HS<V + S0 3 

H 

(2) S0 3 + C 6 H 6 ;^ C 6 frT Slow 



6 

so 3 - 



e/ H 

(3) C 6 H( + HS<V ^= C 6 H 5 S<V + H 2 SO 4 Fast 



(4) C 6 H 5 SO 3 - + H 3 O* J QH 5 SO 3 H + H 2 O Equilibrium far to the left 

Again the first step, which generates the electrophilic sulfur trioxide, is simply 
an ,'icid-base equilibrium, this time between molecules of sulfuric acid. For 



348 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

sulfonation we commonly use sulfuric acid containing an excess of SO 3 ; even if this 
is not done, it appears that SO 3 formed in step (1) can be the electrophile. 



S:O: 



In step (2) the electrophilic reagent, SO 3 , attaches itself to the benzene ring 
to form the intermediate carbonium ion. Although sulfur trioxide is not positively 
charged, it is electron-deficient, and hence an acid, nevertheless. 

Step (3) is the loss of a hydrogen ion to form the resonance-stabilized sub- 
stitution product, this time the anion of benzenesulfonic acid which, being a strong 
acid, is highly dissociated (step 4). 

With some aromatic substrates and at certain acidities, the electrophile may 
be HSO 3 + or molecules that can readily transfer SO 3 or HSO 3 + to the aromatic 
ring. 

Problem 11.2 Write an equation for the formation from H 2 SO 4 of each of the 
following sulfonating electrophiles: (a) H 3 SO 4 + ; (b) HS<V ; (c) H 2 S 2 O 7 . 



UrliO Mechanism of Friedel-Crafts alkylation 

In Friedel-Crafts alkylation, the electrophile is typically a carbonium ion. 
It, too, is formed in an acid-base equilibrium, this time in the Lewis sense: 

(1) RC1 + A1C1 3 Zjz A1C1 4 - + R 

H 



(2) R + C 6 H 6 ;r C 6 u Slow 

R 

e/ H 

(3) C 6 H 5 + A1C1 4 - 5Z C 6 H 5 R + HC1 + A1C1 3 Fast 

R 

In certain cases, there is no free carbonium ion involved. Instead, the alkyl 
group is transferred without a pair of electrons directly to the aromatic ring 
from the polar complex, I, between A1C1 3 and the alkyl halide: 

a H 

Cl At-O R + C 6 H 6 - > C 6 H^ + A1C1 4 - Slow 

a R 

I 

The electrophile is thus either (a) R + or (b) a molecule like I that can readily 
transfer R + to the aromatic ring. This duality of mechanism is common In electro- 
philic aromatic substitution. In either case, the Lewis acid R 4 " is displaced from RC1 
by the other Lewis acid, A1C1 3 . 

We speak of the Friedel-Crafts reaction as electrophilic substitution and, from the 
viewpoint of the aromatic ring, it is. But, just as an acid reacts with a base, so an electro- 
phile reacts with a nucleophile (nucleus-lover), a molecule which can provide the electrons 
that the electrophile seeks. From the opposite point of view, then, this reaction involves 



SEC. 11.11 MECHANISM OF HALOGENATTON 349 

nucleophilic attack by the aromatic ring on the alkyl group of complex I. The A1C1 4 ~ ion 
is a better leaving group than Cl~ would be; the Lewis acid, A1C1 3 , serves the same pur- 
pose here that a Lowry-Br0nsted acid does in protonation of an alcohol (Sec. 5.20). 

As we shall find out when we take up the Friedel-Crafts reaction as a synthetic 
tool (Sec. 12.6), the Friedel-Crafts reaction in its widest sense involves reactants 
other than alkyl halides and Lewis acids other than aluminium chloride: Bp3, 
SnCl 4 , HF, and even H+. 

Problem 11.3 How do you account for the fact that benzene in the presence of 
A1C1 3 reacts: (a) uith //-propyl bromide to give isoprop>lbenzene; (b) with isobutyl 
bromide to yield /r/7-butylbenzene; (c) with neopentyl bromide to yield /err-pehtyl- 
benzenc? (d) By which of the alternative mechanisms for the Friedel-Crafts reaction 
are these products probably formed ? 

Problem 11.4 Write all steps in the most likely mechanism for the reaction of 
benzene: (a) with tert-buiyi alcohol in the presence of H : SLO 4 to yield /v7-butyl benzene; 
(b) with propylene in the presence of H 3 PO 4 to form isopropylbenzene. 



11.1J/ Mechanism of halogcnat ion 

Aromatic halogenation, illustrated for chlorination, involves the following 
steps. 

e & 

(1) C1 2 -f FeCl 3 ^= ChFe-CICl 

II 

( (7, - P H 

(2) a 3 Fc-oi-Cl 4- C 6 H 6 > C 6 H^ + FeCl 4 - Slow 

11 X C1 

H 

(3) C 6 ils + FeCl 4 - v C 6 H 5 C1 + HC1 + FeCl 3 fta 

N C1 

The key step (2) is the attachment of positive chlorine to the aromatic ring, 
ft seems unlikely, though, that an actually free Cl + ion is involved. Instead, ferric 
chloride combines with C1 2 to form complex II, from which chlorine is transferred, 
without its electrons, directly to the ring. 

Addition of halogens to alkenes, we have seen (Sec. 6.13), similarly involves 
attack by positive halogen to form an intermediate carbonium ion. The loosely 
held TT electrons of an alkene make it more reactive, however, and positive halogen 
is transferred from the halogen molecule itself, X 2 , with loss of Cl~. The less 
reactive benzene molecule needs the assistance of a Lewis acid; reaction occurs 
with the loss of the better leaving group, FeCl 4 ~. Indeed, more highly reactive 
aromatic compounds, i.e., those whose it electrons are more available, do react with 
halogens in the absence of any added Lewis acid. 

Problem 11.5 Certain activated benzene rings can be chlorinated by hypo- 
chlorous acid, HOC1, and tlv's reaction is catalyzed by H^. In light of the above dis- 
cussion, can you suggest a possible function of H+ ? 

Problem 11.6 Aromatic bromination catalyzed by the Lewis acid thallium ace- 
tate, Tl(OOCCHjh, gives onl> the para isomer. Suggest an explanation for this regio- 
specificity. (Hint: See Sec. 11.7.) 



350 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

11.12 Desulfonation. Mechanism of protonation 

When an aromatic sulfonic acid is heated to 100-175 with aqueous acid, it 
is converted into sulfuric acid and an aromatic hydrocarbon. This desulfonation 
is the exact reverse of the sulfonation process by which the sulfonic acid was 
originally made. 

C 6 H 6 + H 2 S0 4 ~I C 6 H 5 S0 3 H + H 2 O 
Hydrocarbon Sulfonic acid 

Volatile Non-volatile 

By applying the usual equilibrium principles, we can select conditions that will 
drive the reaction in the direction we want it to go. To sulfonate we use a large 
excess of concentrated or fuming sulfuric acid; high concentration of sulfonating 
agent and low concentration of water (or its removal by reaction with SOj) shift 
the equilibrium toward sulfonic acid. To desulfonate we use dilute acid and often 
pass superheated steam through the reaction mixture ; high concentration of water 
and removal of the relatively volatile hydrocarbon by steam distillation shift the 
equilibrium toward hydrocarbon. 

According to the principle of microscopic reversibility (p. 170), the mechan- 
ism of desulfonation must be the exact reverse of the mechanism of sulfonation. 



(1) C 6 H 5 S0 3 - + H+ 5= CH, 

so 3 - 
ov H 

(2) C 6 H 5 5= C 6 H 6 + S0 3 

X so 3 - 

The reaction is simply another example of electrophilic aromatic substitution. The 
electrophile is the proton, H+, and the reaction is protonation or, more specifically, 
protodesulfonation. 

Sulfonation is unusual among electrophilic aromatic substitution reactions 
in its reversibility. It is also unusual in another way: in sulfonation, ordinary 
hydrogen (protium) is displaced from an aromatic ring about twice as fast as deu- 
terium. These two facts are related to each other and, as we shall see in Sec. 11.16, 
give us a more detailed picture of sulfonation and of electrophilic aromatic substi- 
tution in general. 

Problem 11.7 Predict the product or products of: (a) monobromination of 
toluene; (b) monobromination of /Molucnesulfonic acid followed by treatment with 
acid and superheated steam, (c) Using the principle of (b), and following the guide- 
lines of Sec. 11.7, outline a synthesis from benzene of 0-dibromobenzene; of 
o-bromochlorobenzene. 



11/13 Thallation 

" Treatment of aromatic compounds with thallium trifluoroacetate, 
T1(OOCCF 3 ) 3 , dissolved in trifluoroacetic acid (CF 3 COOH) gives rapidly and in 



SEC. 11.13 



THALLATION 



351 



high yield arylthallium ditrifluoroacetates, stable crystalline compounds. Reaction 
is believed by Taylor and McKillop (p. 345) to involve electrophilic attack on the 
aromatic ring by the (Lewis) acidic thallium. 



ArH + T1(OOCCF 3 ) 3 

Thallium 
trifluoroacetate 



ArTl(OOCCF 3 ) 2 + CF 3 COOH 

Arylthallium 
ditrifluoroacctate 



Thallium compounds are very poisonous, and must be handled with extreme care. 

Although substituent groups affect the reactivity of the aromatic substrate 
as expected for electrophilic substitution, orientation is unusual in a number of 
ways, and it is here that much of the usefulness of thallation lies. Thallation is al- 
most exclusively para to R, Cl, and -OCH 3 , and this is attributed to the bulk 
of the electrophile, thallium trifluoroacetate, which seeks out the uncrowded para 
position. 

Thallation is almost exclusively ortho to certain substituents like COOH, 
COOCH 3 , and CH 2 OCH 3 (even though some of these are normally meta- 
directing), and this is attributed to prior complexing of the electrophile with the 
substituent; thallium is held at just the right distance for easy intramolecular 
delivery to the ortho position. For example: 




CH '\A 



TUOOCCF,), 




TI(OOCCF 3 ) 3 



cocx: H 3 



Methyl benzoate 



0-Carbomcthoxyphenyhhallium 
ditnfluoroacetalte 

Only product 



(In C 6 H 5 CH 2 CH 2 COOH, however, it is evidently held too far from the ring, and 
must leave the substituent before attacking the ring intermolecularly at the para 
position.) 

Thallation is reversible, and when carried out at a higher temperature (73 
instead of room temperature) yields the more stable isomer: usually the meta 
(compare 1,2- and 1,4-addition, Sec. 8.22). For example: 



25 



CH(CH 3 ) 2 



TKOOCCF,), 



73 




Rate-controlled 



JT1(OOCCF 3 ) 2 

85% meta 



Equilibrium-controlled 



352 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. II 

Now, these arylthallium compounds are useful, not in themselves, but as 
intermediates in the synthesis of a variety of other aromatic compounds. Thallium 
can be replaced by other atoms or groups which cannot themselves be introduced 
directly into the aromatic ring - or at least not with the same regiospecificity. In 
this way one can prepare phenols (ArOH, Sec. 24.5) and aryl iodides (Sec. 25.3). 
Direct iodination of most aromatic rings does not work very well, but the process 
of thallation followed by treatment with iodide ion gives aryl iodides in high yields. 

ArH + T1(OOCCF 3 ) 3 CFjCO "> ArTI(OOCCF 3 ) 2 - K -^ Arl 



11.14 Mechanism of electrophilic aromatic substitution: a summary 

Electrophilic aromatic substitution reactions seem, then, to proceed by a 
single mechanism, whatever the particular reagent involved. This can be sum- 
marized for the reagent YZ as follows: 

H 
(1) C 6 H 6 + Y+ - -> C 6 nf Slow 



H 
(2) C 6 Hf + :Z- > C 6 H 5 Y + H:Z Fas. 



Two essential steps are involved: (1) attack by an electrophilic reagent upon the 

H 

ring to form a carbonium ion, C 6 H< , and (2) abstraction of a hydrogen ion from 

Y 

this carbonium ion by some base. In each case there is a preliminary acid-base 
reaction which generates the attacking particle; the actual substitution, however, is 
contained in these two steps. 

Most of the support for this mechanism comes from evidence about the nature 
of the attacking particle in each of these reactions: evidence, that is, that sub- 
stitution is electrophilic. This evidence, in turn, comes largely from kinetics, 
augmented by various other observations: the nitrating power of preformed 
nitronium salts (Sec. 1 1.8), for example, or carbonium ion-like rearrangements in 
some Friedel-Crafts alkylations (Problem 11.3 above). The electrophilic nature 
of these reactions is supported in a very broad way by the fact that other reactions 
which show the same reactivity and orientation features also fit into the same 
mechanistic pattern. 

Problem 11.8 In each of the following reactions, groups on the ring under at- 
tack exert the kinds of effects summarized in Sec. 11.5. Suggest a likely electrophile 
in each case, and write a likely mechanism. 

(a) ArH + R-C-C1 Alcl ' 

il 
O 

(b) ArH + Ar'N 2 + Cr > Ar-N-=N-Ar' 

(c) ArH + HONO ^-> Ar NO + H 2 O 



SEC 11.15 ISOTOPE EFFECTS 353 

Problem 11.9 When phenol is treated with D 2 SC>4 in D Z O (deuterium sulfate in 
heavy water), there is formed phenol containing deuterium instead of hydrogen at 
positions ortho and para to the OH group. Benzene undergoes similar exchange but 
at a much lower rate; under the same conditions benzenesulfonic acid does not under- 
go exchange at all. (a) Outline the most probable mechanism for hydrogen- deuterium 
exchange in aromatic compounds, (b) What is the attacking reagent in each case, and 
to what general class does this reaction belong? 

But this is only part of the mechanism. Granting that substitution is electro- 
philic, how do we know that it involves two steps, as we have shown, and not just 
one! And how do we know that, of the two steps, the first is much slower than 
the second? To understand the answer to these questions, we must first learn 
something about isotope effects. 



11.15 Isotope effects 

Different isotopes of the same element have, by definition, the same electronic 
configuration, and hence similar chemical properties. This similarity is the basis 
of the isotopic tracer technique (Sec. 3.29): one isotope does pretty much what 
another will do, but, from its radioactivity or unusual mass, can be traced through 
a chemical sequence. 

Yet different isotopes have, also by definition, different masses, and because 
of this their chemical properties are not identical: the same reactions can occur but 
at somewhat Different rates (or, for reversible reactions, with different positions of 
equilibrium)/^ difference in rate (or position of equilibrium) due to a difference in 
the isotope present in the reaction system is called an isotope effect. \ 

Theoretical considerations, which we cannot go into, suprforted by much 
experimental evidence, lead to the conclusion: if a particular atom is less tightly 
bound in the transition state of a reaction than in the reactant, the reaction involving 
the heavier isotope of that atom will go more slowly. The hydrogen isotopes have 
the greatest proportional differences i,i mass: deuterium (D) is twice as heavy as 
protium (H), and tritium (T) is three times as heavy. As a result, hydrogen isotope 
effects are the biggest, the easiest to measure, and because of the special impor- 
tance of hydrogen in organic chemistry the most often studied. (If you doubt the 
importance of hydrogen, look at the structure of almost any compound in this 
book.) 

One kind of reaction in which an atom is less tightly bound in the transition 
state than in the reactant is a reaction in which a bond to that atom is being broken. 
Isotope effects due to the breaking of a bond to the isotopic atom are called 
primary isotope effects. They are in general the biggest effects observed for a 
particular set of isotopes. 

In this book we shall be concerned with primary hydrogen isotope effects, 
which amount to this: a bond to protium (H) is broken faster than a bond to deuter- 
ium (D). For many reactions of this kind, 

(1) ~C H + Z ^~> [~C~H~Z] > ~C + H Z 

(2) -C-D + Z -* [~C~D~Z] -C + D-Z 



354 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

in which hydrogen is abstracted as an atom, positive ion, or negative ion, deu- 
terium isotope effects (/e H /& D ) in the range 5 to 8 (at room temperature) have been 
observed ; that is to say, the reaction is 5 to 8 times as fast for ordinary hydrogen 
as for deuterium. (Tritium isotope effects, k H /k T , are about twice as large as 
deuterium isotope effects.) 

These differences in rate can be measured in a variety of ways. In some 
cases, the rates of the two individual reactions (1) and (2) can be measured directly 
and the results compared. Usually, however, it is more feasible, as well as more 
satisfactory, to use our familiar method of competition (Sec. 3.22) in either of two 
ways. 

In intermolecular competition, a mixture of labeled and unlabeled reactants 
compete for a limited amount of reagent; reactions (1) and (2) thus go on in the 
same mixture, and we measure the relative amounts of H Z and D Z r .oduced. 
(Sometimes, larger amounts of the reagent Z are used, and the relative amounts of 
the two reactants ordinary and labeledleft unconswned are measured; the less 
reactive will have been used up more slowly and will predominate. The relative 
rates of reaction can be calculated without much difficulty.) 

In intramolecular competition, a single reactant is used which contains several 
equivalent positions, some labeled and some not: 

Product(D) 



Product(H) 



One can then measure either the relative amounts of H Z and D Z, or the 
relative amounts of the D-containing product formed by reaction (3) and the 
H-containing product formed by reaction (4). 

Problem 11.10 (a) When excess toluene-a-d (C 6 H 5 CH 2 D) was photochemically 
monochlorinated at 80 with 0.1 mole of chlorine, there were obtained 0.0212 mole 
DC1 and 0.0868 mole HC1. What is the value of the isotope effect * H /fc D (per hydrogen 
atom, of course)? (b) What relative amounts of DC1 and HC1 would you expect to get 
fromC 6 H 5 CHD 7 ? 




Mechanism of electrophilic aromatic substitution: the two steps 

Now that we know what isotope effects are and, in a general way, how they 
arise, we are ready to see why they are of interest to the organic chemist. Let us 
return to the questions we asked before : how do we know that electrophilic aromatic 
substitution involves two steps, 



(1) ArH + Y+ - > Ar Slow: rate-determining 

Y 

e/ H 

(2) Ar + :Z > ArY + H:Z Fast 

Y 



SEC. 11.16 MECHANISM OF ELECTROPHILIC AROMATIC SUBSTITUTION 355 

instead of just one, 



(la) ArH + Y + > 



Ar 

VX YJ 



> ArY 



and how do we know that, of these two steps, the first is much slower than the 
second ? 

The answer is found in a series of studies begun by Lars Melander (of the 
Nobel Institute of Chemistry, Stockholm) and extended by many other workers. 
A variety of aromatic compounds labeled with deuterium or tritium were subjected 
to nitration, bromination, and Friedel-Crafts alkylation. It was found that in 
these reactions deuterium or tritium is replaced at the same rate as protium; 
there is no significant isotope effect. 

We have seen that a carbon-deuterium bond is broken more slowly than a 
carbon-prolium bond, and a carbon -tritium bond more slowly yet. How then, 
are we to interpret the fact that there is no isotope effect here? If the rates of 
replacement of the various hydrogen isotopes are the same, it can only mean that 
the reactions whose rates we are comparing do not involve the breaking of a 
carbon- hydrogen bond. 

This interpretation is consistent with our mechanism. The rate of the overall 
substitution is determined by the slow attachment of the electrophilic reagent to 
the aromatic ring to form the carbonium ion. Once formed, the carbonium ion 
rapidly loses hydrogen ion to form the products. Step (1) is thus the rate-deter- 
mining step. Since it does not involve the breaking of a carbon- hydrogen bond, 
its rate- and hence the rate of the overall reaction- -is independent of the particular 
hydrogen isotope that is present. 

If substitution involved a single step, as in (la), this step would necessarily be 
the rate-determining step and, since it involves breaking of the carbon hydrogen 
bond, an isotope effect would be observed. Or, if step (2) of the two-step sequence 
were slow enough relative to step ( 1 ) to affect the overall rate, again we would expect 
an isotope effect. (Indeed, sulfonation does show a small isotope effect and, as 
we shall see, for just this reason. Even in sulfonation, however, the overall rate is 
controlled chiefly by step (1).) 

Thus the absence of isotope effects establishes not only the two-step nature of 
electrophilic aromatic substitution, but also the relative speeds of the steps. 
Attachment of the electrophile to a carbon atom of the ring is the difficult step 
(see Fig. 11.2); but it is equally difficult whether the carbon carries protium or 
deuterium. The next step, loss of hydrogen ion, is easy. Although it occurs more 
slowly for deuterium than for protium, this really makes no difference; slightly 
faster or slightly slower, its speed has no effect on the overall rate. 

Let us look at this matter more closely (Fig. 11.2, insert). Every carbonium 
ion formed, whether I(H) or I(D), goes on to product, since the energy barrier to 

H D 

ov ^ / 

Ar Ar 

X N0 2 X N0 2 

KH) KD) 



356 



ELECTROPHILIC AROMATIC SUBSTITUTION 



CHAP. 11 



the right (ahead of the carbonium ion) whether slightly higher for deuterium or 
slightly lower for protium is still considerably lower than the barrier to the left 
(behind the carbonium ion). But the barrier behind the carbonium ion is the 
ot f r tne reverse of step (1). It is this reverse reaction that must be much slower 
than step (2) if step (1) is to be truly rate-determining (see Sec. 14.12). Summarized 
in terms of the rate constants, k t for the various steps, we have: 

(1) H (2) 

ArH++N0 2 5= Ar' -^ ArNO 2 + H+ k 2 k^ 

*- X N0 2 
I 

We can see why nitration and reactions like it are not reversible. In the reverse 
of nitration, nitrobenzene is protonated (the reverse of reaction 2) to form car- 
bonium ion I; but this is, of course, no different from the ion I formed in the 
nitration process, and it does the same thing: (re)forms nitrobenzene. 



Difficult step 





goes on to product 



ArH + NO? 



ArNO 2 + H f 



Progress of Reaction > 

Figure 11.2. Nitration. Formation of carbonium ion is rate-controlling 
step; occurs equally rapidly whether protium (H) or deuterium (D) at 
point of attack. All carbonium ions go on to product. There is no isotope 
effect, and nitration is irreversible. 



SEC. 11.16 MECHANISM o* iLi^CTROPHILIC AROMATIC SUBSTITUTION 



357 



Unlike most other electrophilic substitution reactions, sulfonation shows a 
moderate isotope effect: ordinary hydrogen (protium) is displaced from an aro- 
matic ring about twice as fast as deuterium. Docs this mean that sulfonation takes 
place by a different mechanism than nitration, one involving a single step? Almost 
certainly not. 

(1) H (2) 



ArH + S0 3 



\ 



H+ 



so 3 - 
11 



Unlike most other electrophilic substitution reactions, sulfonation is revers- 
ible, and this fact gives us our clue. Reversibility means that carbonium ion II 
can lose SO 3 to form the hydrocarbon. Evidently here reaction (2) is not much 





5^A?< ..orA?< - 

go on to product, some revert 
to starting material 



ArH + SO 3 



Progress of Reaction > 

Figure 11.3 Sulfonation. Some carbonium ions go on to product, some 
revert to starting material. There is an isotope effect, and suifonation is 
reversible. 



358 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP, 11 

faster than the reverse of reaction (1). In sulfonation, the energy barriers on either 
side of the carbonium ion II must be roughly the same height; some ions go one 
way, some go the other (Fig. 1 1.3). Now, whether the carbonium ion is II(D) or 
II(H), the barrier to the left (behind it) is the same height. But to climb the barrier 
to the right (ahead), a carbon-hydrogen bond must be broken, so this barrier is 
higher for carbonium ion II(D) than for carbonium ion II(H). More deuterated 
ions than ordinary ions revert to starting material, and so overall sulfonation is 
slower for the deuterated benzene. Thus, the particular shape of potential energy 
curve that makes sulfonation reversible also permits an isotope effect to be 
observed. 

By use of especially selected aromatic substrates highly hindered ones isotope 
effects can be detected in other kinds of electrophilic aromatic substitution, even in 
nitration. In certain reactions the size of the isotope can be deliberately varied by changes 
in experimental conditions- and in a way that shows dependence on the relative rates 
of (2) and the reverse of (1). There can be little doubt that all these reactions follow the 
same two-step mechanism, but with differences in the shape of potential energy curves. 
In isotope effects the chemist has an exceedingly delicate probe for the examination of 
organic reaction mechanisms. 

