UNIT- 4
STEREOCHEMISTRY
4.1. INTRODUCTION
So far, we have understood the composition of the principal functional groups of organic
chemistry. We have mostly viewed molecules in a two-dimensional way and have given
little thought to any consequence that might arise from spatial arrangement of atoms in
molecules. We are ready now to think about three-dimensional structure of organic
compounds, namely Stereochemistry.
Stereochemistry is a branch of chemistry which deals with the three dimensional aspects
of molecules. It explains those properties of molecules that result from the spatial
relationships of their constituent groups. It includes methods for determining and
describing these relationships; the effect on the physical and chemical properties these
relationships impact upon the molecules in question, and the manner in which these
relationships influence the reactivity of the molecules in question.
Central to Stereochemistry is the concept of isomerism. Isomers are sets of chemical
compounds having identical atomic composition but with different structural properties.
With geometric isomers, the differences arise from the atoms being bonded in different
sequences or patterns across a double bond or in a ring. Optical isomers are pairs of
molecules that differ in the same way that a left hand and right hand differ; i.e. they are
mirror images of each other. Such molecules rotate plane polarized light that passes
through them in opposite direction. Constitutional isomers, on the other hand, differ in
the connectivity of atoms in the molecule where as conformational isomers differ by the
structures they have through rotation across a single bond.
Unit -4
Stereochemistry
4.2. Learning objectives of the unit.
At the end of this unit students will be able to:

develop the concept of molecules as three-dimensional objects;

be familiar with stereo chemical principles terms and notations;

predict the existence of different types isomers, represent and designate their
structures; and

determine the stereochemistry of organic molecules.
4.3. Constitutional and Geometric isomers
4.3.1. Introduction
Isomers are different compounds that have the same molecular formula but have
different structures. They have the same numbers and kinds of atoms but differ in the
way the atoms are arranged. Isomers are sub-divided into two groups as constitutional
isomers and stereoisomer. Constitutional isomers are those isomers whose atoms have a
different connectivity. Stereoisomers are those isomers that have the same connectivity
but that differ in the arrangement of their atoms in space. Some of the common
stereoismoers are geometric isomers, diastereomers and enatntiomers.
4.3.2. Learning objectives of the topic
At the end of this topic students will be able to:

write constitutional isomers of common organic compounds;

identify the two types of geometric isomers; and

show the difference between constitutional and geometric isomers.
Unit -4
Stereochemistry
4.3.3. Constitutional isomers
Compounds that have the same molecular formula but different structural formulas due
to different orders of attachment of a toms are called constitutional isomers . They are
said to have a different connectivity. Constitutional isomers occurs widely in different
organic compounds. The common types of constitutional isomers are skeletal, functional
and position isomers. Skeletal isomers have different carbon skeletons . For example, a
compound with molecular formula C4H8 has two Skeletal constitutional isomers as shown
below.
CH3CH2CH2CH3
Butane
and
CH3CHCH3
CH3
Isobutame
In the same way, a compound with molecular formula C5H12 has the following three
skeletal constitutional isomers.
CH3CH2CH2CH2CH3 , CH3CHCH2CH3,
CH3
Isopentane
Pentane
(Straight chain)
and
(one branched chain)
CH3
CH3CCH3
CH3
Neopentane
(two branched chains)
Functional isomers have different functional groups which arise from different
connectivities of atom in the molecule, For example, a compound with molecular formula
C2H5O have the following two functional isomers (alcohol and ether).
CH3CH2OH
Ethyl alcohol
and
CH3OCH3
Dimethyl ether
Unit -4
Stereochemistry
In the same way, a compound with molecular formula C3H6O has the following two
functional isomers ( aldehyde and ketone).
O
O
CH3CH2CH
and
Propanal
CH3CCH3
propanone (acetone).
Position isomers have different positions of the functional group. They have the same
type of functional group at different positions in the molecule. For example, a compound
with molecular formula C3H7NH2 or C3H9N (an amine ) has the following two position
isomers which differ only by the position of the functional group (NH2).
CH3CH2CH2NH2
and
CH3 CH CH3
NH2
Isopropylamine
Propylamine
In the same way a compound with molecular formula of C4 H10O (an alcohol ) has the
following two position isomers which differ only by the position of the functional group (OH).
CH3CH2CH2CH2OH
Butanol
and
CH3 CH CH2CH3
OH
2-Butanol
Unit -4
Stereochemistry
Problem 4-1:-Write all possible constitutional isomers for the following
compounds
1. Alkanes with formula C6H14
2. Alcohols with formula C5H12O
3. Ketones with formula C5H10O
4. Amines with formula
C5H13N
5. A compound with general formula C6 H12O
4.3.4. Geometric isomers (cis - trans isomers)
Geometric isomers or cis-trans isomers are isomers in which substituents are on the
same side or opposite side of double bond or ring. This is the property of disubstituted
alkenes and cycloalkanes. (Disubstituted means that two substituents other than
hydrogen are bonded to the double bond carbons).
When substituents are on the same side of the double bond or the ring, the isomers are
cis- and when substituents are on opposite side, the isomers are trans far example in 2Burene, the two methyl groups can be either on the same side of the double bond or on
apposite side forming the cis- and trans- isomers respectively as shown below.
H3C
CH3
C
H
H3C
H
C
C
H
Cis - 2-Butene
H
C
CH3
trans-2-Butene
Unit -4
Stereochemistry
In the same way 1,2-Dichloroethane has two geometric isomers, the cis-1,2dichloroethane where the two chlorine atoms are on the same side of the double bond
and the trans -1,2- Dichloroethane where the two chlorine atoms are on opposite side of
