Essentials of Organic Chemistry

36 ATOMIC STRUCTURE AND BONDING

the two atoms. Covalent bonds may, therefore, display considerable polarity. This polarity helps us
possess a charge imbalance, with one of the atoms to predict chemical behaviour, and it is crucial to our
taking more than its share of the electrons. This prediction of chemical mechanisms.
is referred to as bond polarity. An atom that is
more electronegative than carbon will thus polarize d+ d− d+ d−
the bond, and we can consider the atoms as being CO CN
partially charged. This is indicated in a structure by
putting partial charges (δ+ and δ−) above the atoms. polarity in C O and C N bonds
It can also be represented by putting an arrowhead
on the bond, in the direction of electron imbalance. We must also modify our thinking of bonding
Alternatively, we use a specific dipole arrow at the as being simply ionic (where there is transfer of
side of the bond. electrons between atoms) or covalent (where there
is equal sharing of electrons). These represent two
bromine is more d+ d− C Br C Br extremes, but bond polarity now provides a middle
electronegative C Br ground where there is sharing of electrons, but an
than carbon dipole on dipole arrow unequal sharing.
partial bond
charges Bond polarity in a molecule can often be measured
by a dipole moment, expressed in Debye units
In general, electronegativities increase from left to (D). However, the physical measurement provides
right across the periodic table, and decrease going only the overall dipole moment, i.e. the sum of the
down a particular column of the periodic table. individual dipoles. A molecule might possess bond
The relative electronegativities of those atoms most polarity without displaying an overall dipole if two
likely to be found in typical organic molecules or more polar bonds are aligned so that they cancel
are included in Table 2.2. The numbers (Pauling each other out. The C–Cl bond is polar, but although
electronegativity values) are on an arbitrary scale chloroform (CHCl3) has a dipole moment (1.02 D),
from Li = 1 to F = 4. carbon tetrachloride (CCl4) has no overall dipole.
Because of the tetrahedral orientation of the dipoles
From the sequence shown, it is readily seen in carbon tetrachloride, the vector sum is zero.
that hydrogen and carbon are among the least
electronegative atoms we are likely to encounter in overall dipole
organic molecules. The relatively small difference H
in electronegativities between hydrogen and carbon
also means there is not going to be much polarity Cl C
associated with a C–H bond. Most atoms other than Cl
hydrogen and carbon when bonded to carbon are Cl
going to be electron rich; therefore, bonds may

Table 2.2 Pauling electronegativity values Cl
tetrahedral orientation of
H
2.1 Cl C dipoles means vector sum
Cl is zero
Li Be B C N O F Cl
1.0 1.6 2.0 2.5 3.0 3.5 4.0
Polarization in one bond can also influence the
Na Mg Al Si P S Cl polarity of an adjacent bond. Thus, in ethyl chloride,
0.9 1.2 1.5 1.8 2.1 2.5 3.0 the polarity of the C–Cl bond makes the carbon more
positive (δ+); consequently, electrons in the C–C
K Br bond are drawn towards this partial positive charge.
0.8 2.8 The terminal carbon thus also experiences a partial
positive charge, somewhat smaller than δ+ and so
I depicted as δδ+.
2.5

CONJUGATION 37

H d− d+ dd+ ddd+ whereas in 1,4-pentadiene they are isolated or non-
d− d+ dd+ X C C CH3 conjugated. The nomenclature ‘diene’ indicates two
Cl C CH3 C=C double bonds, the numbers the position in
inductive effect decreases the molecule (see Section 1.4). Conjugated dienes
H as polar group is located usually display rather different chemical reactivity
inductive effect further away and spectral properties from non-conjugated dienes
(see Section 8.2).

H CH3 Cl CH3 HH HH
Cl C Cl C HCCH HCCH

Cl Cl CCC CCC
H H HH H
inductive effects increase as
number of polar groups increases HHH 1,4-pentadiene

1,3-pentadiene isolated double bonds
conjugated double bonds

This transmission of polarity through the σ bonds The differences arise from the nature of the π
is termed an inductive effect. It is relatively short orbitals in the double bond system. Consider 1,4-
range, decreasing rapidly as the original dipole pentadiene first. We may draw this to show overlap
is located further away. It becomes unimportant of p orbitals to create two separate π bonds, and
after about the third carbon atom. However, the effectively that is all there is that is worthy of note
effects will increase with the number of polar (Figure 2.24). The double bonds are isolated entities
groups, so we see increasing polarization effects that do not interact.
with 1,1-dichloroethane and 1,1,1-trichloroethane.
We shall often need to consider inductive effects In 1,3-pentadiene, however, the p orbitals are
when attempting to predict chemical reactivity. all able to overlap in such a way that a lower
energy molecular orbital can be formed. We have
2.8 Conjugation more physical data available for 1,3-butadiene, so
let us consider this slightly simpler conjugated system
Double bonds, whether they be C=C, C=O, or instead.
C=N, are sites of special reactivity in a molecule.
This reactivity may take on different characteristics We have four 2p orbitals on four adjacent car-
if we have two or more double bonds in the bon atoms, and these can overlap to produce four
same molecule, depending upon whether the double π molecular orbitals. These are as shown, and their
bonds are isolated or conjugated. We use the term relative energies can be visualized from the bond-
conjugated to describe an arrangement in which ing interactions possible (Figure 2.25). Remember,
double bonds are separated by a single bond. Thus, bonding results from overlap of orbitals that have the
in 1,3-pentadiene the double bonds are conjugated, same phase sign of the wave function, whereas anti-
bonding orbitals originate from interaction of orbitals
with different phase signs of the wave function. Thus,
ψ1 has three bonding interactions and no antibonding
interactions, ψ2 has two bonding interactions and one

CC CC CC overlap of p
CCC orbitals

1,4-pentadiene C C CC C

for clarity, all hydrogen
atoms have been omitted

1,3-pentadiene 1,3-butadiene

Figure 2.24 Overlap of p orbitals in dienes

38 ATOMIC STRUCTURE AND BONDING

antibonding y4
interaction

π*

y3

Energy

bonding y2
interaction y1

π

ethylene 1,3-butadiene

Figure 2.25 Energy diagram: molecular orbitals of 1,3-butadiene

antibonding interaction, ψ3 has one bonding interac- not restricted to alkenes. Any system containing two
tion and two antibonding interactions, and ψ4 has or more π bonds may be conjugated, so that we
no bonding interaction and three antibonding inter- can include triple bonds (alkynes), carbonyl groups,
actions. The four electrons will be allocated to ψ1 imines, and nitriles in this description. In its broadest
and ψ2. sense, conjugation refers to a system that has a p
orbital adjacent to a π bond allowing delocalization
Conjugation introduces a number of features. We of electrons. The adjacent p orbital may be a
can consider that the π electrons in a conjugated vacant one, as in a carbocation (see Section 2.6.2),
system are no longer associated with specific bonds, one that contains a single electron, as in a radical
but are delocalized over those atoms constituting the (see Section 2.6.2), or may be part of another π
conjugated system. This has energy implications. The bond, as in a conjugated diene. At first glance, a
overall energy associated with butadiene is actually conjugated anion does not fit the broad definition of
less than we might expect. It is lower than that of conjugation, since we would expect the carbanion
non-conjugated dienes, e.g. 1,4-pentadiene, and less centre to be sp3 hybridized (see Section 2.6.2).
than what we might estimate from figures for the Nevertheless, there is delocalization of electrons and
monounsaturated but-1-ene. Thus, compounds with this system is considered to be conjugated. In this
two conjugated double bonds are thermodynamically system, delocalization results from accommodating
more stable (less reactive) than compounds with the negative charge in a p orbital rather than an sp3
two isolated double bonds. In due course, we shall orbital, so that we again achieve p orbital overlap.
see that the double bond reactivity of butadiene is
also influenced by conjugation: butadiene behaves Conjugated systems also give characteristic spec-
differently from compounds with isolated double tral absorptions, especially in the UV–visible
bonds (see Section 8.2). We also need to appreciate regions. As the extent of conjugation increases, i.e.
that conjugation, and its influence on reactivity, is more than two double bonds separated by single

CONJUGATION 39

HH H H H
CC CCC C C C
CC H CC CC
HH HN HH
conjugated ene–yne
conjugated diene conjugated nitrile conjugated carbocation

H H H H
CO C NR C C
CC CC CC CC
H H HH HH

conjugated carbonyl conjugated imine conjugated radical conjugated carbanion

bonds, these compounds have more intense absorp- levels. This means that, with increasing conjugation,
tions at longer wavelengths (lower energies). This is the characteristic absorption moves from the UV
because the energy difference between bonding and to the visible region and, typically, the compound
antibonding molecular orbitals becomes smaller with becomes coloured. A compound appears coloured to
increasing conjugation. The spectral data arise from the human eye when it removes by absorption some
the transition of an electron between these energy of the wavelengths from white light.

Box 2.1

Carotenoids, vitamin A, and vision

Carotenoids are a group of natural products found predominantly in plants. They are characterized by an extended
chain of conjugated double bonds, giving an extended π electron system. They are highly coloured and contribute
to yellow, orange, and red pigmentations in plants. Lycopene is the characteristic carotenoid pigment in ripe
tomato fruits, and the orange colour of carrots is caused by β-carotene. Capsanthin is the brilliant red pigment
of capsicum peppers.

lycopene OH
β-carotene
O capsanthin
OH

40 ATOMIC STRUCTURE AND BONDING

Box 2.1 (continued)

Carotenoids function along with chlorophylls (see Box 11.4) in photosynthesis as accessory light-harvesting
pigments, effectively extending the range of light that can be absorbed by the photosynthetic apparatus. The
absorption maximum of carotenoids is typically between 450 and 500 nm, which indicates that the energy
difference between bonding and antibonding molecular orbitals is quite small. This absorption maximum
corresponds to blue light, so that with blue light absorbed, the overall impression to the human eye is of a
bright yellow–orange coloration. Recent research suggests that carotenoids are important antioxidant molecules
for humans, helping to remove toxic oxygen-derived radicals, and thus minimizing cell damage (see Box 9.2). The
most beneficial dietary carotenoid in this respect is lycopene, with tomatoes featuring as the predominant source.

Vitamin A1 (retinol) is derived in mammals by oxidative metabolism of plant-derived dietary carotenoids
in the liver, especially β-carotene. Green vegetables and rich plant sources such as carrots help to provide us
with adequate levels. Oxidative cleavage of the central double bond of β-carotene provides two molecules of the
aldehyde retinal, which is subsequently reduced to the alcohol retinol. Vitamin A1 is also found in a number of
foodstuffs of animal origin, especially eggs and dairy products. Some structurally related compounds, including
retinal, are also included in the A group of vitamins.

β-carotene

O2 cleavage of central double bond
generates two molecules of retinal

NADH OH
O
retinal retinol
reduction of aldehyde (vitamin A1)
to alcohol

A deficiency of vitamin A leads to vision defects, including a visual impairment at low light levels, termed
night blindness. For the processes of vision, retinol needs to be converted first by oxidation into the aldehyde
retinal, and then by enzymic isomerization to cis-retinal. cis-Retinal is then bound to the protein opsin in the
retina via an imine linkage (see Section 7.7.1) to give the red visual pigment rhodopsin.

retinol hydrolytic cleavage H
NADP+ of imine opsin
H
11 E N
O H

retinal formation of imine with amino absorption of light energy
enzymic trans–cis group on protein opsin hn restores trans configuration
isomerism of
11,12-double bond H2N opsin of 11,12-double bond

11 Z

11-cis-retinal HO rhodopsin opsin
HN

H

AROMATICITY 41

Rhodopsin is sensitive to light by a process that involves isomerization of the cis-retinal portion back to the

trans form, thus translating the light energy into a molecular change that then triggers a nerve impulse to the
brain. The absorption of light energy promotes an electron from a π to a π∗ orbital, thus temporarily destroying
the double bond character and allowing rotation (see Section 2.6.2). trans-Retinal is then subsequently released

from the protein by hydrolysis, and the process can continue.

2.9 Aromaticity molecular orbital in a conjugated system extending
over all six atoms. The electrons will be distributed
Aromatic compounds constitute a special group evenly, or delocalized, over the whole molecule.
of conjugated molecules; these are cyclic unsatu-
rated molecules with unusual stability and char- The six p atomic orbitals combine to give six
acteristic properties. The term aromatic originates molecular orbitals for the π system. The relative
from the odour displayed by many of the simple energies for these are shown in Figure 2.27. There
examples. is one low-energy bonding molecular orbital and two
degenerate bonding orbitals at higher energy. There
2.9.1 Benzene will be an analogous array of antibonding orbitals at
higher energy.
The parent compound is benzene. Benzene, C6H6,
contains an array of six sp2-hybridized carbons, each The six electrons are assigned to these orbitals
attached by a σ bond to the adjacent carbons, and as we have seen previously, beginning with the
by a third σ bond to a hydrogen atom. The six p lowest energy level. This leads to the six electrons
atomic orbitals from carbon are all aligned so that completely filling the bonding molecular orbitals
they can overlap to form molecular orbitals, and and providing an extremely favourable arrangement,
this is most favourable when the carbons are all in that the overall energy is significantly below
in one plane. The lowest energy molecular orbital that of six electrons in the contributing p atomic
can be considered as an extended ring-like system orbitals. The energy stabilization is considerable,
with a high electron probability above and below the and also much more than could be accounted for
plane of the ring (Figure 2.26). This is a bonding π by simple conjugation. The special stability afforded
by this planar cyclic array is what we understand by
aromaticity. The chemical reactivity associated with
aromatic systems will be covered in Chapter 8.