Problemll.il From the reaction of mesitylene (1,3,5-trimethylbenzene) with 
HF and BF 3 , Olah (see p. 346) isolated at low temperatures a bright-yellow solid whose 
elemental composition corresponds to mesitylene :HF:BF 3 in the ratio 1:1:1. The 
compound was poorly soluble in organic solvents and, when melted, conducted an 
electric current; chemical analysis showed the presence of the BF 4 ~ ion. When heated, 
the compound evolved BF 3 and regenerated mesitylene. 

What is a likely structure for the yellow compound? The isolation of this and 
related compounds is considered to be strong support for the mechanism of electro- 
philic aromatic substitution. Why should this be so? 

Problem 11.12 Dehydrobromination by C 2 H 5 O""Na + of ordinary isopropyl 
bromide and of labeled isopropyl bromide, (CDOaCHBr, at 25" has been studied, and 
the rates found to be in the ratio 1 .76:0.26. (a) What is the value of the isotope effect? 
(b) Is this isotope effect consistent with the mechanism for dehydrohalogcnation given 
in Sec. 5.13? (c) With the following two-step mechanism involving a carbonium ion? 

RBr 2!L> R + Br- -^U alkene 
(d) With the following two-step mechanism involving a carbanion 7 

II I I 

-C C + -OC 2 H 5 -*=-* -C C -^> alkene 




11.17 Reactivity and orientation 

We have seen that certain groups activate the benzene ring and direct sub- 
stitution to 0/7//0 and para positions, and that other groups deactivate the ring and 
(except halogens) direct substitution to meia positions. Let us sec if we can ac- 
count for these effects on the basis of principles we have already learned. 

First of all, we must remember that reactivity and orientation are both matters 
of relative rates of reaction. Methyl is said to activate the ring because it makes 



SEC. 11.18 



THEORY OF REACTIVITY 



359 



the ring read faster than benzene; it causes ortho, para orientation because it makes 
the ortho and para positions react faster than the nieta positions. 

Now, we know that, whatever the specific reagent involved, the rate of electro- 
philic aromatic substitution is determined by the same slow step -attack of the 
electrophile on the ring to form a carbonium ion: 



H 



C 6 H 6 



Y+ 



C 6 H 5 



/ 



Slow: rate-determining 



Any differences in rate of substitution must therefore be due to differences in the 
rate of this step. 

For closely related reactions, a difference in rate of formation of carbonium 
ions is largely determined by a difference in af t , that is, by a difference in stability 
of transition states. As with other carbonium ion reactions we have studied, 
factors that stabilize the ion by dispersing the positive charge should for the same 
rea c on stabilize the incipient carbonium ion of the transition state. Here again 
we expect the more stable carbonium ion to be formed more rapidly. We shall 
therefore concentrate on the relative stabilities of the carbonium ions. 

In electrophilic aromatic substitution the intermediate carbonium ion is a 
hybrid of structures I, II, and III, in which the positive charge is distributed about 
the ring, being strongest as the positions ortho and para to the carbon atom being 
attacked. 



H Y 




H Y 




rep re sen ted as 



III 



H Y 



IV 



A group already attached to the benzene ring should affect the stability of the 
carbonium ion by dispersing or intensifying the positive charge, depending upon its 
electron-releasing or electron-withdrawing nature. It is evident from the structure 
of the ion (I-III) that this stabilizing or destabilizing effect should be especially 
important when the group is attached ortho or para to the carbon being attacked. 



11.18 Theory of reactivity 

To compare rates of substitution in benzene, toluene, and nitrobenzene, we 
compare the structures of the carbonium ions formed from the three compounds: 



H Y 



H Y 



CH 3 



NO 2 



nr 



360 



ELECTROPHILIC AROMATIC SUBSTITUTION 



CHAP. 11 



By releasing electrons, the methyl group (II) tends to neutralize the positive 
charge of the ring and so become more positive itself; this dispersal of the charge 
stabilizes the carbonium ion. In the same way the inductive effect stabilizes the 
developing positive charge in the transition state and thus leads to a faster reaction. 





Transition state. 

developing positive 

chutge 



Carbonium ion: 

full positive 

charge 



The NO 2 group, on the other hand, has an electron- withdrawing inductive 
effect (III); this tends to intensify .the positive charge, destabilizes the carbonium 
ion, and thus causes a slower reaction. 

Reactivity in electrophilic aromatic substitution depends, then, upon the tendency 
of a substituent group to release or withdraw electrons. A group that releases elec- 
trons activates the ring; a group that withdraws electrons deactivates the ring. 



4- Y+ 





Electrophilic Aromatic Substitution 

G releases elections. G 

stabilizes carbonium ion, 

activates 




NH 2 
OH 
OCH} 
NHCOCH 3 



-en. 



G withdraw s electrons G 

destabilizes carbonium ion, 
deactivates 



-NO 2 

-CN 
SO 3 H 
COOH 

-CHO 

-COR 
X 



Like CH 3 , other alkyl groups release electrons, and like CH 3 they activate 
the ring. For example, ferf-butylbenzene is 16 times as reactive as benzene toward 
nitration. Electron release by NH 2 and OH, and by their derivatives OCH 3 
and NHCOCH 3 , is due not to their inductive effect but to resonance, and is' 
discussed later (Sec. 11.20). 

We are already familiar with the electron-withdrawing effect of the halogens 
(Sec. 6.11). The full-fledged positive charge of the N(CH 3 ) 3 + group has, of 
course, a powerful attraction for electrons. In the other deactivating groups 
(e.g., NO 2 , CN, COOH), the atom next to the ring is attached by a multiple 
bond to oxygen or nitrogen. These electronegative atoms attract the mobile n 
electrons, making the atom next to the ring electron-deficient; to make up this 
deficiency, the atom next to the ring withdraws electrons from the ring. 



SEC. 11.19 THEORY OF ORIENTATION 361 

We might expect replacement of hydrogen in CH 3 by halogen to decrease 
the electron-releasing tendency of the group, and perhaps to convert it into an 
electron-withdrawing group. This is found to be the case. Toward nitration, 



H-C-H H-C >C1 H~O>C1 






Activating Weakly Moderately Strongly 

deactivating deactivating deactivating 



toluene is 25 times as reactive as benzene; benzyl chloride is only one-third as 
reactive as benzene. The CH 2 C1 group is thus weakly deactivating. Further 
replacement of hydrogen by halogen to yield the CHC1 2 and the CC1 3 groups 
results in stronger deactivation. 



11.19 Theory of orientation 

Before we try to account for orientation in electrophilic substitution, let us 
look more closely at the facts. 

An activating group activates all positions of the benzene ring; even the posi- 
tions metgj.o it are^noje reactive than any single position in benzene itself. It 
directs ortho and para simply because it activates the ortho and para positions 
much more than it does thojneta % 

A deactivating group deactivates all positions in the ring, even the positions 
meta to it. It directs meta simply because it deactivates the ortho and para positions 
even more than it does the meta. 

Thus both ortho, par a orientation and meta orientation arise in the same way: 
the effect of any group whether activating or deactivating is strongest at the ortho 
and para positions. 

To see if this is what we would expect, let us compare, for example, the car- 
bonium ions formed by attack at the para and meta positions of toluene, a com- 
pound that contains an activating group. Each of these is a hybrid of three struc- 
tures, I-III for para, IV- VI for meta. In one of these six structures, II, the positive 
charge is located on the carbon atom to which CH 3 is attached. Although CH 3 
releases electrons to all positions of the ring, it does so most strongly to the car- 




H Y 

il 

Especially stable: 

charge on carbon 
carrying sitbstituent 




Para attack 



362 



KLECTROPIUUC AROMATIC SUBSTITUTION 



CHAP. 11 



bon atom nearest it; consequently, structure II is a particularly stable one. Be- 
cause of contribution from structure II, the hybrid carbonium ion resulting from 



Meta attack 



attack at the para position is more stable than the carbonium ion resulting from 
attack at a meta position. Para substitution, therefore, occurs faster than meta 
substitution. 

In the same way, it can be seen that attack at an ortho position (VII-IX) 






CH 3 





C'Hj 



VII 



VIII 




On ho attack 



Especially stable: 

charge on carbon 
carrying substituent 

also yields a more stable carbonium ion, through contribution from IX, than attack 
at a meta position. 

In toluene, ortho, para substitution is thus faster than meta substitution because 
electron release by CHj is more effective duringattack at the positions ortho and 
para to it. 

Next, let us compare the carbonium ions formed by attack at the para and 
meta positions of nitrobenzene, a compound that contains a deactivating group. 
Each of these is a hybrid of three structures, X-XII for para attack, XI II -XV for 
meta attack. In one of the six structures, XI, the positive charge is located on the 



Para attack 



Especially unstable: 

charge on carbon 
carrying substituent 

carbon atom to which NO 2 is attached. Although NO 2 withdraws electrons 
from all positions, it does so most from the carbon atom nearest it, and hence this 
carbon atom, already positive, has little tendency to accommodate fc the positive 
charge of the carbonium ion. Structure XI is thus a particularly unstable one and 
does little to help stabilize the ion resulting from attack at the para position. The 
ion for para attack is virtually a hybrid of only two structures, X and XII; the 






SEC. 11.20 ELECTRON RELEASE VIA RESONANCE 363 

positive charge is mainly restricted to only two carbon atoms. It is less stable than 
the ion resulting from attack at a meta position, which is a hybrid of three structures, 
and in which the positive charge is accommodated by three carbon atoms. Para 
substitution, therefore, occurs more slowly than meta substitution. 



Meta attack 





XIV 




In the same way it can be seen that attack at an ortho position (XVI-XVIII) 
yields a less stable carbonium ion, because of the instability of XVIII, than attack 
at a meta position. 






Ortho attack 

XVI XVII XVIII 

Especially unstable: 

charge on carbon 
carrying substituent 

In nitrobenzene, ortho.para substitution is thus slower than meta substitution 
because electron withdrawal by NO 2 is more effective during attack at the 
positions ortho and para to it. 

Thus we see that both ortho.para orientation by activating groups and meta 
orientation by deactivating groups follow logically from the structure of the 
intermediate carbonium ion. The charge of the carbonium ion is strongest at the 
positions ortho and para to the point of attack, and hence a group attached to one 
of these positions can exert the strongest effect, whether activating or deactivating. 

The unusual behavior of the halogens, which direct ortho and para although 
deactivating, results from a combination of two opposing factors, and will be taken 
up in Sec. 11.21. 

11.20 v ^Electron release via resonance 

We have seen that a substituent group affects both reactivity and orientation 
in electrophilic aromatic substitution by its tendency to release or withdraw 
electrons. So far, we have considered electron release and electron withdrawal 
only as inductive effects, that is, as effects due to the electronegativity of the group 
concerned. 

But certain groups (~NH 2 and OH, and their derivatives) act as powerful 
activators toward electrophilic aromatic substitution, even though they contain 
electronegative atoms and can be shown in other ways to have electron-withdrawing 
inductive effects. If our approach to the problem is correct, these groups must 
release electrons in some other way than through their inductive effects; they are 



364 



ELECTROPHILIC AROMATIC SUBSTITUTION 



CHAP. 11 



believed to do this by a resonance effect. But before we discuss this, let us review 
a little of what we know about nitrogen and oxygen. 

Although electronegative, the nitrogen of the NH 2 group is basic and tends 
to share its last pair of electrons and acquire a positive charge. Just as ammonia 
accepts a hydrogen ion to form the ammonium (NH 4 + ) ion, so organic compounds 
related to ammonia accept hydrogen ions to form substituted ammonium ions. 



NH 3 + H+ 



R 2 NH 2 + 



RNH 2 + H+ 
R 3 N + H+ 



RNH 3 + 



The OH group shows similar but weaker basicity; we are already familiar with 
oxonium ions, ROH 2 + . 



H 2 O 



ROH + H* 



ROH 2 



The effects of NH 2 and OH on electrophilic aromatic substitution can be 
accounted for by assuming that nitrogen and oxygen can share more than a pair 
of electrons with the ring and can accommodate a positive charge. 

The carbonium ion formed by attack para to the NH 2 group of aniline, for 
example, is considered to be a hybrid not only of structures I, II, and III, with 
positive charges located on carbons of the ring, but also of structure IV in which the 




:NH 2 




:NH-> 





Para attack 



Especially stable: 
every atom has octet 



:NH 2 





:NH 2 



VI 




Meta attack 



positive charge is carried by nitrogen. Structure IV is especially stable, since in it 
every atom (except hydrogen, of course) has a complete octet of electrons. This 
carbonium ion is much more stable than the one obtained by attack on benzene 
itself, or the one obtained (V-VI1) from attack rneta to the -~NH 2 group of aniline; 
in neither of these cases is a structure like IV possible. (Compare, for example, the 
stabilities of the ions NH 4 + and CH 3 + . Here it is not a matter of which atom, 
nitrogen or carbon, can better accommodate a positive charge; it is a matter of 
which atom has a complete octet of electrons.) 



SEC. 11.21 



EFFECT OF HALOGEN 



365 



Examination of the corresponding structures (VIII-XI) shows that ortho 
attack is much like para attack: 



:NH 2 




VIII 




:NH 2 




NH 2 




Ortho attack 



Especially stable: 
every atom has octet 

Thus substitution in aniline occurs faster than substitution in benzene, and 
occurs predominantly at the positions ortho and para to NH 2 . 

In the same way activation and ortho>para orientation by the OH group is 
accounted for by contribution of structures like XII and XIII, in which every atom 
has a complete octet of electrons: 



OH 




0OH 




XIII 
Ortho attack 



The similar effects of the derivatives of NH 2 and OH are accounted for by 
similar structures (shown only for/wa attack): 



<"N(CH 3 ) 2 




NHCOCHj COOCH 3 

U 

H' Y H' 

-NHCOCH 3 OCH 3 




The tendency of oxygen and nitrogen in groups like these to share more than a 
pair of electrons with an aromatic ring is shown in a number of other ways, which 
will be discussed later (Sec. 23.2 and Sec. 24.7X 



11. 21^ Effect of halogen on electrophilic aromatic substitution 

Halogens are unusual in their effect on electrophilic aromatic substitution: 
they are deactivating yet 0rf/io,/>ara-directing. Deactivation is characteristic of 
electron withdrawal, whereas ortho y para orientation is characteristic of electron 
release. Can halogen both withdraw and release electrons? 

The inswer is yes. Halogen withdraws electrons through its inductive effect, 
and releases electrons through its resonance effect. So, presumably, can the 
NH 2 and OH groups, but there the much stronger resonance effect greatly 



366 



ELECTROPHILIC AROMATIC SLBSTITUTION 



CHAP. 11 




outweighs the other. For halogen, the two effects are more evenly balanced, and 
we observe the operation of both. 

Let us first consider reactivity. Electrophilic attack on benzene yields car- 



Cl \\-iihJra\\s electrons: 
desiahilizcs carhonium ion. 
dcacnralcs ring 



bonium ion I, attack on chlorobenzene yields carbonium ion II. The electron- 
withdrawing inductive effect of chlorine intensifies the positive charge in carbonium 
ion II, makes the ion less stable, and causes a slower reaction. 

Next, to understand orientation, let us compare the structures of the carbonium 
ions formed by attack at the para and nteta positions of chlorobenzene. Each of 





H Y 

IV 
Especially unstable: 

charge on carbon 
bearing substituent 




Para attack 






Met a attack 



VI 



VII 



VIII 



these is a hybrid of three structures, Hl-V for para, VI- VII I for meta. In one of 
these six structures, IV, the positive charge is located on the carbon atom to which 
chlorine is attached. Through its inductive effect chlorine withdraws electrons 
most from the carbon to which it is joined, and thus makes structure IV especially 
unstable. As before, we expect IV to make little contribution to the hybrid, which 
should therefore be less stable than the hybrid ion resulting from attack at the 
meta positions. If only the inductive effect were involved, then, we would expect 
not only deactivation but also meta orientation. 

But the existence of halonium ions (Sec. 7.12) has shown us that halogen 
can share more than a pair of electrons and can accommodate a positive charge. 
If we apply that idea to the present problem, what do we find? The ion resulting 
from para attack is a hybrid not only of structures III-V, but also of structure IX, 
in which chlorine bears a positive charge and is joined to the ring by a double bond. 
This structure should be comparatively stable, since in it every atom (except 
hydrogen, of course) has a complete octet of electrons. (Structure IX is exactly 
analogous to those proposed to account for activation and ort ho, par a direction 
by NH 2 and OH.) No such structure is possible for the ion resulting from 



SEC. 11.21 



EFFECT OF HALOGEN 



367 



meta aitack. To the extent that structure IX contributes to the hybrid, it makes the 
ion resulting from para attack more stable than the ion resulting from meta attack. 

eci 



Para attack 

H Y 

IX 

Comparatively stable: 
every atom has octet 

Although we could not have predicted the relative importance of the two factors 
the instability of IV and the stabilization by IX the result indicates that the con- 
tribution from IX is the more important. 

In the same way it can be seen that attack at an orlho position also yields an 
ion (X-X1II) that can be stabilized by accommodation of the positive charge by 
chlorine. 



Ortho attack 







XII 
Especially unstable' 

charge on carbon 
hearing A itbstitiient 



XII! 

Comparatively stable' 
every atom has octet 



Through its inductive effect halogen tends to withdraw electrons and thus to 
destabilize the intermediate carbonium ion. This effect is felt for attack at all 
positions, but particularly for attack at the positions ortlio and para to the halogen. 

Through its resonance effect halogen tends to release electrons and thus to 
stabilize the intermediate carbonium ion. This electron release is effective only 
for attack at the positions ortho and para to the halogen. 

The inductive eflect is stronger than the resonance effect and causes net electron 
withdrawal and hence deactivation for attack at all positions. The resonance 
effect tends to oppose the inductive effect for attack at the orlho and para positions, 
and hence makes the deactivation less for ortho, para attack than for meta. 

Reactivity is thus controlled by the stronger inductive eflect, and orientation 
is controlled by the resonance effect, which, although weaker, seems to be more 
selective. 

Problem 11.13 Hydrogen iodide adds to vinyl chloride more slowly than. to 
ethylene, and yields 1-chloro-l-iodoethane. (a) Draw the formula of the carbonium 
ion formed in the initial step of the addition to vinyl chloride, (b) Of addition to 
ethylene. (c) Judging from the relative rates of reaction, which would appear to be the 
more stable carbonium ion? (d) Account for the difference in stability. 

(e) Draw the formula for the carbonium ion that would be formed if vinyl chloride 
were to yield I -chloro-2-iodoethane. (f) Judging from the actual orientation of addi- 
tion, which carbonium ion from vinyl chloride is the more stable, (a) or (e)? (g) Account 
for the difference in stability. 

(h) Which effect, inductive or resonance, controls reactivity in electrophilic addi- 
tion to vinyl halides? (i) Which effect controls orientation? 



368 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

Thus we find that a single structural concept partial double-bond formation 
between halogen and carbon helps to account for unusual chemical properties 
of such seemingly different compounds as aryl halides and vinyl halides. The 
structures involving doubly-bonded halogen, which probably make important 
contribution not only to benzenonium ions but to the parent aryl halides as well 
(Sec. 25.6), certainly do not seem to meet our usual standard of reasonableness 
(Sec. 6.27). The sheer weight of evidence forces us to accept the idea that certain 
carbon-halogen bonds possess double-bond character. If this idea at first appears 
strange to us, it simply shows how little, after all, we really know about molecular 
structure. 



11.22 Relation to other carbonium ion reactions 

In summary, we can say that both reactivity and orientation in electrophilic 
aromatic substitution are determined by the rates of formation of the intermediate 
carbonium ions concerned. These rates parallel the stabilities of the carbonium 
ions, which are determined by the electron-releasing or electron-withdrawing 
tendencies of the substituent groups. 

A group may release or withdraw electrons by an inductive effect, a resonance 
effect, or both. These effects oppose each other only for the NH 2 and OH 
groups (and their derivatives) and for the halogens, X. For NH 2 and OH 
the resonance effect is much the more important; for X the effects are more 
evenly matched. It is because of this that the halogens occupy the unusual position 
of being deactivating groups but ortho,para directors. 

We have accounted for the facts of electrophilic aromatic substitution in 
exactly the way that we accounted for the relative ease of dehydration of alcohols, 
and for reactivity and orientation in electrophilic addition to alkenes: the more 
stable the carbonium ion, the faster it is formed; the faster the carbonium ion is 
formed, the faster the reaction goes. 

In al! this \\e have estimated the stability of a carbonium ion on the same 
basis: the dispersal or concentration of the charge due to electron release or electron 
withdrawal by the substituent groups. As we shall see, the approach that has 
worked so well for elimination, for addition, and for electrophilic aromatic sub- 
stitution works for still another important class of organic reactions in which a 
positive charge develops: nucleophilic aliphatic substitution by the S^I mechanism 
(Sec. 14.14). It works equally well for nucleophilic aromatic substitution (Sec. 
25.9), in which a negative charge develops. Finally, we shall find that this approach 
will help us to understand acidity or basicity of such compounds as carboxylic 
acids, sulfonic acids, amines, and phenols. 



PROBLEMS 

1. Give structures and names of the principal products expected from the ring 
monobromination of each of the following compounds. In each case, tell whether bro- 
mination will occur faster or slo\\er than \\ith benzene itself. 

(a) acetanilide(C ft H 5 NHCOCH 3 ) (d) N-methylaniline (C 6 H 5 NHCH,) 

(b) iodobenzene (e) ethyl benzoate (C^COOQHj) 

(c) sec-butylbenzene (f) acetophenone (C 6 H 5 COCH.O 



PROBLEMS 369 

(g) phenetoie (C 6 H 5 OC 2 H 5 ) (j) benzotrifluoride (C 6 H 5 CF 3 ) 

(h) diphenylmethane (C 6 H 5 CH 2 C 6 H5) (k) biphenyl (C 6 H 5 -C 6 H 5 ) 

(i) benzonitrile (C 6 H 5 CN) 

2. Give structures and names of the principal organic products expected from 
mononitration of: 

(a) o-nitrotoluene (g) p-cresol 

(b) /w-dibromobenzene (h) w-nitrotoluene 

(c) /7-nitroacetanilide (i) />-xylene (p-C 6 H 4 (CH 3 ) 2 ) 
(/?-O 2 NC 6 H 4 NHCOCH 3 ) (j) terephthalic acid (p-C 6 H 4 (COOH) 2 ) 

(d) m-dinitrobenzene (k) anilinium hydrogen sulfate 

(e) /w-cresol (w-CH 3 C 6 H 4 OH) (C 6 H 5 NH 3 + HSO 4 -) 

(f) 0-cresol 

3. Give structures and names of the principal organic products expected from the 
monosulfonation of : 

(a) cyclohexyl benzene (g) o-fluoroanisole 

(b) nitrobenzene (h) o-nitroacetanilide 

(c) anisole (C 6 H 5 OCH 3 ) (o-O 2 NC 6 H 4 NHCOCH 3 ) 

(d) benzenesulfonic acid (i) 0-xylene 

(e) salicylaldehyde (0-HOC 6 H 4 CHO) (j) /w-xylene 

(f) Aw-mtrophenol (k) /?-xylene 

4. Arrange the following in order of reactivity toward ring nitration, listing by 
structure the most reactive at the top, the least reactive at the bottom. 

(a) benzene, mesitylene (l,3,5-C ft H 3 (CH3)0, toluene, w-xylene, /?-xylene 

(b) benzene, bromobenzene, nitrobenzene, toluene 

(c) acetanilide (C 6 H 5 NHCOCH 0, acetophenone (C 6 H,COCH,), aniline, benzene 

(d) terephthalic acid, toluene, /7-toluic acid (p-CH 3 C 6 H 4 COOH), /?-xylene 

(e) chlorobenzene, /?-chloronitrobenzene, 2,4-dinitrochlorobenzene 

(f) 2,4-dinitrochlorobenzene, 2,4-dinitrophenol 

(g) /w-dinitrobenzene, 2,4-dinitrotoluene 

5. Even though 1,3,5-trinitrobenzene (TNB) has more shattering power (more 
brisance) and is no more dangerous to handle, 2,4,6-trinitrotoluene (TNT) has always 
been the high explosive in more general use. Can you suggest a reason (connected with 
manufacture) for the popularity of TNT? (Benzene and toluene are both readily avail- 
able materials; for many years benzene was cheaper.) 