the double bond.
Cl
Cl
C
H
Cl
H
C
C
H
Cl
H
Cis -1, 2- Dischloroethane
C
trans-1, 2- Dichloro ethane
In cycloalkanes, there will be cis- trans isomers when they have two-substituents on
different carbons. When these substituents are on the same side of the ring, we have the
trans isomer, for example, 1,2- dimethyl cyclopropane has the following isomers .
CH3
H
CH3
H
cis- 1,2- Dimethylcyclopropane
(the two methyl groups are
on the same side of the ring)
CH3
H
CH3
H
trans- 1,2 - Dumethylcyclopropane
( the two methyl groups are on
opposite side of the ring)
In the same way, 1,3- dimethylcyclopentane have two geometric isomers,
cis - 1,3-
dimethylcyclopentane in which the two methyl groups are on the same side of the
cyclopentane ring and the trans 1,3 -dimethylcyclopentane in which to two methyl groups
are on opposite side of the cyclopentane ring.
Unit -4
Stereochemistry
CH3
H
CH3
H
Cis-1,2- Dimethylcyclopentane
CH3
H
H
CH3
trans-1,2 - Dimethyliyclo pentane
The cis- trans nomenclature for describing the geometry of the double bonds works well
for di substituted double bonds. It fails to work with tri substituted and tetra substituted
double bonds. (Tri substituted means three substituents other than hydrogen are
attached to the double-bond carbon atoms; tetra substituted means four substituents
other than hydrogen are attached to the double bond carbon atoms).
A more general method of describing double-bond stereochemistry is provided by the E,
Z system of nomenclature, which uses a series of sequence rules which is based on an
atomic number criterion to assign priorities to the substituent groups on the double- bond
carbons. The sequence rules are used to decide which of the two groups attached to each
carbon is higher in priority (higher atomic number).The sequence rule which is called the
Cahn-Ingold- Prelog priouty rules are given in table 4.1. If the higher- priority groups on
each carbon are on the same side of the double bond, we say that the double bond has
the Z configuration. Z stands for Zusammen, which means "together". If the higherpriority groups are on the opposite sides of the double bond, we say that
the
configuration is E. Here the symbol E stands for the German word entgegen, which
means "opposite". The assignment is shown below.
Unit -4
Stereochemistry
High
High
C
High
C
Low
C
Low
Low
C
low
High
E - Configuration
Z - Configuration
Example:-.
higher
Cl
C
Cl
C
C
H
H3C
lower
higher
higher
Br
lower
H3C
lower
F
lower
Br
higher
C
Higher-ranked substituents
(Cl and Br) are on opposite
side of double bond.
(E)-1-Bromo-2-Chloro-1-fluoropropene.
Higher-ranked substituents
(Cl and Br) are on the same
side of double bond.
(Z)-1-Bromo-2-Chloropropene
Look at the following example on how E & Z- configuration are assigned
to a double
bond.
Example- 1. Assign E or Z configuration to this compound.
H3C
CH2OH
C
C
CH3
H
Methyl and hydrogen occur as a substituents on the first carbon of the double bond.
According to the rule, methyl has higher priority than hydrogen. The other double bond
carbon bears methyl and CH2OH group. The -CH2OH group is of higher priority than
methyl. Since higher priority substituents
(-CH3 and - CH2OH) are on the same side
of the double bond, the configuration is Z.
higher
H3C
CH2OH
C
Lower
H
higher
C
CH3
Lower
(Z) -1- Hydroxyl-2-methyl-2-butene
Unit -4
Stereochemistry
Example -2- Assign E or Z Configuration to this compound.
H3C
C
CH2CH2CH3
C
CH2CH3
FH2C
On the first carbon -CH3 and -CH2F substituents occur and -CH2F witl be given higher
priority than -CH3. On the 2nd carbon -CH2CH2CH3 and -CH2CH3 group occur. According
to the rule, -CH2CH2CH3 group has the higher priority. Then, the two higher priority
groups (-CH2F and -CH2CH2CH3) are on opposite sides of the double bond, so the
configuration is E.
Lower
higher
H3C
FH2C
C
C
CH2CH2CH3
CH2CH3
higher
Lower
(E) -3- Ethyl -1- fluoro -2- methyl-2- hexene.
Unit -4
Stereochemistry
Table 4-1:- Cahn- Ingold - Prelog Priority Rules.
Rule
Example
1. Higher atomic number takes
higher Br
higher priority
C
CH3 higher
C
H
Lower Cl
Lower
Br(35) receives higher priority than Cl(17),
C(6) of CH3 takes higher priority than H(1).
It has Z-configuration (Br and CH3 on the
same side)
2. When two atoms directly attached to the
double bond are identical, compare the
atoms attached to these two on the basis
of their atomic numbers. Priority is
determined at the first point of difference.
Priority of some groups
C
Multiple-bonded atoms are equivalent
to the same number of single-bonded
atoms.
CH2 is equivalent to
CH
O
O
CH
is equivalent to
CH
O
H
C
C
C
C
C
CH2CH3 higher
Lower Cl
- CH2CH3 have higher priority than –CH3
- It has E configuration (Br and CH2CH3
higher
Cl
CH2OH
C
Lower
H
Lower
CH3
are on opposite sides)
- C(CH3)3 > -CH(CH3)2 > -CH2CH3
3.
higher Br
Lower
C
CH=O
F
higher
has E- configuration
H
higher
HO
CH=CH2
C
Lower
H2N
higher
C
CH(CH3)2
has Z- configuration
Lower
Unit -4
Stereochemistry
Problem 4-2
Assign cis-trans or E-Z- configuration for the following compounds. Write the name of
each compound.
1.
HO
OH
C
H3C
CH2CH3
C
C
CH2CH2OH
Cl
H3CH2C
5. O2N
CN
C
CH2NH2
H3C
C
COOH
C
C
CH2OH
CH3
C
CH2CH3
H3C
H3C
4.
C
C
CH2CH(CH3)2
CH2CH3
(H3C)2HC
H
6.
C
H3C
7. H2C=HC
H3CH2C
C
H3CH2C
3.
2.
C
C
CH3
H3CH2C
8.
H3C
H3CH2C
C
C
CH3
CH2CH3
Unit -4
Stereochemistry
4.4. Optical isomers.
4.4.1 Introduction.
Three dimensional objects such as a sphere, a cube a cone tetrahedron cups, chairs, beds
etc, are all identical with, and can be superimposed on their mirror images and are said
to be symmetric. On the other hand, many other objects cannot be superimposed on
their minor images and are said to be dissymmetric. For example, your left hand and your
right hand are mirror images of each other but cannot be made to superimpose in three
dimensions. In chemistry, compounds like three dimensional objects are symmetric or
dissymmetric which are categorized as achiral and chiral molecules respectively. This
difference gives rise to different isomers with different properties. These isomers are
identified by the spatial arrangement of atoms of the molecule in space.
4.4.2. Learning objectives of the topic
At the end of this topic students will be able to :