H
HCH

CC

CC
HCH

H

benzene H H overlap of p H overlap of p

orbitals H orbitals
H
H HH

H H HH

lowest energy molecular orbital for benzene;
all p orbitals overlapping in phase

Figure 2.26 Lowest energy molecular orbital for benzene

42 ATOMIC STRUCTURE AND BONDING

antibonding
molecular orbitals

Energy six p atomic orbitals with six
unpaired electrons
bonding molecular
orbitals

molecular orbitals in benzene
Figure 2.27 Energy diagram: molecular orbitals of benzene

2.9.2 Cyclooctatetraene 2.9.3 Hu¨ckel’s rule

Let us consider the origins of benzene’s aromatic sta- Cyclooctatetraene has eight π electrons and benzene
bilization. Another cyclic hydrocarbon, cyclooctate- has six. The number of π electrons that confer
traene (pronounced cyclo-octa-tetra-ene), certainly aromaticity is given by Hu¨ ckel’s rule: a planar cyclic
looks conjugated according to our criteria, but chemi- conjugated system will be particularly stable if the
cal evidence shows that it is very much more reactive number of π electrons is 4n + 2, where n is an
than benzene, and does not undergo the same types integer (0, 1, 2, 3, etc). Although the significance
of reaction. It does not possess the enhanced aromatic of this will not become apparent until later (see
stability characteristic of benzene. below), we must stress that 4n + 2 refers to the
number of π electrons, and not the number of
H H H atoms in the ring. Benzene, therefore, with six π
H H electrons (n = 1, 4n + 2 = 6), is aromatic; however,
cyclooctatetraene, with eight π electrons, is not
H H aromatic. Also aromatic would be a system with 10 π
electrons (n = 2), or 14 π electrons (n = 3). The first
H of these would be the compound [10]annulene, and
the second [14]annulene. Annulene is a general term
benzene cyclooctatetraene for a carbon ring system with alternating single and
double bonds; the number in brackets is the number
Further, cyclooctatetraene has been shown to be of carbons in the ring. For example, we could call
non-planar; it adopts a tub shape. This originates from benzene [6]annulene, though in practice, nobody ever
bond angles. A regular octagon has internal bond does.
angles of 135◦, quite far from the optimum angle of
120◦ for sp2 hybridization. In benzene’s hexagon, the Whereas [14]annulene shows aromatic properties,
internal angle is 120◦, a perfect fit for sp2 geometry. [10]annulene, unfortunately, does not, but we know
Cyclooctatetraene thus distorts from the planar to this is a consequence of the molecule adopting a
relieve this strain. A careful consideration of this non-planar shape. The interior angle for a planar
shape may then suggest the immediate consequences. 10-carbon system would have to be 144◦, and this
These are that none of the double bonds are in is too far removed from the sp2-hybridized angle
the same plane; therefore, there is going to be no of 120◦ to be feasible. As ring sizes get larger, it
overlap of p orbitals between the double bonds. We becomes possible to have a cyclic system where all
cannot get any enhanced stability associated with bond angles can be the ideal 120◦. There is a way
conjugation. of drawing a 10-carbon ring system with angles of

AROMATICITY 43

120◦, but we must realize that this attempts to place cyclopropenyl cation cyclopentadienyl anion
two hydrogens in the same space. This is clearly two π electrons (n = 0) six π electrons (n = 1)
not feasible; as the hydrogens are pushed away from
each other, therefore, this must lead to a non-planar with them being in atomic orbitals. We have seen
molecule. how this originates in benzene by allocating electrons
to the bonding orbitals. We can apply the same
looks OK, but procedure to other annulene compounds, and there
cannot be planar exists a very neat way of finding the relative
because of energies of molecular orbitals without resource to
locating hydrogen mathematical calculations. This device, the Frost
atoms circle, inscribes the appropriate polygon in a circle,
with one vertex pointing vertically downwards. The
[10]annulene [10]annulene intersections of other vertices with the circle then
mark the positions of the molecular orbitals. The
10 π electrons (n = 2) position of the horizontal diameter represents the
energy of the carbon p orbital; intersections below
[14]annulene [18]annulene this are bonding, those above are antibonding, and
14 π electrons (n = 3) 18 π electrons (n = 4) nonbonding orbitals are on the diameter line. Frost
circles for benzene and cyclooctatetraene are drawn
Structures that are also aromatic are the in Figure 2.28.
cyclopropenyl cation (2 π electrons; n = 0) and
the cyclopentadienyl anion (6 π electrons; n = 1). We can immediately see that allocating six elec-
Although we do not wish to pursue these examples trons into the benzene molecular orbitals fills all three
further, they are representative of systems where the bonding orbitals (a closed shell structure) and there is
number of π electrons is not the same as the number substantial aromatic stabilization, in that the energy
of carbon atoms in the ring. associated with electrons in the molecular orbitals is
greatly reduced compared with that of electrons in the
The stabilization conferred by aromaticity results six atomic orbitals. For cyclooctatetraene, allocating
primarily from the much lower energy associated with eight electrons to the molecular orbitals leads to three
a set of electrons in molecular orbitals compared filled orbitals, but then the remaining two electrons
are put singly into each of the degenerate nonbonding
orbitals. Cyclooctatetraene does not have a filled shell

Energy antibonding
molecular orbitals

nonbonding
molecular orbitals

bonding
molecular orbitals

benzene cyclooctatetraene

Figure 2.28 Relative energies of benzene and cyclooctatetraene molecular orbitals from Frost circles

44 ATOMIC STRUCTURE AND BONDING

structure like benzene, but has two nonbonding elec- to give a dibromo product (see Section 8.1.2). This
trons; it does not have the special stability we see in reaction destroys the π bond. When it comes to com-
benzene. As we have seen in Section 2.9.2, cyclooc- pounds such as annulenes, it is not always easy to
tatetraene also adopts a non-planar shape, lacks the synthesize sufficient material to demonstrate typical
stabilization associated with conjugation, and behaves chemical reactivity, and a simple spectroscopic anal-
like four separate normal alkenes. ysis for aromaticity is infinitely preferable. Nuclear
magnetic resonance (NMR) spectroscopy has pro-
2.9.4 Kekule´ structures vided such a probe.

Benzene is usually drawn as a structure with The proton NMR signals for hydrogens on a double
alternating single and double bonds. We can draw bond are found in the region δ 5–6 ppm. In contrast,
it in two ways. those in benzene are detected at δ 7.27 ppm. This
substantial difference is ascribed to the presence
Kekulé representations benzene; circle represents of a ring current in benzene and other aromatic
of benzene delocalized π electrons compounds. A ring current is the result of circulating
electrons in the π system of the aromatic compound.
These two forms are so-called Kekule´ structures; Without entering into any discussion on the origins
but neither is correct, in that benzene does not have of NMR signals, the ring current creates its own
single and double bonds. This immediately follows magnetic field that opposes the applied magnetic
from a measurement of C–C bond lengths. For sp2- field, and this affects the chemical shift of protons
hybridized carbons, we expect the C=C bonds to bonded to the periphery of the ring. Signals are
be about 1.34 A˚ , whereas the C–C bond length shifted downfield (greater δ) relative to protons in
would be about 1.47 A˚ . Measurements show that all alkenes. Proton NMR spectroscopy can, therefore,
of the carbon–carbon bond lengths are the same, be used as a test for aromaticity. In this way,
at 1.40 A˚ . This length is between that of single [14]annulene and [18]annulene have been confirmed
and double bonds, and suggests that we have C–C as aromatic.
bonds that are somewhat between single and double
bond in character. From the point of stability, and 2.9.6 Aromatic heterocycles
now also bond lengths, we must view benzene as
quite different from cyclohexatriene. To emphasize In due course we shall see that unsaturated cyclic
this, a different representation for the benzene ring compounds containing atoms other than carbon, e.g.
has been proposed, i.e. a circle within a hexagon. nitrogen, oxygen, or sulfur, can also be aromatic. For
The circle represents the six π-electron system, and example, pyridine can be viewed as a benzene ring
this, therefore, highlights the special nature of the in which one CH has been replaced by a nitrogen.
aromatic ring. As we shall see in due course, this It is aromatic and, like benzene, displays enhanced
representation has considerable limitations, and most stability. Pyrrole is a five-membered heterocycle, but
chemists, ourselves included, do not use it. also displays aromaticity. Like the cyclopentadienyl
anion (see Section 2.9.3), the number of π electrons
is not the same as the number of atoms. In
pyrrole, nitrogen provides two of the six π electrons.
Examples of these molecules are discussed under
heterocycles in Chapter 11.

2.9.5 Aromaticity and ring currents benzene N N
pyridine H
One can demonstrate the particular stability of aro- pyrrole
matic compounds by their characteristic chemical
reactions. For example, benzene reacts with bromine 2.9.7 Fused rings
only with difficulty and gives bromobenzene, a
substitution product (see Section 8.4). This leaves the We may also encounter aromatic hydrocarbons that
aromatic ring intact. By contrast, a typical alkene feature fused rings. Thus, naphthalene is effectively
reacts readily with bromine by an addition process

RESONANCE STRUCTURES AND CURLY ARROWS 45

two benzene rings fused together, and anthracene 18 π electrons. Note that Hu¨ckel’s rule applies only to
has three fused rings. The heterocycle quinoline (see monocyclic compounds, and although 10 π electrons
Section 11.8.1) is a fusion of benzene and pyridine. (naphthalene) and 14 π electrons (anthracene) seem
These ring systems are undoubtedly aromatic, and to be meet the criteria for aromaticity, there is good
they display the enhanced stability and reactivity evidence to suggest we should consider the aromatic
associated with simple aromatic compounds like system not as a combination of benzene rings, but as
benzene. a single ring involving the periphery of the molecule.

naphthalene anthracene N 2.10 Resonance structures and curly
quinoline
arrows
naphthalene anthracene
(10 π electrons) (14 π electrons) The molecular orbital picture of benzene proposes
that the six π electrons are no longer associated
these structures are strictly incorrect if with particular bonds, but are effectively delocalized
the circle represents six π electrons over the whole molecule, spread out via orbitals
that span all six carbons. This picture allows us to
naphthalene anthracene appreciate the enhanced stability of an aromatic ring,
(10 π electron system) (14 π electron system) and also, in due course, to understand the reactivity
of aromatic systems. There is an alternative approach
the π electron system may involve just based on Lewis structures that is also of particular
the periphery of the molecule value in helping us to understand chemical behaviour.
Because this method is simple and easy to apply, it is
Molecular orbital calculations suggest that the π an approach we shall use frequently. This approach
electrons in naphthalene are delocalized over the is based on what we term resonance structures.
two rings and this results in substantial stabilization.
These molecules are planar, and all p orbitals are Let us go back to the two Kekule´ representations
suitably aligned for overlap to form π bonding for benzene. The Lewis structure for benzene has
molecular orbitals. Although we can draw Kekule´ alternating single and double bonds, but there are two
structures for these compounds, it is strictly incorrect ways of writing this. In one form, a particular bond
to use the circle in hexagon notation since the is single; in the other form, this bond has become
circle represents six π electrons. Naphthalene has 10 double. Resonance theory suggests that these two
carbons, and therefore 10 π electrons, and anthracene structures are both valid representations, and that each
has 14 π electrons. The circle notation suggests 12 or contributes to the structure of benzene, but the true
structure is something in between, a lower energy
hybrid of the two Kekule´ forms, a resonance hybrid.
If this is the case, then each bond is neither single
nor double, but, again, something in between. As we
have already seen, all C–C bond lengths in benzene
are 1.40 A˚ , which is in between the bond lengths for
single (1.47 A˚ ) and double (1.34 A˚ ) bonds.

double-headed arrow is used to
indicate resonance structures

Kekulé representations curly arrow represents the we could also have written
of benzene movement of two electrons curly arrows like this

46 ATOMIC STRUCTURE AND BONDING

To indicate resonance forms, we use a double- but the atom involved is still part of a molecule,
headed arrow between the contributing structures. and the molecule consequently also carries a formal
This arrow is reserved for resonance structures and charge. We have already met a few such entities
never used elsewhere. The difference between the in this chapter, e.g. the ammonium and hydronium
two structures is that the electrons in the π bonds cations, looking specifically at the molecular orbital
have been redistributed, and we can illustrate this by descriptions (see Section 2.6.3). As indicated above,
use of another type of arrow, a curly arrow. This the use of curly arrows may involve species with
arrow is used throughout chemistry to represent the positive or negative charges.
movement of two electrons. In the benzene case,
a cyclic movement of electrons accounts for the As simple examples, ammonia and water are neu-
apparent relocation of double bonds, though there tral molecules. Nitrogen has five valence electrons,
are two ways we might show this process; both are and it acquires a stable octet of electrons in making
equally satisfactory. three bonds to hydrogen atoms. Each hydrogen has
its stable arrangement of two electrons. The nitrogen
Benzene is a nice example to choose to illustrate in ammonia also carries a lone pair of electrons. Oxy-
the concept of resonance. It is not the best example gen, with six valence electrons, makes two bonds to
for explaining the rules governing the use of curly hydrogen atoms. Its octet of electrons will carry two
arrows, so we must move to some simpler com- lone pairs.
pounds. Being able to draw curly arrows is an essen-
tial skill for an organic chemist, and you will see HNH HOH H H
from a cursory glance at the following chapters just H water HOH
how frequently they are employed. We shall use the HNH
same curly arrows and precisely the same principles ammonia hydronium
for predicting the outcome of chemical reactions (see H cation
Section 5.1). They allow us to follow bond making ammonium
and bond breaking processes, and provide us with a
device we can use to keep track of the electrons. cation

• The curly arrow represents the movement of two We can deduce the charge associated with ammo-
electrons. nia and water by simply considering that the com-
ponent atoms are neutral, that all we have done is
• The tail of the arrow indicates where the electrons share the electrons, so the molecules must also be
are coming from, and the arrowhead where they neutral. The formal charge on an individual atom
are going to. can be assessed more rigorously by subtracting the
number of valence electrons assigned to an atom in
• Curly arrows must start from an electron-rich its bonded state from the number of valence elec-
species. This can be a negative charge, a lone pair, trons it has as a neutral free atom. Electrons in bonds
or a bond. are considered as shared equally between the atoms,
whereas unshared lone pairs are assigned to the atom
• Arrowheads must be directed towards an electron- that possesses them.
deficient species. This can be a positive charge,
the positive end of a polarized bond, or a formal = number of valence – number of
suitable atom capable of accepting electrons, i.e. charge electrons as neutral valence electrons
an electronegative atom.
free atom assigned in
In our brief introduction to Lewis structures (see bonded state
Section 2.2), we paid particular attention to valency,
the number of bonds an atom could make to other Hence, for nitrogen, the number of valence elec-
atoms via the sharing of electrons. We must now trons in a free atom is five. In ammonia, the number
broaden this idea to consider atoms in a molecule of assigned electrons is also five (three in bonds plus
that are no longer neutral, but which carry a formal a lone pair). Therefore, the formal charge on nitrogen
positive or negative charge. This means we are is zero. For hydrogen, the formal charge is also zero,
considering cations and anions, as in ionic bonding, since the number of valence electrons is one, and the
number of assigned electrons is one. For oxygen, the
number of valence electrons in a free atom is six. In
water, the number of assigned electrons is also six

RESONANCE STRUCTURES AND CURLY ARROWS 47

formal charge +1 formal charge 0 formal charge −1

C CC C CC C CC

N NN N NN NN
OO OO O

X X X
X = F, Cl, Br, I

Figure 2.29 Formal charges of common atoms and ions

(two in bonds plus two lone pairs). Therefore, the five (three bonds plus a lone pair). Therefore, the
formal charge on oxygen is zero. The hydrogens are formal charge on carbon is 4 − 5 = −1. We must
also uncharged, as in ammonia. always indicate the charge in structures pictured as
shown in the right-hand representation; the left-hand
Now consider the ammonium and hydronium representation is incomplete and, therefore, wrong.
cations. In the ammonium system, for nitrogen The most common formal charges we shall meet are
the formal charge is now +1. This follows from summarized in Figure 2.29.
the number of valence electrons, i.e. five, minus
the number of assigned electrons, i.e. four (four in Now let us return to curly arrows and resonance
bonds). In the hydronium system, the formal charge structures.
on oxygen is also +1. This is assessed from the
number of valence electrons, i.e. six, minus the • Resonance structures differ only in the position of
number of assigned electrons, i.e. five (three in bonds the electrons; the positions of the atoms do not
plus a lone pair). Of course, we already knew that change.
ammonium and hydronium cations were the result
of bonding neutral ammonia or water with a proton • Resonance structures can be interconverted by the
(charge +1), so an overall charge of +1 comes as no movement of electrons indicated by curly arrows.
particular surprise (see Section 2.6.3).
• Three main types of electron movement can be
Other systems are less familiar, and will therefore implicated:
have to be assessed carefully. For example, what
charge is associated with the structure shown on the bonding to nonbonding
left below?