6. For each of the following compounds, indicate which ring you would expect to 
be attacked in nitration, and give structures of the principal products. 



(a) 2 N< 



/7-Nitrobiphenyl /;/-Nitrodiphen\lmethanc Phcn\l ben/oate 



7. Arrange the compounds of each set in order of reactivity toward electrophilic 
substitution. Indicate in each set which would yield the highest percentage of meta 
isomer, and which would yieldLthe lowest. 

(a) C 6 H 5 N(CH 3 ) 3 +, C 6 H 5 CH 2 N(CH 3 ) 3 +, C 6 H 5 CH 2 CH 2 N(CH 3 ) 3 +, 

C 6 H 5 CH 2 CH 2 CH 2 N(CH 3 V 

(b) C 6 H 5 N0 2 , C 6 H 5 CH 2 N0 2 , C 6 H 5 CH 2 CH 2 NO 2 

(c) C 6 H 5 CH 3 , C 6 H 5 CH 2 COOC 2 H 5 , C 6 H 5 CH(COOC 2 H 5 ) 2 , C 6 H 5 C(COOC 2 H 5 ) 3 

8. There is evidence that the phenyl group, C 6 H 5 -, has an electron-withdrawing 
inductive effect. Yet each ring of biphenyl, C 6 H 5 -C 6 H 5 , is more reactive than benzene 





370 ELECTROPHILIC AROMATIC SUBSTITUTION CHAP. 11 

toward electrophilic substitution, and the chief products are ortho and para isomers. 
Show how reactivity and orientation can be accounted for on the basis of resonance. 

9. When ^-phenylethyl alcohol, C 6 H 5 CH2CH 2 OH, is treated with thallium trifluoro- 
acetate followed by potassium iodide, there is obtained predominantly one aryl iodide, 
the particular isomer depending upon the conditions of thallation: (a) 25, ortho\ (b) 75, 
meta\ (c) prior conversion to the ester, C 6 H 5 CH 2 CH 2 OCOCH3, then 25, para. Suggest 
an explanation for each case of regiospecificity. 

10. There is evidence that the reaction between HNO 3 and H 2 SO 4 to generate + NO 2 
(which we have summarized in one equation, Sec. 1 1 .8) actually involves tjiree steps, 
the second of which is the slowest one and the one that actually produces - + NO 2 . Can 
you suggest a reasonable sequence of reactions? (Hint: See Sec. 5.20.) 

11. Treatment of sulfanilic acid (p-H2NC 6 H 4 SO 3 H) with 3 moles of bromine yields 
2,4,6-tribromoaniline. Treatment of 4-hydroxy-l,3-benzenedisuIfonic acid with nitric 
acid yields picric acid, 2,4,6-trinitrophenol. (a) Outline the most probable mechanism 
for the replacement of SO 3 H by Br and by NO 2 . (b) To what general class of 
organic reactions do those reactions belong? 

12. Using only individual steps with which you are already familiar, outline a likely 
mechanism for the following reaction. 

C 6 H 5 C(CH 3 ) 3 4- Br 2 (AlBr 3 ) > C 6 H 5 Br + HBr + (CH 3 ) 2 O-CH 2 

13. In light of what you have learned in this chapter, predict the major products of 
each of the following reactions. 

(a) (CHO^NCH-CH, + HI 

(b) CH 2 --CHCF.1 + HBr(AlBr 3 ) 

(c) What is the function of AlBr 3 in (b)? Why is it needed here? 

14. You are trying to find out whether or not there is an isotope effect in a particular 
kind of substitution in which the electrophile Y replaces a hydrogen of an aromatic ring. 
I.i each of the following cases, tell what you would do^ and what you would expect to 
observe if there were an isotope effect. (You can quantitatively analyze mixtures of 
isomers. Your mass spectrometer will tell you what percentage of the hydrogen in a 
compound is deuterium, but not the location of deuterium in a molecule.) 

(a) C 6 Hfi and C () D 6 are allowed to react separately but under identical conditions. 

(b) A 50:50 mixture of C 6 H 6 and C 6 D 6 is allowed to react with a limited amount of the 
reagent. * 

(c) Anisole and amsole-4-d arc allowed to react separately. (Both your watch and your 
mass spectrometer arc under repair when this particular experiment is carried out.) 

(d) Benzene- 1, 3, 5-d 3 (1,3,5-trideuteriobenzene) is allowed to react. 

15. Outline all steps in the laboratory synthesis of the following compounds from 
benzene and/or toluene, using any needed aliphatic or inorganic reagents. (Review the 
general instructions on p. 224. Assume that a pure para isomer can be separated from 
an ortho,para mixture.) 

(a) /Miitrotoluene (i) 1,3,5-trjnitrobenzene 

(b) p-bromonitrobcnzene (j) 2-bromo-4-nitrotoluenc 

(c) /?-dichloroben/ene (k) 2-bromo-4-nitrobenzoic acid 

(d) w-bromobcnzencsulfonic acid (1) 4-bromo-3-nitrobenzoic acid 

(e) /7-bromobenzenesulfonic acid (m) 3,5-dinitrobenzoic acid 

(f ) /7-bromobenzoic acid (n) 4-nitro-l,2-dibromobenzene 

(g) /H-bromobenzoic acid (o) 2-nitro-M-dichlorobenzene 
(h) 0-iodobenzoic acid (p) w-iodotoluene 



PROBLEMS 371 

16. Outline all steps in the following laboratory syntheses, using any needed aliphatic 
or inorganic reagents. (Follow the other instructions in Problem 15). 

(a) 4-nitro-2,6-dibromoanisole from anisole (C 6 H 5 OCH3) 

(b) 4-bromo-2-nitrobenzoic acid from 0-nitrotoluene 

(c) 2,4,6-tribromoaniline from aniline 

(d) 2,4-dinitroacetanilide from acetanilide (C 6 H 5 NHCOCH 3 ) 

(e) 5-nitroisophthalic acid from w-xylene 

(f ) 4-nitroisophthalic acid from /w-xylene 

(g) 2-nitroterephthalic acid from /;-xylene (two ways) 
(h) Which way in (g) is preferable? Why? 



Chapter 

12 



Arenes 



12.1 Aliphatic-aromatic hydrocarbons 

From our study so far we know what kind of chemical properties to expect of 
an aliphatic hydrocarbon, that is, of an alkane, alkene, or alkyne. We know what 
kind of chemical behavior to expect of the parent aromatic hydrocarbon, benzene. 
Many important compounds are not just aliphatic or just aromatic, however, but 
contain both aliphatic and aromatic units; hydrocarbons of this kind are known 
collectively as arenes. Ethylhenzene, for example, contains a benzene ring and an 
aliphatic side chain. 

1CH 2 CH 3 



Ethylbcn/cne 

What kind of chemical properties might we expect of one of these mixed 
aliphatic-aromatic hydrocarbons? First, we might expect it to show two sets of 
chemical properties. The ring of ethylbenzcne should undergo the electrophilic 

Mote tettJily 
benzene 



0-Nitroethylbcn/cnc 



/J-Nitrocthylbcn/cnc 



CH 2 CH 3 



Br 

1 -Bromo-1 -phenylethane 

(a-Phenylethyl bromide) 

Only product 

372 



More readily 
than for 
ethane 



SEC. 12.2 



STRUCTURE AND NOMENCLATURE 



373 



substitution characteristic of benzene, and the side chain should undergo the free- 
radical substitution characteristic of ethane. Second, the properties of each portion 
of the molecule should be modified by the presence of the other portion. The ethyl 
group should modify the aromatic properties of the ring, and the ring should 
modify the aliphatic properties of the side chain. 

These predictions are correct. Treatment of ethylbenzcne with nitric acid and 
sulfuric acid, for instance, introduces a nitro group into the ring; treatment with 
bromine in the presence of light introduces a bromine atom into the side chain. 
But because of the ethyl group, nitration takes place more readily than with benzene 
itself, and occurs chiefly at the positions ortho and para to the ethyl group; and 
because of the ring, bromination takes place more readily than with ethane, and 
occurs exclusively on the carbon nearer the ring. Thus each portion of the molecule 
affects the reactivity of the other portion and determines the orientation of attack. 

In the same way we may have a molecule that is part aromatic and part alkene, 
or part aromatic and part alkyne. Again each portion of such a molecule shows 
the properties characteristic of its particular structure, although these properties 
are modified by the other portion of the molecule. 

We shall examine most closely the compounds made up of aromatic and alkane 
units, the alkylbenzenes. We shall look much more briefly at the aromatic-alkene 
compounds (alkenylbenzencs) and aromatic-alkyne compounds (alkynylbenzenes). 

We shall encounter the benzyl free radical and the benzyl carbonium ion, 
which pretty much complete our lists of these reactive particles, and shall see 
how their relative stabilities can be accounted for. 



12.2 Structure and nomenclature 

The simplest of the alkylbenzenes, methylbenzene, is given the special name of 
toluene. Compounds containing longer side chains are named by pjefixing the 
name of the alkyl group to the word -benzene, as, for example, in ethylbenzene, 
n-propylbenzene, and isobittylbenzene. 



CH 2 CH 2 CH 3 

(O) 

/i-Propy I benzene Isobutylbenzene 






Toluene 



Ethylbenzene 



The simplest of the dialkylbenzenes, the dimethylbenzenes, are given the 
special names of xylenes; we have, then, o-xylene, m-xylene, and p-xylene. Dialkyl- 
benzenes containing one methyl group are named as derivatives of toluene, while 
others are named by prefixing the names of both alkyl groups to the word -benzene. 
A compound containing a very complicated side chain might be named as a 





o-Xylene 



w-Xy!ene 




374 ARENES CHAP. 12 

CHj 






jTHCHj 
-Hj 

/>-Ethyltoluene m-Ethylisopropylbenzene 2-Methyl-3-phenylpentane 

phenylalkane (C 6 H 5 = phenyl). Compounds containing more than one benzene 
ring are nearly always named as derivatives of alkanes. 

)CH 2 CH 2 < 

Diphenylmethane 1 ,2-DiphcnyIethane 

The simplest alkenylbenzene has the special name styrene. Others are generally 
named as substituted alkenes, occasionally as substituted benzenes. Alkynyl- 
benzenes are named as substituted alkynes. 

CH 3 

Styrene All>lhen/cne 2-Phen>l-2-butenc Phen> lacetylcne 

(Vinylben/ene) (3-Phen>lpropene) 

(Phenylethylcne) 

12.3 Physical properties 

As compounds of low polarity, the alkylbenzenes possess physical properties 
that are essentially the same as those of the hydrocarbons we have already studied. 
They are insoluble in water, but quite soluble in non-polar solvents like ether, 
carbon tetrachloride, or ligroin. They are almost always less dense than water. 
As we can see from Table 12.1, boiling points rise with increasing molecular weight, 
the boiling point increment being the usual 20-3u 3 for each carbon atom. 

Since melting points depend not only on molecular weight but also on mo- 
lecular shape, their relationship to structure is a very complicated one. One 
important general relationship does exist, however, between melting point and 
structure of aromatic compounds: among isomeric disubstituted benzenes, the 
para isomer generally melts considerably higher than the other two. The xylenes, 
for example, boil within six degrees of one another; yet they differ widely in melting 
point, the o- and w-isomers melting at -25 and -48, and the/MSomer melting 
at -f 13. Since dissolution, like melting, involves overcoming the intermolecular 
forces of the crystal, it is not surprising to find that generally the para isomer is also 
the least soluble in a given solvent. 

The higher melting point and lower solubility of a para isomer is only a special 
example of the general effect of molecular symmetry on intracrystalline forces. 
The more symmetrical a compound, the better it fits into a crystal lattice and hence 
the higher the melting point and the lower the solubility. Para isomers are simply 
the most symmetrical of disubstitut r d benzenes. We can see (Table 12.1) that 



SEC. 12.4 INDUSTRIAL SOURCE OF ALKYLBENZENES 

Table 12.1 ALIPHATIC-AROMATIC HYDROCARBONS 



375 



Name 


Formula 


M.p., 
C 


c' 


Density 
(20C) 


Benzene 


C 6 H 6 


5.5 


80 


0.879 


Toluene 


C 6 H 5 CH 3 


- 95 


111 


.866 


o-Xylene 


1,2-C 6 H 4 (CH 3 ) 2 


- 25 


144 


.880 


w-Xylene 


1,3-C 6 H 4 (CH 3 ) 2 


- 48 


139 


.864 


p-Xylene 


1,4-C 6 H 4 (CH 3 ) 2 


13 


138 


.861 


Hemimellitene 


1,2,3-C 6 H 3 (CH 3 ) 3 


- 25 


176 


.895 


Pseudocumene 


1,2,4-C 6 H 3 (CH 3 ) 3 


- 44 


169 


.876 


Mesitylene 


1,3,5-C 6 H 3 (CH 3 ) 3 


- 45 


165 


.864 


Prehnitene 


1,2,3,4-C 6 H 2 (CH 3 ) 4 


- 6.5 


205 


.902 


Isodurene 


1,2,3,5-C 6 H 2 (CH 3 ) 4 


- 24 


197 




Durene 


1,2,4>C 6 H 2 (CH 3 ) 4 


80 


195 




Pentamethylbenzene 


C 6 H(CH 3 ) 5 


53 


231 




Hexamethyl benzene 


C 6 (CH 3 ) 6 


165 


264 




Ethylbenzene 


C 6 H 5 C 2 H 5 


- 95 


136 


.867 


w-Propylbenzene 


C 6 H 5 CH 2 CH 2 CH 3 


- 99 


159 


.862 


Cumene 


C 6 H S CH(CH 3 ) 2 


- 96 


152 


.862 


w-Butylbenzene 


C 6 H 5 (CH 2 ) 3 CH 3 


- 81 


183 


.860 


Isobutylbenzene 


C 6 H 5 CH 2 CH(CH 3 ) 2 




171 


.867 


sec- Butyl benzene 


C 6 H 5 CH(CH 3 )C 2 H 5 


- 83 


173.5 


.864 


ten- Butyl benzene 


C 6 H 5 C(CH 3 ) 3 


- 58 


169 


.867 


p-Cymene 


1,4-CH 3 C 6 H 4 CH(CH 3 ) 2 


- 70 


177 


.857 


Biphenyl 


C 6 H 5 C 6 H5 


70 


255 




Diphenylmethane 


C6H 5 CH 2 C6H 5 


26 


263 




Triphenylmethane 


(C 6 H 5 ) 3 CH 


93 


360 




1,2-Diphcnylethane 


C6H 5 CH 2 CH 2 C 6 H 5 


52 


284 




Styrene , 


C 6 H 5 CH^CH 2 


- 31 


145 


.907 


//w/s-Stilbene 


tronS'Cf) H sC H : = :: CHC'6 H5 


124 


307 




cw-Stilbene 


c/y-CgHsCH^CHCgHf 


6 






unsym- Di pheny let hy lene 


(C 6 H 5 ) 2 C-CH 2 , % 


9 


277 


1.02 


Triphenylethylene 


(C 6 H 5 ) 2 C =CHCe,H 5 


73 






Tetraphenylethylene 


(C 6 H 5 ) 2 0-C(C 6 H 5 ) 2 


227 


425 




Phenylacetylene 


C 6 H 5 C^CH 


- 45 


142 


0.930 


Diphenylacetylene 


C 6 H 5 C-CC 6 H 5 


62.5 


300 





1,2,4,5-tetramethylbenzene melts 85 to 100 higher than the less symmetrical 
1,2,3,5- and 1,2,3,4-isomers. A particularly striking example of the effect of 
symmetry on melting point is that of benzene and toluene. The introduction of a 
single methyl group into the extremely symmetrical benzene molecule lowers the 
melting point from 5 to -95. 



12.4 Industrial source of alkylbenzenes 

It would be hard to exaggerate the importance to the chemical industry and to 

our entire economy of the large-scale production of benzene and the alkylbenzenes. 

, Just as the alkanes obtained from petroleum are ultimately the source of nearly all 

our aliphatic compounds, so benzene and the alkylbenzenes are ultimately the 

source of nearly all our aromatic compounds. When a chemist wishes to make a 



376 ARENES CHAP. 12 

complicated aromatic compound, whether in the laboratory or in industry, he does 
not make a benzene ring ; he takes a simpler compound already containing a benzene 
ring and then adds to it, piece by piece, until he has built the structure he wants. 

Just where do the enormous quantities of simple aromatic compounds come 
from? There are two large reservoirs of organic material, coal and petroleum, 
and aromatic compounds are obtained from both. Aromatic compounds are 
separated as such from coal tar, and are synthesized from the alkanes of petroleum. 

By far the larger portion of coal that is mined today is converted into coke, 
which is needed for the smelting of iron to steel. When coal is heated in the 
absence of air, it is partly broken down into simpler, volatile compounds which are 
driven out; the residue is coke. The volatile materials consist of coal gas and a 
liquid known as coal tar. 

From coal tar by distillation there are obtained a number of aromatic com- 
pounds. Upon coking, a ton of soft coal may yield about 120 pounds of coal tar. 
From this 120 pounds the following aromatic compounds can be separated: 
benzene, 2 pounds; toluene, 0.5 pound; xylenes, 0.1 pound; phenol, 0.5 pound; 
cresols, 2 pounds; naphthalene, 5 pounds. Two pounds of benzene from a ton of 
coal does not represent a very high percentage yield, yet so much coal is coked every 
year that the annual production of benzene from coal tar is enormous. 

Still larger quantities of aromatic hydrocarbons are needed, and these are 
synthesized from alkanes through the process of catalytic reforming (Sec. 9.3). 
This can bring about not only dehydrogenation, as in the formation of toluene from 
methylcyclohexane, but also cyclization and isomerization, as in the formation of 
toluene from //-heptane or 1,2-dimethylcyclopentane. In an analogous way, ben- 
zene is obtained from cyclohexane and methylcyclopentane, as well as from the 
hydrodealkylation of toluene. 

Today, petroleum is the chief source of the enormous quantities of benzene, 
toluene, and the xylenes required for chemicals and fuels. Half of the toluene and 
xylenes are utilized in high-test gasoline where, in a sense, they replace the aliphatic 
compounds inferior as fuels from which they were made. (A considerable 
fraction even of naphthalene, the major component of coal tar distillate, is now 
being produced from petroleum hydrocarbons.) 

12.5 Preparation of alkylbenzenes 

Although a number of the simpler alkylbenzenes are available from industrial 
sources, the more complicated compounds must be synthesized in one of the ways 
outlined below. 



PREPARATION OF ALKYLBENZENES 
1. Attachment of alkyl group: Friedel-Crafts alkylation. Discussed in Sees. 12.6-12,8. 



+ HX R may rearrange 



Lewis acid: AlClj, BF 3 , HF, etc. 
Ar-X cannot be used in place o/R-X 




SEC. 12.6 FRIEDEL-CRAFTS ALKYLATION 377 

2. Conversion of side chain. Discussed in Sec. 19.10. 

^ 

)CH 2 R Clemmensen or Wolff-Kishner 




re 



)CH 2 CH 2 R 



Friedel-Crafts alkylation is extremely useful since it permits the direct attach- 
ment of an alkyl group to the aromatic ring. There are, however, a number of 
limitations to its use (Sec. 12.8), including the fact that the alkyl group that 
becomes attached to the ring is not always the same as the alkyl group of tha parent 
halide; this rearrangement of the alkyl group is discussed in Sec. 12.7. 

There are frequently available aromatic compounds containing aliphatic side 
chains that are not simple alkyl groups. An alkylbenzene can be prepared from 
one of these compounds by converting the side chain into an alkyl group. Al- 
though there is an aromatic ring in the molecule, this conversion is essentially the 
preparation of an alkanc from some other aliphatic compound. The methods used 
are those that we have already learned for the preparation of alkanes: hydro- 
genation of a carbon carbon double bond in a side chain, for example. Many 
problems of the alkylbenzenes are solved by a consideration of simple alkane 
chemistry. 

The most important side-chain conversion involves reduction of ketones either 
by amalgamated /inc and HC1 (Clemmensen reduction) or by hydrazine and strong 
base (Wolff-Kishner reduction). This method is important because the necessary 
ketones are readily available through a modification of the Friedel-Crafts reaction 
that involves acid chlorides (see Sec. 19.6). Unlike alkylation by the Friedel- 
Crafts reaction, this method does not involve rearrangement 

Problem 12.1 How might you prepare ethylbenzene from: (a) benzene 
and ethyl alcohol; (b) acetophenone, C 6 H 5 COCH 3 ; (c) styrene, C 6 H 5 CH=CH 2 ; 
(d) a-phenylethyl alcohol, C 6 H 5 CHOHCH 3 ; and (e) 0-phenylethyl chloride, 
C 6 H 5 CH 2 CH 2 CI? 

Problem 12.2 How might you prepare 2,3-diphenylbutane from a-phenylethyl 
alcohol, QH 5 CHOHCH 3 ? * 



12.6 Friedel-Crafts alkylation 

If a small amount of anhydrous aluminum chloride is added to a mixture of 
benzene and methyl chloride, a vigorous reaction occurs, hydrogen chloride gas is 

f CH 3 C1 -42*+ (Q) " 3 + HC1 
Toluene 



378 ARENES CHAP. 12 

evolved, and toluene can be isolated from the reaction mixture. This is the 
simplest example of the reaction discovered in 1877 at the University of Paris by 
the French- American team of chemists, Charles Friedel and James Crafts. Con- 
sidered in its various modifications, the Frie del-Crafts reaction is by far the most 
important method for attaching alkyl side chains to aromatic rings. 

Each of the components of the simple example just given can be varied. The 
alkyl halide may contajnan alkyl group more complicated than methyl, and a 
halogen atom other thancHToTmc; in some cases alcohols are used or especially 
in industry alkenes. Substituted alkyl halides, like benzyl chloride, C h H 5 CH 2 Cl, 
also can be used. Because of the low reactivity of halogen attached to an aromatic 
ring (Sec. 25.5), aryl halides (Ar X, e.g., bromo- or chlorobenzene) cannot be used 
in plaCe of alkyl halides. 

The aromatic ring to \vhich the side chain becomes attached may be that of 
benzene itself, certain substituted benzenes (chiefly alkylbenzenes and haloben- 
zenes), or more complicated aromatic ring systems like naphthalene and anthracene 
(Chap. 30). 

, In place of aluminum chloride, other Lewis acids can be used, in particular 
BF 3 , HF, and phosphoric acid. 

The reaction is carried out by simply mixing together the three components; 
usually the only problems are those of moderating the reaction by cooling and of 
trapping the hydrogen halide gas. Since the attachment of an alkyl side chain 
makes the ring more susceptible to further attack (Sec. 1 1.5), steps must be taken 
to limit substitution to /?70/70alkylation. As in halogenation of alkanes (Sec. 2.8), 
this is accomplished by using an excess of the hydrocarbon. In this way an alkyl 
carbonium ion seeking an aromatic ring is more likely to encounter an unsub- 
stituted ring than a substituted one. Frequently the aromatic compound does 
double duty, serving as solvent as well as reactant. 

From polyhalogenated alkanes it is possible to prepare compounds containing 
more than one aromatic ring: 



2C 6 H 6 + CH 2 CI 2 -^> C 6 H 5 CH 2 C 6 H 5 + 2HC1 
Diphenylmethane 



2C 6 H 6 + C1CH 2 CH 2 C1 - C 6 H 5 CH 2 CH 2 C 6 H 5 + 2HC1 

1 ,2-Diphenylethane 



3C 6 H 6 + CHC1 3 -^2i* C 6 H 5 - C-C 6 H 5 + 3HC1 

H 

Triphenylmethane 



3C 6 H 6 + CC1 4 -^* C 6 H5-C-C 6 H 5 + 3HCI 

CJ 
Triphenylchloromethane 



SEC. 12.7 MECHANISM OF FRIEDEL-CRAFTS ALKYLATION 379 

12.7 Mechanism of Friedel-Crafts alkylation 

In Sec. 11.10 we said that two mechanisms are possible for Friedel-Crafts 
alkylation. Both involve electrophilic aromatic substitution, but they differ as to 
the nature of the electrophile. 