identify chiral and achiral molecules ;

define enantiomers and diastercomers; and

assign configuration of chiral molecules.
Unit -4
Stereochemistry
4.4.3. Chirality
A)
Enatiomers
Molecules that are not super imposable on their mrror images are said to be chiral. That
is they show handedness. Your left and right hands are Chiral. On the other hand, if the
molecule is super imposable on its mirror image, it is said to be achiral. The molecule
and its mirror image have the same constitution. That is, the atoms are connected to
each other in the same order. But the molecules differ in the arrangement of their atoms
in space, they are stereoisomers. Stereoisomers that are related as an object and its non
superimposable mirror image are classified as enantiomers. That is, enantiomers are the
chiral molecule and its nonsuperimposable mirror image. The word enantiomer describes
a particular relationship between two objects (the chiral molecule and its mirror image),
The chirality of an organic molecule can be illustrated using a relatively simple compound,
2- butanol.
CH3 CH CH2 CH3
OH
If we draw the three dimensional representation of2- butanol (moldel- I) and keep it
infront of a mirror, model- II is seen in the mirror. Models I and II are not superimposable
on each other (this can be demonstrated using a model), therefore, they represent
different, but isomeric molecules. Because models I and II are non super imposable mirror
images of each other, the molecules they represent are enantiomers.
Unit -4
Stereochemistry
HO
C
CH2
CH3
H
------------------------------------
CH3
H3C
H
OH
CH3
H2C
H3C
mirror
I
II
Model - III and model - IV are enantromers of lactic acid, CH3CHCOOH
OH
COOH
HO
C
CH3
III
H
HOOC
H
C
OH
H3 C
IV
We can predict whether a given molecule is chiral or achiral by examining its structures
. If a molecule contains a plane of symmetry it is achiral and if it doesn’t, it is chiral. A
plane of symmetry is an imaginary plane passing through the molecule which divides the
molecule in such away that one half of the molecule is an exact mirror image of the other
half. Thus, propanoic acid has a plane of symmetry when it is lined up as shown below
and is
chiral.
therefore, achiral. Lactic acid, however, has no plane of symmetry and is thus
Unit -4
Stereochemistry
No plane of symmetry
Plane of symmetry
------------
------------
CH3
CH3
H
C
H
H
C
OH
COOH
COOH
------------
------------
CH3CHCOOH
CH3CH2COOH
OH
P ropanoic acid (achiral)
Lactic acid (Chiral).
Noting the presence of stereogenic center in a given molecule is a simple and rapid way
to determine that the molecule is chiral. For example, C–2 is a stereogenic center in 2butanol and bears a hydrogen atom, methyl, ethyl, and hudroxyl groups as its four
different substituents. By way of contrast, none of the carbon atoms bear four different
groups is the alcohol
1
CH3
H
2
C
3
2- propanol.
H
4
CH2CH3
OH
2- Butanol : Chiral
four different substituents at C-2
CH3
C
CH3
OH
2- Propanol: achiral
two of the substituents at C-2 are the same.
Here are some more examples of chloral molecules. The stereogenic carbon is indicated
by an asterisk (*). Carbons that are part of a double bond or a triple bond can not be
stereogenic centers.
Unit -4
Stereochemistry
*
CH3CHCOOH
,
*
CH3CHCH2CH3
OH
Br
*
CH3CH2CH2CH2CHCH2CH2CH3
,
*
CH3CH2CHCH2OH
,
CH3CH2*
CHCH=CH2
CH3
OH
CH3
*
(CH3)2C=CHCH2CH2CCH=CH2
OH
NO2
A carbon atom in a ring can be stereogenic center if it bears two different substituents
and the path traced around the ring from that carbon in one direction is different from
the one traced in the other. The carbon atom that bears the methyl group in 1,2– epoxy
propane, for example, is a stereogenic center. The sequence of the group is –CH2-O as
one proceeds clockwise around the ring from that atom, but is –O- CH2 in the
anticlockwise direction. On the other hand, methylcyclohexane is achiral because there
is no stereogenic center in the molecule. The carbon atom that bears the methyl group
is not stereogenic because as one proceeds clockwise and anticlockwise around the ring
from that atom the sequence of groups are the same –CH2 – groups on both sides.
H3C
*
HC
CH2
O
1-2- Epoxypropane ( Chiral)
H2
C
H2C
H2C
CH
C
H2
CH3
CH2
Methylcyclohecane(achiral).
Unit -4
Stereochemistry
Additional examples of chiral molecules are given below. Check for yourself that the
labeled centers (ones with asterisk ) are stereogenic,
CH3
O
O
CH3
CH3
*
*
C
*
OH
CH2
OH
CH3
CH3
*
CH2CH3
*
*
O
CH3CH2
As we have already seen, enantiomers are different compounds, and you must expect,
therefore, that they differ in some properties. One property that differs enantiomers is
their effect on the plane of polarized light. A plane polarized light is light vibrating in only
one plane in one direction. Each member of a pair of enantiomers
rotate the plane of
polarized light, and for this reason, enantiomers are said to be optically active. While each
enatiomer is optically active the two rotate the plane of polarized light in opposite
directions and are also known as optical isomers. This physical property of enantiomers
is called optical activity. Optical activity is the ability of a chiral substance to rotate the
plane of polarized light. A substance that does not rotate the plane of polarized light is
said to be optically inactive, that is, have no optical acivity. All achiral molecules are
optically inactive.
Unit -4
Stereochemistry
The ability of molecules to rotate the plane of polarized light can be observed using a
polarimeter. Polarimeter is an instrucment that is used to measure
the degree of rotation of plane polarized light as it passes through the
sample of a
molecule. Enantiomers rotate the plane polarized light either to the right or to the left. If
the enantiomer rotates the plane prlarized light to the left, it is said to be levorotatory () (Latin: laevus , on the left side). If it rotates the plane prlarized light to the right, it is
said to be dextrorotatory (+) (Latin : dexter , on the right side).
The magnitude of the observed rotation for a particular compound depends on the
concentration, the temperature and wave length of the light used. To standardize optical
rotation data, the term specific rotation. [x], is defined as the observed rotation at specific
concentration, temperature and wave length of the light used. Mostly specific rotation is
given for 1 molar concentration at 250c(T)using sodium lamp ( = 589nm) which are
designated as superscript (T) and subscript (D)=  =589nm) to the square brackets, [
], Which are used to denote 1 molar concentration.
Specific rotation = [X]T = [x]25
(
)
For example the specific rotation of lactic acid CH3CHCOOH
is designated as
[x]25D = + 3.8
of cholestrol is,
[x]25D = -31.5
OH
In the same way, the specific rotation
In reporting specific rotation, it is common to indicate a dextrorotatory compound with a
plus sign in parentheses,(+) or with a letter 'd' in parenthesis(d) and a levorotatory
compound with a minus sign in parentheses, (-) or a letter 'l' in parenthesis (l). Most
commonly the (+) and
Unit -4
Stereochemistry
(-) indications are used. For any pair of enantiomers, one enantiomer is dextrorotatory,
and the other is levorotatory. For each enantiomer of equal
concentration, the value of the specific rotation is exactly the same, but the sign is
opposite. Following are the specific rotations of the enantiomers of
2-butanol at 250 C
using the D line of sodium.
OH
HO
C
H
C
CH3
CH3CH2
CH3
(+) -2-Butanol
x
H
CH2CH3
(-) -2- Butanol
25
= + 13.52
D
x
25
= -13.52
D
O
The specific rotations for enantiomers of glyceraldehyde
HCCHCH2OH
OH
H
H
C
OHC
CH2OH
HOH2C
OH
(-)- Glyceraldehyde
x
C
25
D
= - 8.70
CHO
OH
(+) - Glyceraldehyde
x
25
D
= + 8.70
are
Unit -4
Stereochemistry
B. Specification of configuration of Enantiomers.
Although drawing provides a pictorial representation of stereochemistry, they are difficult
to translate into words.The three dimensional structure of molecules are often shown
pictorially by using a solid wedge (
the plane), dashed wedge (
)and a single line (
) to show a bond projecting out words (above
) to show a bond projecting inwards (below the plane
) represent a bond that lies in the plane of the paper .In a
wedge –and- dash representation of the structure of lactic
CH3
CH3 CHCOOH
OH
,
HO
C

COOH
- OH group is below (behind) the plane, the –COOH group is above the plane while CH3
and H groups are on the plane of paper.
A verbal method for indicating the three dimensional arrangement of atoms (the
configuration) at stereogenic center is necessary. A system for designating the
configuration of a stereogenic center was devised by R.S. Cahn, C.K, Ingold and V.Prelog.
The system is named as the Cahn-Ingold-Prelog convention of the R,S convention. The
orientation of groups about a stereogenic center is specified as R or S using the following
rules.
1. Identify the stereogenic center in the molecule. Look at the four atoms directly
attached to the stereogenic center and assign priorities in order of decreasing
atomic number. The atom with highest atomic number is ranked first (1) the atom
with lowest atomic number is ranked fourth (4) (refer to Table 4-1 for priority
rules).
Unit -4
Stereochemistry
1. Orient the molecule in space such that the group of lowest priority(4) points away
from you in which the three groups of higher priority (1-3) project toward you.
2. Look at the three groups projecting towards you in order from highest priority(1)
to the lowest priority(3). Then,
i) If they are in a clockwise direction, the configuration at the
stereogenic
center is designated as R (Latin: Rectus, right).
ii) If they are in a counter clockwise direction, the configuration at the
stereogenic center is designated as S (Latin: Sinster, left)
Examples:
1. Absolute comfiguration of 2-butanol
CH2CH3
C
H
OH
CH3
i) . Assign priority of the four groups on the stereogenic center.
-OH > - CH2CH3 > - CH3 > -H
1st
2
nd
3rd
(higher)
4th
(lowest)
ii) Orient the molecule such that the lowest priority group is away from you
and the remaining three groups are toward you. (From the wedge-and-dash
drawing given above, the molecule is in an appropriate orientation because
the lowest priority group, hydrogen, is below the plane, that is away from
you).
H3CH2C
H
OH
C
CH3
Unit -4
Stereochemistry
iii)
Look at the three groups towards you from the highest to the lowest
priority (1-3). The three groups are arranged in counter clockwise
direction. Therefore, the configuration of the stereogenic center is S.
( 2nd) H3CH2C
H
st
OH ( 1 )
C
CH3
(3rd)
2. Absolute configuration of lactic acid
CH3
HO
C