HCH HCH CO CO
H H

methyl anion two electrons are moved from the π bond to the
electronegative oxygen;
This is the methyl anion, and carries one negative
charge. Carbon has four valence electrons, and in carbon now has formal charge +1, oxygen has formal
this structure the number of assigned electrons is charge −1;
the molecule still has overall charge of zero

48 ATOMIC STRUCTURE AND BONDING

nonbonding to bonding • Separation of charge in a resonance structure
decreases stability.
OC OC
• Structures with charge separation are more stable if
the two lone pair electrons are used to make a π bond the negative charge is located on an electronegative
to carbon; this effectively assigns one electron to atom.
each atom;
C+ was electron deficient but now gains one electron • Structures with adjacent like charges are dis-
from the lone pair; favoured, as are those with multiple isolated
oxygen now loses one electron from the lone pair and charges.
carries formal charge +1;
the new structure still has overall charge +1 • The σ bond framework and steric factors must
permit a planar relationship between contributory
bonding to new bonding resonance structures.

This is illustrated for a carbonyl compound and an
alkene.

C C CO CO
CC CC
most favourable charge separation;
resonance form less favourable;
negative charge on more
π electrons from the double bond are used to make a electronegative atom
new double bond;
C+ was electron deficient but now gains one electron CO CO
from the pair of π electrons;
the carbon at the donor end of the arrow is now least important;
electron deficient and carries formal charge +1; negative charge on less
the new structure still has overall charge +1 electronegative atom

• All structures must be valid Lewis structures. An CC CC
atom may become electron deficient, but, on the
other hand, it must never be shown with more most favourable charge separation;
valence electrons than it can accommodate. For resonance form less favourable;
example, it is not possible to have pentavalent positive and negative charges
carbon. on same type of atom

• The overall charge must remain the same. CC CC

• There should be the same number of unpaired also unfavourable
electrons in each structure.
We then consider the potential relative importance
This redistribution of electrons provides us with of the resonance structures we have drawn.
one or more new resonance structures (also
called canonical structures, limiting structures, or • Equivalent resonance structures contribute equally
mesomers). However, some structures are more real- to the hybrid.
istic than others.

• The more covalent bonds a structure has, the more
stable it is.

• A structure in which all the atoms have the noble
gas structure is particularly stable.

HYDROGEN BONDING 49

• Structures that are not equivalent do not contribute H H
equally; the more stable a structure is, the more it CO CO
is likely to contribute.
H H
• Highly unstable structures make little contribution
and may be ignored. formaldehyde

Acceptable resonance structures can then be imag- CH3 CH3
ined as contributing to the overall electronic distribu- OC OC
tion in the molecule. By considering the properties of H CH3 H CH3
contributing structures, we can also predict some of
the properties of the molecule. We imagine that the conjugate acid
molecule is not fully represented by a single structure, of acetone
but is better represented as a hybrid of its contributing
resonance forms. It is likely that the energy associ- H H
ated with the molecule is actually lower than that of H CH H CH
any contributing resonance form; therefore, the delo-
calization of electrons that resonance represents is CC CC
a stabilizing feature. The larger the number of sta- HH HH
ble resonance structures we can draw, the greater
the extent of delocalization. The difference in energy allylic cation
between the actual molecule and that suggested by
the best of the resonance structures is termed the res- note that this resonance is only
onance energy or resonance stabilization energy. possible if the atoms are coplanar
This can usually only be an estimated amount.
find that electron-rich reagents (nucleophiles) attack
The resonance terminology and the double-headed the carbon atom (see Section 7.1). This is reasonable,
arrow may give the impression that the structures since we can show this carbon as positively charged
are rapidly interconverting. This is not true. We in the right-hand resonance structure. The third
must appreciate right from the start that resonance example is the allylic cation. This is a reasonably
structures are entirely hypothetical. They are our stable carbocation, and we attribute this to resonance
(sometimes clumsy) attempt to write down on paper stabilization; this is particularly favourable in this
what the bonding in the molecule might be like, case, since both contributing resonance forms are
and they may depict only the extreme possibilities. identical. We can visualize the allylic cation as an
The molecule is presumably happily going about its entity in which the positive charge is delocalized over
business in a form that we cannot easily depict. the whole structure (strictly, it is the electrons that are
Nevertheless, resonance structures are extremely delocalized, but we are one short of a full complement
useful and do help us to explain chemical behaviour. and it is the positive charge that dominates the
representation).
Let us look again at the simple examples shown
above and the consequences of our hypothetical 2.11 Hydrogen bonding
resonance structures.
Hydrogen bonds (H-bonds) describe the weak attrac-
We shall see that most of the reactions of tion of a hydrogen atom bonded to an electronega-
simple carbonyl compounds, like formaldehyde, are tive atom, such as oxygen or nitrogen, to the lone
a consequence of the presence of an electron-deficient pair electrons of another electronegative atom. These
carbon atom. This is accounted for in resonance bonds are different in nature from the covalent bonds
theory by a contribution from the resonance structure we have described; they are considerably weaker than
with charge separation (see Section 7.1). The second covalent bonds, but turn out to be surprisingly impor-
example shows the so-called conjugate acid of tant in chemistry and biochemistry.
acetone, formed to some extent by treating acetone
with acid (see Section 7.1). Protonation in this way Let us consider a molecule possessing an O–H σ
typically activates acetone towards reaction, and we bond. This bond is polar because hydrogen is less
electronegative than oxygen (see Section 2.7), and

50 ATOMIC STRUCTURE AND BONDING

this allows the partially positive hydrogen atom to are known to form hydrogen-bonded ‘dimers’ in solu-
associate with a centre in another molecule carrying a tion through two quite strong intermolecular hydro-
partial negative charge. This is likely to be the oxygen gen bonds. Hydrogen bonds involving N–H are also
atom in another molecule. common, and we can also meet hydrogen bonding
between alcohols and amines.

d− d+ d− d+ H
OH OH OO

hydrogen bond CC
H3C C CH3
If we consider the interaction between two water
molecules, then the partial positive charge on hydro- H
gen induced by the electronegativity of oxygen is
attracted to the high electron density of the oxy- intramolecular hydrogen bonding
gen lone pair in another molecule. This hydrogen is in enol form of acetylacetone
now linked to its original oxygen by a σ bond, and
to another oxygen by an electrostatic attraction. The O HO
bond length of the H–O hydrogen bond is typically
about twice the length of the H–O covalent bond, RC CR
and the hydrogen bond is very much weaker than the
covalent bond, though stronger than other interactions OH O
between molecules. In water, further hydrogen bonds
involving other molecules are formed, leading to a intermolecular hydrogen bonding
network throughout the entire sample. The extensive in carboxylic acids
hydrogen bonding in water is responsible for some of
water’s unusual properties, its relatively high boiling Box 2.2
point, and its high polarity that makes it a partic-
ularly good solvent for ionic compounds. Alcohols Hydrogen bonds and DNA
also exhibit hydrogen bonding; but, with only a sin-
gle O–H, the network of bonds cannot be as extensive The nucleic acids known as deoxyribonucleic acid
as in water. (DNA) are the molecules that store genetic informa-
tion. This information is carried as a sequence of
Hydrogen bonds connecting different molecules bases in the polymeric molecule. Remarkably, the
are termed intermolecular, and when they connect interpretation of this sequence depends upon simple
groups within the same molecule they are called hydrogen bonding interactions between base pairs.
intramolecular. A simple example of an intramolec- Hydrogen bonding is fundamental to the double helix
ular hydrogen bond is seen in the enol form of arrangement of the DNA molecule, and the transla-
acetylacetone (see Section 10.1). Carboxylic acids tion and transcription via ribonucleic acid (RNA) of
the genetic information present in the DNA molecule.

HH

O OH CH3 CH3
OH O
d+ H H H
d− H H O H O
OH CH3 CH3
O d− OH O
d+ H d+ H H

H OH

d+

extensive hydrogen H hydrogen bonding in methanol
bonding in water

MOLECULAR MODELS 51

In DNA, the base pairs are adenine–thymine Another pyrimidine base, uracil, is found in RNA
and guanine–cytosine. Adenine and guanine are instead of thymine. Base pairing between adenine
purine bases, and thymine and cytosine are and uracil involves two hydrogen bonds and resem-
pyrimidines (see Section 14.1). bles the adenine–thymine interaction. This type of
base pairing is of importance in transcription, the
HH CH3 synthesis of messenger RNA (see Section 14.2.5).
N O
2.12 Molecular models
NN NN
NN H We soon come to realize that molecules are not two-
dimensional objects as we draw on paper; they are three-
O dimensional and their overall size and shape can have a
thymine profound effect on some of their properties, especially
biological properties. We have seen that four single
adenine bonds to carbon are distributed in a tetrahedral array, an
arrangement that minimizes any steric or electrostatic
H interactions (see Section 2.6.2). Atoms around double
bonds are in a planar array, and angles are 120◦.
N Again, this trigonal arrangement minimizes interactions.
H A triple bond creates a linear array of atoms. Now,
careful measurements of bond angles and also bond
O NN lengths in a wide variety of molecules have convinced
us that these features are sufficiently constant that we
N H O can use them to predict the shape and size of other
N molecules.

N N H cytosine Bond lengths in molecules usually correlate with one
N of five kinds, and their typical measurements are shown
in Table 2.3. The five types of bond are:
H
• single bonds between atoms, one of which is hydro-
guanine gen;

HH O • single bonds between atoms, neither of which is
N hydrogen;
NN
NN H • double bonds;
NN
O • triple bonds;
uracil
• bonds in aromatic rings.
adenine
Bond angles can be related to hydridization, and
Thus, each purine residue is specifically linked so can bond lengths. Thus, electrons in sp hybrid
by hydrogen bonding to a pyrimidine residue. orbitals are held closer to the nucleus than electrons
This may involve either two or three hydrogen in sp2 orbitals, which are correspondingly closer than
bonds, with hydrogen of N–H groups bonding electrons in sp3 orbitals (see Section 2.6.2). The bond
to oxygen or to nitrogen. The result of these lengths below follow this generalization. The shortest
interactions is that each base can hydrogen bond bond lengths involve bonds to hydrogen, the smallest
only with its complementary partner. The specific atom that utilizes an s orbital in bonding. Note also
base-pairing means that the two strands in the that aromatic carbon–carbon bonds have a bond length
DNA double helix are complementary. Wherever between that of single and double bonds, a feature of
adenine appears in one strand, thymine appears aromatic bonding (see Section 2.9.4).
opposite it in the other; wherever cytosine appears
in one strand, guanine appears opposite it in the
other.

52 ATOMIC STRUCTURE AND BONDING

Table 2.3 Typical bond lengthsa

Bond type Bond Length (A˚ ) Bond Length (A˚ ) Bond Length (A˚ )

Single H–X H–C 1.06 – 1.10 H–N 1.01 H–O 0.96
Single C–X 1.54 1.47 C–O 1.43
Double C–C 1.34 C–N 1.30 C=O 1.23
Triple C=C 1.20 C=N 1.16
Aromatic C≡C 1.40 C≡N

C–C aromatic

a 1 A˚ (Angstrom unit) = 10−10 m.

With these five typical bond lengths, and the typical nitrogen. The resultant model tells us about the bonding,
bond angles for tetrahedral, trigonal, and linear arrays, it and the overall size and shape of the molecule, but gives
becomes possible to construct molecular models to pre- us little indication about the volume taken up by the
dict a molecule’s size and shape. This may be achieved atoms and electrons. Nevertheless, this type of model
via a molecular model kit, or by computer graphics. is probably the most popular, and it provides a lot of
information. It is the three-dimensional equivalent of our
Many different molecular model kits have been two-dimensional line drawings of structures.
produced over the years, each varying in their approach
to atoms and bonds, and also in their cost. However, In ball-and-stick models, balls with holes drilled at
there are three main types, which can provide us appropriate angles are used to represent the atoms, and
with three main types of information. These are the sticks or springs are used for the bonds to link them. The
framework, ball-and-stick, and space-filling versions resultant model is rather like the framework model, but
(Figure 2.30). has representations of the atoms. Models tend to look
better than the framework type, but in practice tend to
Framework models concentrate on the bonds in be bigger and less user friendly.
the molecule. In the least expensive kits, these are
represented by narrow tubes, joined by linker pieces Space-filling model kits are even less user friendly.
that signify the atom position. The linkers have four, They employ specially shaped atomic pieces that clip
three, or two stubs to fit into the tubes, according to the together, each representing the volume taken up by the
number of bonds required, and these are also arranged atom and its bonding electrons. This system produces
at appropriate angles (tetrahedral 109◦, trigonal 120◦, a rather more globular model that indicates the whole
linear 180◦). The linkers are also coloured differently bulk of the molecule, including the electron clouds that
to show the atom they represent, typically black for are involved in bonding. The value of this type of model
carbon, white for hydrogen, red for oxygen, and blue for is that it shows just how big the molecule really is, and

CO2H 4,4-dimethylcyclohexane-

carboxylic acid

(a) (b) (c)

Figure 2.30 Molecular models depicting 4,4-dimethylcyclohexanecarboxylic acid: (a) framework; (b) ball-and-stick;
(c) space-filling. Note that the size of atoms reflects the electronic charge associated with the atom. Therefore, as seen
in models (b) and (c), a hydrogen atom attached to electronegative oxygen appears smaller than a hydrogen atom
attached to carbon

MOLECULAR MODELS 53

not just the atoms and/or bonds. When one is faced with Whilst many people still like to handle and view a
interpreting how a molecule might interact with, say, a model, computer programs have allowed us to cre-
receptor protein, then space-filling models are essential. ate and manipulate representations of three-dimensional
There is a downside, however, because it becomes very molecules rapidly with varying degrees of accuracy and
difficult indeed to visualize the structure and bonding in sophistication. We can easily view the image from any
the molecule. angle, and can see the effect that any molecular mod-
ifications might have. Further, although interactions of
These three types of model are illustrated in groups in a molecule can lead to changes in bond angles,
Figure 2.30, though we have employed computer graph- and to a lesser extent in bond lengths, ordinary mod-
ics to generate these pictures. Significantly, computer els may not show this. Computer graphics programs are
modelling programs all allow generation of images able to carry out quite sophisticated energy calculations
according to the three main types of model; each pro- to show the most favourable arrangement of atoms, the
vides us with subtly different information, so they interatomic distances, and the bond angles.
are not alternatives, but are actually complementary.