One mechanism for Friedel-Crafts alkylation involves the following steps, 

(1) RC1 + A1C1 3 ^= A1C1 4 - + R^ 

R 

(2) R J + C$H$ < > C6H5 

X H 

R 

(3) C 6 H^ + A1C1 4 - ^= C 6 H 5 R + HC1 + A1C1 3 

N H 

in which the electrophile is an alkyl carbonium ion. The function of the aluminum 
chloride is to generate this carbonium ion by abstracting the halogen from the 
alkyl halide. It is not surprising that other Lewis acids can function in the same 
way and thus take the place of aluminum chloride: 



R:X: + A1:CJ: ^I R3+:X:AI:CJ:O 
:C:" ":C: " 



R:X: + B:F: J= R;+:X:B:F:3 _ , . . 

. P. ' " . p. " " Carbonium ions 

: .v : *! : from alkyl 

halides 



R:X: + Fe:Cl: ^= R3 + :X:Fe::0 



K:X: + H:F: jz RO + :X:~H:F:9 

Judging from the mechanism just described, we might expect the benzene 
ring to be attacked by carbonium ions generated in other ways: by the action of 
acid on alcohols (Sec. 5.20) and on alkenes (Sec. 6.10). 

ROH + H + ^Z ROH 2 O T""*' R -H H 2 O Carbonium ions 

from alcohols 

|| || and from 

-C=C- + H + ^Z -C-C0 alkenes 



This expectation is correct: alcohols and alkenes, in the presence of acids, alkylate 



380 



ARENES 



CHAP. 12 



aromatic rings in what we may consider to be a modification of the Friedel-Crafts 
reaction. 



C 6 H 6 + (CH 3 ) 3 COH 

fer/-Butyl alcohol 



H : SO 4 



C 6 H 5 -C(CH 3 ) 3 

fer/- Butyl benzene 



C 6 H 6 -f (CH 3 ) 2 G=CH 2 n2 4 > C 6 H 5 -C(CH 3 ) 3 

^hobutylcne terr-Butylbenzene 

Also judging from the mecrTnism, we might expect Friedel-Crafts alkylation to 
be accompanied by the kind of rearrangement that is characteristic of carbonium 
ion reactions (Sec. 5.22). This expectation, too, is correct. As the 'following 
examples show, alkylbenzenes containing rearranged alkyl groups are not only 
formed but are sometimes the sole products. In each case, we see that the 



C 6 H 6 



CH 3 CH 2 CH 2 Cl 

w-Propyl chloride 



AlClj 



-18 to 8 



C 6 H 5 CH 2 CH 2 C S H 3 

//-Propyl benzene 

35-31% 



CH 3 

and C 6 H 5 CHCH 3 
Isopropyl benzene 
65-69% 



C 6 H 



66 



CH 3 CH 2 CH 2 CH 2 C1 

n-Butyl chloride 



AIC1 3 

o 



C 6 H 5 CH 2 CH 2 CH 2 CH 3 

/i-Butyl benzene 

34% 



and 



CH, 

C 6 H 5 CHCH 2 CH 3 

sec-Butylbenzene 
66% 



C 6 H 6 



CH 3 

CH 3 CHCH 2 C1 

Isobutyl chloride 



V CH, 
+ CttjCCHjOH 

V CH 3 

Neopentyl alcohol 



AlClj 



-18 c to80 



CH 3 
C 6 H 3 CCH, 

CH 3 

ter/-Butylbenzene 
Only product 

CH, 



_BPi 

~ 



6 H 5 C 



:CH 2 CH 3 

CH 3 

/erf-Pentylbenzene 
Only product 



particular kind of rearrangement corresponds to what we would expect if a less 
stable (1) carbonium ion were to rearrange by a 1,2-shift to a more stable (2 or 
3) carbonium ion. 

We can now make another addition to our list of carbonium ion reactions 
(Sec. 6.16). A carbonium ion may: 

(a) eliminate a hydrogen ion to form an alkene; 

(b) rearrange to a more stable carbonium ion; 

(c) combine with a negative ion or other basic molecule; 

(d) add to an alkene to form a larger carbonium ion; 

(e) abstract a hydride ion from an alkane; 

(f) alkylate an aromatic ring. 

A carbonium ion formed by (b) or (d) can subsequently undergo any of the 
reactions. 



SEC. 12.8 LIMITATIONS OF FRIEDEL-CRAFTS ALKYLATION 381 

In alkylation, as in its other reactions, the carbonium ion gains a pair of 
electrons to complete the octet of the electron-deficient carbonthis time from the 
TI cloud of an aromatic ring. 

Problem 12.3 ter/-Penty I benzene is the major product of the reaction of benzene 
in the presence of BFj with each of the following alcohols: (a) 2-methyl-l-butanol, 
(b) 3-methyl-2-butanol, (c) 3-methyl-l-butanol, and (d) neopentyl alcohol. Account 
for Us formation in each case. 

In some of the examples given above, we see that part of the product is made 
up of unrearranged alkylbenzenes. Must we conclude that part of the reaction 
does not go by way of carbonium ions? Not necessarily. Attack on an aromatic 
ring is probably one of the most difficult jobs a carbonium ion is called on to do; 
that is to say, toward carbonium ions an aromatic ring is a reagent of low reactivity 
and hence high selectivity. Although there may be present a higher concentration 
of the more stable, rearranged carbonium ions, the aromatic ring may tend to seek 
out the scarce unrearranged ions because of their higher reactivity. In some cases, 
it is quite possible that some of the carbonium ions react with the aromatic ring 
before they have time to rearrange; the same low stability that makes primary 
carbonium ions, for example, prone to rearrangement also makes them highly 
reactive. 

On the other hand, there is additional evidence (of a kind we cannot go into 
here) that makes it very likely that there is a second mechanism for Friedel-Crafts 
alkylation. In this mechanism, the electrophile is not an alkyl carbonium ion, 
but an acid-base complex of alkyl halide and Lewis acid, from which the alkyl 
group is transferred in one step from halogen to the aromatic ring. 



Cl 



ol ^ 

Cl-Al-Cl-R + C 6 H 6 > 



c 6 H 5 ; 



H 
/ 



8. 



H 
C 6 H/ + AK 



\ 






~R-C1A1C1 3 J R 



This duality of mechanism does not reflect exceptional behavior, but is usual 
for electrophilic aromatic substitution. It also fits into the usual pattern for 
nucleophilic aliphatic substitution (Sec. 14.16), which from the standpoint of the 
alkyl halide is the kind of reaction taking place. Furthermore, the particular 
halides (1 and methyl) which appear to react by this second mechanism are just 
the ones that would have been expected to do so. 

12.8 Limitations of Friedel-Crafts alkylation 

We have encountered three limitations to the use of Friedel-Crafts alkylation: 
(a) the danger of polysubstitution ; (b) the possibility that the alkyl group will 
rearrange; and (c) the fact that aryl halides cannot take the place of alkyl halides. 
Besides these, there are several other limitations. 

(d) An aromatic ring less reactive than that of the halobenzenes does not 
undergo the Friedel-Crafts reaction; evidently the carbonium ion, R+, is a less 
powerful nucleophile than NO 2 * and the other electron-deficient reagents that 
bring about electrophilic aromatic substitution. 



382 ARENES CHAP. U 

Next, (e) aromatic rings containing the NH 2 , NHR, or NR 2 group do 
not undergo Friedel-Crafts alkylation, partly because the strongly basic nitrogen 
ties up the Lewis acid needed for ionization of the alky] halide: 

OA1C1 3 
C 6 H 5 NH 2 + A1CI 3 > C 6 H 5 NH 2 

I 

Problem 12.4 Tying up of the acidic catalyst by the basic nitrogen is not the 
only factor that prevents alkylation, since even when excess catalyst is used, reaction 
does not occur. Looking at the structure of the complex (I) shown for aniline, can you 
suggest another factor? (Hint: See Sec. 11.18.) 

Despite these numerous limitations, the Friedel-Crafts reaction, in its various 
modifications (for example, acylation, Sec. 19.6), is an extremely useful synthetic 
tool. 



12.9 Reactions of alkylbenzenes 

The most important reactions of the alkylbenzenes are outlined below, with 
toluene ana ethylbenzene as specific examples; essentially the same behavior is 
shown by compounds bearing other side chains. Except for hydrogenation and 
oxidation, these reactions involve either electrophilic substitution in the aromatic 
ring or free-radical substitution in the aliphatic side chain. 

In following sections we shall be mostly concerned with (a) how experimental 
conditions determine which portion of the molecule aromatic or aliphatic 
is attacked, and (b) how each portion of the molecule modifies th'- reactions of the 
other portion. 



REACTIONS OF ALKYLBENZENES 



1. Hydrogenation. 

Example: 



(g)CH 2 CH 3 + 3Hj N|p . Mt QCH 2 CH 3 

Ethylbenzene Ethylcyclohexane 



2. Oxidation. Discussed in Sec. 12.10. 
Example: 



KMn0 4 (/-x-iCOOH (+CO 2 ) 

(or K 2 Cr 2 7 , 
ordil. HNO,) 

Ethylbenzene Benzoic acid 



SEC 12.9 



REACTIONS OF ALKYLBENZENES 



383 



3. Substitution in the ring. ElectrophiUc aromatic substitution. Discussed in Sec. 
12.11. 



Examples: 



CH 3 



Toluene 



CH 3 



N 2 



and 



CH 3 

y 

N0 2 

0-Nitrotoluene p-Nitrotoluene 

Chief products 



H 2 so 4 , so, 



CH 3 

[6 



0-Toluenesulfonic 
acid 



CH 3 

and (O) 

S 3H R: activates 
p-Toluenesulfonic and directs 
acid ortho, para 



CH,X, A1C1 3 



CH 3 

and O 

CH 3 

o-Xylene />-Xylene 

Temperature may affect orientation 



, FcX 3 



X = Cl, Br 



CH 3 

[6l x 



and 



(QJ 

X 



4. Substitution in the side chain. Free-radical halogenation. Discussed in Sees. 
12.12-12.14. 



Examples: 

CH 3 

(6) 

Toluene 



CH 2 C1 CHC1 2 

(6) -HSterlSl 



CC1 3 



Benzyl chloride 



Benzal chloride 



Benzotrichloride 



*|CH 2 CH 3 Cla 



[QJJHC8, 



:H 2 CH 2 C1 



heat of light 
Ethyl benzene 

Chief product 
Note: Competition between ring and side chain. Discussed in Sec. 12.12. 



-Phenylethyl 
chloride 



0-PhenyIethyl 
chloride 



384 ARENES 


CHAP. 12 


CH 2 C1 




CH 3 

frYl Cl2 


hcalorlight^ rV-O 


Fre^ radical 
substitution 


CH 3 CH 3 


lUJ 




LFCC.^ (gjCl ^ |gj 


Electrophilic 
substitution 


Cl 






12.10 Oxidation of alkylbenzenes 

Although benzene and alkanes are quite unreactive toward the usual oxidizing 
agents (KMnO 4 , K 2 Cr 2 O 7 , etc.), the benzene ring renders an aliphatic side chain 
quite susceptible to oxidation. The side chain is oxidized down to the ring, only a 
carboxyl group ( COOH) remaining to indicate the position of the original side 
chain. Potassium permanganate is generally used for this purpose, although 
potassium dichromate or dilute nitric acid also can be used. (Oxidation of a side 
chain is more difficult, however, than oxidation of an alkene, and requires pro- 
longed treatment with hot KMnO 4 .) 

jgjCH 2 CH 2 CH 2 CH 3 ho.KMn 0<) 

//-Bu(\lhcn/cnc Bcn/oic acid 

This reaction is used for two purposes: (a) synthesis of carboxylic acids, and 
(b) identification of alkylbenzenes. 

(a) Synthesis of carboxylic acids. One of the most useful methods of preparing 
an aromatic carboxylic acid involves oxidation of the proper alkylbenzene. For 
example: 




Terephthalic acid 
(1,4-Benzenedicarboxylic acid) 



CH 3 


COOH 


fr^i Cri< 


>7 "~ >H \ Ir^l 


IL)J 


'tyj 


NO 2 


N0 2 


/7-Nitrotoluene 


p-Nitroberizoic acid 



(b) Identification of alkylbenzenes. The number and relative positions of side 
chains can frequently be determined by oxidation to the corresponding acids. 



SEC. 12.11 ELECTROPHILIC AROMATIC SUBSTITUTION IN ALKYLBENZENES 



385 



Suppose, for example, that we are trying to identify an unknown liquid of formula 
C 8 H 10 and boiling point 137-139 that we have shown .in other ways to be an 
alkylbenzene (Sec. 12.22). Looking in Table 12.1 (p. 375), we find that it could be 
any one of four compounds: o- 9 m~, or p-xylene, or ethylbenzene. As shown below, 
oxidation of each of these possible hydrocarbons yields a different acid, and these 
acids can readily be distinguished from each other by their melting points or the 
melting points of derivatives. 





o-Xylene Phthalic acid, m.p. 231 

(b.p. 144) (p-nitrobcnzyl ester, m.p. 155) 



OOH 



w-Xylene Isophthalic acid, m.p. 348 
(b.p. 139) (p-nitrobenzyl ester, m.p. 215) 



COOH 

COOH 

Terephthalic acid, m.p. 300 subl. 
(/j-nitrobenzyl ester, m.p. 263 ) 





Ethylbenzene Benzoic acid, m.p. 122 
(b.p. 136 ) (/;-nitrobenzyl ester, m.p. 89 ) 



12.11 Electrophilic aromatic substitution in alkylbenzenes ^/ 

Because of its electron-releasing effect, an alkyl group activates a benzene 
ring to which it is attached, and directs ortho and para (Sees. 11.18 and 11.19). 

Problem 12.5 Treatment with methyl chloride and AIC1 3 at converts toluene 
chiefly into <? and p-xylenes; at 80, however, the chief product is w-xylene. Further- 
more, either o- or p-xylene is readily converted into m-xylene by treatment with AlClj 
and HC1 at 80. 

How do you account for this effect of temperature on orientation? Suggest a 
role for the HC1. 



Problem 12.6 Why is polysubstitution a complicating factor in Friedel-Crafts 
alkylation but not in aromatic nitration, sulfonation, or halogenation? 



386 ARENES CHAP. 12 

12*12 Halogenation of alkylbenzenes: ring vs. side chain 

Alkylbenzenes clearly offer two main areas to attack by halogens: the ring and 
the side chain. We can control the position of attack simply by choosing the 
proper reaction conditions. 

Halogenation of alkanes requires conditions under which halogen atoms are 
formed, that is, high temperature or light. Halogenation of benzene, on the other 
hand, involves transfer of positive halogen, which is promoted by acid catalysts 
like ferric chloride. 

CH 4 + C1 2 heat rllght > CH 3 C1 + HC1 
C 6 H 6 + C\ 2 F ^ l3>c ld > C 6 H 5 C1 + HC1 

We might expect, then, that the position of attack in, say, toluene would be 
governed by which attacking particle is involved, and therefore by the conditions 
employed. This is so: if chlorine is bubbled into boiling toluene that is exposed to 

Atom: attacks side chain 



.C\+ Ion: attacks ring 



ultraviolet light, substitution occurs almost exclusively in the side chain; in the 
absence of light and in the presence of ferric chloride, substitution occurs mostly in 
the ring. (Compare the foregoing with the problem of substitution vs. addition in 
the halogenation of alkenes (Sec. 6.21), where atoms bring about substitution and 
ions or, more accurately, molecules that can transfer ions bring about 
addition.) 

Like nitration and sulfonation, ring halogenation yields chiefly the o- and 




CH 3 




Cl z . Fe, or feCl 3> 



Toluene o-Chlorotoluene /?-Chlorotoluene 

M a/ o 42' { , 

/j-isomers. Similar results are obtained with other alkylbenzenes, and with bromine 
as well as chlorine. 

Side-chain halogenation, like halogenation of alkanes, may yield polyhalo- 
genated products; even when reaction is limited to monohalogenation, it may yield 
a mixture of isomers. 

Side-chain chlorination of toluene can yield successively the mono-, di-, and 
trichloro compounds. These are known as benzyl chloride, benzal chloride, and 





light ^-^ Hght ^^ light 

Toluene Benzyl chloride Benzal chloride Benzotrichloride 



SEC. 12.13 SIDE-CHAIN HALOGENATION OF ALKYLBENZENES 38 / 

benzotrichloride\ such compounds are important intermediates in the synthesis of 
alcohols, aldehydes, and acids. 



12.13 Side-chain halogenation of alkylbenzenes 

Chlorination and bromination of side chains differ from one another in 
orientation and reactivity in one very significant way. Let us look first at bromina- 
tion, and then at chlorination. 

An alkylbenzene with a side chain more complicated than methyl offers more 
than one position for attack, and so we must consider the likelihood of obtaining 
a mixture of isomers. Bromination of ethylbenzene, for example, could theoretically 
yield two products: 1-bromo-l-phenylethane and 2-bromo-l-phenylethane. Despite 





1 -Bromo- 1 -phenylethane 

Only product 
heat, 
light 

Ethylbenzene 

t- * 

2-Bromo- 1 -phenylethane 

a probability factor that favors 2-bromo-l -phenylethane by 3:2, the 

found is 1-bromo-l-phenylethane. Evidently abstraction of the hydrogens aiiacned 

to the carbon next to the aromatic ring is greatly preferred. 

Hydrogen atoms attached to carbon joined directly to an aromatic ring are called 
benzylic hydrogens. 




Benzylic hydrogen : 

easy to abstract 

The relative ease with which benzylic hydrogens are abstracted is shown not 
only by orientation of bromination but also and in a more exact way by com- 
parison of reactivities of different compounds. Competition experiments (Sec. 
3.22) show, for example, that at 40 a benzylic hydrogen of toluene is 3.3 times as 
reactive toward bromine atoms as the tertiary hydrogen of an alkane -and nearly 
100 million times as reactive as a hydrogen of methane! 

Examination of reactions that involve attack not only by halogen atoms but 
by other free radicals as well has shown that this is a general rule: benzylic hydro- 
gens are extremely easy to abstract and thus resemble allylic hydrogens. We can 
now expand the reactivity sequence of Sec. 6.22: 



Ease of abstraction allylic 
of hydrogen atoms benzylic 



> y > r> > r > CHj vinv i ic 

> 3 >2 >l >I4 , vmylic 



Side-chain halogenation of alkylbenzenes proceeds by the same mechanism as 




388 ARENES CHAP. 12 

halogenation of alkanes. Bromination of toluene, for xample, would include the 
following steps: 

:r\ _ s\ 

CH 2 

Benzyl radical Benzyl bromide 

The fact that benzylic hydrogens are unusually easy to abstract means that 
benzyl radicals are unusually easy to form. 

Ease of formation allyl y T , CH , , 

of free radicals benzyl > 3 > 2 > ' > CH 3 vmvl 

Again we ask the question: are these findings in accord with our rule that the 
more stable the radical, the more rapidly it is formed*! Is the rapidly formed benzyl 
radical relatively stable? 

The bond dissociation energies in Table 1.2 (p. 21) show that only 85 kcal is 
needed for formation of benzyl radicals from a mole of toluene, as compared 
with 91 kcal for formation of terf-butyl radicals and 88 kcal for formation of allyl 
radicals. Relative to the hydrocarbon from which each is formed, then, a benzyl 
radical contains less energy and is more stable than a tert-bulyl radical. 

We can now expand the sequence of radical stabilities (Sec. 6.22). Relative to 
the hydrocarbon from which each is formed, the relative stability of free radicals is: 

Stability of allyl ^ *o ^ r> -* i* >> ru vinv , 

free radicals benzyl > 3 > 2 > 1 > CH 3 -, vinyl 

Orientation of chlorination shows that chlorine atoms, like bromine atoms, 
preferentially attack benzylic hydrogen ; but, as we see, the preference is less marked: 

CH 2 CH 3 

~hi "'" I W I I 

:i 

1 -Chloro- 1 -phenylethane 2-Chloro- 1 -phenylet hane 
Ma for product, 9l n/ 9 

Furthermore, competition experiments show that, under conditions where 3, 2, 
and P hydrogens show relative reactivities of 5.0:3.8:1.0, the relative rate per 
benzylic hydrogen of toluene is only 1.3. As in its attack on alkanes (Sec. 3.28), 
the more reactive chlorine atom is less selective than the bromine atom: less selec- 
tive between hydrogens in a single molecule, and less selective between hydrogens 
in different molecules. 

In the attack by the comparatively unreactive bromine atom, we have said 
(Sec. 2.23), the transit. !>n state is reached late in the reaction process: the carbon- 
hydrogen bond is largely broken, and the organic group has acquired a great deal 
of free- radical character. The factors that stabilize the benzyl free radical stabilize 
the incipient benzyl free radical in the transition state. 

In contrast, in the attack by the highly reactive chlorine atom, the transition 
state is reached early in the reaction process: the carbon-hydrogen bond is only 
slightly broken, and the organic group has acquired little free-radical character. 
The factors that stabilize the benzyl radical have little effect on this transition state. 





SEC. 12.14 RESONANCE STABILIZATION OF THE BENZYL RADICAL 



389 



Just why benzylic hydrogens are less reactive toward chlorine atoms than even 
secondary hydrogens is not understood. It has been attributed to polar factors (Sec. 32.4), 
but this hypothesis has been questioned. 

12.14 Resonance stabilization of the benzyl radical 

How are we to account for the stability of the benzyl radical? Bond dissocia- 
tion energies indicate that 19 kcal/mole less energy (104 - 85) is needed to form 
the benzyl radical from toluene than to form the methyl radical from methane. 



C 6 H 5 CH 3 

Toluene 



C 6 H 5 CH 2 - + H- 
Benzyl radical 



A//- +85kcal 



As we did for the allyl radical (Sec. 6.24), let us examine the structures in- 
volved. Toluene contains the benzene ring and is therefore a hybrid of the two 
Kekule structures, I and II: 



i 11 

Similarly, the benzyl radical is a hybrid of the two Kekule structures, III and IV: 

CH 2 - CH 2 - 



This resonance causes stabilization, that is, lowers the energy content. However, 
resonance involving Kekule structures presumably stabilizes both molecule and 
radical to the same extent, and hence does not affect the difference in their energy 
contents. If there were no other factors involved, then we might reasonably expect 
the bond dissociation energy for a benzylic hydrogen to be about the same as that 
of a methane hydrogen (see Fig. 12.1). 

Considering further, however, we find that we can draw three additional 
structures for the radical: V, VI, and VII. In these structures there is a double 
bond between the side chain and the ring, and the odd electron is located on the 
carbon atoms ortho and para to the side chain. Drawing these pictures is, of 



equivalent to 







VII 



course, our way of indicating that the odd electron is not localized on the side 
chain but is delocalized, being distributed about the ring. We cannot draw com- 
parable structures for the toluene molecule. 



390 



ARENES 



CHAP. 12 



Contribution from the three structures, V-VII, stabilizes the radical in a way 
that is not possible for the molecule. Resonance thus lowers the energy content of 
the benzyl radical more than it lowers the energy content of toluene. This extra 
stabilization of the radical evidently amounts to 19 kcal/mole (Fig. 12.1). 



+H- 



.CH... {-H- 




Extra stabilization of 
ben; vl radical. 19 kcal 



CH, 



QH,CH 3 



Progress of reaction > 

Figure 12.1. Molecular structure and rate of reaction. Resonance- 
stabilized ben/yl radical formed faster than methyl radical. (Plots aligned 
with each other for easy comparison.) 

We say, then, that the benzyl radical is stabilized by resonance. When we use 
this expression, we must always bear in mind that we actually mean that the benzyl 
radical is stabilized by resonance to a greater extent than the hydrocarbon from 
which it is formed. 

In terms of orbitals, delocalization results from overlap of the p orbital occu- 
pied by the odd electron with the TT cloud of the ring. 




Figure 12.2. Benzyl radical. The p orbital occupied by the odd electron 
overlaps TT cloud of ring. 



Problem 12.7 It is believed that the side-chain hydrogens of the benzyl radical 
lie in the same plane as the ring. Why should they? 

Problem 12.8 The strength of the bond holding side-chain hydrogen in 
m-xylene is the same as in toluene; in o- and p-xylene it is 3-4 kcal lower. How do 
you account for these differences? 