COOH
Priorities
–OH > -COOH > -CH3 > H
1st
2nd
3rd
4th
When the lowest Priority group (H) is behind the plane the remaining groups have the
following orientation.
(3rd) H3C
H
st
OH ( 1 )
C
COOH ( 2nd)
The groups are arranged in clockwise direction. Therefore, the configuration the molecule
is R.
Unit -4
Stereochemistry
3. Absolute configuration of
OH
C
H3C
CH=CH2
H
Priorities:-OH > CH=CH2 >CH3 > H (remember that double bond is considered to be equivalent to
two single bonded atoms).
Orientation:In the compound given when the lowest priority group (H) is behind the plane the
orientation of the other groups will be as shown below.
OH ( 1st)
H
C
H2C=HC
( 2nd)
CH3 (3rd)
Since the groups are arranged in counter clockwise direction, the configuration is S.
Compounds in which a stereogenic center is part of a ring are handled is a similar way.
For example, look at how the configuration of
4- methyl cyclohexene
is determined.
CH3
H
4- Methylcylohexene.
Identify the stereoginic center and give priorities to the groups by treating the right-and
left-hand paths around the ring as if they were independent substituents. Therefore, the
molecule will be treated as follows.
Unit -4
Stereochemistry
3rd
th
H4
CH3
CH3
C
H
is treated as
2
nd
CH2
CH2
CH2
CH
1st
C C
C
The priorities of groups around the stereogenic center are :C
-CH2CH
> - CH2CH2 - C > - CH3 > - H.
C
The orientation of the three highest priority groups are:CH3 (3rd)
H
C CH2CH2
C
C
CH2CH
( 2nd)
( 1st)
C
The groups are arranged in clockwise direction. Therefore, the configuration of the
molecule is R.
Problem 4-3:Assign priorities to these set of Substituents
a) -CH2CH2OH,-Br,-H, -CH2CH3
b) -COOH,-COOCH3,-CH2OH,-OH
c) –CH2NHCH3,-CH2NH2,-NH2,-CN
d) –Br,-CH2Br,-Cl,-CH2Cl
Unit -4
Stereochemistry
Problem 4-4
Assign absolute configurations as R or S to each of the following compounds.
a)
C
H
H
b)
CH3
CH3
C
CH3CH2
COOH
CH2OH
Br
NH2
CH3
C)
C
CHO
d)
CN
H
e)
CH2OH
OH
H
H
C
OH
f)
Br
CH3
O
CH3
CH3
CH3
C) Diastereomers:When a molecule contains two or more sterogenic centers, it can have more than two
stereoisomers. For a molecule with n stereogenic centers, the maximum number of
stereoisomers possible are 2n. For molecules, with one sterogenic center, 21=2
stereoisomes,
with
two
stereogenic
center,
22=4
stereoisomers etc. For example, 2,3-dihydroxybutanoic acid (CH3CHCHCOOH)
OH OH
stereogenic centers have 4 stereoisomers.
with two
Unit -4
Stereochemistry
The absolute configuration at C-2 may be R or S. Likewise, C-3 may have ether the R or
the S-configuration. The four possible combinations of these two stereogenic center are:stereoisomer I:- (2R,3R),
stereoisomer II:- (2S,3S)
stereoisomer III:- (2R,3S),
stereoisomer IV:- (2S,3R).
Stereoisomers I and II are enantiomers of each other; the enantiomer of (R,R) is (S,S).
Likewise stereoisomers III and IV are enantiomers of each other, the enantiomers of
(R,S) being (S,R). Look at the structure formulas of these four stereoisomers given below.
1 COOH
1 COOH
H
H
C
2
C
3
HO
OH
OH
enantiones
C2
HO
C
H
3
H
4
4
CH3
CH3
I (2R,3R)
II( 25, 35 )
diastereomers
diastereomers
diastereomers
1 COOH
H
HO
C2
C
1 COOH
OH
3
Enantiomers
H
4 CH3
III ( 2R, 35 )
HO
H
C2 H
C
3
H
4 CH3
IV (25, 3R)
Unit -4
Stereochemistry
Stereoisomer I is not a mirror image of III or IV, and so it is not an enantiomer of either
one. Stereoisomers that are not related as an object and its mirror image are called
diastereomers. Diastereomers are stereoisomers that are not mirror images of each other.
Thus, stereoisomer I is a diastereomer of III and a diastereomer of IV. Similarity, II is a
diasteremer of III and IV.
Diastereomers have opposite configurations at some (one or more) stereogenic centers,
but have the same at others. Emantiomers, on the other hand, have opposite
configurations at all stereogenic center. For instance; in order to convert a molecule with
two stereogenic centers to its enantiomer, the configuration at both centers must be
changed. Reversing the configuration at only one stereogenic center converts it to a
disteromeric
structure.For example, the amino acid threonine (2-amino-3-hydroxybutonoic acid),
CH3CHCHCOOH with two stereogenic centers (C2&C3),
the Stereoisomer with
OH NH2
configuration 2R, 3R, will have enantomer with configuration 2R,3S and 2S,3R. The
following table shows the complete description of the stereoisomers of threonine.
Table 4.2. The relationships between the four stereoisomers of threonine.
Stereoisomer
Enantiomeric with
Diastereomeric with
2R,3R
25,3S
2R.3S and 2S,3R
2S,3S
2R,3R
2R,3S amd 2S,3S
2S,3R
2R,3S
2R,3R and 2S,3S
Unit -4
Stereochemistry
Certain molecules containing two or more stereo centers have special symmetry
properties that reduce the number of stereoisomers to fewer than what is predicted by
the 2n rule. One such molecule is 2,3-dihydroxybutanoic acid, more commonly named
tartaric acid.
1 2 3 4
HOOCCHCHCOOH
OHOH
2,3-Dihydroxybutanoic acid (Tartaric acid).
In tartaric acid, Carbons 2 and 3 are stereogenic centers, and using the 2 n rule, the
maximum number of stereoisomers possible are four as shown below.
1COOH
H
HO
C
OH
C
H
2
3
Mimor
4
COOH
I(2R,3R)
H
H
1COOH
1COOH
C
2
3
C
4
COOH
II(2S,3S)
OH
OH
H
C
2
3
C
H
Mirror
OH
OH
4
COOH
III (2R,3S)
HO
1
COOH
2
C
3
C
HO
H Plane of
symmetry
H
4
COOH
IV (2S, 3R)