3

Stereochemistry

3.1 Hybridization and bond angles drawing stereostructures:

From our discussions of bonding, we have learnt C might also be drawn as
something about the arrangement of bonds around
various atoms (see Chapter 2). These concepts are CC
fundamental to our appreciation of the shape of
molecules, i.e. stereochemistry. Before we delve into note that whilst these are OK
these matters, let us recap a little on the disposition
of bonds around carbon. this could lead to confusion

Bonding at four-valent carbon is tetrahedral, with always try to keep a tetrahedral appearance
four sp3-hybridized orbitals mutually inclined at
109.5◦. Remember that the tetrahedral array is demon- use these bond angles for nice structures
strated by experimental measurements, and that 120° 90°
hybridization is the mathematical model put forward 120° 90°
to explain this observation (see Section 2.6.2). We can
conveniently represent the tetrahedral arrangement in Bonding at three-valent carbon is trigonal planar
two dimensions by using a wedge–dot convention. In with bond angles of 120◦, an observation that
this convention, single bonds written as normal lines we account for through sp2 hybridization plus
are considered to be in the plane of the paper. Bonds formation of a π bond by overlap of p orbitals (see
in front of this plane, i.e. coming out from the paper, Section 2.6.2). Thus, an alkene double bond involves
are then drawn as a wedge, whilst bonds behind the electrons in sp2 hybrid orbitals making σ single
plane, i.e. going into the paper, are drawn as a broken bonds, and the remaining electrons in p orbitals
or dotted bond (see Section 2.6.2). overlapping to produce the π-bond component of
the double bond. We can draw this as a planar
in plane behind plane representation, all single bonds in the plane of the

sp3 hybridization C wedge−dot
angle 109.5° representation
tetrahedral
in front of plane

As we get more familiar with this representation,
we may begin to abbreviate it by showing either the
wedge or the dotted bond, rather than both. Of course,
it is important to remember that these abbreviated
forms actually represent a tetrahedral array, and not
something with three bonds planar plus one other.

Essentials of Organic Chemistry Paul M Dewick
 2006 John Wiley & Sons, Ltd

56 STEREOCHEMISTRY

paper, or show the π bonding in the plane of the Bonding at nitrogen and oxygen approximates to that at
paper, so that some bonds now require to be drawn carbon via lone pairs:
in wedge form and others in dotted form.

sp2 hybridization CC N O sp3 tetrahedral
angle 120°
planar single bonds in plane of
paper, π bond
perpendicular to plane N O sp2 trigonal

CC overlap of p orbitals N sp linear
generates π bond
in cyanide, will dictate a linear arrangement, with a
π bond in plane nitrogen lone pair occupying a nonbonding sp orbital
of paper (see Section 2.6.3).

Bonding at two-valent carbon is linear, i.e. bond Bond angles depend upon the type of hybridization
angles are 180◦, and the triple bond comprises two as just described, but in most molecules they appear
π bonds and a σ single bond formed from sp hybrid to be very similar. There can often be a small
orbitals (see Section 2.6.2). The two π bonds are at degree of variation because of the nature of the
right angles to each other. precise atoms being bonded, and the presence of
lone pair electrons (see Section 2.6.3), but the level
sp hybridization CC of consistency is very high. Similarly, bond lengths
angle 180° are also remarkably consistent, depending mainly
linear on the nature of the atoms bonded and whether
bonds are single, double, aromatic, or triple (see
overlap of p orbitals Section 2.12). With bond lengths and bond angles
generates two π bonds being sufficiently consistent between molecules, it is
possible to predict the shape and size of a molecule
Although most of the atoms in the framework of an using simple molecular models or computer graphics
organic molecule tend to be carbon, other atoms, such (see Section 2.12).
as oxygen and nitrogen, are routinely encountered.
We can consider the arrangement of bonds around 3.2 Stereoisomers
these atoms as approximately the same as the sp3-
hybridized tetrahedral array seen with carbon (see For a given molecular formula there is often more
Section 2.6.3). One (nitrogen) or two (oxygen) of the than one way of joining the atoms together, whilst
sp3 orbitals will be occupied by lone pair electrons. still satisfying the rules of valency. Such variants
The consequences of this include the fact that the two are called structural isomers or constitutional iso-
single bonds to oxygen are not linear, but are inclined mers – compounds with the same molecular formula
at about 109◦ (see Section 2.6.3), and the three bonds but with a different arrangement of atoms. A simple
to nitrogen are similarly not planar. example is provided by C4H10, which can be accom-
modated either by the straight-chained butane, or by
When oxygen or nitrogen are linked to another the branched-chain isobutane (2-methylpropane).
atom, e.g. carbon, by double bonds, the arrangement
will be equivalent to the trivalent carbon, i.e. trigonal structural isomers
planar with a π bond perpendicular to the plane (see constitutional isomers
Section 2.6.3). Lone pair electrons (one lone pair for
nitrogen, two in the case of oxygen) will occupy HH CH3
nonbonding sp2 orbitals. A triple bond to nitrogen, as H3C CH3
H3C H CH3
HH
isobutane
butane (2-methylpropane)

CONFORMATIONAL ISOMERS 57

Stereoisomers, on the other hand, are compounds draw a similar Newman projection for the second
with the same molecular formula, and the same wedge–dot representation, but the C–H bonds of the
sequence of covalently bonded atoms, but with front and rear methyls will appear to be on top of each
a different spatial orientation. Two major classes other. We therefore draw a slightly modified version
of stereoisomers are recognized, conformational showing all bonds, but must remember that this really
isomers and configurational isomers. represents a system where the bonds at the rear are
obscured by the bonds at the front.
Conformational isomers, or conformers, intercon-
vert easily by rotation about single bonds. Configura- In the sawhorse representation, the molecule is
tional isomers interconvert only with difficulty and, viewed from an oblique angle, and all bonds can be
if they do, usually require bond breaking. We shall seen.
study these in turn.
H H
3.3 Conformational isomers HH
HH
3.3.1 Conformations of acyclic compounds HH HH
H
Let us consider first the simple alkane ethane. Since H sawhorse
both carbons have a tetrahedral array of bonds, representations
ethane may be drawn in the form of a wedge–dot
representation. H HH
HH H HH
Now let us consider rotation of the right-hand H
methyl group about the C–C bond, and we eventually H
get to a different wedge–dot representation as shown. HH
This is more easily visualized by looking at the
molecule from one end down the C–C bond, and The two representations shown here are actually
this gives us what is termed a Newman projection. two different conformers of ethane; there will be
The Newman projection shows the hydrogen atoms an infinite number of such conformers, depending
and their bonds, but the carbons are represented by upon the amount of rotation about the C–C bond.
a circle; since we are looking down the C–C bond, Although there is fairly free rotation about this
we cannot see the rear carbon. A further feature bond, there does exist a small energy barrier to
is that the C–H bonds of the methyl closest to rotation of about 12 kJ mol−1 due to repulsion of
us are shown drawn to the centre of this circle, the electrons in the C–H bonds. By inspecting
whilst those of the rear methyl are partially obscured the Newman projections, it can be predicted that
and drawn only to the edge of this circle. We can this repulsion will be a minimum when the C–H
bonds are positioned as far away from each other

view H H rotation of right H H wedge–dot
from end H C hand methyl H H representation

C H about C C bond C C
H H
H H

view from end gives Newman projection HH HH H
generally HH
60˚ drawn as H
H
HH H HH
H
HH eclipsed conformer shows both bonds,
H high energy but angle is
assumed to be 0˚
staggered conformer
low energy

58 STEREOCHEMISTRY

as is possible. This is when the dihedral angle take the form of a sine wave, because rotations of
between the C–H bonds of the front and rear methyls either 120◦ or 240◦ will produce an indistinguishable
is 60◦, as exists in the left-hand conformer. This conformer of identical energy. This is shown in
conformation is termed the staggered conformation. Figure 3.1.
On the other hand, electronic repulsion will be
greatest when the C–H bonds are aligned, as in the It follows that the preferred conformation of ethane
right-hand conformer. This conformation is termed is a staggered one; but, since the energy barrier to
the eclipsed conformation. In between these two rotation is relatively small, at room temperature there
extremes there will be other conformers of varying will be free rotation about the C–C bond.
energies, depending upon the degree of rotation.
Energies for these will be greater than that of Let us now consider rotation about the central C–C
the staggered conformer, but less than that of the bond in butane. Rotation about either of the two
eclipsed conformer. Indeed, if one considers a gradual other C–C bonds will generate similar results as with
rotation about the C–C bond, the energy diagram will ethane above. Wedge–dot, Newman, and sawhorse
representations are all shown; use the version that
appears most logical to you.

wedge–dot representations

view HH CH3 HH C HCH3 HH H HH HH HH H HH CHH3
C C C H C C C C C CH C C

H3C HH H3C H3C HCH3 H3C CH3 H3C CH3 H3C H

rotation of right-hand group

Newman projections H H H H CHH3
H3C H HH H
rotation of H
rear groups

CH3 H3C H H H H3C CH3 H H3C H
HH eclipsed conformer* H3C H HH eclipsed conformer*
eclipsed conformer
HH H highest energy H CH3 * equal energies
CH3 CH3 CH3

staggered conformer staggered conformer** staggered conformer**
anti gauche gauche

lowest energy ** equal energies

sawhorse representations H3C H H HH H H CH3
rotation of H HH HH3CH H H HCH3 HH
rear groups CH3 H3C H3C H H CH3 H
H
HH
HH

H3C H3C H3C H3C

C CH3 bonds shown in bold

As we rotate the groups, we shall get a series a methyl–hydrogen interaction, which in turn will
of staggered and eclipsed conformers. The energy be larger than that arising from hydrogen–hydrogen
barrier to rotation will be larger than the 12 kJ mol−1 interactions. Logically then, we predict that the
seen with ethane. This is because, in addition to the energy of the eclipsed conformer in which the
similar electronic repulsion in the bonds, there is methyl groups are aligned will be higher than that
now a spatial interaction involving the large methyl in which there are methyl–hydrogen alignments, and
groups. It follows that the repulsive energy associated that there will be two equivalent versions of the
with a methyl–methyl interaction will be larger than latter.

CONFORMATIONAL ISOMERS 59

ethane eclipsed eclipsed
12 kJ mol−1

Relative energy 0 staggered staggered
staggered

60 120 180 240 300

Dihedral angle (degrees)
Figure 3.1 Energy diagram: ethane conformations

Similarly, of the low-energy staggered conform- conformers, where there must be at least some
ers, there will be two equivalent ones where the spatial interaction between the methyl groups. The
carbon–methyl bonds are inclined at 60◦ to each staggered conformer with maximum separation of
other, and one in which the carbon–methyl bonds methyl groups is termed the anti conformer (Greek:
are inclined at 180◦. We can also predict that anti = against), whilst the two other ones are termed
the latter conformer, which has the methyl groups gauche conformers (French: gauche = left). The
as far away from each other as possible, will energy diagram observed (Figure 3.2) reflects these
be of lower energy than the alternative staggered predictions, and the energy difference between the

butane
18.8 kJ mol−1

15.9

Relative energy 3.8 gauche
gauche
anti
0
anti

180 120 60 0 60 120 180

Me Me dihedral angle (degrees)
Figure 3.2 Energy diagram: butane conformations

60 STEREOCHEMISTRY

low-energy staggered anti conformer and the highest cyclopropane • angle 60˚
energy eclipsed conformer is about 18.8 kJ mol−1. • highest ring strain
• must be planar
There will still be free rotation about C–C bonds • bonds eclipsed
in butane at room temperature, but the larger energy
barrier compared with that for ethane means that the HH CC CC
staggered conformers are preferred, and calculations HH
show that, at room temperature, about 70% of maximum poor orbital
molecules will be in the anti conformer and about HH orbital overlap overlap
15% in each gauche conformer. poor orbital overlap
in C C bonds
3.3.2 Conformations of cyclic compounds
HH CH2
Cyclopropane, cyclobutane, cyclopentane, HH
cyclohexane
A further feature of three-membered rings is that
The practical consequences of conformational iso- they must be planar, and a consequence of this is
merism become much more significant when we con- that, in cyclopropane, all C–H bonds are in the high-
sider cyclic compounds. The smallest ring system will energy eclipsed state. There can be no conformational
contain three atoms; in the case of hydrocarbons this mobility to overcome this.
will be cyclopropane.
In cyclobutane, the internal angle is 90◦. Conse-
Now, simple geometry tells us that the inside angle quently, there is high ring strain, but this is not so
in cyclopropane must be 60◦. This is considerably great as in cyclopropane.
less than the 109.5◦ of tetrahedral carbon, and the
consequences are that the amount of overlap of If cyclobutane were planar, all C–H bonds would
the sp3 orbitals in forming the C–C bonds must be in the high-energy eclipsed state. It transpires
be considerably less than in an acyclic system like that cyclobutane is not planar, since it can adopt a
ethane. With poorer overlap, we get a potentially
weaker bond that can be broken more easily. We term
this ring strain, and although three-membered rings
exist and are quite stable, they are frequently subject
to ring-opening reactions (see Section 6.3.2).

cyclobutane • angle 90°
• high ring strain
• non-planar conformer minimizes eclipsing

equilibrium arrow is used to indicate
interconversion of structures

H H H H H H
HH H HH H H

H H HH H HH H CH2
H H H H CH2
HH
interconversion of equal energy H
non-planar conformers through
planar intermediate note how we can show
perspective by having a full
bond at the front, and an
incomplete bond at the rear

CONFORMATIONAL ISOMERS 61

more favourable conformation in which eclipsing is are ‘pushed’ or ‘pulled’. Both conformers will be
reduced, and the ring appears puckered. This appears produced equally, and can interconvert at room
to be achieved by pushing pairs of opposite carbons temperature because the energy barrier is fairly small
in different directions; but, in reality, it is only a at about 5.8 kJ mol−1. The interconversion of the
combination of rotations about C–C bonds as we two forms is depicted by the equilibrium arrow,
have seen with the simpler acyclic compounds. It comprised of two half arrows. At equilibrium, both
is not possible to achieve the ideal 60◦ staggered conformers coexist, and in this case in equal amounts
arrangement, but it does produce a lower energy since they have the same energy.
conformer. Of course, there are two alternative ways
of doing this, depending on whether pairs of carbons The planar form of cyclobutane will be the energy
maximum in the interconversion of conformers.