SEC. 12.15 TRIPHENYLMETHYL: A STABLE FREE RADICAL 391 

12.15 Triphenylmethyl: a stable free radical 

We have said that benzyl and allyl free radicals are stabilized by resonance; 
but we must realize, of course, that they are stable only in comparison with simple 
alkyl radicals like methyl or ethyl. Benzyl and allyl free radicals are extremely 
reactive, unstable particles, whose fleeting existence (a few thousandths of a second) 
has been proposed simply because it is the best way to account for certain experi- 
mental observations. We do not find bottles on the laboratory shelf labeled 
"benzyl radicals" or "allyl radicals." Js there, then, any direct evidence for the 
existence of free radicals? 

In 1900 a remarkable paper appeared in the Journal of the American Chemical 
Society and in the Berichte der deutschen chemischen Gesellschaft\ its author was 
the young Russian-born chemist Moses Gomberg, who was at that time an 
instructor at the University of Michigan. Gomberg was interested in completely 
phenylated alkanes. He had prepared tetraphenyl methane (a synthesis a number 
of eminent chemists had previously attempted, but unsuccessfully), and he had now 
set himself the task of synthesizing hexaphenylethane. Having available triphenyl- 
chloromethane (Sec. 12.6), he went about the job in just the way we might today: 
he tried to couple together two triphenylmethyl groups by use of a metal (Sec. 9.4). 
Since sodium did not work very well, he used instead finely divided silver, mercury, 
or, best of all, zinc dust. He allowed a benzene solution of triphenylchloromethane 
to stand over one of these metals, and then filtered the solution free of the metal 
halide. When the benzene was evaporated, there was left behind a white crystalline 
solid which after recrystallization melted at 185; this he thought was hexaphenyl- 
ethane. 



Hexaphenylethane -f ZnCl2 
Expected product 



Triphenylchloromethane 
2 moles 

As a chemist always does with a new compound, Gomberg analyzed his 
product for its carbon and hydrogen content. To his surprise, the analysis showed 
88% carbon and 6% hydrogen, a total of only 94%. Thinking that combustion 
had not been complete, he carried out the analysis again, this time more carefully 
and under more vigorous conditions; he obtained the same results as before. 
Repeated analysis of samples prepared from both triphenylchloromethane and 
triphenylbromomethane, and purified by recrystallization from a variety of sol- 
vents, finally convinced him that he had prepared not a hydrocarbon not hexa- 
phenylethane but a compound containing 6% of some other element, probably 
oxygen. 

Oxygen could ha've come from impure metals; buj extremely pure samples of 
metals, carefully freed of oxygen, gave the same results. 




ARENES 



CHAP. 12 



Oxygen could have come from the air, although he could not see how molec- 
ular oxygen could react at room temperature with a hydrocarbon. He carried 
out the reaction again, this time under an atmosphere of carbon dioxide. When 
he filtered the solution (also under carbon dioxide) and evaporated the solvent, 
there was left behind not his compound of m.p. 185 but an entirely different 
substance, much more soluble in benzene than his first product, and having a 
much lower melting point. This new substance was eventually purified, and on 
analysis it gave the correct composition for hexaphenylethane: 93.8'% carbon, 
6.2% hydrogen. 

Dissolved in benzene, the new substance gave a yellow solution. When a 
small amount of air was admitted to the container, the yellow color disappeared, 
and then after a few minutes reappeared. When more oxygen was admitted, the 
same thing happened: disappearance of the color and slow reappearance. Finally 
the color disappeared for good; evaporation of the solvent yielded the original 
compound of m.p. 185. 

Not only oxygen but also halogens were rapidly absorbed by ice-cold solutions 
of this substance; even solutions of normally unreactive iodine were instantly 
decolorized. 

The compound of m.p. 185 was the peroxide, 

(C 6 H 5 ) 3 C-0-0-C(C 6 H 5 )3 

as Gomberg showed by preparing it in an entirely different way. The products of 
the halogen reactions were the triphenylhalomethanes, (C 6 H 5 ) 3 C X. 

If this new substance he had made was indeed hexaphenylethane, it was be- 
having very strangely. Cleavage of a carbon-carbon bond by such mild reagents 
as oxygen and iodine was unknown to organic chemists. 




Triphenylmethyl 



> Dimer 
Colorless 



Triphenylmethyl 
Yellow 

"The experimental evidence presented above forces me to the conclusion that 
we have to deal here with a free radical, triphenylmethyl, (C 6 H 5 ) 3 C. On this 
assumption alone do the results described above become intelligible and receive 



SEC. 12.15 TRIPHENYLMETHYL: A STABLE FREE RADICAL 393 

an adequate explanation." Com berg was proposing that he had prepared a stable 
free radical. 

It was nearly ten years before Gomberg's proposal was generally accepted. 
It now seems clear that what happens is the following: the metal abstracts a 
chlorine atom from triphenylchloromethane to form the free radical triphenyl- 
methyl; two of these radicals then combine to form a dimeric hydrocarbon. But 
the carbon-carbon bond in the dimer is a very weak one, and even at room tem- 
perature can break to regenerate the radicals. Thus an equilibrium exists between 
the free radicals and the hydrocarbon. Although this equilibrium tends to favor 
the hydrocarbon, any solution of the dimer contains an appreciable concentration 
of free triphenylmethyl radicals. The fraction of material existing as free radicals' 
is about 2% in a 1 M solution, 10% in a 0.01 M solution, and nearly 100 in 
very dilute solutions. We could quite correctly label a bottle containing a dilute 
solution of this substance as "triphenylmethyl radicals." 

Triphenylmethyl is yellow; both the dimer and the peroxide are colorless. 
A solution of the dimer is yellow because of the triphenylmethyl present in the 
equilibrium mixture. When ox>gen is admitted, the triphenylmethyl rapidly 
reacts to form the peroxide, and the yellow color -disappears. More dimer dis- 
sociates to restore equilibrium and the yellow color reappears. Only when all the 
dimer-triphenylmethyl mixture is converted into the peroxide does the yellow 
color fail to appear. In a similar way it is triphenylmethyl that reacts with iodine. 



-^ (C 6 H 5 ) 3 C-0~0-C(C 6 H 5 ) 3 
Dimer *=- 2(C 6 H 5 ) 3 O 

L^ 2(C 6 H 5 ) 3 C-l 

Triphenylmethyl 
radical 

Thus the dimer undergoes its surprising reactions by first dissociating into 
triphenylmethyl, which, although unusually stable for a free radical, is nevertheless 
an exceedingly reactive particle. 

Now, what is this dimer? For nearly 70 years it was believed to be hexaphenyl- 
ethane. It and dozens of analogs were studied exhaustively, and the equilibria 
between them and triarylmethyl radicals were interpreted on the basis of the hexa- 
arylethane structure. Then, in 1968, the dimer was shown to have the structure I. 

<C 6 H 5 ) 3 C /=> 



Gomberg's original task is still unaccomplished: hexaphenylethane, it seems, has 
never been made. 

The basic significance of Gomberg's work remains unchanged. Many dimers 
have been prepared, and the existence of free triarylmethyl radicals has been 
substantiated in a number of ways; indeed, certain of these compounds seem to 
exist entirely as the free radical even in the solid state. The most convincing evi- 
dence for the free-radical nature of these substances lies in properties that arise 
directly from the odd electron that characterizes a free radical. Two electrons 



394 ARENES CHAP. 12 

that occupy the same orbital and thus make up a pair have opposite spins (Sec. 
1.6); the magnetic moments corresponding to their spins exactly cancel each other. 
But, by definition (Sec. 2.12), the odd electron of a free radical is not paired, and 
hence the effect of its spin is not canceled. This spin gives to the free radical a net 
magnetic moment. This magnetic moment reveals itself in two ways: (a) the com- 
pound is paramagnetic: that is, unlike most matter, it is attracted by a magnetic 
field; and (b) the compound gives a characteristic paramagnetic resonance absorp- 
tion spectrum (or electron spin resonance spectrum, Sec. 13.14) which depends upon 
the orientation of the spin of an unpaired electron in a changing external magnetic 
field. This latter property permits the detection not only of stable free 'radicals 
but of low concentrations of short-lived radical intermediates in chemical reactions, 
and can even give information about their structure. (See, for example, Sec. 6.17). 

The remarkable dissociation to form free radicals is the result of two factors. 
First, triphenylmethyl radicals are unusually stable because of resonance of the 
sort we have proposed for the benzyl radical. Here, of course, there are an even 
larger number of structures (36 of them) that stabilize the radical but not the 
hydrocarbon; the odd electron is highly delocalized, being distributed over three 
aromatic rings. 

Second, crowding among the large aromatic rings tends to stretch and weaken 
the carbon-carbon bond joining the triphenylmethyl groups in the dimer. Once 
the radicals are formed, the bulky groups make it difficult for the carbon atoms to 
approach each other closely enough for bond formation: so difficult, in fact, 
that hexaphenylethane is not formed at all, but instead dimer I even with the 
sacrifice of aromaticity of one ring. Even so, there is crowding in the dimer, and 
the total effect is to lower the dissociation energy to only 1 1 kcal/mole, as compared 
with a dissociation energy of 80-90 kcal for most carbon-carbon single bonds. 

It would be hard to overestimate the importance of Gomberg's contribution to 
the field of free radicals and to organic chemistry as a whole. Although triphenyl- 
methyl was isolable only because it was not a typical free radical, its chemical 
properties showed what kind of behavior to expect of free radicals in general; 
most important of all, it proved that such things as free radicals could exist. 

Problem 12.9 The A// for dissociation of the dimer I has been measured as 
11 kcal/mole, the " act as 19 kcal/mole. (a) Draw the potential energy curve for the 
reaction, (b) What is the energy of activation for the reverse reaction, combination of 
triphenylmethyl radicals? (c) How do you account for this unusual fact? (Compare 
Sec. 2.17.) 

Problem 12.10 When 1.5 g of "diphenyltetra(0-tolyl)ethane" is dissolved in 50 g 
of benzene, the freezing point of the solvent is lowered 0.5 (the cryoscopic constant 
for benzene is 5). Interpret these results. 



12.16 Preparation of alkenylbenzenes. Conjugation with ring 

An aromatic hydrocarbon with a side chain containing a double bond can be 
prepared by essentially the same methods as simple alkenes (Sees. 5.12 and 5.19). 
In general, these methods involve elimination of atoms or groups from two ad- 
jacent carbons. The presence of the aromatic ring in the molecule may affect the 
orientation of elimination and the ease with which it takes place. 



SEC. 12.16 



PREPARATION OF ALKENYLBENZENES 



395 



On an industrial scale, the elimination generally involves dehydrogenation. 
For example, styrene, the most important of these compounds and perhaps the 
most important synthetic aromatic compound can be prepared by simply heating 
ethylbenzene to about 600 in the presence of a catalyst. The ethylbenzene, in 



CH 2 CH 3 



Cr 2 O 3 -Al 2 O 3 . 600 
90% yield > 




Ethylbcn/cne 



St>rene 



turn, is prepared by a Friedel-Crafts reaction between two simple hydrocarbons, 
benzene and ethylene. 

In the laboratory, however, we are most likely to use dehydrohalogenation or 
dehydration. 



-CH-CH3 - 



Cl 
1-Phenyl-l-chloroethane 



Styrene 



-CH--CH 3 
OH 

1-Phenylethanol 



Styrene 



Dehydrohalogenation of l-phenyl-2-chloropropane, or dehydration of 
l-phenyl-2-propanol, could yield two products: 1-phenylpropene or 3-phenyl- 
propene. Actually, only the first of these products is obtained. We saw earlier 
(Sees. 5.14 and 5.23) that where isomeric alkenes can be formed by elimination, the 



KOH 



>CH 2 CHCH 3 

Cl 

1 -Phenyl-2-chloropropanc 



1 -Phenylpropene 
Only product 



3-Phenylpropenc 






heat 



H 2 CHCH 

OH 

I -Phenyl-2-propano 



preferred product is the more stable alkene. This seems to be the case here, too. 
That 1-phenylpropene is much more stable than its isomer is shown by the fact that 
3-phenylpropene is rapidly converted into 1-phenylpropene by treatment with hot 
alkali. 



>CH 2 -CH=CH 2 

3-Phenylpropene 
(Allylbcnzenc) 



>CH=CH-CH 3 

1- Phenylpropene 




396 ARENES CHAP. 12 

A double bond that is separated from a benzene ring by one single bond is 
said to be conjugated with the ring. Such conjugation confers unusual stability on 

Double bond conjugated with ring: 

unusually stable system 

a molecule. This stability affects not only orientation of elimination, but, as we 
shall see (Sec. 21.6), affects the ease with which elimination takes place. 

Problem 12.11 Account for the stability of alkenes like styrene on the basis of: 
(a) delocalization of rr electrons, shew ing both resonance structures and orbital overlap; 
and (b) change in hybfidization. 



12.17 Reactions of alkenylbenzenes 

As we might expect, alkenylbenzenes undergo two sets of reactions: substitution 
in the ring, and addition to the double bond in the side chain. Since both ring and 
double bond are good sources of electrons, there may be competition between the 
two sites for certain clectrophilic reagents; it is not surprising that, in general, the 
double bond shows higher reactivity than the resonance-stabilized benzene ring. 
Our main interest in these reactions will be the way in which the aromatic ring 
affects the reactions of the double bond. 

Although both the benzene ring and the carbon carbon double bond can be 
hydrogenated catalytically, the conditions required for the double bond are much 

CH 2 CH 3 CH 2 CH 3 




H ;> Ni. 20. 2-3 atm f/^\] H 2 , N., 125, 110 

75 minutes 



atm r^^l 



Styrene Kthylben/cne fthylcyclohcxanc 

milder; by proper selection of conditions it is quite easy to hydrogenate the side 
chain without touching the aromatic ring. 

Mild oxidation of the double bond yields a glycol; more vigorous oxidation 
cleaves the carbon- carbon double bond and generally gives a carboxylic acid in 
which the COOH group is attached to the ring. 



OH OH 

Sivrcnc A gl\col Ben/oic acid 

Both double bond and ring react with halogens by ionic mechanisms that have 
essentially the same first step: attack on the n cloud by positively charged halogen. 
Halogen is consumed by the double bond first, and only after the side chain is 
completely saturated does substitution on the ring occur. Ring-halogenated 
alkenylbenzenes must be prepared, therefore, by generation of the double bond 
after halogen is already present on the ring. For example: 



SEC. 12.18 



STABILITY OF THE BENZYL CATION 



397 




HCICH 3 



CH=CH 2 



Cl 

/>-("hiorost>rCne 

In a similar way, alkenyl benzenes undergo the other addition reactions 
characteristic of the carbon- carbon double bond. Let us look further at the 
reactions of conjugated alkenylbenzenes, and the way in which the ring affects 
orientation and reactivity. 



12.18 Addition to conjugated alkeny (benzenes: orientation. Stability of the 
benzyl cation 

Addition of an unsymmetrical reagent to a double bond may in general yield 
two different products. In our discussion of alkenes (Sees. 6.11 and 6.17), we 
found that usually one of the products predominates, and that we can predict 
which it will be in a fairly simple way: in either electrophilic or free-radical addition, 
the first step takes place in the way that yields the more stable particle, carbonium ion 
in one kind of reaction, free radical in the other kind. Does this rule apply to 
reactions of alkenylbenzcnes? 

The effect of the benzene ring on orientation can be well illustrated by a single 
example, addition of HBr to 1-phenylpropene. In the absence of peroxides, 
bromine becomes attached to the carbon adjacent to the ring; in the presence of 
peroxides, bromine becomes attached to the carbon once removed from the ring. 
According to the mechanisms proposed for these two reactions, these products are 
formed as follows: 



C 6 H 5 CH--CHCH 3 



C 6 H 5 CH-CHCH 3 



C b H 5 CHCH 2 CH 3 

A benzyl cation 
C 6 ri 5 CHCHCHj 

Br 
A benzyl free radical 



C 6 H 5 CHCH 2 CH 3 

i 

Rr 
DI 



C 6 H 5 CH 2 CHCH 3 
Br 



No 
peroxides 



Peroxides 
present 



The first step of each of these reactions takes place in the way that yields the benzyl 
cation or the benzyl free radical rather than the alternative secondary cation or 
secondary free radical. Is this consistent with our rule that the more stable particle 
is formed faster? 

Consideration of bond dissociation energies has already shown us that a benzyl 
free radical is an extremely stable one. We have accounted for this stability on 
the basis of resonance involving the benzene ring (Sec. 12.14). 

What can we say about a benzyl cation? From the ionization potential (179 
kcal) of the benzyl free radical, we can calculate (Sec. 5.18) that the benzyl cation 
is 69 kcal more stable than the methyl cation, and just about as stable as the 
ter/-butyl cation. We can now expand our sequence of Sec. 8.21 to include the 
benzyl cation : 



Stability of 
carbonium ions 



benzyl aliyl 
3 > 2 



CH 3 



398 ARENES CHAP. 12 

The stability of a benzyl cation relative to the compounds from which it is 
made is also accounted for by resonance involving the benzene ring. Both the 
carbonium ion and the compound from which it is made are hybrids of Kekule 
structures. In addition, the carbonium ion can be represented by three other 
structures, I, II, and III, in which the positive charge is located on the oriho and 
para carbon atoms. Whether we consider this as resonance stabilization or simply 



CHCH 2 CH 3 CHCH 2 CH 3 CHCH 2 CH 3 

" " 






in 

as dispersal of charge, contribution from these structures stabilizes the carbonium 
ion. 

The orbital picture of the benzyl cation is similar to that of the benzyl free 
radical (Sec. 12.14) except that the p orbital that overlaps the *r cloud is an empty 
one. Thep orbital contributes no electrons, but permits further delocalization of the 
n electrons to include the carbon nucleus of the side chain. 

Problem 12.12 How do you account for the following facts? (a) Triphenylchloro- 
methane is completely ionized in certain solvents (e.g., liquid SO 2 ); (b) triphenyl- 
carbinol, (C 6 H 5 ) 3 COH, dissolves in concentrated H 2 SO 4 to give a solution that has 
the same intense >cllow color as triphenylchloromethane solutions. (Note: This yellow 
color is different from that of solutions of tnphenylniethyl.) 

Problem 12.13 In light of Problem 12.12, can you suggest a possible reason, 
besides steric hindrance, why the reaction of CCI 4 \sith benzene stops at triphenyl- 
chloroinethane? (See Sec. 12.6.) 



12.19 Addition to conjugated alkenylbenzenes: reactivity 

On the basis of the stability of the particle being formed, we might expect 
addition to a conjugated alkenylbenzcne, which yields a stable benzyl cation or 
free radical, to occur faster than addition to a simple alkene. 

On the other hand, we have seen (Sec. 12.17) that conjugated alkenylbenzenes 
are more stable than simple alkenes. On this basis alone, we might expect addition 
to conjugated alkenylbenzenes to occur more slowly than to simple alkenes. 

The situation is exactly analogous to the one discussed for addition to con- 
jugated dienes (Sec. 8.24). Both reactant and transition state are stabilized by 
resonance; whether reaction is faster or slower than for simple alkenes depends 
upon which is stabilized more (see Fig. 8.9, p. 275). 

The fact is that conjugated alkenylbenzenes are much more reactive than 
simple alkenes toward both ionic and free-radical addition. Here again as in 
most cases of this sort resonance stabilization of the transition state leading to a 
carbonium ion or free radical is more important than resonance stabilization of the 
reactant. We must realize, however, that this is not always true. 



SEC. 12.21 ANALYSIS OF ALKYLBENZENES 399 

Problem 12.14 Draw a potential energy diagram similar to Fig. 8.9 (p. 275) to 
summarize what has been said in this section. 

Problem 12.15 Suggest one reason why tetraphenylethylene does not react with 
bromine in carbon tetrachloride. 

12.20 Alkynylbenzenes 

The preparations and properties of the alkynylbenzenes are just what we might 
expect from our knowledge of benzene and the alkynes. 

. Problem 12.16 Outline all steps in the conversion of: (a) ethylbenzene into 
phenylacetylene; (b) /ra//$-l-phenylpropene into c/s-1-phenylpropene. 

12.21 Analysis of alkylbenzenes 

Aromatic hydrocarbons with saturated side chains are distinguished from 
alkenes by their failure to decolorize bromine in carbon tetrachloride (without 
evolution of hydrogen bromide) and by their failure to decolorize cold, dilute, 
neutral permanganate solutions. (Oxidation of the side chains requires more 
vigorous conditions; see Sec. 12.10.) 

They are distinguished from alkanes by the readiness with which they are 
sulfonated by and thus dissolve in cold fuming' sulfuric acid (see Sec. 11.4). 

They are distinguished from alcohols and other oxygen-containing com- 
pounds by their failure to dissolve immediately in cold concentrated sulfuric acid, 
and from primary and secondary alcohols by their failure to give a positive chromic 
anhydride test (Sec. 6.30). 

Upon treatment with chloroform and aluminum chloride, alkylbenzenes give 
orange to red colors. These colors are due to triarylmethyl cations, Ar 3 C+, 
which are probably produced by a Friedel-Crafts reaction followed by a transfer 
of hydride ion (Sec. 6.16): 

ArH CHCi " A1C S ArCHQ 2 ArH> AlCli > Ar 2 CHCl ArH - A1C S Ar 3 CH 
Ar 2 CHCl &*+ Ar 2 CHMlCl 4 -- 



Ar 3 CH- 



Ar 2 CH 2 + Ar 3 C< A1C 



Orange to red color 



This test is given by any aromatic compound that can undergo the Friedel-Crafts 
reaction, with the particular color produced being characteristic of the aromatic 
system involved: orange to red from halobenzenes, blue from naphthalene, purple 
from phenanthrene, green from anthracene (Chap. 30). 

Problem 12.17 Describe simple chemical tests (if any) that would distinguish 
between: (a) w-propyl benzene and 0-chlorotoluene; (b) benzene and toluene; 
(c) w-chlorotoluene and m-dichlorobenzene; (d) bromobenzene and bromocyclo- 
hexane; (e) bromobenzene and 3-bromo-l-hexene; (f) ethylbenzene and benzyl 
alcohol (C 6 H 5 CH 2 OH). Tell exactly what you would do and see. 

The number and orientation of side chains in an alkylbenzene is shown by 
the carboxylic acid produced on vigorous oxidation (Sec. 12.10). 



400 ARENES CHAP. 12 

Problem 12.18 On the basis of characterization tests and physical properties, 
an unknown compound of b.p. 182 is believed to be either m-diethylbenzene or 
//-butylbenzene. How could you distinguish between the two possibilities? 

(Analysis of alkylbenzenes by spectroscopic methods will be discussed in 
Sees. 13.15-13.16.) 

12.22 Analysis of alkenyl- and alkynylbenzenes 

Aromatic hydrocarbons with unsaturated side chains undergo the reactions 
characteristic of aromatic rings and of the carbon-carbon double or triple bond. 
(Their analysis by spectroscopic methods is discussed in Sees. 13.15-13.16.) 

Problem 12.19 Predict the response of allylbenzene to the following test reagents: 
(a) cold concentrated sulfuric acid; (b) Br^ in CC1 4 ; (c) cold, dilute, neutral perman- 
ganate; (d) CHC1 3 and A1CV, (e) CrO 3 and H 2 SO 4 . 

Problem 12.20 Describe simple chemical tests (if any) that would distinguish 
between : (a) styrene and ethylbenzene ; (b) styrene and phenylacetylene; (c) allylbenzene 
and 1-nonene; (d) allylbenzene and allyl alcohol (CH 2 =^CH CH 2 OH). Tell exactly 
what you would do and see. 



PROBLEMS 

1. Draw the structure of: 

(a) w-xylene (g) isopropylbenzene (cumene) 

(b) mesitylene (h) /raws-stilbene 

(c) o-ethyltoluene (i) l,4-diphenyl-l,3-butadiene 

(d) /7-di-/er/-butyl benzene (j) p-dibenzylbenzene 

(e) cyclohexy I benzene (k) /w-bromostyrene 

(f) 3-phenylpentane (1) diphenylacet>lene 

2. Outline all steps in the synthesis of ethylbenzene from each of the following 
compounds, using any needed aliphatic or inorganic reagents. 