Structures (I) and (II) are nonsuperimposable mirror images and are, therefore, a pair
of enantiomers. Structures (III) and (IV) are also mirror images, but they
are
superimposable because the molecules have a plane symmetry which cuts through the
C2 – C3 bond , making one half of the molecule a mirror image of the other half. Therefore,
the two structures are not different molecules; they are the same molecule, just oriented
differently which can be seen by rotating one structure 1800 which also gives the other
structure. The structure represented as III or IV is achiral because it is superimposable
on IV. Such a molecule with two or more stereogenic
Unit -4
Stereochemistry
centers which are identically substituted and is achiral is called meso compound. They
are achiral molecules that have stereogenic centers. Thus, tartaric acid exists in three
stereo isomeric forms; two enantiomers (I & II) and one meso form (III or IV). In the
same way, 2,3 – butanediol
CH3CHCHCH3
OHOH
have three isomers, two enantiomers and one meso form. Write the enantiomers and the
meso form for this compound.
Problem 4-5.
Using R and S descriptors, write all the possible combination (Pair) of enantiomers,
diastereomers and meso forms.
i) 2 R-2- methyl butanol
(
(
CH3CH2CHCH2OH
CH3
CH3CHCHCH2OH
ii) 2S,3S -1,2,3 - butanetriol
OHOH
iii)
2S,3R-3- Chloro-2- Pentanol
iv) 2R, 3R 2,3- Dibromobutane
V) 3S,4S -3,4-hexanediol
(
(
(
)
)
)
)
)
CH3CHCHCH2CH3
OHCl
CH3CHCHCH3
Br Br
CH3CH2CHCHCH2CH3
OHOH
Unit -4
Stereochemistry
D) Fischer Projections
In writing structures for Chiral molecules, we have thus far used lines, solid wedges and
broken wedges to indicate configuration about a stereogenic center. The use of these
drawing conventions a accurately describes
the stereochemistry and helps to treat
molecules as three- dimensional objects. It is possible, however, to convey
stereochemical information in standard and an abbreviated form using a method devised
by the German chemist
Emil Fisher.
The Fisher projection is a two dimensional representation showing the configuration of a
chiral molecule in which a tetrahedral carbon atom is represented as the center of two
perpendicular lines crossing each other. The horizontal lines represent bonds projecting
forward (above the plane) and Vertical lines represent bond projecting to the back
(behind the plane) where the stereogenic carbon is on the plane.
bond behind the plane
Stereogenic
Carbon.
Corresponds to
C
bond above the plane
To write a Fischer projection, orient the stereogenic center of the chiral molecule so that
two horizontal bonds are facing you (above the plane) and two vertical bonds are moving
back (behind the plane). Then write the molecule as a two dimensional figure with the
stereogenic carbon indicated
as the intersection of the two crossed lines. For example, we can convert the three
dimensional formula of (S)-2- Butanol into fischer projection as follows.
Unit -4
Stereochemistry
OH
C
H
CH3
CH2CH3
(S) -2- Butanol
Turn the molecule so that the two groups on a plane (-OH and –CH3) will be behind the
plane (away from you ) where both the other groups, -H (behind the plane) and –CH2
CH3 (above the plane) both face towards you (above the plane). Then write the Fisher
projection for the molecule.
C
H
OH
OH
OH
H
CH3
C
H
CH2CH3
When truned
CH2CH3
CH2CH3
In Fischer
Projection
CH3
CH3
Cl
In the same way, the Fisher progection for
CH3
CH3
H
is
CH2OH
Cl
Cl
C
C
H
HOCH2
CH2OH
C
Cl
H
CH3
when turned
H
HOCH2
CH3
Fisher projection
Unit -4
Stereochemistry
Problem:- 4.6.
Write Fischer projection for the following molecules
CH3
a)
H
C
OH
b)
C
CH2Br
H
OH
C)
H
NH2
COOH
CH2CH3
OH
CH2OH
d)
C
Br
CH2CH2CH3
C
CHO
H
4.5. Conformational isomerism
4.5.1. Introduction
Structural formulas are useful to show the order of attachment of atoms in a molecule.
However, they usually don’t show three- dimensional shapes. hence, it becomes
increasingly important to understand more about the three- dimensional shapes of
molecules. We know that an SP3- hybridized carbon atom has tetrahedral geometry and
that the Carbon-Carbon bonds result from overlapping
of carbon SP3 orbitals. Free
rotation can occur around Carbon- Carbon single bonds due to cylindrical symmetry of
sigma bond. The rotation about carbon-carbon single bond gives spatial relationships
between hydrogen on one carbon and the hydrogens on
The different arrangements of atoms that result from
a neighboring carbon.
Unit -4
Stereochemistry
rotation about a single bond are called conformations, and a specific conformation is
called a conformer (Conformational isomer) .
In this section, conformations that are adopted by individual molecules of alkanes and
cycloalkanes will be examined. Unlike constitutional isomers, different conformers can't
usually be isolated, because they interconvert too rapidly. The particular conformation
that a molecule adopts can exert a profound influence on its properties. There is an
energy change involved during the rotation of the single bond. An analysis of the energy
changes that a molecule undergoes as groups rotate about single bonds is called a
conformational analysis. It shows the stability of the different conformers. Conformational
analysis of some simple alkanes will be considered in our discussion below.
4.5.2. Learning objectives of the topic.
At the end of this topic students will be able to :