Box 3.1

Compounds with cyclopropane or cyclobutane rings

A cyclopropane ring has the highest level of ring strain in the carbocycles. This means that they are rather
susceptible to ring-opening reactions, but it does not mean that they are unstable and cannot exist. Indeed, there
are many examples of natural products that contain cyclopropane rings, and these are perfectly stable under
normal conditions.

One group of natural cyclopropane derivatives of especial importance is the pyrethrins, insecticidal components
of pyrethrum flowers, and widely used in agriculture and in the home. These compounds have very high toxicity
towards insects without being harmful to animals and man, and are rapidly biodegraded in the environment. The
pyrethrins are esters of two acids, chrysanthemic acid and pyrethric acid, with three alcohols, pyrethrolone,
cinerolone, and jasmolone, giving six major ester structures. The acids contain the cyclopropane ring, and this
appears essential for the insecticidal activity.

OR2 acids: CO2H CO2H
R1 O chrysanthemic acid MeO2C pyrethric acid
HO
pyrethrins alcohols: HO HO
general structure
O O O
semi-synthetic pyrethrins pyrethrolone cinerolone jasmolone

O Cl O O
O Cl O O

O O O
bioresmethrin
permethrin phenothrin

Many semi-synthetic esters, e.g. bioresmethrin, permethrin, and phenothrin, have been produced and these
have increased toxicity towards insects and also extended lifetimes. All such esters retain a high proportion of
the natural chrysanthemic acid or pyrethric acid structure.

62 STEREOCHEMISTRY

Box 3.1 (continued)

The drugs naltrexone and nalbuphine are semi-synthetic analogues of the analgesic morphine. Morphine is
a good painkiller, but has some unpleasant side effects, the most serious of which is the likelihood of becoming
addicted.

HO HO HO

O O O

H NMe HO N HO N
HO H H O H

HO

morphine nalbuphine naltrexone

Nalbuphine is a modified structure containing a cyclobutane ring as part of the tertiary amine function.
Extending the size of the nitrogen substituent makes the drug larger and allows it to exploit extra binding
sites on the receptor that morphine cannot interact with. Nalbuphine is found to be a good analgesic with
fewer side effects than morphine. Naltrexone incorporates a cyclopropane ring in the nitrogen substituent.
This, together with the other structural modifications, produces a drug that has hardly any analgesic effects,
but is a morphine antagonist. Accordingly, it can be used to assist in detoxification of morphine and heroin
addicts.

Let us move on to cyclopentane, where geometry The energy barrier to this conformational change
tells us the internal angle is 108◦. This is so close is about 22 kJ mol−1. There is no reason why any one
to the tetrahedral angle of 109.5◦ that cyclopentane particular carbon should be out of the plane, and at
can be considered essentially free of ring strain. room temperature there is rapid interconversion of all
However, planar cyclopentane would have all its C–H possible variants. Again, a planar form would feature
bonds eclipsed, which is obviously not desirable. as the energy maximum in the interconversions. The
Accordingly, it adopts a lower energy conformation conformation with four carbons in plane and one out
in which one of the carbon atoms is out of planarity. of plane is termed an envelope conformation. This
‘Pushing’ this carbon out of the plane is achieved by terminology comes from the similarity to an envelope
rotation about C–C bonds, and it reduces eclipsing with the flap open.
along all but one of the C–C bonds.
For cyclohexane, the calculated internal angle is
cyclopentane • angle 108° 120◦ if the molecule were to be planar, but the
• little ring strain tetrahedral angle of 109.5◦ turns out to be perfect if
• non-planar conformer the molecule is non-planar. It is possible to construct
a cyclohexane ring from tetrahedral carbons without
minimizes eclipsing introducing any strain whatsoever. The ring shape
formed in this way is termed a chair conformation.
HH H There is considerable resemblance to a folding chair
H having a back rest and leg rest, though the open seat
H might be regarded as a distinct disadvantage. Not
H H only is the bond angle perfect, but it also turns out
H that all C–H bonds are in a staggered relationship
with adjacent ones. The chair conformation cannot
HH be improved upon.

envelope conformation

CONFORMATIONAL ISOMERS 63

cyclohexane • angle 109.5° if non-planar
• no ring strain
• no eclipsing in chair conformation

axis

H CH2 H Hax Hax hydrogens are axial
Newman projection of chair H H Heq Hax
conformation looking along
Heq Heq

two opposite C C bonds; H CH2 H Heq Heq or equatorial
all bonds are staggered H H Hax Heq

Hax Hax

chair conformation

The total ring strain in various cycloalkanes com- Table 3.1 Ring straina in cycloalkanes
pared with their strain-free acylic counterparts has
been estimated, as shown in Table 3.1. Thus, small Number of Total ring Number of Total ring
rings like cyclopropane and cyclobutane have con- atoms atoms
siderable ring strain, and cyclohexane is effectively in ring strain in ring strain
strain free. Larger rings (8–11 atoms) have more (kJ mol−1) (kJ mol−1)
ring strain than might be predicted, certainly much
more than cyclohexane, but any puckering that 3 115 8 41
reduces ring strain actually creates eclipsing. We shall
meet rings containing more than six carbons only 4 110 9 53
infrequently.
5 26 10 51

6 0 11 47

7 26 12 17

a Values relative to strain-free acyclic analogue, e.g. cyclobu-
tane and butane.

Box 3.2

How to draw chair conformations of cyclohexane

You can only appreciate stereochemical features if you can draw a representation that correctly pictures the
molecule. One of the most challenging is the chair conformation of cyclohexane. Practice makes perfect; so this
is how it is done.

Draw two inclined bonds Draw two further Add the two remaining Put in the axial substituent
bonds ensuring they are bonds, up from top points,
of the same length parallel bonds parallel to existing bonds down from bottom points

ensure top points these bonds are
are level parallel to each other

these bonds these bonds are
are parallel parallel to each other
to each other

64 STEREOCHEMISTRY

Box 3.2 (continued) The end result − perfect!
Put in wedges and bold bond for
Add three pairs of equatorial substituent bonds perspective if required; the lower
ensuring they are parallel to existing bonds part of the ring is always at the front

these four bonds are all
parallel to each other

these four bonds are all these four bonds are all

parallel to each other parallel to each other

Note that the wedges and bold bonds help to show how we are looking at the cyclohexane chair. In practice,
particularly to speed up the drawing of structures, we tend to omit these. Then, by convention, the lower bonds
represent the nearest part of the ring.

for ease of drawing, we the lower bonds always
usually omit bold bonds represent the nearest part
and wedges of the ring

When one looks at the hydrogens in the chair This interconversion may be considered as the
conformation of cyclohexane, one can see that they simultaneous pushing down/pulling up of carbons on
are of two types. Six of them are parallel to the central opposite sides of the ring, as indicated in the left-
rotational axis of the molecule, so are termed axial. hand structure. As a result, the ring ‘flips’ into an
The other six are positioned around the outside of alternative conformation, also a chair, as in the right-
the molecule and are termed equatorial. One might hand structure. This ring flip is actually achieved
imagine, therefore, that these two types of hydrogen by rotation about several of the C–C bonds at the
would have some different characteristics, and be same time. The ring flip can be demonstrated with
detectable by an appropriate spectral technique. Such suitable molecular models, and it is possible to feel
a technique is NMR spectroscopy; but, at room the resistance in the model to this rotation, which
temperature, only one type of proton is detectable. represents the energy barrier to the change. Both
At room temperature, all hydrogens of cyclohexane conformers have the same energy, but the energy
can be considered equivalent; this is a consequence barrier is about 42 kJ mol−1. The energy barrier looks
of conformational mobility, and the interconversion high compared with those in ethane or butane, but
of two chair conformations. this is because the interconversion involves rotations
about several C–C bonds at the same time.
*Hax * Heq
•Heq •Hax Look at the hydrogen atoms shown labelled in
the left-hand structure. Note particularly that, after
interconversion of conformers via ring flip ring flip, the axial hydrogen becomes equatorial,
changes axial / equatorial relationship; whilst the equatorial hydrogen becomes axial. Similar
the conformers have the same energy changes occur at all other positions. With rapidly
interconverting conformers, the hydrogens cannot be
distinguished by NMR spectroscopy and they all
merge to give a single signal. However, as one cools

CONFORMATIONAL ISOMERS 65

the sample, the energy available to overcome the not planar, but has a chair conformation. We shall fre-
interconversion energy barrier diminishes, until at quently want to use the hexagon representation, and
a sufficiently low temperature, the interconversion it will be necessary to assign hydrogens or other sub-
stops, and two types of hydrogen are detectable in stituents onto the chair representation with the correct
the NMR spectrum. This temperature is −89 ◦C. stereochemistry. At this stage it is salutary to look at
Measurement of this temperature allows the energy both the two-dimensional hexagon and the chair rep-
barrier to be calculated. resentations of cyclohexane. Note particularly that we
must not confuse ‘up’ with axial, and ‘down’ with
If we look at the two-dimensional hexagon repre- equatorial. As the structures show, ‘up’ hydrogens or
sentation for cyclohexane, we could put in the bonds substituents will alternate axial and equatorial as we
to hydrogens as wedges (up bonds) or dotted lines go round the ring positions.
(down bonds). We now know the cyclohexane ring is

up up

HH H H H up up H
HH
HH H H HH down hydrogens shown in bold are 'up'
H H H up and alternate axial−equatorial
HH H H H
HH down H H around the ring; they are not all
HH H H
up HH axial or all equatorial
H down downH
HH down
incorrect planar down
representation

The chair is not the only conformation that folding; sea-worthiness is rather questionable. Again,
cyclohexane might adopt. An alternative boat there is no ring strain in this conformation, but
conformation is attained if the ring flip-type process it turns out that some of the C–H bonds are
is confined to just one carbon. The name boat eclipsed, as seen in the accompanying Newman
comes from the similarity to boats formed by paper projection.

flagpole interaction

HH • no ring strain H H
• eclipsing in boat conformation H HH
HH H • flagpole interaction in boat
H H • eclipsing and flagpole interaction HH H
HH H HH
reduced in twist-boat HH
H • both higher energy than chair
H

chair boat twist-boat

CH2
H CH2 H
HH

H H HH

Newman projection of boat
conformation looking along
the two horizontal bonds;
some bonds are eclipsed

66 STEREOCHEMISTRY

conformations of cyclohexane half-chair
half-chair

42 kJ mol−1

boat

27
21

Relative energy twist-boat twist-boat

0

chair chair

Figure 3.3 Energy diagram: cyclohexane conformations

In addition, the hydrogens at the top of the In practice, only the chair conformation is impor-
structure are getting rather close to each other, tant for cyclohexane, since the energy differences
and there is some interaction, termed a flagpole between it and the other conformations make them
interaction, again from the nautical analogy. Both much less favourable. However, there are plenty
the eclipsing and the flagpole interactions can be of structures where cyclohexane rings are forced
minimized when the boat conformation undergoes into the boat or twist-boat conformation because of
further subtle changes by rotation about C–C bonds other limiting factors. For example, bornane is a
to form the twist-boat. This is a result of twisting the terpene hydrocarbon where opposite carbons in a
flagpole hydrogens apart. Making a molecular model cyclohexane ring are bridged by a methylene group.
of the boat conformation immediately shows how This is stereochemically impossible to achieve with
easy it is to modify it to the twist-boat variant; the a chair – the carbons are too far apart. However,
boat conformation is quite floppy compared with the it is possible with a boat conformation. In such a
chair, which is very rigid. An energy diagram linking structure, there are no further possibilities for confor-
the chair, boat and twist-boat conformations is shown mational mobility – the conformation is now fused
in Figure 3.3. The boat conformation is represented in and no further changes are possible, even though
by an energy maximum. there may be unfavourable eclipsing interactions.

H3C CH3 H

HH CH3 HH H
H H H H
H
H HH

HH H H
H

bornane boat conformation cyclohexene half-chair conformation tetrahydro-
(planar around double bond, naphthalene
non-planar elsewhere
removes eclipsing)

CONFORMATIONAL ISOMERS 67

In cyclohexene, the double bond and adjacent Let us look at a simple example, namely methyl-
carbons must all be planar. The remainder of the cyclohexane.
molecule avoids unfavourable eclipsing interactions
by adopting what is termed a half-chair conforma- Ring flip in the case of methylcyclohexane
tion. This would also be found in a cyclohexane ring achieves interconversion of one conformer where the
fused onto an aromatic ring (tetrahydronaphthalene) methyl group is equatorial into a conformer where
or fused to a three-membered ring (see Section 3.5.2). this group is axial (compare the hydrogens in cyclo-
The half-chair conformation in cyclohexane (without hexane). It turns out that the conformer with the
the double bond) is thought to be equivalent to the equatorial methyl group is favoured over the con-
energy maximum in Figure 3.3 that must be overcome former where the methyl group is axial. The energy
in the chair–twist-boat interconversion. difference of these two conformers is estimated to
be about 7.1 kJ mol−1; this is the energy difference,
Substituted cyclohexanes not the barrier to interconversion. Because of this
energy difference, the equilibrium mixture at room
The ring flipping conformational mobility in the un- temperature has about 95% of conformers with the
substituted compound cyclohexane has little practical equatorial methyl and only 5% where the methyl is
significance; but, when the ring is substituted, we axial. We can account for the difference in energy
have to take ring flip into account, because one partic- between the two conformers quite easily using the
ular conformation is usually favoured over the other. reasoning we applied earlier for the acyclic hydro-
carbon butane.

methylcyclohexane

Newman projection interconversion of conformers via ring flip Newman projection
down 2,1-bond down 2,1-bond
changes axial methyl to equatorial methyl;
CH3
H CH2 equatorial conformer favoured H

3 CH3 H 65 4 H3C CH2

H CH2 1 6H 5 H3C 1 6
H6 H
H 2

3 3 H CH2
24 H3

gauche methyl axial methyl equatorial anti

higher energy conformer lower energy conformer

1,2-gauche interaction and 1,2-anti and

1,3-diaxial interactions no 1,3-diaxial interaction

We need to consider a Newman projection looking down the 6,1 bond. Now, in the conformer where
down the 2,1 bond. When the methyl is axial, it the methyl is equatorial the Newman projection
can be seen that there will be a gauche interaction shows the most favourable anti arrangement for the
between this methyl and the ring methylene (C-3); a methyl and methylene(s); there will be a similar anti
second, similar interaction will be seen if we looked interaction if we looked down the 6,1 bond. On

68 STEREOCHEMISTRY

this basis alone, we can predict that the equatorial Note that it is not necessary to consider both forms
conformer is of lower energy and, thus, more of cyclohexane, where the methyl is either wedged
favoured. However, there is a further feature that (up) or dotted (down). If the cyclohexane ring were
destabilizes the axial conformer, and that is the planar, the two structures would be the same, since
spatial interaction between the axial methyl and the one merely has to turn the structure over to get the
axial hydrogens at positions 3 and 5, termed a other. Although the cyclohexane ring is not planar,
1,3-diaxial interaction. Together, they account for it turns out that the two structures are still identical,
the equilibrium mixture consisting mainly of the because of the ring flip process. This is shown below.
equatorial conformer. We can indicate this by using One set of conformers is simply the upside-down
arrows of unequal size in the equilibrium equation. version of the other.