(a) benzene (f) 1-chloro-l-phenylethane 

(b) styrene (g) 2-chloro-l-phenylethane 

(c) phenylacetylene (h) /?-bromoetJhylbenzene 

(d) a-phenylethyl alcohol (C 6 H 5 CHOHCH,) (i) acetophenone (C 6 H 5 CCH 3 ) 

(e) -phenylethyl alcohol (C H 5 CH 2 CH 2 OH) "\ 

3. Give structures and names of the principal organic products expected from 
reaction (if any) of //-propy I benzene with each of the following. Where more than one 
product is to be expected, indicate which will predominate. 

(a) H 2 , Ni, room temperature, low (k) C1 2 , Fe 
pressure (1) Br 2 , Fe 

(b) H : , Ni, 200% 100 atm. (m) 1 2 , Fe 

(c) cold dilute KMnO 4 (n) Br 2 , heat, light 

(d) hot KMnO 4 (o) CH 3 CI, AlClj, 

(e) K 2 Cr 2 O 7 , H 2 SO 4 , heat (p) C 6 H 5 CH 2 CI, A1CI 3 , (Note: A benzyl 

(f) boiling NaOH(aq) halide is not an aryl halide.) 

(g) boiling HCI(aq) (q) C 6 H 5 C1, A1C1 3 , 80 
(h) HNO 3 , H,SO 4 (r) isobutylene, HF 

(i) H,SO 4 , SO 3 (s) /m-butyl alcohol, H 2 SO 4 

(j) Tl (OOCCF 3 ) 3 (t) cyclohexene, HF 



PROBLEMS 401 

4. Give structures and names of the principal organic products expected from 
reaction (if any) of mw.y-1-phenyl-l-propene with: 

(a) H 2 , Ni, room temperature, low pressure (i) Br 2 , H 2 O 

(b) H 2 , Ni, 200, 100 atm. (j) cold dilute KMnO 4 

(c) Br 2 inCa 4 (k) hot KMnO 4 

(d) excess Br 2 , Fe (1) HCO 2 OH 

(e) HCI (m) O 3 , then H 2 O/Zn 

(f) HBr (n) Br 2 , 300 

(g) HBr (peroxides) (o) CHBr 3 , f-BuOK 

(h) cold cone. H 2 SO 4 (p) product (c), KOH(alc) 

5. Give structures and names of the principal organic products expected from each 
of the following reactions:. 

(a) benzene + cyclohexene + HF 

(b) phenylacetylene 4- alcoholic AgNO 3 

(c) w-nitroben/yl chloride f K 2 Cr 2 O 7 + H 2 SO 4 + heat 

(d) allylbenzene + HCI 

(e) 1 />-chlorotoluene + hot KMnO 4 

(f) *w<?w0/tCi H 12 O 2 , 2-methoxy-4-allylphenol) -f hot KOH 
> isoeugenol (C| H 12 O 2 ) 

(g) benzyl chloride 4- Mg + dry ether 
(h) product of (g) + H 2 O 

(i) /j-xylene + Br 2 -f Fe 

(j) l-phenyl-l,3-butadiene + one mole H 2 + Ni, 2 atm., 30 

(k) //ww-stilbene + O 3 , then H 2 O/Zn 

(1) 1,3-diphenylpropyne + H 2 , Pd *C, 5 H 14 

(m) 1,3-diphenylpropyne -H Li, NH 3 (Jiq) >C I5 H 14 

(n) p-CH 3 OC 6 H 4 CH- CHC 6 H 5 + HBr 

6. Treatment of ben/yl alcohol (C 6 H 5 CH 2 OH) with cold concentrated H 2 SO 4 
yields a high-boiling resinous material. What is a likely structure for this material, and 
how is it probably formed ? 

7. Label each set of hydrogens in, each of the following compounds in order of 
expected ease of abstraction by bromine atoms. Use (1) for the most reactive, (2) for 
the next, etc. 

(a) l-phenyl-2-hexene 



(c) 1,2,4-trimethylbenzene (Hint: See Problem 12.8, p. 390.) 

(d) What final monobromination product or products would abstraction of each kind 
of hydrogen in (a) lead to?. 

8. Give structures and names of the products expected from dehydrohalogenation 
of each of the following. Where more than one product can be formed, predict the 
major product. 

(a) 1-chloro-l-phenylbutane (c) 2-chloro-2-phenylbutane 

(b) l-chloro-2-phenylbutane (d) 2-chloro-l-phenylbutane 

(e) 3-chloro-2-phenylbutane 

9. Answer Problem 8 for dehydration of the alcohol corresponding to each of 
the halides given. (Hint: Do not forget Sec. 5.22.) 

10. Arrange in order of ease of dehydration: (a) the alcohols of Problem 9; 
(b) C 6 H 5 CH 2 CH 2 OH, C 6 H 5 CHOHCH 3> (C 6 H 5 ) 2 C(OH)CH 3 . 



402 ARENES CHAP. 12 

11. Arrange the compounds of each set in order of reactivity toward the indicated 
reaction. 

(a) addition of HC1 : styrene, /7-chlorostyrene, p-methylstyrene 

(b) dehydration : a-phenylethyl alcohol (O 6 H 5 CHOHCH 3 ), a-(p-nitrophenyi)ethyl alcohol, 
a-(/?-aminophenyl)ethyl alcohol. 

12. (a) Draw structures of all possible products of addition of one mole of Br 2 to 
I-phenyl-l,3-butadiene. (b) Which of these possible products are consistent with the 
intermediate formation of the most stable carbonium ion? (c) Actually, only 1-phenyl- 
3,4-dibromo-l-butene is obtained. What is the most likely explanation of this fact? 

13. (a) The heats of hydrogenation of the stereoisomeric stilbenes (1,2-diphenyl- 
ethenes) are: cis- t 26.3 kcal; trans-, 20.6 kcal. Which isomer is the more stable? (b) cis- 
Stilbene is converted into fra/w-stilbene (but not vice versa) either (i) by action of a very 
small amount of Br 2 in the presence of light, or (ii) by action of a very small amount of 
HBr (but not HC1) in the presence of peroxides. What is the agent that probably brings 
about the conversion ? Can you suggest a way in which the conversion might take place? 

(c) Why is rro/ij-stilbene not converted into c/s-stilbene ? 

14. One mole of triphenylcarbinol lowers the freezing point of 1000 g of 100% 
sulfuric acid twice as much as one mole of methanol. How do you account for this? 

15. Can you account for the order of acidity : triphenylmethane > diphenylmethane > 
toluene > w-pentane (Hint: See Sec. 12.19.) 

16. When a mixture of toluene and CBrCl 3 was irradiated with ultraviolet light, 
there were obtained, in almost exactly equimolar amounts, benzyl bromide and CHC1 3 . 
(a) Show in detail all steps in the most likely mechanism for this reaction, (b) There 
were also obtained, in small amounts, HBr and C 2 C1 6 ; the ratio of CHC1 3 to HBr was 
20:1. How do you account for the formation of HBr? Of C 2 C1 6 ? What, specifically, 
does the 20:1 ratio tell you about the reaction? (c) When the reaction was carried out 
on a series of ^-substituted toluenes, G C 6 H 4 CH 3 , the following order of reactivity 
was observed 

G = CH 3 O > CH 3 > H > Br 

How do you account for this order of reactivity? 

17. When the product of the HF-catalyzed reaction of benzene with l-dodecene, 
previously reported to be pure 2-phenyldodecane, was analyzed by gas chromatography, 
five evenly-spaced peaks of about the same size were observed, indicating the presence of 
five components, probably closely related in structure. What five compounds most 
likely make up this mixture, and how could you have anticipated their formation? 

18. The bond dissociation energy for the central C C bond of hexacyclopropyl- 
ethane is only 45 kcal/mole. Besides steric interaction, what is a second factor that may 
contribute to the weakness of this bond? (Hint: See Sec. 9.9.) 

19. On theoretical grounds it is believed that a primary isotope effect is greatest if 
bond breaking and bond making have proceeded to an equal extent in the transition 
state, (a) In free-radical halogenation of the side chain of toluene, k. H /k D is about 2 in 
chlorination and about 5 in bromination. There are two possible interpretations of 
this. What are they? (b) In light of Sec. 2.23, which interpretation is the more likely? 

20. The three xylenes are obtained as a mixture from the distillation of coal tar; 
further separation by distillation is difficult because of the closeness of their boiling points 
(see Table 12.1, p. 375), and so a variety of chemical methods have been used. In each 
case below tell which isomer you would expect to react preferentially, and why. 

(a) An old method : treatment of the mixture at room temperature with 80% sul- 
furic acid. 

(b) Another old method: sulfonation of all three xylenes, and then treatment of the 
sulfonic acids with dilute aqueous acid. 



PROBLEMS 403 

(c) A current method : extraction of one isomer into a BFj/HF layer. 

(d) A proposed method: 

C 6 H 5 C(CH 3 ) 2 -Na + + xylenes 7- C 6 H 5 CH(CH 3 ) 2 -f A 4- two xylenes 
(Hint to part (d): See Sees. 8.10, 5.17, and 11,18.) 

21. Upon ionic addition of bromine, c/s-1-phenyl-l-propene gives a mixture of 17% 
erythro dibromide and 83%-' threo\ /raws- 1-pheny 1-1 -propene gives 88% erythro, 12% 
threo\ and /ra/w-l-(p-methoxyphenyl)propene gives 63% erythro, 37% threo. 

Ar Ar 

-Br 



~Br H 

-Br Br 



-H 



CH 3 

and enantiomer and enantiomcr 

Ervthro Threo 

How do these results compare with those obtained with the 2-butenes (Sec. 7.11)? 
Suggest a possible explanation for the difference. What is the effect of the p-methoxy 
group, and how might you account for this? 

22. Outline all steps in a possible laboratory synthesis of each of the following 
compounds from benzene and/or toluene, using any necessary aliphatic or inorganic 
reagents. Follow instructions on p. 224. Assume a pure para isomer can be separated 
from an ortho.para mixture. 

(a) ethylbenzene (i) c/s-1-phenylpropene 

(b) styrene (j) p-/m-butyltoluene 

(c) phenylacetylene (k) p-nitrostyrene 

(d) isopropylbenzene (1) p-bromobenzyl bromide 

(e) 2-phenylpropene (m) /?-nitrobenzal bromide 

(f ) 3-phenylpropene (allylbenzenc) (n) p-bromobenzoic acid 

(g) 1-phenylpropyne (two ways) (o) w-bromobenzoic acid 
(h) mr/fj-1-phenylpropcne (p) 1,2-diphenylethane 

(q) /?-nitrodiphenylmethane (/7-O 2 NC 6 H 4 CH 2 C 6 H 5 ) (Hint: See Problem 3(p).) 

23. Describe simple chemical tests that would distinguish between: 

(a) benzene and cyclohcxane 

(b) benzene and 1 -hexene 

(c) toluene and //-heptane 

(d) cyclohexylbenzene and 1-phenylcyclohexcne 

(e) benzyl alcohol (C () H 5 CH 2 OH) and /f-pentylben/ene 

(f) cinnamyl alcohol (C 6 H 5 CH--CHCH,OH) and 3-phenyl-l-propanol 
(C 6 H 5 CH 2 CH 2 CH 2 OH) 

(g) chlorobenzene and ethylbenzene 

(h) nitrobenzene and m-dibromobenzene 

24. Describe chemical methods (not necessarily simple tests) that would enable 
you to distinguish between the compounds of each of the following sets. (For example, 
make use of Table 18.1, page 580. 

(a) 1-phenylpropene, 2-phenylpropene, 3-phenylpropene (allylbenzene) 

(b) all alkylbenzenes of formula G>Hi 2 

(c) w-chlorotoluene and benzyl chloride 

(d) /Miivinylbenzene (p-C 6 H 4 (CH--^CH 2 ) 2 ) and l-phenyl-l,3-butadiene 

(e) C 6 H 5 CHaCH 3f p-CH 3 C 6 H 4 CH 2 Cl, and 



404 ARENES CHAP. 12 

25. An unknown compound is believed to be one of the following. Describe how 
you would go about finding out which of the possibilities the unknown actually is. 
Where possible, use simple chemical tests; where necessary, use more elaborate chemical 
methods like quantitative hydrogenation, cleavage, etc. Where necessary, make use of 
Table 18.1, page 580. 

b.p. b.p. 

bromobenzene 156 p-chlorotoluene 162 

3-phenylpropene 157 0-ethyltoluene 162 

m-ethyltoluene 158 p-ethyltoluene 163 

/i-propylbenzene 159 mesitylene 165 

0-chlorotoluene 159 2-phenylpropene 165 

m-chlorotoluene 162 

26. The compound indene, C 9 H 8 , found in coal tar, rapidly decolorizes Br 2 /CCl 4 
and dilute KMnO 4 . Only one mole of hydrogen is absorbed readily to form indane, 
C 9 Hi . More vigorous hydrogenation yields a compound of formula C 9 Hi 6 . Vigorous 
oxidation of indene yields phthalic acid. What is the structure of indene? Of indane? 
(Hint: See Problem 9.17, p. 313.) 

27. A solution of 0.01 mole tert-buly] peroxide (p. 114) in excess ethylbenzene was 
irradiated with ultraviolet light for several hours. Gas chromatographic analysis of 
the product showed the presence of nearly 0.02 mole of tert-butyl alcohol. Evaporation 
of the alcohol and unreacted ethylbenzene left a solid residue which was separated by 
chromatography into just two products: X (1 g) and Y (1 g). X and Y each had the em- 
pirical formula C 8 H 9 and m.w. 210; each was inert toward cold dilute KMnO 4 and 
toward Br 2 /CCl 4 . 

When isopropyl benzene was substituted for ethylbenzene in the above reaction, 
exactly similar results were obtained, except that the single compound Z (2.2 g) was 
obtained instead of X and Y. Z had the empirical formula C 9 H H , m.w. 238, and was 
inert toward cold, dilute KMnO 4 and toward Br 2 /CCl 4 . 

What are the most likely structures for X, Y, and Z, and what is the most likely 
mechanism by which they are formed? 



Chapter 

Spectroscopy and Structure 



13.1 Determination of structure: spectroscopic methods 

Near the beguiling of our study (Sec. 3.32), we outlined the general steps an 
organic chemist takes when he is confronted with an unknown compound and sets 
out to find the answer to the question: nhat is it? We have seen, in more detail, 
some of the ways in which he carries out the various steps: determination of mo- 
lecular weight and molecular formula; detection of the presence or absence of 
certain functional groups; degradation to simpler compounds; conversion into 
derivatives; synthesis by an unambiguous route. 

At every stage of structure determination from the isolation and purification 
of the unknown substance to its final comparison with an authentic sample 
the use of instruments has, since World War II, revolutionized organic chemical 
practice. Instruments not only help an organic chemist to do what he does/<w/er 
but, more important, let him do what could not be done at all before: to analyze 
complicated mixtures of closely related compounds; to describe the structure of 
molecules in detail never imagined before; to detect, identify, and measure the 
concentration of short-lived intermediates whose very existence was, not so long 
ago, only speculation. 

By now, we are familiar with some of the features of the organic chemical 
landscape; so long as v^e do not wander too far from home, we can find our way 
about without becoming lost. We are ready to learn a little about how to interpret 
the kind of information these modern instruments give, so that they can help us 
to see more clearly the new things we shall meet, and to recognize them more 
readily when we encounter them again. The instruments most directly concerned 
with our primary interest, molecular structure, are the spectrometers measurers o 
spectra. Of the various spectra, we shall ^ual^^ 

and nucleajjna&w^^^ since they are thc_workhorsesjpf the organic 

chemical laboratory today ; of these, we shall spend most of our time with nmr. 



406 SPKCTROSCOPY AND STRl CTl RK CHAP. 13 



We shall look \ery briefly at three other kinds of spectra: ma\v. ultraviolet 
and eject rott \j?in,resongucc_(e&L}-~-' 

In all this, \\e must constantly keep in mind that \\hat xsc learn at this stage must 
be greath simplified. There are many exceptions to the general i/ations \\e shall 
learn: there are main pitfalls into \\hich \\e can stumble Our ability to apply 
spectroscopic methods to the determination of organic structure is limited by 
our understanding of organic chemistry as a \vhole and in this \\e are, of course, 
onl> beginners. But so long as \se are aware of the dangers of a little learning, 
and are \\iiling to make mistakes and profit from them, it is \\ortlnvhile for us 
to become beginners in this area of organic chemistry, too. 

Let us look first at the mass spectrum, and then at the others, which, as we 
shall see, are all parts different ranges of \\avelengths-ofa single spectrum: 
/that of electromagnetic radiation. 

13.2 The mass spectrum 

In the mass spectrometer, molecules are bombarded with a beam of energetic 
electrons. The molecules are ionized and broken up into many fragments, some 
of \\hich are positive ions. Each kind of ion has a particular ratio of mass to 
charge, or /;/ e Value. For most ions, the charge is I, so that m/e is simply the 
mass of the ion. Thus, for neopentane: 

CH, 

CHj-C-CH, 
CH 3 



2e' + (C 5 H, 2 )t mje = 72 
Molecular ion 



and others 



The set of ions is analyzed in such a way that a signal is obtained for each 
value of m/e that is represented; the intensity of each signal reflects the relative 
abundance of the ion producing the signal. The largest peak is called the base 
peak: its intensity is taken as 100, and the intensities of the other peaks arc ex- 
pressed relative to it. A plot or even a list showing the relative intensities of 
signals at the various m/e values is called a mass spectrum, and is highly character- 
istic of a particular compound. Compare, for example, the spectra of two isomers 
shown in Fig. 13.1. 

JjrtasjLspectra can, be used in two general ways: (a) to prpvejjie identity of 
two cjornpounds^.and (b) toJveJpLestjahlisJillie .suucturc of a^jieyv^^mpoMTKt ^ 

"Two compounds are shown to be identical by the fact that they have identical 
physical properties: melting point, boiling point, density, refractive index, etc. 
The greaterjhe number _ofj^ysical.prqperties measured, the stronger the evidence. 





1 


I 


I 






(C 4 H 9 ) + 






(C : H 3 


m'e: 


57 


41 


29 


27 


Relative 










intensity: 


100 


41.5 


38.5 


15.7 




Base peak 









SEC. 13 

100 


2 THE MASS SPECTRUM 40 


- 


(a) 


80 


- 




| 60 






c 






o 


" 




1 40 


- 




OL 


- 






20 


- 




1 



1 


l 




II .1 i . J 


i ' I ^^ T ' i ' r ' T 1 1 ' 1 ' j ' i i 
20 30 40 50 60 70 80 90 100 110 12 



m/e 



100 



80 



I > 

C 

s 

'5 40 



20 







X 



10 20 30 40 50 60 70 80 90 100 110 120 

m/e 

Figure 13.1. Mass spectra of two isomeric alkanes. (a) //-Octane; 
(ft) 2,2,4-trimethylpentane. 



Now,_a single mass spectrum amounts to dozens of physical properties, since it 
'sTfows the relative abundances of dozens of different fragments. If we measure the 
mass spectrum of an unknown compound and find it to be identical with the spec- 
trum of a previously reported compound of known structure, then we can conclude 
that- almost beyond the shadow of a doubt -the two compounds are identical. 
JP.ec.t ru m hcl ps^ tojesta bli sh thc^stnic^re^qf jijieir comjxwnd in 
: it can give an exact niolecularjyeight; it cajn_g[ve a molecular 
^or at least najro^11ie"po^M bill lies Jo $ very fe\v; and it can indicate the 

certain structural units. 

If one electron is removed from the parent molecule, there is produced the 
molecular ion (ur parent ion), whose m\e value is, of course, the molecular weight of 



408 SPECTROSCOPY AND STRUCTURE CHAP. 13 

the compound. Sometimes the M+ peak is the base peak, and is easily recognized; 
often, though, it is not the base peak it may even be very small and consider- 
able work is required to locate it. Once identified, it gives the most accurate 
molecular weight obtainable. 

M + e~ > M* + 2e- 

Moiecular ion 

(Parent ion) 

m/e SB mol. wt. 

We might at first think that the M + peak would be the peak of highest m/e 
value. This is not so, however. Most elements occur naturally as several isotopes; 
generally the lightest one greatly predominates, and the heavier ones occur to 
lesser extent. Table 13.1 lists the relative abundances of several heavy isotopes. 

Table 13.1 ABUNDANCE OF SOME HEAVY ISOTOPES 



Heavy 
isotope 


Abundance relative 
to isotope of lowest 
atomic weight 


2H 


0.015% 


13C 


1.11 


15N 


0.37 


18Q 


0.20 


33S 


0.78 


34 S 


4.4 


37C1 


32.5 


8iBr 


98.0 



The molecular weight that one usually measures and works with is the sum of 
the average atomic weights of the elements, and reflects the presence of these 
heavy isotopes. This is. not true, however, of the molecular weight obtained from 
the mass spectrum; here, the M+ peak is due to molecules containing only the 
commonest isotope of each element. 

Consider benzene, for example. The M f peak, m/e 78, is due only to ions of 
formula C 6 H 6 +. There is a peak at m/e 79, the M + 1 peak, which is due to 
C 5 13 CH 6 + and C 6 H 5 D + . There is an M + 2 peak at m/e 80, due to C 4 13 C 2 H 6 % 
C 5 13 CH 5 D% and C 6 H 4 D 2 +. Now, because of the low natural abundance of most 
heavy isotopes, these isotopic peaks are generally much less intense than the 
M+ peak; just how much less intense depends upon which elements they are due 
to. In the case of benzene, the M + 1 and M + 2 peaks are, respectively, 6.58% 
and 0.18% as intense as the M* peak. (Table 13.1 shows us, however, that a 
monochloro compound would have an M -f 2 peak about one-third as intense 
as the M * peak, and a monobromo compound would have M and M -f 2 peaks 
of about equal intensity.) 

It is these isotopic peaks that make it possible for us to determine the molecular 
formula of the compound. Knowing the relative natural abundances of isotopes, 
one can calculate for any molecular formula the relative intensity to be expected 
for each isotopic peak: M + 1, M + 2, etc. The results of such calculations are 
available in tables. Consider, for example, a compound for which 'M* is 44. 



SEC. 13.3 THE ELECTROMAGNETIC SPECTRUM 409 

The compound might be (among other less likely possibilities) N 2 O, CO 2 , C 2 H 4 O, 
or C 3 H 8 . By use of Table 13.2, we clearly could pick out the most likely formula 
from the mass spectral data. 

Table 13.2 CALCULATED INTENSITIES OF ISOTOPIC PEAKS 

M M + l M + 2 



N 2 O 100 0.80 0.20 

CO 2 100 1.16 0.40 

C 2 H 4 100 1.91 0.01 

C 3 H 8 100 3.37 0.04 



Finally, study of compounds of known structure is beginning to reveal the 
factors that determine which fragments a particular structure is likely to break into. 
In this we can find much that is familiar to us: the preferential formation of car- 
bonium ions that we recognize as being relatively stable ones; elimination of small, 
stable molecules like water, ammonia, and carbon monoxide. Under the energetic 
conditions, extensive rearrangement can occur, complicating the interpretation; 
but here, too, patterns are emerging. The direction of rearrangement is, as we 
would expect, toward more stable ions. As this knowledge accumulates, the process 
is reversed: from the kind of fragmentation an unknown compound gives, its 
structure is deduced. 

Problem 13.1 (a) Referring to the neopcntane fragmentation (p. 406), what is a 
likely structure for C 4 H 9 + ; C 3 H 5 + ; C 2 H 5 + ; QH 3 + ? (b) Write a balanced equation 
for the formation of C 4 H 9 + from the molecular ion CjH^*. 



13.3 The electromagnetic spectrum 

We ^rQ^ready^^l^rjw^^vdinQU^^ kinds_of electromagnetic radiation: 
light visible, ultraviolet^ infrared x-rays, radio and radar waves. These are 
simply Different parts of a broad spectrum that stretches from g^rnj^ra)^whose 
wyekngths are measured in fractions of an Angstrom unit, to radio wayes^ whose 
wayeleugthsjire measuredjiri meters or even kilometers. AlMhese wave? have. 
the ^sam^yg[ocijty y 3 x 10 l j:enti imeters. pe second. Their Jfrequency js related 
to the wavelength~by the exgression_ ~~~~ "' 



where v = frequency, in Hz (Hertz, cycles/sec) 

A wavelength, in cm 
c = velocity, 3 x 10 10 cm/sec 

Theshortcr the wavelength^ the higher the frequency. 