explain conformational isomerism;

write different conformers of simple alkane and cycloalkane; and

analyze conformations of simple alkanes .
4.5.2. Conformations of open Chain alkanes
Alkanes of two or more carbons can be twisted into a number of different three
dimensional arrangements of their atoms by rotating about a carbon-carbon bond or
bonds which results in the formation of different conformational isomers.
Unit -4
Stereochemistry
Conformational isomers are represented using Newman projections. In a Newman
projection, a molecule is viewed along the axis of a C-C bond. The three atoms or groups
of atoms on the front carbon are shown on three lines extending from the center of the
circle at an angle of 1200. The three atoms or group of atoms on the back carbon are
shown on three lines extending from the circumference of the circle, also at angles of
1200.
A
B
A
C
View
C
A
B
B
A
B
back carbon
B
front carbon
A
A
B
Molecule
Newman Proojection
There are two extreme conformations, the staggered conformation and th eclipsed
conformation. In the staggered conformation when viewed along
C - C axis, each atom
or group of atoms on the first carbon (front carbon) is seen to be positioned perfectly
between two atoms or groups of atoms on the second carbon (back carbon). The
staggered conformation is a conformation where the atoms or groups of atoms on one
carbon are as far apart as possible from atoms or groups of atoms on an adjacent carbon.
It is the lowest-energy, most stable conformation. The degree of rotation in which the
angle between C - H(or other atoms) bonds on the front and back carbons goes full circle
from 00 to 3600 is called a
dihedral angle (). It is the angle formed by the two
intersecting planes of the bonds on the front and back carbons. Staggered conformations
are observed at dihedral angles () of 600, 1800 and 3000. This is shown by the Newman
projections below.
Unit -4
Stereochemistry
A
B
B
A
A
600
B
B
B
B
A
3000
A
1800
A
A
A
A
B
B
B
 =600
 = 1800
 = 3000
In the eclipsed conformation when viewed along the C-C axis, all atoms or groups of
atoms on the first carbon (front carbon ) are directly opposite to those on the second
carbon (back carbon ) that is, those on the first eclipse those on the second. Eclipsed
conformation is a conformation where atoms or groups of atoms on one carbon are as
close as possible to the atoms or groups of atoms on an adjacent carbon. It is the
highest-energy, least stable conformation due to non bond interaction that arises when
atoms or groups of atoms not bonded to each other are forced into close proximity.
Eclipsed conformations are observed at dihedral angles ( ) of 00, 1200, 2400 and 3600.
This is shown by the New man projections below.
B A
B A
B A
B A
120o
360o
240o
B
A
B
A A
B
 = 00
B
 = 1200
A
B
A
B
A
B
A
 = 240o
BA
 = 3600
When each staggered conformation is rotated by 600 it gives a new but equivalent
eclipsed conformation. There is a difference in energy between
Unit -4
Stereochemistry
these extreme conformations. Eclipsed conformation is at highest energy state due to
strong non bonded interaction and the staggered conformation is at the lowest energy
state due to no or very weak non bonded interaction. These energy differences between
the eclipsed and staggered conformation is known as torsional or rotational energy (strain
). It is the force that opposes the rotation of one part of a molecule about a bond while
the other part is held past from the staggered conformation to the eclipsed conformation.
B A
A
B
B
+
A
Energy
(Torsional energy)
A
rotate
600
B
Staggered conformation
A.
B
A
B
A
Eclipsed conformation
Conformations of Ethane
If we build a molecular model of ethane (CH3CH3), we can see that the hydrogen atoms
on the two methyl groups are readily rotated with respect to each other. These rotational
movement gives the staggered and eclipsed conformations. All staggered conformations
of ethane ( = 600, 1800 and 2400 ) are equivalent and at the lowest energy state. While
all eclipsed conformations of ethane  = 0, 120, 2400 and 3600) are also equivalent but
at the highest energy state. The torsional energy between these conformations is only 3
K Cal. mol-1(figure 4-1). The torsional strain in eclipsed ethane is due to the slight
repulsion of electron pairs of adjacent C-H bonds when they rotate past each other in
converting from one staggered conformation to another.
Unit -4
Stereochemistry
H
Figure 4-1. Conformations of ethane
H
C
H
C
H
H
H
Staggered conformation
H
H
H
H
H
600
H
H
H
H
H
3000
H
1800
H
H
H
H
H
H
H
 = 600
 = 1800
 = 300
Eclipsed conformations
H H
HH
H H
120
H H
240o
o
360o
H
H
H
H H
H
H
=0
H
 = 1200
H
H
H
H
H
H
 = 2400
HH
 = 3600
Equilibrium between conformations of ethane
HH
H
H
H
rotate
+ 3K cal. mol
600
H
H
-1
H
H
H
Staggered conformation
 = 60
o
, 180 o , 300
o
H
H
Eclipsed conformation
=0
o
, 120 o , 240 o , 360
o
Unit -4
Stereochemistry
Problem 4-7
Cl
Following is the structural formula of 1,2– dichloroethane.
H
H
C
C
H
H
Cl
a) Draw Newman projections for all staggered and eclipsed conformations formed
by rotation from 00 to 3600 about the
carbon- carbon single bond.
b) Which staggered conformation(s) has the lowest energy; which has the highest
energy?
c) Which eclipsed conformation(s) has the lowest energy which has the highest
energy?
B.
Conformation of Propane.
Propane is the next higher member in the alkane series. Let us see how the potential
energy change when a substituent is added to ethane. Consider propane (CH 3CH2CH3)
whose structure is similar to that of ethane, except that a methyl group replaces one of
ethane hydrogen atoms.
H
H
H
C1
C
H
H
2
H
Ethane (CH3CH3)
H
;
H
H
H
C
C
H
H
1
2
CH3
3
Propane (CH3CH2CH3)
H
H
H
H
CH3
H
H
H
H
H
H
Unit -4
Stereochemistry
The Newman projections of propane differ from those of ethane only by the substituted
methyl group. Again, the extreme conformations are staggered and eclipsed. However,
the torsional energy between the two conformations is 3.4 K cal mol-1, slightly higher
than for ethane. This difference is due to the steric interaction between the methyl
substituent and the eclipsing hydrogen atom, known as steric hindrance. This effect can
be attributed to the bulk (size) where two molecular fragments can not occupy the same
region in space. Steric hindrance in propane due to the methyl substitution raises the
energy, not only of the eclipsed conformation, but also of the staggered (lowest energy,
or ground state) one, the latter to a lesser extent because of less steric interaction.
H CH3
H
H
CH3
rotate
+ 3.4 K cal. mol
H
H
H
Staggered propane
-1
Torsional energy
600
H
H
H
Eclipsed propane
H
Unit -4
Stereochemistry
Problem: 4. 8.
Following is the Structural formal of Chloropropane
Cl
a)
H
H
C1
C
H
H
H
2
C3 H
H
Draw Newman projections for all staggered and eclipsed conformation formed by
rotation about C1 – C2 bond from 00 to 3600.
b) Which Staggered conformation(s) has the lowest energy; which has the highest
energy?
c) Which eclipsed conformation(s) has the lowest energy, which has thhe highest
energy?
C. Conformation of Butane
The conformational Situation becomes more complex as alkane becomes larger, Let
as look at conformations of butane. Butane can be consider as disubstituted ethane
in which one hydrogen each on ethane carbon atoms are substituted by methyl group.
H
H
H
C1
C
H
H
2
H
Ethane (CH3CH3)
H
H
H
C
C
H
H
1
H
2
H
C3 C
H
H
H
Butane(CH3CH2CH2CH3)
When the conformations of butane are viewed along the bond between carbons 2 and
3, there are two types of staggered conformations and two types of eclipsed
conformations for butane. The staggered conformation
Unit -4
Stereochemistry
in which the methyl groups are at the maximum distance apart ( = 1800) is called the
( = 600) is called the
anti conformation; and that in which they are closer together
gauche conformation. In one eclipsed conformation
( = 0), methyl is eclipsed by
methyl. In the other ( = 1200), methyl is eclipsed by hydrogen. Anti conformation is the
most stable conformation of butane while the eclipsed conformation ( = 0 = 3600) is
the most unstable conformation of butane.
CH3
CH3
H CH3
H
H
H
CH3
H
HH
H
HH
CH3
Relative
torstional 0.0 Kcal. mol-1
(reference)
energy
= 180
Anti
(Most stable)
H
0.9 Kcal.mol-1
H
H
CH3 CH3
H
CH3
3.8 Kcal.mol-1
H
H
HH
4.5 Kcal.mol-1