H3C CH3

≡ HA H

H H B A=D
H3C B=C
CH3 C
D
Now, as the substituent gets bigger, the proportion
of axial conformer will diminish even further. With a is such that essentially all molecules are in the
substituent as big as a tert-butyl group, the equilibrium equatorial conformation; in general terms, we can
consider that a tert-butyl group will never be axial.

H3C CH3 H
H3C H

H

H
tert-butyl group never axial

Although analysis of the consequences of cis. The terms trans and cis are used to describe
ring flip in a monosubstituted cyclohexane is the configuration, not conformation, of the isomers;
pretty straightforward, the presence of two or in the trans isomer, the two methyl substituents are
more substituents requires careful consideration to on opposite sides (faces) of the ring (Latin: trans =
decide which conformer, if any, is the more across), whereas in the cis isomer they are on the
favoured. Let us illustrate the approach using 1,4- same side of the ring (Latin: cis = on this side).
dimethylcyclohexane. Now, two configurational These concepts will become clear when we reach
isomers of this structure can exist, namely trans and Section 3.4.

CONFORMATIONAL ISOMERS 69

1,4-dimethylcyclohexane H CH3 ax
trans eq CH3
H3C H
eq H
H
lower energy − both methyl
substituents equatorial CH3
ax
higher energy − both methyl
substituents axial

ax 180º ax ax

CH3 CH3 CH3

cis H CH3 H3C H ≡ H CH3
eq eq
HH H eq

both conformations have same energy −
one axial methyl and one equatorial methyl

In the trans isomer, one methyl is written down have one equatorial methyl and one axial methyl; they
(dotted bond) whilst the other is written up (wedged must, therefore, be of the same energy, so form a
bond). If we transform this to a chair conformation, 50 : 50 equilibrium mixture. In fact, it is also easy
as shown in the left-hand structure, the down methyl to see that rotation of either structure about its central
will be equatorial and the up methyl will also be axis produces the other structure, a clear illustration
equatorial. With ring flip, both of these substituents that they must be energetically equivalent. Note that
then become axial as in the right-hand conformer. the cis isomer with both methyls down is actually the
From what we have learned about monosubstituted same compound viewed from the opposite side.
cyclohexanes, it is now easily predicted that the
diequatorial conformer will be very much favoured This type of reasoning may be applied to other
over the diaxial conformer. dimethylcyclohexanes, as indicated in the figure.
There is no easy way to predict the result; it must
In the cis isomer, both methyls are written with be deduced in each case. One conformer is of much
wedges, i.e. up. In the left-hand chair conformation, lower energy in the cases of trans-1,2-, cis-1,3-,
one methyl is therefore axial and the other is equatorial. and trans-1,4-dimethylcyclohexane; both conformers
With ring flip, the axial methyl becomes equatorial and have equal energy in the cases of cis-1,2-, trans-1,3-,
the equatorial methyl becomes axial. Both conformers and cis-1,4-dimethylcyclohexane.

cis-1,2-dimethylcyclohexane methyls eq and ax ax ax
trans-1,2-dimethylcyclohexane methyls both eq or both ax 1 eq
cis-1,3-dimethylcyclohexane methyls both eq or both ax eq 2 3
trans-1,3-dimethylcyclohexane methyls eq and ax
cis-1,4-dimethylcyclohexane methyls eq and ax ax eq
trans-1,4-dimethylcyclohexane methyls both eq or both ax eq

4

ax

Should the two substituents be different, and espe- the most favoured conformer is going to be the
cially of different sizes, then the simple reasoning one with the maximum number of equatorial sub-
used above with two methyl substituents will need stituents, or perhaps where we have the large sub-
adapting; the larger substituent will prefer to be equa- stituents equatorial. This is seen in the following
torial. Where we have three or more substituents, examples.

70 STEREOCHEMISTRY

HO HO OH
menthol
all substituents equatorial; all substituents axial
favoured
OH OH
HO OH HO
HO OH HO HO
HO OH HO OH
OH
HO OH OH five substituents axial,
myo-inositol OH one equatorial

HO five substituents equatorial, CO2H
OH one axial; OH
favoured HO OH
HO OH
CO2H HO three substituents axial, two
HO CO2H equatorial
HO OH
quinic acid OH

three substituents equatorial,
two axial;
favoured

HO HO OH
neoisomenthol
two substituents equatorial two substituents axial with large
but the large isopropyl group isopropyl group equatorial;
is axial favoured

Box 3.3

How to draw conformational isomers and to flip cyclohexane rings

Interpreting a two-dimensional stereochemical structure, converting it into a conformational drawing, and
considering the consequences of ring flip can cause difficulties. The process can be quite straightforward if
you approach it systematically.

We saw early in Section 3.3.2 that, if we draw cyclohexane in typical two-dimensional form, the bonds to the
ring could be described as ‘up’ or ‘down’, according to whether they are wedged or dotted. This is how we would
see the molecule if we viewed it from the top. When we look at the molecule from the side, we now see the
chair conformation; the ring is not planar as the two-dimensional form suggests. Bonds still maintain their ‘up’
and ‘down’ relationship, but this means bonds shown as ‘up’ alternate axial–equatorial around the ring; they are

CONFORMATIONAL ISOMERS 71

not all axial or all equatorial. Whilst the ring flip process changes equatorial bonds to axial bonds, and vice versa,
it does not change the ‘up’–‘down’ relationship.

HH HH HH
HH HH
HH ring flip HH
HH
HH HH H H
HH HH H H
HH HH
HH

HH

hydrogens shown in bold are 'up' and ring flip changes equatorial to axial,
alternate axial−equatorial around the ring; and axial to equatorial; it does not
they are not all axial or all equatorial change the 'up'−'down' relationship

Let us consider the trimethylcyclohexane isomer shown below. All three substituents are ‘up’. We need to use
one of the carbons as a reference marker; let us choose the top one. I like to make this the left-hand carbon in
the chair; to make the process more obvious, we could turn the structure so that our reference carbon is also on
the left. It is most important to have this reference carbon, so that as we put the various substituents in we put
them on the correct carbons.

rotate structure 90º this is the top view of the
chair conformation

draw bonds at relevant carbons put in substituents

up CH3 up up up up up H3C CH3
H3C CH3 up
1 3
12 CH3 23 1 23 CH3
3 1 H H
≡ H3C down down 2
up
CH3
up becomes front down H

part of chair

= reference carbon

up CH3 3 CH3
up
2
draw flipped ring; down 23 H
align left-hand carbons
and right-hand carbons up H3C

1 1 H

down H
down

Now draw the two chair conformations of cyclohexane, both having the reference carbon on the left. The
carbons opposite our reference point must be furthest right. If we draw the structures one above the other, left-
hand carbons and right-hand carbons should be aligned. Draw axial and equatorial bonds at the relevant carbons
where we have the substituents and identify them as ‘up’ or ‘down’. Since we are interpreting the structure as
though we are looking down on it from the top, the lower part of the ring represents the nearmost part of the
conformational drawing. It can also help to number the carbons. Then fill in the substituents as necessary. In
this example, our three methyl groups are all ‘up’, which means that in one conformer the groups will be axial,
equatorial, and axial, whereas in the other they will be equatorial, axial, and equatorial. The latter conformer,
with the most equatorial substituents, will be the favoured one.

A word of warning is appropriate here. As we shall see in due course (see Box 3.11), merely changing a
substituent from, say, equatorial to axial without flipping the ring changes the configuration, and can produce a
different molecule. It would also destroy the ‘up’ or ‘down’ identifier.

72 STEREOCHEMISTRY

To take this general principle to its extreme, we instead the ring system adopts the less favourable
noted above that tert-butyl groups are sufficiently twist-boat conformation. It follows, therefore, that
large that they never occupy an axial position. It there must be a greater energy difference between
is possible to make di-tert-butylcyclohexanes where chair conformations carrying axial and equatorial
conformational mobility would predict that one of tert-butyl substituents than there is between chair
these groups would have to be axial, namely cis- and twist-boat conformations. These conformational
1,2-, trans-1,3- or cis-1,4-derivatives. As a result, in changes are shown for trans-1,3-di-tert-butyl-
these cases, we do not see an axial tert-butyl, but cyclohexane.

H H each conformer has one
trans-1,3-di-tert-butylcyclohexane H tert-butyl group axial

H

H
tert-butyl group never axial,
so chair forced into twist-boat
conformation

H

We noted earlier that bonds around nitrogen and on the heteroatom is considered to be larger than
oxygen atoms occupied some of the tetrahedral array, the lone pair electrons. Some common examples are
lone pairs taking up other orbitals. This means that shown below. As we shall see in Section 12.4, the
we can use essentially the same basic principles for heteroatom may have other influences, and there are
predicting the shape and conformation of heterocy- sometimes unexpected effects involving a substituent
cles as we have used for carbocycles. A substituent adjacent to the heteroatom.

O

O N NH O
tetrahydrofuran
H ethylene oxide
piperidine (planar)

O CH2OH OO
HO O
N
O HO OH H NH
tetrahydropyran OH morpholine

glucose (cyclic hemiacetal form)

CONFIGURATIONAL ISOMERS 73

Box 3.4

Conformation of lindane

Chlorination of benzene gives an addition product that is a mixture of stereoisomers known collectively as
hexachlorocyclohexane (HCH). At one time, this was incorrectly termed benzene hexachloride. The mixture has
insecticidal activity, though activity was found to reside in only one isomer, the so-called gamma isomer, γ-HCH.
γ-HCH, sometimes under its generic name lindane, has been a mainstay insecticide for many years, and is about
the only example of the chlorinated hydrocarbons that has not been banned and is still available for general
use. Although chlorinated hydrocarbons have proved very effective insecticides, they are not readily degraded in
the environment, they accumulate and persist in animal tissues, and have proved toxic to many bird and animal
species.

180º

Cl Cl Cl Cl Cl Cl Cl Cl
Cl Cl Cl
Cl Cl Cl Cl ≡
Cl Cl Cl Cl
Cl Cl Cl Cl
lindane
γ-HCH Cl

The stereochemistry of the γ-isomer is shown in the diagram, and when converted into a conformational
stereodrawing it can be seen that there are three axial chlorines and three equatorial ones. Ring flip produces an
alternative conformation of equal energy, but it can be seen that this is identical to the first structure; rotation
through 180◦ produces an identical and, therefore, superimposable structure. It can be seen that conformational
change will not stop the compound interacting with the insect receptor site.

3.4 Configurational isomers mirror
AA
As we have now seen, conformational isomers inter-
convert easily by rotation about single bonds. Con- E B B E
figurational isomers, on the other hand, are isomers D D
that interconvert only with difficulty, and it usually
requires bond breaking if they do interconvert. chiral centre

3.4.1 Optical isomers: chirality and optical Four different groups on tetrahedral carbon can
activity be arranged in two ways − non-superimposable
molecules with a mirror image relationship
If tetrahedral carbon has four different groups
attached, it is found that they can be arranged in two Arrangement is described as chiral
different ways. These molecules are not superimpos-
able and they have a mirror image relationship to The two arrangements (non-superimposable
each other. This is most easily seen with models. mirror images) are called enantiomers

Such an arrangement is called chiral (Greek:
cheir = hand), and the carbon atom is termed a
chiral centre or stereogenic centre. Look at your

74 STEREOCHEMISTRY

two hands. You will see that they appear identical enantiomers are not going to interconvert readily, and
(allowing for minor blemishes or broken fingernails). to achieve interconversion we would have to break
However, do what you will, it is not possible one of the bonds then remake it so as to get the other
to superimpose them, and you should be able to configuration.
appreciate the mirror image relationship. The two
different arrangements – non-superimposable mirror Note that the enantiomer of a particular compound
images – are called enantiomers (Greek: enantios = can be drawn by reversing two of the substituents;
opposite), and we say that enantiomers have different this is actually much easier than drawing the mir-
configurations. The configuration is thus the spatial ror image compound, especially in more complicated
sequence about a chiral centre. It is also apparent that structures. As an alternative, the wedge–dot relation-
ship could be reversed.

A mirror rotation about ≡ A enantiomer can also be
E BB C−A axis E obtained by reversing the
D AA B D wedge−dot relationship

E ≡D B

DE

enantiomer is obtained by
reversing two substituents

Molecules that are superimposable on their mirror attached groups are the same. This introduces a plane of
images are said to be achiral. With tetrahedral carbon, symmetry into the molecule; molecules with a plane of
this is typically the case when two or more of the symmetry can be superimposed on their mirror images.

mirror DB D B plane of
AA molecule with a plane of symmetry

A B B A symmetry is achiral AA AA
D D

molecules that are superimposable
on their mirror images are achiral

Note that chirality is not restricted to tetrahe- tetrahedral systems, such as quaternary nitrogen
dral carbon; it can also be associated with other compounds.

mirror C
AA NB

EN NE NB A rapid nitrogen inversion
D D C means individual enantiomers
B B A are not isolated

quaternary N can also be chiral

However, non-quaternary nitrogen, although tetra- occurs readily at room temperature. This usually
hedral, is not chiral. There is a rapid inver- makes it impossible to obtain neutral amines in opti-
sion that converts one enantiomer into the other; cally active form; quaternization stops this inversion.
effectively, the lone pair does not maintain its posi-
tion. The energy barrier to interconversion is about We shall later need to introduce a related term,
25 kJ mol−1, which is sufficiently low that inversion prochiral. The concept of prochirality is discussed
in Section 3.7.