" When a beam of electromagnetic radiation is passed through a substance, the 
radiation can be either absorbed or transmitted, depending upon its frequency 
and the structure of the molecules it encounters. Electromagn^c jadjatioiL ia. 
^nergy, and hejice ,when a molecule absorbs radiationfu gams energy. Just how 



410 SPECTROSCOPY AND STRUCTURE CHAP. 13 

much energy it gains depends upon the frequency of the radiation: the higher the 
frequency (the shorter the wavelength), the greater the gain in energy. 

A = hv 

where A = gain in energy, in ergs 

h = Planck's constant, 6.5 x 10~ 27 erg-sec 
v = frequency, in Hz 

The energy joined byjjiejiplecule jn this way may bring about increased 
vibration or rotation of the atoms, or may raise electron^ to higher energy levels. 
The particular 'Frequency of radiation that a given molecule can absorb~de"pends 
upon the changes in vibrations or rotations or electronic slates that are permitted 
to a molecule of that structure. The spectrum of a compound is a plot that shows 
how much electromagnetic radiation is absorbed (or transmitted) at each frequency. 
It can be highly characteristic of the compound's structure. 



13.4 The infrared spectrum 

Of all the properties of an organic compound, the one that, by itself, gives the 
.most jn form at jon about the compound's structure is its infrared spectrum!""*" 

A molecule is cor^sTanjj^viFrajingj i tsjonds^rr^// (aHd co n tract), and Lfefl</ 
with^ respect lo 'each, othejv Qianges^in vibrations, of a_ molecule , are caused by 
ataorpijpn. of infraredjjght; light lyjn&_ beyond (lower frequency, longer wave- 
length, less energy) thejred end ofjthe visible spectrum., 

A particular part of the infrared spectrum is referred to either by its wavelength 
or and this is considered preferable by its frequency. Wavelength is expressed 
in microns, |JL (1 tx = 10 ~ 4 cm or 10 4 A). Frequency is expressed, not in Hertz, but 
in \cavenumbersi cm" 1 , often called reciprocal ceiwmeters; the wavenumber is 
simply the number of waves per centimeter, a~nd is equal to the reciprocal of the 
wavelength in centimeters. 

Like the mass spectrum, an infrared spectrum is a highly characteristic property 
of an organic compound see, for example, the spectra in Fig. 13.2, p. 411 and 
can be used both to establish the identity of two compounds and to reveal the 
structure of a new compound. 

Two substances that have identical infrared spectra are, in effect, identical in 
thousands of different physical properties the absorption of light at thousands of 
different frequencies and must almost certainly be the same compound. (One 
region of the infrared spectrum is called, appropriately, the fingerprint region.) 

The infrared spectrum heipsto rcveaijhc structure ofa new compound by _. 
tcljjng us what^ groups arejpresent in or absent fromthe molecule., A particular 
group of atoms jjvcs rise to characteristic absorption bands', that is to say, ajjarlicu.- , 
lafgroup absorbs light of certain frequencies that are much the same from com- 
pound to compound. For example, the --OH group of alcohols absorbs strongly 
at 3200-3600 cm - *jjhc C_Ojroup of keioneTglTl'O crrrj ;" the j-CsN group,* 
^^^J^Tt^J^ grojjpatJ450and l^cnr}^ * 

Interpretation of an infrared spectrurn is not a simple matter. Bands may be 
obscured by the overlapping of other bands. Overtones (harmonics) may appear 
at just twice the frequency of the fundamental band. The absorption band of a 



SEC. 13.4 



THE INFRARED SPECTRUM 



411 




4000 3500 

Sadder 8334 K 



2000 1800 1600 

Frequency, cm" 1 



Wavelength, 




4000 3500 

Sadder 10952 K 



2000 1800 1600 

Frequency, cm"' 



A-H2. 



Wavelength, M 




mono- 
subsld. 
H 2 CH Z CH 3 benzene 




Sadlltr 8457 K 



90 1800 1600 

Frequency, cm" 1 



Figure 13.2. Infrared spectra, (a) 1-Octene; (b) isopropyl bromide; 
(c) rt-butylbenzene. 



412 



SPECTROSCOPY AND STRUCTURE 



CHAP. 13 



particular group may be shifted by various structural features conjugation, 
electron withdrawal by a neighboring substituent, angle strain or van der Waals 
strain, hydrogen bonding and be mistaken for a band of an entirely different 
group. (On the other hand, recognized for what they are, such shifts reveal the 
structural features that cause them.) 

In our work we shall have modest aims: to learn to recognize a few of the more 
striking absorption bands, and to gain a little practice in correlating infrared data 
with other kinds of information. We must realize that we shall be taking from an 
infrared spectrum only a tiny fraction of the information that is there, and which 
can be gotten from it by an experienced person with a broad understanding of 
organic structure. 

Table 13.3 lists infrared absorption frequencies characteristic of various groups. 
We shall look more closely at the infrared spectra of hydrocarbons in Sec. 13.15 
and, in following chapters, at the infrared spectra of other families of compounds. 

Table 13.3 CHARACTERISTIC INFRARED ABSORPTION FREQUENCIES'* 

Frequency range, 



Bond 


Compound type 


cm" 1 


Reference 


C H 


Alkanes 


2850-2960 


Sec. 13.15 






1350-1470 




C-H 


Alkenes 


3020-3080 (m) 


Sec. 13.15 






675-1000 




CH 


Aromatic rings 


3000-3 100 (m) 


Sec. 13.15 






675-870 




C H 


Alkynes 


3300 


Sec. 13.15 


ox: 


Alkenes 


1640-1680 (0) 


Sec. 13.15 


feC 


Alkynes 


2100-2260 (v) 


Sec. 13.15 


G-C 


Aromatic rings 


1500, 1600< 


Sec. 13.15 


C-vO 


Alcohols, ethers, carboxyiic acids, esters 


1080-1300 


Sec. 16.13 








Sec. 17.17 








Sec. 18.22 








Sec. 20.25 


O=O 


Aldehydes, ketones, carboxyiic acids, esters 


1690-1760 


Sec. 19.17 








Sec. 18.22 








Sec. 20.25 


O H 


Monomeric alcohols, phenols 


3610-3640 (v) 


Sec. 16.13 








Sec. 24.13 




Hydrogen-bonded alcohols, phenols 


3200-3600(&r<?a</) 


Sec. 16.13 








Sec. 24.13 




Carboxyiic acids 


2500-3000 (broad) 


Sec. 18.22 


N H 


Amines 


3300-3500 (m) 


Sec. 23.20 


CN 


Amines 


1180-1360 


Sec. 23.20 


C=~N 


Nitriles 


2210-2260 (v) 




NO 2 


Nitro compounds 


1515-1560 








1345-1385 





* All bands strong unless marked: m, moderate; u\ weak; v, variable. 



13.5 JThe ultrayio lel^spgctruni 

Light of wavelength between about 4000 A and 7500 A (400-750 mpO is visible. 
Just beyond the red end of the visible spectrum (A greater than 750 m^) lies the 



SEC. 13.5 THE ULTRAVIOLET SPECTRUM 413 

infrared region which we have just discussed. Just beyond the violet end of the 
visible spectrum (A less than 400 mjji) lies the ultraviolet region. 

The ultrayiojet spectrometers commonly used measure absorption_of.lightin 
the visible and "near" ultraviolet region, that is, in the 200-750 mjx range.lThis 
Fight is of higher 'frequency (and greater energy) than infrared light and, wh^n it is 
absorbed by a molecule, the changes it produces are, naturally, ones that require 
greater energy : changes in electronic states. 

In a transition to a higher electronic level, a molecule can go from any of a 
number of sub-levels corresponding to various vibrational and rotational states 
to any of a number of sub-levels; as a result, ultraviolet absorption bands are broad. 
Where an infrared spectrum shows many sharp peaks, a typical ultraviolet spectrum 
shows only a few broad humps. One can conveniently describe such a spectrum 
in terms of the position of the top of the hump (h mAX ) and the intensity of that 
absorption (e mA1t , the extinction coefficient). 

When we speak of a molecule as being raised to a higher electronic level, 
we mean that an electron has been changed from one orbital to another orbital of 
higher energy. This electron can be of any of the kinds we have encountered: 
a a electron, a TT electron, or an n electron (a non-bonding electron that is, one 
of an unshared pair). A a electron is held tightly, and a good deal of energy is 
required to excite it: energy corresponding to i,Hraviolet light of short wavelength, 
in a region "far" ultraviolet outside the range of the usual spectrometer. It is 
chiefly excitations of the comparatively loosely held n and TT electrons that appear 
in the (near) ultraviolet spectrum, and, of these, only jumps to the lower more 
stable excited states. 

The electronic transitions of most concern to the organic chemist are: 
(a) // -> TT*, in which the electron of an unshared pair goes to an unstable (and- 
bonding) IT orbital, as, for example, 



and (b) TT -> TT*, in which an electron goes from a stable (bonding) n orbital to an 
unstable TT orbital, as, for example, 

\=6: > \-6: TT > TT* 

/ / 

A TT -> TT* transition can occur for even a simple alkene, like ethylene, but 
absorption occurs in the far ultraviolet. Conjugation of double bonds, however, 
lowers the energy required for the transition, and absorption moves to longer 
wavelengths, where it can be more conveniently measured. If there are enough 
double bonds in conjugation, absorption will move into the visible region, and the 
compound will be colored. ^-Carotene, for example, is a yellow pigment found 
in carrots and green leaves, and is a precursor of vitamin A; it contains eleven 
carbon-carbon double bonds in conjugation, and owes its color to absorption 
at the violet end of the visible spectrum (A max 451 m^). 

How does conjugation bring about this effect? We have seen (Sec. 8.17) 
that 1,3-butadiene, for example, is stabilized by contribution from structures 
involving formal bonds. Stabilization is not very great, however, since such 
structures and additional, ionic structures are not very stable and make only 



414 SPECTROSCOPY AND STRUCTURE CHAP. 13 

small contribution to the hybrid. Similar structures contribute to an excited state 
of butadiene, too, but here, because of the instability of the molecule, they make 
much larger contribution. Resonance stabilizes the excited state more than it 
stabilizes the ground state, and thus reduces the difference between them. 

In contrast to the infrared spectrum, the ultraviolet spectrum is not used 
primarily to show the presence of individual functional groups, but rather to 
show relationships between functional groups, chiefly conjugation: conjugation 
between two or more carbon-carbon double (or triple) bonds; between carbon- 
carbon and carbon-oxygen double bonds; between double bonds and an aromatic 
ring; and even the presence of an aromatic ring itself. It can, in addition, reveal 
the number and location of substituents attached to the carbons of the conjugated 
system. 

Problem 13.2 In Problem 9.19, page 313, you calculated the number of rings in 
^-carotene. Taking into account also the molecular formula, the number of double 
bonds, conjugation, its natural occurrence, and its conversion into vitamin A (p. 277), 
what possible structure for 0-carotene occurs to you ? 

Problem 13.3 Compounds A, B, and C have the formula C S H 8 , and on hydro- 
genation all yield //-pentane. Their ultraviolet spectra show the following values of 
Ama: A, 176 m,u; B, 211 mjx; C, 215 m^. (1-Pentene has A max 178 mjx.) (a) What is a 
likely structure for A? For B and C? (b) What kind of information might enable you 
to assign specific structures to B and C? 



13.6 The nuclear magnetic resonance (nmr) spectrum 

Like electrons, the nuclei of certain atoms are considered to spin. The spin- 
ning of these charged particles the circulation of charge generates a magnetic 
moment along the axis of spin, so that these nuclei act like tiny bar magnets. 
One such nucleus and the one we shall be mostly concerned withis the proton, 
the nucleus of ordinary hydrogen, 1 H. 

Now, if a proton is placed in an external magnetic field, its magnetic moment, 
according to quantum mechanics, can be aligned in either of two ways: with or 
against the external field. Alignment with the field is the more stable, and energy 
must be absorbed to "flip" the tiny proton magnet over to the less stable align- 
ment, against the field. 

Just how much energy is needed to flip the proton over depends, as we might 
expect, on the strength of the extcn.il field: the stronger the field, the greater the 
tendency to remain lined up with it, and the higher the frequency (Remember: 
A = hv) of the radiation needed to do the job. 



where v = frequency, in Hz 

H = strength of the magnetic field, in gauss 
y = a nuclear constant, the gyromagnetic ratio, 
26,750 for the proton 

In a field of 14,092 gauss, for example, the energy required corresponds to electro- 
magnetic radiation of frequency 60 MHz (60 megahertz or 60 million cycles per 



SEC 13.6 THE NUCLEAR MAGNETIC RESONANCE (NMR) SPECTRUM 



415 



second): radiation in the radiofrequcncy range, and of much lower energy (lower 
frequency, longer wavelength) than even infrared light. 

In pnncjle^we^ojLiildj)lace^a substance in a magnetic field of constant strength, 
and then oTStajnTi^^ injraredjor arTultraviolet 

spectrum: pass radiation ofjste^ijyj^iju^ 



andjabservc the frcq ucncy a t vvh ichjgd iat ion is absorbed. In practice, 
Tfhas been foun^niore coiiv'ementjo_keeplhe radiatloji fre.qj4nc_y coastant t a.rjd 
to vary the strength of the magnetic fieldj at some value oH he field -Strength the 
energy required to flip "(TuT proton" niatches the energy^of the radration, absorption 
occurs/aTrd^lTsTgnaT is observed. Such a spectrum is called a nuclear magnetic 
resonance (nmr) spectrum (Fig. 13.3). ~~~ 

Since the nucleus involved is the proton, the spectrum is sometimes called a pmr 
(proton magnetic resonance) spectrum, to differentiate it from spectra involving such 
nuclei as 13 C (called r/Mrjypectra) orJ^R_~ ~ 



Downfield direction 



Signal 




Low field 



High field 



Magnetic field sweep > 
Figure 13.3. The nmr spectrum. 



Now, if the situation were as simple as we have so far described it, all the 
protons in an organic molecule would absorb at exactly the same field strength. 
and the spectrum would consist of a single signal that would tell us little about the 
structure of the molecule. But the frequency at which a proton absorbs depends 
on the magnetic field which that proton feels, and this effective field strength is not 
exactly the same as the applied field strength. The effective field strength at each 
proton depends on the environment of that proton on, among other things, the 
electron density at the proton, and the presence of other, nearby protons. Each 
protonor, more precisely, each set of equivalent protons will have a slightly 
different environment from every other set of protons, and hence will require a 
slightly different applied field strength to produce the same effective field strength: 
the particular field strength at which absorption takes place. 

At a given radjcfrequcncy, then, all protons absorb at the same effective field 
ffe 



^ ajjdjfferent applied field strengths. It is this applied field 

strygtlnhatjsjTieasured, and against which the absorption is plotted. " 



416 SPECTROSCOPY AND STRUCTURE CHAP. 13 

The result is a spectrum showing many absorption peaks, whose relative 
positions, reflecting as they do differences in environment of protons, can give 
almost unbelievably detailed information about molecular structure. 

In the following sections, we shajMookjrt various aspects of the nmr spec- 
trum; 

(a) the number of signals^ which tells us how many different "kinds" of pro- 
tons there are in a molecule; * ~~ "" """ ~~ 

~" (b) the positions of the signals, which tell us something about the electronic 
kind of proton ; " *""" " ~" 



(c) the intensities ofjhjjgha[s 9 which tell us how many protons of_each kind 
there are ^and *" ~ 

(d) the splitting of a^ ^/^^intosey^ml^peaks^jwhich tells usjabout the en- 

n with respect to jpther, nearby protons. 



13.7 Nmr. Number of signals. Equivalent and non-equivalent protons 

In a given molecule, protons with the same environment absorb at the same 
(applied) field strength; protons with different environments absorb at different 
(applied) field strengths. A set of protons with the same environment are said to 
be equivalent; the number of signals in the nmr spectrum tells us, therefore, how 
many sets of equivalent protons how many "kinds" of protons a molecule 
contains. 

For our purposes here, equivalent protons are simply chemically equivalent 
protons, and we have already had considerable practice in judging what these are. 
Looking at each of the following structural formulas, for example, we readily 
pick out as equivalent the protons designated with the same letter: 

CH 3 -CH 2 -C1 CH 3 CHC1 CH 3 CH 3 CH 2 -CH 2 -Ci 

a b aba a b c 

2 nmr signals 2 nmr signals 3 nmr signals 

Ethyl chloride Isopropyl chloride w-Propyl chloride 

Realizing that, to be chemically equivalent, protons must also be stereo- 
chemically equivalent, we find we can readily analyze the following formulas, too: 



a b a b a b 

CH 3 H CH 3 H H U 

x x >< 

CH 3 H Br H Cl H 

a b c c 

2 nmr signals 3 nmr signals 3 nmr signals 4 nmr signals 

Isobutytene 2-Bromopropenc Vinyl chloride Methylcyclopropane 

1,2-Dichloropropane (optically active or optically inactive) gives four nmr 




SEC. 13.7 



NMR. NUMBER OF SIGNALS 



417 




signals, and it takes only a little work with models or stereochemical formulas 
to see that this should indeed be so. 



c 

H 

CH.I CHCI- c; ci 

i 
H 

a h (I 



4 nmr signals 
\ ,2-DichIoropropane 

The environments of the two protons on C-l are not the same (and no amount of 
rotation about single bonds will make them so); the protons are not equivalent, 
and will absorb at different field strengths. 

We can tell from a formula which protons are in different environments and hence 
should give different signals. We cannot always tell -particularly \\ith stereochemically 
different protons just how different these environments are; they may not be different 
enough for the signals to be noticeably separated, and we may seejewer signals than we 
predict. 

Now, just how did we arrive at the conclusions of the last few paragraphs? 
Most of us perhaps without realizing it judge the equivalence of protons by 
following the approach of isomcr number (Sec. 4.2). This is certain!) the easiest 
\vay to do it. We imagine each proton in turn to be replaced by some other atom 
Z. If replacement of either of t\\o protons by Z would yield the same product 
or enantiomeric products then the t\vo protons are chemically equivalent. We 
ignore the existence of conformational isomcrs and, as we shall see in Sec. 13.13, 
this is just \vhat we should do. 

Take, for example, ethyl chloride. Replacement of a methyl proton would 
give CH 2 Z CH 2 C1; replacement of a methylene proton would give CH 3 CHZC1. 
These are, of course, different products, and we easily recognize the methyl protons 
as being non-equivalent to the methylene protons. 

The product CH 2 Z CH-.C1 is the same regardless of which one of the three 
methyl protons is replaced. The (average) environment of the three protons is 
identical, and hence we expect one nmr signal for all three. 

Replacement of either of the two methylene protons would give one of a pair 
of enantiomers: 



CHj 



CI 

Enanliotopic 
protons 

Ethyl chloride 



H--0-2 

a 




418 



SPECTROSCOPY AND STRUCTURE 



CHAP. 13 



Such pairs of protons are called enantiotopic protons. The environments of these 
two protons are mirror images of each other; these protons are equivalent, and 
we see one nmr signal for the pair. (Like any other physical property except 
rotation of polarized light the nmr spectrum does not distinguish between mirror 
images.) 

Turning to 2-bromopropene, we see that replacement of either of the vinylic 
protons gives one of a pair of diastereomers (geometric isomers, in this case): 



CH 3 



H 







Br 

Diastereotopic 
protons 

2-Bromopropene 

Such pairs of protons are called diastereotopic protons. The environments of these 
two protons are neither identical nor mirror images of each other; these protons 
are non-equivalent, and we expect an nmr signal from each one. 

Similarly, in 1,2-dichloropropane the two protons on C-l are diastereotopic, 
non-equivalent, and give separate nmr signals. 






H 



Diastereotopic 
protons 

1 ,2-Dichloropropane 

In Sec. 13.13, we shall take a closer look at equivalence. The guidelines we 
have laid down here, however based on rapid rotation about single bonds 
hold for most spectra taken under ordinary conditions, specifically, at room 
temperature. 



Problem 13.4 Draw the structural formula of each of the following compounds 
(disregarding enantiomcrism), and label all sets of equivalent protons. How many 
nmr signals would you expect to see from each? 

(a) the two isomers of formula C 2 H 4 CI 2 

(b) the four isomers of C 3 H 6 Br 2 

(c) ethylbcnzcnc and p-xylene 

(d) mesitylene, p-cthyltoluenc, isopropylbenzene 
(c) CH 3 CH 2 OH and CH 3 OCH, 

(f) CH 3 CH 2 OCH 2 CH 3 , CH 3 OCH 2 CH 2 CH 3 , CH 3 OCH(CH,) 2 , CH 3 CH 2 CH 2 CH 2 OH 

(g) CH 2 -CH 2 , CH 3 -CH-CH 2 (Hint: Make Models.) 

V 



SEC. 13.8 



NMR. POSITIONS OF SIGNALS. CHEMICAL SHIFT 



419 



(h) CH 3 CH 2 C-H, CH 3 CCH 3 , and C 

A A 

Problem 13.5 Three isomeric dimethylcyclopropanes give, respectively, 2, 3, and 
4 nmr signals. Draw a stereoisomeric formula for the isomer giving rise to each num- 
ber of signals. 

Problem 13.6 How many nmr signals would you expect from cyclohexane? 
Why? 



13.8 >JnirJP_Qsitioiis of signals. Chemical shift 

Just as the number of signals in an nmr spectrum tells us how many kinds 
of protons a molecule contains, so the positions of the signals help to tell us what 
kinds of protons they are: aromatic, aliphatic, primary, secondary, tertiary; 
benzylic, vinylic, acetyienic; adjacent to halogen or to other atoms or groups. 
These different kinds of protons have different electronic environments, and it is 
the electronic environment that determines just where in the spectrum a proton 
absorbs. 

When a molecule is placed in a magnetic field as it is when one determines an 
nmr spectrum its electrons are caused to circulate and, in circulating, they 
generate secondary magnetic fields: induced magnetic fields. 

Circulation of electrons about the proton itself generates a field aligned in 
such a way that at the proton it opposes the applied field. The field felt by the 
proton is thus diminished, and the proton is said to be shielded. 

Circulation of electrons specifically, n electrons about nearby nuclei 
generates a field that can either oppose or reinforce the applied field at the proton, 
depending on the proton's location (Fig. 13.4). If the induced field opposes the 

Induced 
fields 




Acetylene 
(b) 



Figure 13.4. Induced field (0) reinforces applied field at the aromatic 
protons, and (b) opposes applied field at the acetyienic protons. Aromatic 
protons are deshielded; acetyienic protons are shielded. 



420 SPECTROSCOPY AND STRUCTURE CHAP. 13 

applied field, the proton is shielded, as before. If the induced field reinforces the 
applied field, then the field felt by the proton is augmented, and the proton is 
said to be deshieldc * 

Compared with a naked proton, a shielded proton requires a higher applied 
field strength and a deshielded proton requires a lower applied field strength 
to provide the particular effective field strength at which absorption occurs. 
Shielding thus shifts the absorption upfiekt anddeshielding shifts the absorption 
downfield^ Such shifts in the position of nmr afeb/ptioirarising from shielding 
and deshieiding by electrons, are called chemical shifts. 

" How are the direction and magnitude the value of a particular chemical 
shift to be measured and expressed ? 

The unit in which a chemical shift is most conveniently expressed is parts per 
rnillipn (pptrQ of the total applied magnetic field. Since shielding and deshieiding 
arise from induced secondary fields, the magnitude of a chemical shift is propor- 
tional to the strength of the applied field or, what is equivalent, proportional to 
the radiofrequency the field must match. If, however, it is expressed as a fraction 
of the applied field that is, if the observed shift is divided by the particular 
radiofrequency used then a chemical shift has a constant value that is independent 
of the radiofrequency and the magnetic field that the nmr spectrometer employs. 

The reference point from which chemical shifts are measured is, for practical 
reasons, not the signal from a naked proton, but the signal from an actual com- 
pound: usually tetramethylsilane, (CH 3 ) 4 Si. Because of the low electronegativity 
of silicon, the shielding of the protons in the silane is greater than in most other 
organic molecules; as a result, most nmr signals appear in the same direction from 
the tetramethylsilane signal: downfield. 

The most commonly used scale is the B (delta) scale. The position of the 
tetramethylsilane signal is taken as 0.0 ppm. Most chemical shifts have S values 
between and 10 (minus 10, actually). A small 8 value represents a small downfield 
shift, and a large 8 value represents a large downfield shift. 