 = 600
Gauche
 = 120
eclipsed
 = 360
eclipsed
(Least stable)
Note that both gauche and anit conformations of butane are staggered conformations,
yet the gauche conformations are approximately 0.9 K cal mol-1 higher in energy than
the anti conformation. The difference in energy between these conformations is due
to non bonded interactions between hydrogen atoms and methyl groups. The non
bonded interaction in the case of gauche butane arises because the two methyl groups
are close to each other than they are in the anti conformation. You should also note
that, even though the two staggered conformations with methyl group gauche
(dihedral angles 600 and 3000) have equal energies, they are not identical. They are
related by reflection; one is the reflection of the other. Notice that the conformations
with eclipsed –CH3 and –H groups (dihedral angles of 1200 and 2400 ) are also related
by reflection.
Unit -4
Stereochemistry
If we assign the energy values for the different eclipsing interaction, we can predict the
torsional energy in each eclipsed conformation. The torsional energy of eclipsed
conformation with dihedral angle ()= 00
(4.5 K. cal. mol-1) is the sum of the energy of
interaction of the two methyl groups (-CH3/-CH3 eclipsing interaction ) which is 2.5 K cal
mol- and the energy of interactions of the two hydrogen atoms(Hydrogen /Hydrogen
interactions ) which is 2 K cal mol-1(2 X 1k cal mol-1) as shown below .
2.5 K cal mol-1
CH3 CH3
Total:- 2.5 + (2x1) = 4, 5 K cal mol-1
-1
1 K cal mol
H
H
H
H
1 K cal mol-1
In
the
same way the torsional energy of eclipsed conformation with dihedral angle () 1200 (3.8
K cal mol-1), is the sum of the energy of interaction for the two methyl /Hydrogen
interactions which is 2.4 Kcalmol-1(2 x 1.4 K cal mol-1) and the energy of interaction for
Hydrogen /Hydrogen eclipsing interaction which is 1 K cal mol-1 as shown below.
1.4 K cal mol-1
H CH3
Total = 1 + (2X1.4 ) = 3.8 Kcalmol-1
H
H
1 K cal mol-1
H
CH3
1.4 K cal mol-1
Unit -4
Stereochemistry
The notion of assigning definite energy values to specific interactions with in a molecule
is very useful because it helps for conformational analysis of other alkanes. A summary
of what we have seen thus far is given in Table 4.3.
Table 4-3. Energy costs for interactions in alkane conformers
Interaction
H
Cause
H eclipsed
H
CH3 eclipsed
CH3
Energy cost (Kcal mol-1)
Torsional Strain
1.0
Mostly torsional strain
1.4
CH3 eclipsed Torsional
plus
Steric
2.5
strain
CH3
CH3 gauche
Steric Strain
0.9
Problem 4.9
Sight along the C2 – C3 bond of 2,3 –dimethylbutane, and draw a Newman projections for
all the staggered and eclipsed conformation. Draw the expected potential energy diagram
for this molecule.
1
2
CH3 CH2
CH3
3
CH
4
CH3
CH3
2, 3 - dimethyl butane
Unit -4
Stereochemistry
4.5.3. Conformations of Cycloalkanes
The smallest cycloalkane is cyclopropane. It is much less stable than expected for three
methylene groups (-CH2- ). This is also the same for the other cycloalkanes such as
cyclobutane and cyclopentane; why should this be ? A theoretical interpretation of this
observation was proposed by Adolf
Von Baeyer. Baeyer suggested that, since carbon
prefers to have tetrahedral geometry with bond angles of approximately 1090, ring sizes
other than five and six may be strained to exist and be very much unstable. Baeyer based
his hypothesis on the simple
geometric
notion that a three- membered ring
(cyclopropane ) should be an equilateral triangle with bond angles of 600, a four
membered ring (cylclobutane ) should be a square with bond angle of 900, a fivemembered ring (cyclopentane) should be a regular pentagon with bond angles of 108 0,
and so on.
According to baeyers analysis, cyclopropane, with a bond- angle compression of 1090 –
600 = 490, Should have a large amount of angle strain and must therefore be highly
reactive. Cyclobutane (1090 – 900) =190 angle strain) must be some what strained, but
cycloheptane (1090 – 1280 = 190 angle strain) and higher cycloalkanes that have very
large rings should be strain-free. The concept of angle strain- the strain that arises when
a bond angle is either compressed or expanded compared to its normal value (tetrahedral
value) is very useful.
Several factors in addition to angle strain are involved in determining the shape and total
strain energy of cycloalkanes. One such factor is the barrier to bond rotation (torsional
strain) encountered earlier in the discussion of alkane conformations. We said at that
time that open chain alkanes are most
Unit -4
Stereochemistry
stable in a staggered conformation and least stable in an eclipsed conformation. A similar
conclusion holds for cycloalkanes. Torsional strain is present in cycloalkanes if any
neighboring C-H bonds eclipse each other. For example, Cyclopropane must have
considerable torsional strain( in addition to angle strain), because C-H bonds on
neighboring carbon atoms are eclipsed. Larger cycloalkanes minimize torsional strain by
adopting puckered non planar conformations.In addition to angle strain and torsional
strain, steric strain is yet a third factor that contributes to the overall strain energy of
cycloalkanes.It is the strain due to repulsive interactions when atoms approach each other
too closely. As in gauche butane, two nonbonded groups repeal each other if they
approach too closely and attempt to occupy the same space. Such non-bonded steric
interactions are important in determining the minimum energy conformations of medium
– ring (C7 –C11) Cycloalkanes. Generally, cycloalkanes adopt their minimum-energy
conformations for a combination of these three factors namely, Angle strain, Torsional
strain and steric strain.
A.
Cyclopropane.
The observed C-C-C bond angles in cyclopropane are 600, a value considerably smaller
than that of the 109.50 predicted for sp3- hybridzed carbon atoms. Furthermore, hydrogen
atoms on adjacent carbons are forced into an eclipsed relationship; cyclopropane is a
strained molecule due to both angle strain and torsional strain. Cyclopropane's three
carbon atoms are of geometric necessity, coplanar, and rotation about, its carbon- carbon
bonds is impossible. Torsional strain in cyclopropane arises because of the eclipsed C-H
bonds around the ring. The total strain energy in cyclopropane is approximately 28K cal
mol-1. It is because of its extreme degree of
Unit -4
Stereochemistry
intermolecular strain that cyclopropane and its derivatives undergo several ring-opening
reactions not shown by larger cycloalkanes.
H
H
All adjacent pairs of bonds are eclipsed
H
600
H
H
H
B. Cyclobutane .
In all cycloalkanes larger than cyclopropane, non planar or puckered conformations are
favored. Cyclobutane is not quite flat but is slightly bent so that one carbon atom lies
about 250 above the plane of the other three which gives it a puckered conformation. If
cyclobutane were planar, all C-C-C bond angles would be 900, and there would be eight
pairs of eclipsed hydrogen interactions which were not observed. Puckering of the ring
changes the energy of the ring by decreasing the torsional strain associated with eclipsed
hydrogen interactions but it increase further the angle strain due to compression of C-CC bond angles. Since the decrease in torsional strain is greater than the increase in angle
strain, puckered cyclobutane. is more stable than planar cyclobutane. In the conformation
of lowest energy(pucked conformation),the measured bond angle is 88o. The strain
energy in cyclobutane is approximately 26 Kcal mol-1.
25o
............................
Planar conformation
of cyclobutane
Puckered conformation
of cyclobutane
Unit -4
Stereochemistry
C. Cyclopentane.
Cyclopentane might be expected to be in a planar conformation because the angles in a
regular pentagon are 108o close to tetrahedral angle(109.5o) with a little angle strain.
However, such a planar arrangement would have ten
H - H eclipsing interactions
which creates a torsional strain of approximately 10 Kcal mol-1. The puckering of the ring
reduces part of this eclipsed torsional strain by twisting the ring into an " envelop"
conformation. In this conformation, four carbon atoms are in a plane and the fifth is bent
out of the plane, rather like an envelope with its flap bent upward. In the envelope
conformation, C-C-C bond angles are reduced (increasing angle strain), but the number
of eclipsed hydrogen interactions is reduced (Decreasing torsional stain). Overall, the
molecule is more stable in the envelope conformation than in the planar conformation.
The average C-C-C bond angle in cyclopentane is 1050, indicating that in its conformation
of lowest energy, cyclopentane is slightly puckered. The stain energy in cyclopentane is
approximately 6.5 K cal mol-1.
Planar Conformation
of cyclopentane
Envelope Conformation
of cyclopentane
D) Cyclophexane.
Planar cyclohexane would have twelve H - H eclipsing interactions and six fold bondangle strain(a regular hexagon requires 1200 bond angles) which will make it unstable.
However, it is puckered into a three-dimensional conformation that relieves all strain in
cyclohexane which adopts a number of
Unit -4
Stereochemistry
puckered conformations, t he most stable of which is the chair conformation. In a Chair
conformation, all C-C-C bond angles are 109.50 (Strain free) and hydrogen’s on
adjacent carbons are staggered with respect to one another (removing torsional
strain). Also no two atoms are close enough to each other for any non bonded
interaction strain to exist. Thus, there is no strain of any kind in a chair conformation
of cyclohexane.
H
H
H
1200
H
H
H
H
H H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
1200
Planar cyclohecane
0
(120 bond angles, 12 eclipsing hydrogens)
Chair cyclohexane
(Nearly tetrahedral bond angles,
no eclipsing hydrogens).