CONFIGURATIONAL ISOMERS 75

Box 3.5

Manipulating stereostructures

It is not always easy to look at stereostructures – two-dimensional representations of three-dimensional
molecules – and decide whether two separate representations are the same or different. To compare structures, it is
usually necessary to manipulate one or both so that they can be compared directly. Here are a few demonstrations
of how to approach the problem on paper. Of course, constructing models for comparison is the easiest method,
but there will always be occasions when we have to figure it out on paper.

Question: A molecule is represented by the Newman projection:

OH
H CH3

HH
OH

Which of the following are equivalent to the above Newman projection, or to its enantiomer?

H H H HO HO
H OH Me H
OH
CH3 H OH HO OH OH D
A B C

Answer: A and B are same as original; C and D are enantiomers of original

OH turn sideways H CH3 ≡ HO H CH3 look towards CH2OH
H CH3 HO OH sequence Me → H → OH is clockwise
OH
HH HH Note: use any sequence. For this purpose we
OH do not need to obey any priority rules
not a chiral
centre turn round
180°

H CH3 OH CH2OH and OH still in plane of paper;
methyl was at front, now at rear;

hydrogen was at rear, now at front

HO

turn sideways rotate 60° about
H central bond

H H OH ≡ H OH look towards CH2OH
H OH H3C OH sequence Me → H → OH is clockwise;

CH3 H OH H3C OH H this is the same as original
A

H H ≡ MeH look towards CH2OH
Me H sequence Me → H → OH is clockwise;

HO OH HO OH this is the same as original
B

HO turn 30° H OH look towards CH2OH
sequence Me → H → OH is anticlockwise;
OH ≡ this is enantiomer of original
C
HO look towards CH2OH
sequence Me → H → OH is anticlockwise;
HO turn 30° HO H this is enantiomer of original

OH ≡

H3C OH
D

76 STEREOCHEMISTRY

Optical activity is the ability of a compound to could be used to determine the proportions of
rotate the plane of polarized light. This property each (see Box 3.6). Note that it is not possible to
arises from an interaction of the electromagnetic predict the sign or magnitude of the optical activity
radiation of polarized light with the unsymmetric for a particular enantiomer; it must be measured
electric fields generated by the electrons in a chiral experimentally. The presence of more than one chiral
molecule. The rotation observed will clearly depend centre in a molecule results in an optical rotation that
on the number of molecules exerting their effect, i.e. reflects a contribution from each centre, though this
it depends upon the concentration. Observed rotations is unlikely to be a simple summation. It must also
are thus converted into specific rotations that are be appreciated that a positive contribution from one
a characteristic of the compound according to the centre may be reduced, countered, or cancelled out by
formula below. a negative contribution arising from another centre or
centres (see Section 3.4.5).
temperature
Box 3.6
observed rotation
Optical purity and enantiomeric excess
specific [a]Dt (solvent) = a (degrees)
rotation lc concentration A racemic mixture contains equal amounts of the
two enantiomeric forms of the compound and has an
(g ml−1) optical rotation of zero: the optical rotations arising
from each of the two types of molecule are cancelled
wavelength of length of sample tube out. It follows that a mixture of enantiomers in
unequal proportions will have a rotation that is
monochromatic light (decimetres) numerically less than that of an enantiomer. Here,
we see how to use the measured optical activity
D = Na 'D' line 589 nm to determine the proportions of each enantiomer in
solvent used must be quoted: the mixture, and therefore its optical purity. Optical
purity is a measure of the excess of one enantiomer
rotation is solvent dependent over the other in a sample of a compound.

The observed rotation in degrees is divided by the There are a number of occasions when optical
sample concentration (g ml−1) and the sample tube purity is of interest. We shall see later that many
length (decimetres). The unusual units used transform drugs are chiral compounds, and that biological activ-
the measured small rotations into more manageable ity often resides in just one enantiomer (see Box 3.7).
numbers. The specific rotation is then usually in To minimize potential side effects, it is desirable to
the range 0–1000◦; the degree units are strictly supply the drug in a single enantiomeric form. This
incorrect, but are used for convenience. The polarized might be achieved by devising a synthetic procedure
light must be monochromatic, and for convenience that produces a single enantiomer, an enantiospecific
and consistency the D line (589 nm) in the sodium synthesis. However, syntheses that are enantiospe-
spectrum is routinely employed. Both the temperature cific can be difficult to achieve, and it is more likely
and solvent may influence the rotation somewhat, so that the procedure is only enantioselective, i.e. it
must be stated. produces both enantiomers but with one predomi-
nating. Alternatively, it is possible to separate the
Enantiomers have equal and opposite rotations. racemic mixture into the two enantiomers (resolu-
The (+)- or dextrorotatory enantiomer is the one tion; see Section 3.4.8). This might not be achieved
that rotates the plane of polarization clockwise (as in a single step. In both cases, it is usually necessary
determined when facing the beam), and the (−)- or to monitor just how much of the desired enantiomer
laevorotatory enantiomer is the one that rotates the is present in the product mixture.
plane anticlockwise. In older publications, d and l
were used as abbreviations for dextrorotatory and To illustrate the calculation of optical purity,
laevorotatory respectively, but these are not now we shall consider another type of reaction of
employed, thus avoiding any possible confusion with interest, racemization. This is the conversion of
D and L (see Section 3.4.10).

An equimolar mixture of enantiomers is optically
inactive, since the individual effects from the two
types of molecule are cancelled out. This mixture is
called a racemic mixture or racemate, and can be
referred to as the (±)-form. A mixture of enantiomers
in unequal proportions has a rotation numerically
less than that of either enantiomer; this measurement

CONFIGURATIONAL ISOMERS 77

a single enantiomer into a racemic mixture of the the sample is 56.2% racemic, and contains 71.9%
two enantiomers. It depends upon the chemical laevorotatory enantiomer and 28.1% dextrorotatory
nature of the compound whether this is easily enantiomer. These figures are derived as follows:
achievable (see Sections 10.1.2 and 10.8). One
compound that racemizes readily is hyoscyamine, the optical purity(%)
a natural alkaloid found in deadly nightshade,
which is used as an anticholinergic drug (see = specific rotation of sample × 100
Box 3.7). The natural compound is laevorotatory, specific rotation of pure enantiomer
[α]2D0 − 21◦ (EtOH), and the enantiomer is almost
devoid of biological activity. = −9.2◦/ − 21◦ × 100 = 43.8%

Upon heating with dilute base such as 1% The sample thus contains 43.8% of laevorotatory
NaOH for about an hour, hyoscyamine racem- enantiomer and 100 − 43.8% = 56.2% of racemate,
izes, and the solution becomes optically inactive the latter contributing no overall optical activity. The
(see Box 10.9). At shorter times, racemization is racemate contains equal amounts of laevorotatory
incomplete and the solution will still be optically and dextrorotatory enantiomers, i.e. it contributes
active. Consider first a very simple situation in 28.1% of each isomer to the overall mixture. There-
which exactly half of the material has racemized. fore, we have 43.8 + 28.1 = 71.9% of laevorotatory
Half of the material is now optically inactive, enantiomer, and 28.1% of dextrorotatory enantiomer
consisting of equal amounts of each enantiomer, in the partially racemized mixture.
whilst the other half is still unchanged. Since
the concentration of the unchanged part is half Many workers use the equivalent term percentage
of the original concentration, the optical rotation enantiomeric excess rather than optical purity:
will also have dropped to half its original value.
The solution will contain 50% laevorotatory iso- % Enantiomeric excess
mer and 50% racemate. However, the racemate is moles of one enantiomer−
itself a 50 : 50 mixture of the two enantiomers, so
the solution actually contains 25% dextrorotatory = moles of other enantiomer × 100
and 25 + 50% = 75% laevorotatory enantiomers. total moles of both enantiomers

Now let us consider when measurements indi- but this is exactly equivalent to optical purity.
cate [α]D20 − 9.2◦. Calculations now tell us that From the above calculations, one can see that the
laevorotatory enantiomer (71.9%) is in excess of the
dextrorotatory enantiomer (28.1%) by 43.8%.

The physical properties of enantiomers and race- usually the same. The melting points of (+)- and (−)-
mates, except for optical rotation and melting points, are enantiomers are the same, though that of the racemate is

Box 3.7
Pharmacological properties of enantiomers

Although most physical properties of enantiomers are identical, pharmacological properties may be different.
There are examples of compounds where:

• only one enantiomer is active;
• both enantiomers show essentially identical activities;
• both enantiomers have similar activity, but one enantiomer is more active;
• enantiomers show different pharmacological activities.

These observations may reflect the proximity of the chiral centre to the part of the molecule that binds with the
receptor site.

78 STEREOCHEMISTRY

Box 3.7 (continued) chiral receptor for chiral
centre part of molecule
AA
X X
Y Y

Z Z

receptor for achiral
part of molecule

If binding to the receptor involves the chiral centre, then we may see activity in only one enantiomer, but if binding
does not involves the chiral centre, then there may be similar activities for each enantiomer. Binding close to
the chiral centre may cause the same type of activity but of a different magnitude. A different pharmacological
activity for each enantiomer almost certainly reflects different receptors.

Further, drug absorption, distribution, and elimination from the body may vary due to differences in protein
binding, enzymic modification, etc, since proteins are also chiral entities (see Chapter 13).

Thus, the anticholinergic activity of the alkaloid hyoscyamine is almost entirely confined to the (−)-isomer, and
the (+)-isomer is almost devoid of activity. The racemic (±)-form, atropine, has approximately half the activity
of the laevorotatory enantiomer. An anticholinergic drug blocks the action of the neurotransmitter acetylcholine,
and thus occupies the same binding site as acetylcholine. The major interaction with the receptor involves that
part of the molecule that mimics acetylcholine, namely the appropriately positioned ester and amine groups. The
chiral centre is adjacent to the ester, and also influences binding to the receptor.

the descriptors R, S, and RS
are defined in Section 3.4.2

Me N Me N Me N

H CH2OH H CH2OH CH2OH
O O
O
S R RS
O
O O

(–)-hyoscyamine (+)-hyoscyamine (±)-hyoscyamine (atropine)

relative

anticholinergic 100 0 50

activity (%)

The major constituent of caraway oil is (+)-carvone, and the typical caraway odour is mainly due to this
component. On the other hand, the typical minty smell of spearmint oil is due to its major component, (−)-
carvone. These enantiomers are unusual in having quite different smells, i.e. they interact with nasal receptors
quite differently. The two enantiomeric forms are shown here in their half-chair conformations.

H O
OO

O

(+)-carvone (caraway) half-chair conformation; (–)-carvone (spearmint) H
isopropenyl group
nearly equatorial half-chair conformation;
isopropenyl group
nearly equatorial

CONFIGURATIONAL ISOMERS 79

One of the most notorious and devastating examples of a drug’s side effects occurred in the early 1960s, when
thalidomide was responsible for many thousands of deformities in new-born children. Thalidomide was marketed
in racemic form as a sedative and antidepressant, and was prescribed to pregnant women. Although one enantiomer,
the (R)-form, has useful antidepressant activity, it was not realized at that time that the (S)-form, thought to be
inactive, actually has mutagenic activity and causes defects in the unborn fetus. Furthermore, the (S)-isomer also
has antiabortive activity, facilitating retention of the damaged fetus in the womb, so that any natural tendency to
abort a damaged fetus was suppressed.

O O O O O
H NH HN H H

N N ≡ NO

OO OO NH
OO
(R)-thalidomide
(S)-thalidomide
useful antidepressant activity
mutagenic activity
antiabortive activity − retained damaged fetus

It is now general policy in the pharmaceutical industry to release new drugs as optically pure isomers, rather
than as racemates. It is desirable to minimize the amount of foreign chemical a patient is subjected to, since even
the inactive portion of a drug has to be metabolized and removed from the body. Such tragedies as occurred
with thalidomide may also be avoided. Where a drug is supplied as a single enantiomer, the optical isomer is
often incorporated into the drug name, e.g. dexamfetamine, dexamethasone, levodopa, levomenthol, levothyroxine.
Nevertheless, many racemic compounds are currently used as drugs, including atropine, mentioned above, and
the analgesic ibuprofen.

ibuprofen H
CO2H

(R)-(−)-isomer inactive

metabolic
conversion

H
CO2H

(S)-(+)-isomer active

Ibuprofen is an interesting case, in that the (S)-(+)-form is an active analgesic, but the (R)-(−)-enantiomer
is inactive. However, in the body there is some metabolic conversion of the inactive (R)-isomer into the active
(S)-isomer, so that the potential activity from the racemate is considerably more than 50%. Box 10.11 shows a
mechanism to account for this isomerism.

There are two approaches to producing drugs as a single enantiomer. If a synthetic route produces a racemic
mixture, then it is possible to separate the two enantiomers by a process known as resolution (see Section 3.4.8).
This is often a tedious process and, of course, half of the product is then not required. The alternative approach,
and the one now favoured, is to design a synthesis that produces only the required enantiomer, i.e. a chiral
synthesis.

Note, the descriptors R and S for enantiomers and RS for racemates are defined in Section 3.4.2.

80 STEREOCHEMISTRY

usually different and can be greater or less than document in order to encompass all possibilities. Here is
the melting point of the enantiomers. Most spectral a very short version suitable for our requirements. Note
properties, e.g. NMR, mass spectrometry, etc., of (+)-, that it applies to both acyclic and cyclic compounds.
(−)-, and (±)-forms are indistinguishable. However,
pharmacological properties are frequently different, • Higher atomic number precedes lower,
because they may depend upon the overall shape of the e.g. Br > Cl > S > O > N > C > H.
compound and its interaction with a receptor.
• For isotopes, higher atomic mass precedes lower,
3.4.2 Cahn–Ingold–Prelog system to e.g. T > D > H.
describe configuration at chiral centres
• If atoms have the same priority, then secondary
The arrangement of groups around a chiral atom is groups attached are considered. If necessary, the
called its configuration, and enantiomers have different process is continued to the next atom in the chain.
configurations. Therefore, it is necessary for us to have
a means of describing configuration so that we are in e.g. CH2 CH3 > CH2 H
no doubt about which enantiomer we are talking about.
Although enantiomers have equal and opposite optical first atom is carbon in both cases;
rotations, the sign of the optical rotation does not tell us consider the second atom:
anything about the configuration. The system adopted carbon as second atom has higher
by IUPAC for describing configuration was devised by priority than hydrogen
Cahn, Ingold, and Prelog, and is often referred to as
the R,S convention. CH3

The approach used is as follows:

• Assign an order of priority, 1, 2, 3, and 4, to the CH2 CH > CH2 CH2 CH2 CH3
substituents on the chiral centre.
CH3

• View the molecule through the chiral centre towards first atom is carbon in both cases;
the group of lowest priority, i.e. priority 4. consider the second atom:
second atom is carbon in both cases;
• Now consider the remaining groups in order of consider the next atom(s):
decreasing priority. If the sense of decreasing priority carbon directly bonded to two further
1 → 2 → 3 gives a clockwise sequence, then the carbons has higher priority than carbon
configuration is described as R (Latin: rectus = directly bonded to just one further carbon
right); if the sequence is anticlockwise, then the
configuration is described as S (Latin: sinister = left). • Double and triple bonds are treated by assuming each
atom is duplicated or triplicated.

numbers indicate e.g. C C is considered to be equivalent to C C
assigned priorities CC

1 1

4 view 2 C O is considered to be equivalent to C O
2 3 OC

3 clockwise: R

1 1 CC is considered to be equivalent to CC
CC
4 view 3 CC
3 2

2 anticlockwise: S

The remaining part of the procedure is to assign the As simple examples of the approach, let us consider
priorities. The IUPAC priority rules form a rather long the amino acid (−)-serine and the Krebs cycle interme-
diate (+)-malic acid.