One commonly encounters another scale: the r (tan) scale, on which the tetra- 
nethylsilane signal is taken as 10.0 ppm. Most r values lie between and 10. The two 
scales are related by the expression r = 10 8. 

An nmr signal from a particular proton appears at a different field strength 
han the signal from tetramethylsilane. This difference the chemical shift is 
measured not in gauss, as we might expect, but in the equivalent frequency units 
Remember: v = y// /27r), and it is divided by the frequency of the spectrometer 
ised. Thus, for a spectrometer operating at 60 MHz, that is, at 60 x 10 6 Hz: 

. observed shift (Hz) x 10* 
60 x 10 (HZ) 

The chemical shift for a proton is determined, then, by the electronic environ- 
nent of the proton. In a given molecule, protons with different environments 
ion-equivalent protons have different chemical shifts. Protons with the same 
jnvironm-^t equivalent protons have the same chemical shift; indeed, for nmr 
wrposes, equivalent protons are defined as those with the same chemical shift. (We 
lave already seen what the equivalence of protons means in terms of molecular 
tructure.) 



SEC. 13.8 



NMR. POSITIONS OF SIGNALS. CHEMICAL SHIFT 

Table 13.4 CHARACTERISTIC PROTON CHEMICAL SHIFTS 
Type of proton Chemical shift, ppm 



421 







8 


Cyclopropane 




0.2 


Primary 


RCH 3 


0.9 


Secondary 


R 2 CH 2 


1.3 


Tertiary 


R 3 CH 


1.5 


Vinylic 


C=C H 


4.6-5.9 


Acetylenic 


C^C H 


2-3 


Aromatic 


Ar-H 


6-8.5 


Benzylic 


Ar C- H 


2.2-3 


Allylic 


O=C CH 3 


1.7 


Fluorides 


HC-F 


4-4.5 


Chlorides 


HC Cl 


3-4 


Bromides 


HC Br 


2.5-4 


Iodides 


HC I 


2-4 


Alcohols 


HC OH 


3.4-4 


Ethers 


HC OR 


3.3-4 


Esters 


RCOO-CH 


3.7^.1 


Esters 


HC COOR 


2-2.2 


Acids 


HC COOH 


2-2.6 


Carbonyl compounds 


HC C=O 


2-2.7 


Aldehydic 


RCHO 


9-10 


Hydroxylic 


ROH 


1-5.5 


Phenolic 


ArOH 


4-12 


Enolic 


C=C OH 


15-17 


Carboxylic 


RCOOH 


10.5-12 


Amino 


RNH 2 


1-5 



Furthermore, it has been found that a proton with a particular environment 
shows much the same chemical shift, whatever the molecule it happens to be part 
of. Take, for example, our familiar classes of hydrogens: primary, secondary, 
and tertiary. In the absence of other nearby substituents, absorption occurs at 
about these values: 

RCH 3 8 0.9 
R 2 CH 2 5 1.3 
R 3 CH 8 1.5 

All these protons, in turn, differ widely from aromatic protons which, because of 
the powerful deshielding due to the circulation of the * electrons (see Fig. 13.4, 
p. 419), absorb far downfield: 

Ar-H 8 6-8.5 

Attachment of chlorine to the carbon bearing the proion causes a downfield 
shift. If the chlorine is attached to the carbon once removed .from the carbon 
bearing the proton, there is again a downward shift, but this time much weaker. 



CHj~Cl 8 3.0 

R-CH 2 -CI 8 3.4 

R 2 CH-C1 8 4.0 



CH 3 -C -Cl 

R-CH 2 ~C~C1 

R 2 CH-C-O 



81.5 
81.7 
8 1.6 



422 



SPECTROSCOPY AND STRUCTURE 



CHAP. 13 



Toluene 



r-n: 



Tin 



o* 





-:'. !> ; : :: M ;::i: ;-.'i; : i. ;;: 


J f. - | 

;; iw> ^H* 




, ; . , . . | , . ' , 1 . ***.,},,,.. 


j , ' ' - 






hn.....;iih..j;.ji.4n. 



oa 



1 
300 



(c) Mesitylene 




CH, 

a 



b 

i 



OHi 



i t t 1 . lit i ^ f ,f 



Figure 13.5. Nmr spectra: chemical shift, (a) Toluene; (b) p-xylenc; 
(c) mesitylene. 



SEC. 13.9 NMR. PEAK AREA AND PROTON COUNTING 423 

Two chlorines cause a greater downfield shift. Other halogens show similar 
effects. 

The downfield shift caused by chlorine is what we might have expected from its 
inductive effect: electron withdrawal lowers the electron density in the vicinity of the 
proton and thus causes deshielding. The effect of a substituent on the chemical shift 
is unquestionably the net result of many factors; yet we shall often observe chemical 
shifts which strongly suggest that an inductive effect is at least one of the factors at work. 

Table 13.4 lists chemical shifts for protons in a variety of environments. 

The nmr spectra (Fig. 13.5, p. 422) of the alkylbenzenes toluene, p-xylene, 
and mesitylene illustrate the points we have just made. In each spectrum there are 
two signals: one for the side-chain protons, and one for the ring protons. (Here, 
as in some though not most aromatic compounds, the ortho, meta, and para 
protons have nearly the same chemical shifts, and hence for nmr purposes are 
nearly equivalent.) 

In each spectrum, the ring protons show the low-field absorption we have 
said is characteristic of aromatic protons. Absorption is not only at low field, 
but at nearly the same field strength for the three compounds: at 8 7.17, 7.05, and 
6.78. (These values are not exactly the same, however, since the environments of 
the aromatic protons are not exactly the same in the three compounds.) x 

In each compound, side-chain protons benzylic protons are close enough 
to the ring to feel a little of the deshielding effect of the TT electrons (Fig. 13.4, p. 
419), and hence absorb somewhat downfield from ordinary alkyl protons: at 
8 2.32, 2.30, and 2.25. In all three compounds, the environment of the side-chain 
protons is almost identical, and so are the chemical shifts. 

The similarity in structure among these three alkylbenzenes is thus reflected 
in the similarity of their nmr spectra. There is, however, a major difference in their 
structures a difference in numbers of aromatic and side-chain protons and, as 
we shall see in the next section, this is reflected in a major difference in nmr spectra. 

The chemical shift is fundamental to the nmr spectrum since, by separating the 
absorption peaks due to the various protons of a molecule, it reveals all the other features 
of the spectrum. The numerical values of chemical shifts, although significant, do not 
have the overriding importance that absorption frequencies have in the infrared spectrum. 
In our work with nmr, we shall escape much of the uncertainty that accompanies the 
beginner's attempts to identify precisely infrared absorption bands; at the same time, \\e 
have a greater variety of concepts to learn about but these, at our present level, \\e may 
find more satisfying and intellectually more stimulating. 

Problem 13.7 What is a possible explanation for the following differences in 
chemical shift for aromatic protons? Benzene 5 7.37; toluene 8 7.17; /?-xylene S 7.05; 
mesitylene 8 6.78. 



13.9 Nmr. Peak area and proton counting 

Let us look again at the nmr spectra (Fig. 13.5, p. 422) of toluene, />-xylene, 
and mesitylene, and this time focus our attention, not on the positions of the 
signals, but on their relative intensities, as indicated by the sizes of the absorption 
peaks. 

Judging roughly from the peak heights, we see that the (high-field) peak for 



424 



SPECTROSCOPY AND STRUCTURE 



CHAP. 13 



side-chain protons is smaller than the (low-field) peak for aromatic protons in 
the case of toluene, somewhat larger in the case of p-xylene, and considerably 
larger in the case of mesitylene. More exact comparison, based on the areas under 
the peaks, shows that the peaks for side-chain and aromatic protons have sizes in 
the ratio 3 : 5 for toluene ; 3 : 2 (or 6 : 4) for /?-xylene ; and 3:1 (or 9 : 3) for mesitylene. 
This illustrates a general quality of all nmr spectra. The area under an nmr 
signal is directly proportional to the number of protons giving rise to the signal* 

It is not surprising that this is so. The absorption of every quantum of energy is 
due to exactly the same thing: the flipping over of a proton in the same effective magnetic 
field. The more protons flipping, the more the energy absorbed, and the greater is the 
area under the absorption peak. 

Areas under nmr signals are measured by an electronic integrator, and are 
usually given on the spectrum chart in the form of a stepped curve; heights of 
steps are proportional to peak areas. Nmr chart paper is cross-hatched, and we 
can conveniently estimate step heights by simply counting squares. We arrive 
at a set of numbers that are in the same ratio as the numbers of different kinds of 
protons. We convert this set of numbers into a set of smallest whole numbers just 
as we did in calculating empirical formulas (Sec. 2.27). The number of protons 



I 

200 




A 







a 9H 



O 

.J. 






JU 



Figure 13.6. Nmr spectrum of p-ferr-butyltoluene. Proton counting. 
The ratio of step heights a:b:c is 

8.8:2.9:3.8 = 3.0:1.0:1.3 = 9.0:3.0:3.9 



Alternatively, since the molecular formula 



is known, 



16 H 



1.03 x 8.8 = 9.1 



f _ u 

__ - 1.03 H per unit 
15.5 units 

b = 1.03 x 2.9 = 3.0 c = 1.03 x 3.8 = 3.9 

Either way, we find: a, 9H; 6, 3H; c, 4H. 

The 4H of c (8 7.1) are in the aromatic range, suggesting a disubsti- 
tuted benzene C 6 H 4 . The 3H of b (5 2.28) have a shift expected for 
ben zy lie protons, giving CH 3 C 6 H 4 . There is left C 4 H 9 which, in view 
of the 9H of a (S 1.28) must be C(CH 3 ) 3 ; since these are once removed 
from the ring their shift is nearly normal for an alkyl group. The compound 
is /ir/-butyltoluene (actually, as shown by the absorption pattern of the 
aromatic protons, the />-isomer). 



SEC. 13.10 NMR. SPLITTING OF SIGNALS. SPIN-SPIN COUPLING 425 

giving rise to each signal is equal to the whole number for that signal or to some 
multiple of it. See, for example, Fig. 13.6. 

We take any shortcuts we can. If we know the molecular formula and hence the 
total number of protons, we can calculate from the combined step heights the number of 
squares per proton. If we suspect a particular structural feature that gives a characteristic 
signal --an aldehydic (--CHO) or carboxylic ( COOH) proton, say, which gives a far- 
downfield peakwe can use this step height as a starting point. 

Working the following problems will give us some idea of the tremendous help 
"proton counting" by nmr can be in assigning a structure to a compound. 

Problem 13.8 Go back to Problem 13.4 (p. 418), where you predicted the num- 
ber of nmr signals from several compounds. Tell, where you can, the relative positions 
of the signals (that is, their sequence as one moves downfield) and, roughly, the 5 
value expected for each. For each signal tell the number of protons giving rise to it. 

Problem 13.9 Give a structure or structures consistent with each of ihe nmr 
spectra shown in Fig. 1 3. 7 (p. 426). 



13.10 Nmr. Splitting of signals. Spin-spin coupling 

An nmr spectrum, we have said, shows a signal for each kind of proton in a 
molecule; the few spectra we have examined so far bears this out. If we look 
much further, however, we soon find that most spectra are or appear to be 
much more complicated than this. Figure 13.8 (p. 427), for example, shows the 
nmr spectra for three compounds, 

CH 2 Br-CHBr 2 CH 3 -CHBr 2 CH 3 -CH 2 Br 

1,1,2-Tribromoethane 1,1-Dibromoethane Ethyl bromide 

each of which contains only two kinds of protons; yet, instead of two peaks, 
these spectra shower*?, six, and seven peaks, respectively 

What docs this multiplicity of peaks mean? How does it arise, and what can 
it tell us about molecular structure? 

The answer is that we are observing the splitting _of mrir^signals caused J>y^ 
spin -spin coupling. The signal we expect from each set of equivalent protons is 
*Sppearing, not as a single peak, but as a group of peaks. Splitting reflects the 
environment of the absorbing protons: not with respect to electrons, but with 
respect to other, nearby protons. It is as though we were permitted to sit on a 
proton and look about in all directions: we can see and count the protons attached 
to the carbon atoms next to our own carbon atom and, sometimes, even see protons 
still farther away. 

Let us take the case of adjacent carbon atoms carrying, respectively, a pair 
of secondary protons and a tertiary proton, and consider first the absorption by 
one of the secondary protons: 

CH-CHj- 

The magnetic field that a secondary proton feels at a particular instant is slightly 
increased or slightly decreased by the spin of the neighboring tertiary proton: 
increased if the tertiary proton happens at that instant to be aligned uv7/i the applied 




(ib) 



I 
too 



1" 



itiiiiiiiTiiiiinr-niiiiiiiiii 



Till i i t i IT! 





i . , * t .. 


-.4* ' 4 . . - . I . 


, > 1 ! 1 










.;. ; . 


' , . } : .' * . i i ' . . 


:,'*,. , ; . wo 


' ; *i 


. 




""""*. 


~* "~~!~i r* 


. :-:--]- .-.-/TT- ;: . 


<c) CjHs& v v ! , 












: : , : I . . : , 

. , , * : ( * . ' ' : '. 4 , 


'*''"'* i 1 ' 




' 4 t 




; . 
l 


1 ~ 


. t l:.. , ;., . '.:.!:i f i 4 .. ..!'.: 


i t r i i * . * * . 
' : 1 , * 1 ', ' . i 1 ! 














uji^.n;:;;; '""j? ' 


i 


, ' " 


1 






' -' ' r H ' 




, 








i * " . ' .j 


} :';:!i: j - l !: "jj,:;: 


iJVT:. jj' .: [, ; {. *: 


( . ' ! 1 

1 r | i I 


: H J 


,ii' 


i i i 


ririti 


1 1 ; ri i 1 lit i 1 1 i tt ii f \ 1 t i i i i . i i i 


MMi-i(:!iiiii f iii!ilii!ii 


iliiii 


.liiiii 


1 , 


1 


\ 


7*54 


I S 2 


1 


c 


> 



Figure 13.7. Nmr spectra for Problem 13.9, p. 425. 



m44 } M-U 1} ! ''iliiH-M *t4J-M j ; * ; ' j- 
: ! ! U I JB* ' \ \ 1 M : t r r : an t 1 ' ' ' 






^iiiijiiil 




i j;iii iii tmf^ 

OJ 



n r 1 1 1 i 1 1 i i i 1 1 i i i i ' . 1 1 1 f r//< i a i i n 1 1 n i i : 1 1 i i i t i n 1 1 



ii..i.ii'ii.-t.-rminifm 
2 i 05 




:!(c) Ethyl ibromide |.- ; :; 

';W^^br 



:|; : r;r\;i{|;i::;:ri 




5 4 3 2 I 

Figure 13.8. Nmr spectra : splitting of signals, (a) 1 , 1 ,2-Tribromoethane ; 
(b) l,l~dibromoelhane; (c) ethyl bromide. 



428 SPECTROSCOPY AND STRUCTURE CHAP. 13 

field; or decreased if the tertiary proton happens to be aligned against the applied 
field. 

For half the molecules, then, absorption by a secondary proton is shifted 
slightly down field, and for the other half of the molecules the absorption is shifted 
slightly upfield. The signal is split into two peaks: a doublet, with equal peak in- 
tensities (Fig. 13.9). 

Applied 
field 

I 



Signal from \^ 

uncoupled proton "~~ I 

Spin combinations ^ 

for adjacent CH 

Figure 13.9. Spin-spin coupling. Coupling with one proton gives a 1 : 1 
doublet. 

Next, what can we say about the absorption by the tertiary proton? 

-CH-CH 2 - 

It is, in its turn, affected by the spin of the neighboring secondary protons. But 
now there are two protons whose alignments in the applied field we must consider. 
There are four equally probable combinations of spin alignments for these two 
protons, of which two are equivalent. At any instant, therefore, the tertiary 
proton feels any one of three fields, and its signal is split into three equally spaced 
peaks: a triplet, with relative peak intensities 1:2:1, reflecting the combined 
(double) probability of the two equivalent combinations (Fig. 13.10). 



Applied 
f ' c!d 



Signal from 
uncoupled proton 



if 

* ' 



Spin combinations 
for adjacent CH 2 



Figure 13.10. Spin-spin coupling. Coupling with two protons gives a 
1:2:1 triplet. 

Figure 13.11 (p. 429) shows an idealized nmr spectrum due to the grouping 
CH CH 2 . We see a 1:1 doublet (from the CH 2 ) and a 1:2:1 triplet 
(from the CH ). The total area (both peaks) under the doublet is twice as big 
as the total area (all three peaks) of the triplet, since the doublet is due to absorption 
by twice as many protons as the triplet. 

A little measuring shows us that the separation of peaks (the coupling constant, 



SEC. 13.10 NMR. SPLITTING OF SIGNALS. SPIN-SPIN COUPLING 



429 



CH 2 Br CHBr a 
a b 




Figure 13.11. Spin- spin splitting. Signal a is split into a doublet by coup- 
ling with one proton ; signal b s split into a triplet by two protons. Spacings 
in both sets the same (/ a iJ. 



/, Sec. 13.11) in the doublet is exactly the same as the separation of peaks in the 
triplet. (Spin-spin coupling is a reciprocal affair, and the effect of the secondary 
protons on the tertiary proton must be identical with the effect of the tertiary 
proton on the secondary protons.) Even if they were to appear in a complicated 
spectrum of many absorption peaks, the identical peak separations would tell us 
that this doublet and triplet were related : that the (two) protons giving the doublet 
and the (one) proton giving the triplet are coupled, and hence are attached to 
adjacent carbon atoms. 

We have seen that an nmr signal is split into a doublet by one nearby proton, 
and into a triplet by two (equivalent) nearby protons. What splitting can we 
expect more than two protons to produce? In Fig. 13.12 (p. 430), we see that three 
equivalent protons split a signal into four peaks a quartet with the intensity 
pattern 1:3:3:1. 

It can be shown that, in general, a set ofn equivalent protons will split an nmr 
signal into n + 1 peaks. 

If we turn once more to Fig. 13.8 (p. 427), we no longer find these spectra so 
confusing. We now see not just five or six or seven peaks, but instead a doublet 
and a triplet, or a doublet and a quartet, or a triplet and a quartet. We recognize 
each of these multiplets from the even spacings within it, and from its symmetrical 



430 



SPECTROSCOPy AND STRUCTURE 



CHAP. 13 



intensity pattern (1:1, or 1:2:1, or 1:3:3:1). Each spectrum does show % absorp- 
tion by just two kinds of protons; but clearly it shows a great deal more than that. 



Applied 
field 



Signal from 
uncoupled proton 



uutuu 
.tutu, lit 
N tit ' * 

Spin combinations 
for adjacent CHa 

Figure 13.12. Spin-spin coupling. Coupling with three protons gives a 
1:3:3:1 quartet. 

If we keep in mind that the peak area reflects the number of absorbing protons, 
and the multiplicity of splittings reflects the number of neighboring protons, we 
find in each spectrum just what we would expect. 

In the spectrum of CHBr 2 CH 2 Br we see 

Downfield triplet and Up field doublet 
Area: 1 Area: 2 



C CH 2 

In the spectrum of CH 3 CHBr 2 we see 

Downfield quartet and 
Area: 1 

C-CH 3 
H 

and in the spectrum of CH 3 CH 2 Br we see 

Downfield quartet and 
Area: 2 

-CH 2 -CH 3 



C CH 2 
H 



Vpfield doublet 
Area: 3 

-C CH 3 
H 



Upfie'd triplet 
Area: 3 

CH 2 -CH, 



We see chemical shifts that are consistent with the deshielding effect of halo- 
gens: in each spectrum, the protons on the carbon carrying the greater number of 
halogens absorb farther downfield (larger 8). 

In each spectrum, we see that the spacing of the peaks within one multiplet 
is the same as within the other, so that even in a spectrum with many other peaks, 
we could pick out these two multiplets as being coupled. 

Finally, we see a feature that we have not yet discussed: the various multi- 
plets do not show quite the symmetry we have attributed to them. In spectrum 
(a), v>e see 



SEC. 13.10 



NMR. SPLITTING OF SIGNALS. SPIN-SPIN COUPLING 



431 



not 
and in spectrum (b) 



but something like 



not 
and in spectrum (c) 



but something like 



not 



I 



but something like \ 



In each case, the inner peaks the peaks nearer the other, coupled multiplets are 
larger than the outer peaks. 

Perfectly symmetrical multiplets are to be expected only when the separation 
between multiplets is very large relative to the separation within multiplets 
that is, when the chemical shift is much larger than the coupling constant (Sec. 
13.11). The patterns we see here are very commonly observed, and are helpful 
in matching up multiplets: we know in which direction upfield or downfield 
to look for the second multiplet. 

We have not yet answered a very basic question: just which protons in a 
molecule can 'be coupled? We may expect to observe spin spin splitting only be- 
tween non-equivalent neighboring protons. By " non-equivalent" protons we mean 
protons with different chemical shifts, as we have already discussed (Sec. 13.8). 
By "neighboring" protons we mean most commonly protons on adjacent carbons, 
as in the examples we have just looked at (Fig. 13.8, p. 427); sometimes protons 
further removed from each other may also be coupled', particularly if TT bonds 
intervene. (If protons on the same carbon are non-equivalent as they sometimes 
are they may show coupling.) 

We do not observe splitting due to coupling between the protons making up 
the same CH 3 group, since they are equivalent. We do not observe splitting 
due to coupling between the protons on C-l and C-2 of 1,2-dichloroethane 



CH 2 CH 2 

Cl Cl 
1 ,2-Dichloroethane 



No splitting 



since, although on different carbons, they, too, are equivalent. 



432 SPECTROSCOPY AND STRUCTURE CHAP, 13 

In the spectrum of 1 ,2-dibromo-2-methylpropane, 

CH 3 
CH 3 C CH 2 Br No splitting 

Br 

1 ,2-Dibromo-2-methylpropane 

we do not observe splitting between the six methyl protons, on the one hand, and 
the two CH 2 protons on the other hand. They are non-equivalent, and give 
rise to different nmr signals, but they are not on adjacent carbons, and their 
spins do not (noticeably) affect each other. The nmr spectrum contains two' singlets, 
with a peak area ratio of 3: 1 (or 6:2). For the same reason, we do not observe 
splitting due to coupling between ring and side-chain protons in alkylbenzenes 
(Fig. 13.5, p. 422). 

We do not observe splitting between the two vinyl protons of isobutylene 

CH 3 H 



No splitting 
CH 3 X X H 

- Isobutylene 

since they are equivalent. On the other hand, we may observe splitting between 
the two vinyl protons on the same carbon if, as in 2-bromopropene, they are non- 
equivalent. 

CH 3 H a 



B/ V 

2-Bromopropene 

The fluorine ( 19 F) nucleus has magnetic properties of the same kind as the proton. 
It gives rise to nmr spectra, although at a quite different frequency-field strength com- 
bination than the proton. Fluorine nuclei can be coupled not only with each other, but 
also with protons. Absorption by fluorine does not appear in the proton nmr spectrum 
it is far off the scale but the splitting by fluorine of proton signals can be seen. The 
signal for the two protons of l,2-dichloro-l,l-difluoroethane, for example, 

H F 

I I 
Cl-C-C Cl 



appears as a 1 : 2 : 1 triplet with peak spacings of 1 1 Hz. (What would you expect to see 
in the fluorine nmr spectrum?) 

Figures 13.13 and 13.14, p. 433, and Fig. 13.15, p. 434, illustrate some of the 
kinds of splitting we are likely to encounter in nmr spectra. 



SEC. 13.10 



NMR. SPLITTING OF SIGNALS. SPIN-SPIN COUPLING 



433 



1 1 1 

400 300 200 1 


I 
00 Ocpi 


Br 


/ 


CHj-CH CH 3 \ 


1 ^s'^tto 


v aba 


j 


b 




/. 




10111! , . 




8765432 





Figure 13.13. Nmr spectrum of isopropyl bromide. Absorption by the 
six methyl protons H appears upfield, split into a doublet by the single 
adjacent proton H b . Absorption by the lone proton H b appears down- 
field (the inductive effect of bromine) split into a septet by the six ad- 
jacent protons with the small ou