CONFIGURATIONAL ISOMERS 81

NH2 OH

H CH2OH H CO2H
CO2H CH2CO2H

(−)-serine (+)-malic acid

priorities NH2 > CO2H > CH2OH > H priorities OH > CO2H > CH2CO2H > H

highest O H lowest atomic highest O H lowest atomic
atomic CO CO number atomic CO CC number
number number O H
OH

same atomic number; same atomic number;
next atom then considered, next atom then considerd O > C;
both have oxygen; carbonyl double bond considered
carbonyl double bond to be duplicated
considered to be duplicated;
three oxygens > one oxygen

priority 1 view from this priority 1 priority 1 view from priority 1
side this side OH

NH2 NH2 OH priority 2

H CH2OH HO2C CH2OH H CO2H HO2CH2C CO2H
priority 4 CO2H priority 3 priority 2 priority 3 priority 4 CH2CO2H priority 3 priority 2
priority 2
priority 3

anticlockwise: S clockwise: R

nomenclature showing

optical activity and

configuration: (−)-(S)-serine (+)-(R)-malic acid

It is now possible to incorporate the configuration descriptor (RS ) is used to indicate a (±) racemic mixture
of the compound into its nomenclature to give more (see Section 3.4.1).
detail. (−)-Serine becomes (−)-(S)-serine, whilst (+)-
malic acid becomes (+)-(R)-malic acid. Because there Note also that the configuration (R) or (S) is defined
is no relationship between (+)/(−) and configuration by the priority rules, and configuration (R) could easily
(R)/(S), it is necessary to quote both optical activity become (S) merely by altering one substituent. For
and configuration to convey maximum information. The instance, all the amino acids found in proteins can be
represented by the formula

R CO2H
α-amino acids in proteins

H NH2

priority 3 priority 2 HO CO2H R = CH2OH (−)-(S)-serine
R CO2H H NH2 R = CH3 (+)-(S)-alanine
R = (CH2)4NH2 (+)-(S)-lysine
NH2 CO2H R = CH2Ph (−)-(S)-phenylalanine
priority 1 H NH2

anticlockwise: S H2N CO2H

priorities: H NH2
NH2 > CO2H > alkyl > H CO2H

H NH2

82 STEREOCHEMISTRY

priority 2 priority 3 HS CO2H R = CH2SH (+)-(R)-cysteine
HSH2C CO2H H NH2

NH2
priority 1

clockwise: R

priorities: the case of cysteine, R = CH2SH, and since S has a
higher atomic number than any of the other atoms under
NH2 > CH2SH > CO2H> H consideration, this group will have a higher priority than
the carboxyl. The net result is that cysteine is (R)-
Now all these amino acids that are chiral (glycine, cysteine.
R = H is achiral) have the (S) configuration except for
cysteine, which is (R). Just looking at the structures, Configurations in cyclic compounds are considered
one might imagine that they would all have the same in the same way as for acyclic compounds. If you cannot
configuration, and indeed one can consider that they get an answer with the first atom, move on to the
have; they differ only in the nature of the R group, but next, even though this may mean working around the
are all arranged around the chiral centre in the same ring system. Consider, for example, the stereoisomer of
manner. But since (R) and (S) are only descriptors of 3-methylcyclohexanol.
configuration, the designation depends upon the nature
of the R group. In most cases, R is an alkyl or substituted
alkyl, so it has a lower priority than the carboxyl. In

anticlockwise: 1S clockwise: 3R

3H H priority 2 priority 3
3 H priority 4
12 priority 3 1 priority 2

H OH

(1S,2R)-3-methylcyclohexanol H OH priority 1
priority 4 priority 1 H OH

priorities: priorities:

OH > CH2CH(CH3)CH2 > CH2CH2CH2 > H CH2CH(OH)CH2 > CH2CH2CH2 > CH3 > H

This has two chiral centres, C-1 and C-3. It can Box 3.8
readily be deduced that this isomer is actually (1S,2R)-
3-methylcyclohexanol. Configurations in 6-aminopenicillanic acid

At both centres, two of the groups under consideration Let us look at the common substructure of the
for priority assignment are part of the ring system. penicillin antibiotics, namely 6-aminopenicillanic
These are only differentiable when one comes acid, to illustrate some aspects of working out whether
to the ring substituent, the methyl group when a chiral centre is allocated the R or S configuration.
one considers C-1 and the hydroxyl when one
considers C-3. In each case, the substituted arm First of all, there are three chiral centres in this
is going to take precedence over the unsubstituted molecule, carbons 3, 5 and 6; note that carbon 2
arm. A more interesting example (6-aminopenicillanic is not chiral, since two of the groups attached are
acid) containing heterocyclic rings is discussed methyls. Only the three carbons indicated have four
in Box 3.8. different groups attached.

CONFIGURATIONAL ISOMERS 83

* chiral centre Lastly, suppose one is asked to draw a particular
configuration at C-6, namely 6R. There is no way
H2N H one can visualize a particular configuration, so the
approach is to draw one and see if it is correct; if it is
* * S not correct, then change it by reversing wedged/dotted
bonds. And which to try first? Well, always put
6 5 1 the group of lowest priority, usually H, away
7 2 from you, i.e. dotted/down. Then you can see the
N4 clockwise/anticlockwise relationship easily from the
3 front. In this case, the version with H down gave the
6R configuration; but, if it were to be wrong, then the
O* alternative configuration at this centre would be the
required one, i.e. a wedged bond to the hydrogen.
CO2H

6-aminopenicillanic acid

priority 4

priority 3 H S priority 1
C

5 clockwise: 5R

N to draw (6R)-configuration:
priority 2
first try this:
priority 1 S C priority 2 viewed from the S priority 4
N 3 CC front clockwise H2N H H CO2H H priority 2

HCO therefore, if viewed 6 H
priority 4 O O priority 3 from rear, must be N CS
anticlockwise: 3S N priority 1 6 N
O
C
ON

O priority 3

clockwise: R

The chirality at C-5 is assigned in the usual way. it is always easier to see clockwise /
The groups attached have easily assigned priorities, anticlockwise if the group of lowest
with S > N > C > H. The configuration is thus priority is at the rear (dotted)
5R. For the chirality at position 3, the priorities
are assigned N > C–S > C–O > H. Now a it turns out to be R;
very useful hint. Since the group of lowest priority if it were incorrect, then the
is wedged/up, it is rather difficult to imagine the required isomer would be:
sequence when viewed from the rear. Accordingly,
view the sequence from the front, which is easy, and H2N H H H2N H
reverse it. From the front, the sequence for C-3 looks O
clockwise, so if viewed from the rear, it must be 6 S S
anticlockwise, and the descriptor is 3S. Note how CO2H 6 51
we consider substituents in the standard way even N 2
if they are part of a ring system. If you cannot get O 7 N4 3
an answer with the first atom, move on to the next
around the ring system. CO2H

6S configuration (3S,5R,6R)-6-aminopenicillanic
acid

3.4.3 Geometric isomers With a double bond, rotation would destroy the π
bond that arises from overlap of p orbitals; conse-
Restricted rotation about double bonds or due to the quently, there is a very large barrier to rotation. It
presence of ring systems leads to configurational isomers is of the order of 263 kJ mol−1, which is very much
termed geometric isomers. Thus, we recognize two higher than any of the barriers to rotation about single
isomers of but-2-ene, as shown below, and we term these bonds that we have seen for conformational isomerism.
cis and trans isomers. We have met these terms earlier Accordingly, cis and trans isomers do not intercon-
(see Section 3.3.2). vert under normal conditions. Ring systems can also
lead to geometric isomerism, and cis and trans isomers

84 STEREOCHEMISTRY

of cyclopropane-1,2-dicarboxylic acid similarly do not priority 1 priority 1 priority 1 priority 2
interconvert; interconversion would require the breaking
of bonds.

but-2-ene priority 2 priority 2 priority 2 priority 1
H3C E
CH3 H3C H Z

HH H CH3 priority 1 H3C CH3 priority 2
cis trans priority 2 priority 1
2 3
cyclopropane-1,2-
dicarboxylic acid H CH2CH3

HO2C CO2H HO2C H (E)-3-methylpent-2-ene

HH H CO2H priority 1 H3C CH2CH3 priority 1
cis trans priority 2 priority 2
2 3
The terms cis and trans are used to describe the
configuration, which is considered to be the spatial H CH3
sequence about the double bond or the spatial sequence
relative to a ring system. The cis isomer has substituents (Z)-3-methylpent-2-ene
on the same side of the double bond or ring system
(Latin: cis = on this side), whereas the trans isomer has higher priority. Thus, the high-priority groups are on
substituents on opposite sides (Latin: trans = across). opposite sides of the double bond, and this isomer has
the E configuration. The alternative arrangement with
With simple compounds, like the isomers of but-2- high-priority substituents on the same side of the double
ene, the descriptors cis and trans are quite satisfactory, bond has the Z configuration.
but a compound such as 3-methylpent-2-ene causes
problems. Do we call the isomer below cis because the Box 3.9
methyls are on the same side, or trans because the main
chain goes across the bond? Configurations of tamoxifen, clomifene and
triprolidine
3-methylpent-2-ene H3C CH3
The oestrogen-receptor antagonist tamoxifen is used
H CH2CH3 in the treatment of breast cancer, and is highly
is this cis or trans? successful. Clomifene is also an oestrogen-receptor
antagonist, but is principally used as a fertility drug,
For double bonds, the configuration is now usually interfering with feedback mechanisms and leading
described via the non-ambiguous E,Z nomenclature, to ova release, though this often leads to multiple
assigned using the Cahn–Ingold–Prelog priority rules pregnancies.
for substituents on each carbon. First, consider each
carbon of the double bond separately, and assign priority 2 priority 1
priorities to its two substituents. Then consider the O
double bond with its four substituents. If the two NMe2
substituents of higher priority are on the same side
of the double bond, the configuration is Z (German: configuration (Z)
zusammen = together), whereas if they are on opposite
sides, the configuration is E (German: entgegen = priority 2
across).
priority 1
Thus, for the 3-methylpent-2-ene isomer we can see tamoxifen
that, for C-2, the substituents are methyl and hydrogen
with priorities methyl > hydrogen. For C-3, we have
substituents methyl and ethyl, with ethyl having the

CONFIGURATIONAL ISOMERS 85

priority 2 priority 1 Starting with two chiral centres, there should, there-
O fore, be four stereoisomers, and this is nicely exem-
NMe2 plified by the natural alkaloid (−)-ephedrine, which is
employed as a bronchodilator drug and decongestant.
configuration (Z) Ephedrine is (1R,2S)-2-methylamino-1-phenylpropan-
Cl 1-ol, so has the structure and stereochemistry shown.

priority 1 mirror

HO CH3 H3C OH
H H
priority 2
clomifene Ph 1 2 HNHCH3 H3CHNH 2 1

Ph

As can be deduced from application of the 1R,2S 1S,2R
Cahn–Ingold–Prelog priority rules, high-priority (–)-ephedrine (+)-ephedrine
groups are positioned on the same side of the double
bond in each case. Note that the substituted aromatic H CH3 H3C H
ring has higher priority than the unsubstituted ring. HO OH
Both tamoxifen and clomifene thus have the Z con-
figuration. Ph 1 2 HNHCH3 H3CHNH 2 1

Ph

1S,2S 1R,2R
(+)-pseudoephedrine (–)-pseudoephedrine

priority 1 priority 2 1R,2S 1S,2R
N (–)-ephedrine (+)-ephedrine
H enantiomers
N
configuration (E)

diastereoisomers diastereoisomers diastereoisomers

Me priority 1
priority 2

1S,2S enantiomers 1R,2R

triprolidine (+)-pseudoephedrine (–)-pseudoephedrine

The antihistamine drug triprolidine has the E Now the other three of the possible four stereoisomers
configuration; note that the heterocyclic pyridine ring are the (1S,2S), (1R,2R), and (1S,2R) versions. These
takes priority over the benzene ring, even though are also shown, and mirror image relationships are
the latter has a substituent. Priority is deduced by emphasized. The (1S,2R) isomer is the mirror image
working along the carbon chain towards the first atom of (−)-ephedrine, which has the (1R,2S) configuration.
that provides a decision, in this case the nitrogen Therefore, it is the enantiomer of (−)-ephedrine, and can
atom in the pyridine. be designated (+)-ephedrine. Note that the enantiomeric
form has the opposite configuration at both chiral
3.4.4 Configurational isomers with several centres.
chiral centres
The other two isomers are the (1S,2S) and (1R,2R)
Configurational isomerism involving one chiral centre isomers, and these two also share a mirror image rela-
provides two different structures, the two enantiomers. tionship, have the opposite configuration at both chi-
If a structure has more than one chiral centre, then there ral centres, and are, therefore, a pair of enantiomers.
exist two ways of arranging the groups around each From a structure with two chiral centres, we thus have
chiral centre. Thus, with n chiral centres in a molecule, four stereoisomers that consist of two pairs of enan-
there will be a maximum number of 2n configurational tiomers. Stereoisomers that are not enantiomers we term
isomers. Sometimes, as we shall see in Section 3.4.5, diastereoisomers, or sometimes diastereomers. Thus,
there are less. the (1S,2S) and (1R,2R) isomers are diastereoisomers
of the (1R,2S) isomer. Other enantiomeric or diastere-
omeric relationships between the various isomers are
indicated in the figure.