In a chair conformation of cyclohexane, the C-H bonds are arranged in two different
orientations; six C-H bonds are parallel to the principal molecular axis and are called axial
hydrogen’s and the other six are perpendicular to this axis and are located approximately
along the equator of the molecule and are called equatorial hydrogens. Three axial
hydrogens point straight up; the other three axial bonds point straight down. Notice also
that axial bonds alternate, first up and then down as you move from one carbon of the
ring to the next. Equatorial bonds also alternate first slightly up and then slightly down
as you move from one carbon of the ring to the next, Notice further
Unit -4
Stereochemistry
that if the axial bond of a carbon point upwards, then the equatorial bond on that carbon
points slightly downward. Conversely, if the axial bond on a particular carbon points
downward, then the equatorial bond on that carbon points slightly upward.
H
H
H
H
H
H
H
H
H
H
H
H
Equatorial hydrogen's
(three slightly up & three slightly down )
Axial hydrogens
(three upward & three downward)
H
H
H
H
H
H
H
H
H
H
H
H
Axial and equatorial hydrogens
(Six axial and six equatorial).
There are other non planar conformations of cyclohexane, two of which are, the boat
conformation and twist-boat conformation. A boat conformation is considerably less
stable than a chair conformation because of the torsional
strain associated with four
sets of eclipsed hydrogen and the non bonded interaction strain between the two flagpole
hydrogens. The difference in potential energy between chair and boat conformations is
approximately
6.5 K cal mol-1. Some of the strain in the boat conformation can be
relieved by a slight twisting of the ring to form a twist-boat conformation . It is estimated
that a twist-boat is favored over a boat conformation by approximately 1.5 K calmol-1.
Unit -4
Stereochemistry
H
H
H
H
H
H
H
H
Boat
Twist Boat
H
H
H
H
H
H
H
H
H
H
H
Chair
H
Boat
Cyclohexane is not conformationally rigid. It is capable of interconverting one chair
conformation to the other chair conformation through the boat conformation. The two
equivalent chair conformations can be interconverted by twisting one chair first to a boat
and then to the other chair. The two chair conformations readily interconvert, resulting
in the exchange of axial and equatorial positions. This interconversion of chair
conformations is usually referred to a ring-flip. In this process (“flipping"), all axial
hydrogens in one chair become equatorial in the other and vice versa. The energy for
this conversion is 10.8 K Cal mol-1 and is rapid at room temperature.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Boat
H
H
H
Chair
H
Chair
H
Unit -4
Stereochemistry
If a hydrogen atom of cyclohexane is replaced by a methyl group or other alkyl group,
the group occupies an equatorial position in one chair and an axial position in the other
chair. This means that the two chairs are no longer of equal stability. Structural studies
have established that approximately
95 % of the molecules of methylcyclohexane are
in the chair conformation that has an equatorial methyl group while only 5 % of the
molecules have an axial methyl group at room temperature.
CH3
H
CH3
5%
95 %
H
What is the structural basis for the observation that equatorial methylcyclohexane is more
stable that axial methylcyclohexane? A convenient way to describe the relative stabilities
of chair conformations with equatorial or axial substituents is in terms of a type of non
bonded interaction called diaxial interaction. It refers to the repulsion between an axial
substituent and an axial hydrogen (or other group) on the same side of the ring. In
methylcyclohexane, when the –CH3 is axial, it is parallel to the axial C-H bonds on carbons
3 and 5 where the –CH3 group is crowded with two unfavorable methyl - hydrogen diaxial
interactions. No such unfavorable 1, 3- diaxial interactions exist when the methyl group
is in an equatorial position. For methylcyclohexane, the equatorial methyl conformation
is favored over the axial methyl conformation by about 1.74 K col mol-1. The greater
stability of an equatorial group, compared with axial group is due to steric effect where
the axial group is crowded
due to 1,3 – diaxial repulsion between itself and the other
two axial substituents located on the same side of the ring.
Stereochemistry
....
...
C
H
H
....
.
H
H
H
....
...
...
...
...
.
...
Unit -4
H
H
C
H
1,3- diaxial interactions
between hydrogen of axial
CH3 and axial hydrogens
At C-3 and C-5. (less stable)
H
H
No-1,3-diaxcal interactions
between hydrogen of equatorial
CH3 and axial hydrogen's at
C-3 and C-5 (more stable)
Other substituted cyclohexanes are similar to methyl cyclohexane. The two chair
conformations exist in rapid equilibrium, and the one in which the substituent is equatorial
is more stable. The relative amounts of the two conformations depend on the effective
size of the substituent. As the size of the alkyl substituent increases, preference for
conformations with the group equatorial increases. A branched alkyl group such as
isopropyl and tert -butly groups show greater preference for the equatorial orientation
than does methyl.
CH(CH3)2
H
CH(CH3)2
3%
H
97 %
C(CH3)3
H
C(CH3)3
0.1 %
99.9 %
H
Unit -4
Stereochemistry
Problem 4-10
Draw two different chair conformations of bromocyclohexane, showing all hydrogen
atoms. identify
each substituent as axial or equatorial. Indicate the most stable
conformation.
Summary
Our concern in the chapter has been with the spatial arrangement of atoms and groups
in molecules. Chemistry in three dimensions is known as stereochemistry. It deals with
molecular structures. Compounds that have the same molecular formula but different
structures are known as isomers. There are different types of isomers. The common ones
are constitutional isomers and stereoisomer.
Constitutional isomers are compounds whose atoms are connected differently.
Constitutional isomers may have different carbon skeletons (Skeletal isomers), different
functional groups (functional isomers) or different locations of a functional group along
the chain (positional isomers). Stereo isomers are compounds which have the same
connection (same order of attachment) of atoms in their molecules but a different three
dimensional orientation of their atoms in space (different geometry). Stereo isomers can
be divided into enantiomers, diasteromers and geometric isomers.
nonsuperimposable
mirror
–
image
stereoisomers.
Enantiomers are
Diastereomers
are
nonsuperimposatble, non-mirror image stereo isomers and geometric isomers are those
which have substituents on the same side or opposite side of a double bond or ring.
Unit -4
Stereochemistry
A molecule is chiral if it can’t be superimposed on its mirror image. The most common
kind of chiral molecule contains a carbon atom that bears four different substituent. Such
a carbon is called a stereogenic center (asymmetric center). Chirality is a property of a
molecule as a whole, not of a particular atom. An achiral molecule is that which can be
superimposed on its mirror image. It has a plane of symmetry that bisects the molecule
into two mirror image halves.
Optical activity, the capacity to rotate plane polarized light, is a physical property of chiral
molecules. Enantiomeric forms of the same molecule rotate plane polarized light an equal
amount but in opposite directions. The enantiomer that rotates plane polarized light in
the clockwise (positive) direction is said to be dextrorotatory and that which rotates in
anticlockwise (negative) direction is said to be levorotatory. In order to be optically active
a substance must be chiral, and one enantiomer must be present in an amount greater
than the other. A racemic mixture is optically inactive and contains equal quantities of
enantiomers. A ploarimeter in an instrument used to detect and measure the magnitude
of optical activity. Specific rotation is the number of degrees the plane polarized light is
rotated by a molecule measured at 1 molar concentration and 250 C using sodium lamp
(589 nm). It is used to express the optical activity of a molecule.
The configuration at any stereocenter can be specified by the Cahn- Ingold-Prelog
convention, known alternatively as the R.S convention. A Fischer projection is a two –
dimensional representation showing the configuration of a chiral molecule.Horizontal lines
represent bonds projecting forward and vertical lines represent bond projecting to
forward and vertical lines represent bond projecting to the back where the stereogenic
center is at the
Unit -4
Stereochemistry
intersection of the two perpendicular lines. Meso compounds contain two or more
stereogenic centers, but are achiral ( optically inactive ) overall because they have plane
of symmetry due to identical substituents on the stereogenic centers. A Conformation is
any three dimensional arrangement of atoms of a molecule that results by rotation about
one or more single bonds. One convention for showing conformations is the Newman
projection. A dihedral angle is the angle created by the two intersecting planes. Staggered
conformations occur at dihedral angles of 600, 1800, and 3000. Eclipsed conformations
occur at dihedral angles of 00, 1200, and 2400. Staggered conformations are more stable
than eclipsed conformations due to less intermolecular strain (torsional strain, angle strain
and non-bonded interaction strain). The lowest- energy conformation of cyclopentane is
an envelope conformation. The lowest-energy conformations of cyclohexane are two
interconvertible chair conformations. In a chair conformation,six bonds are axial and six
bonds are equatorial. Bonds which are axial in one chair are equatorial in the alternative
chair. Boat and twist- boat conformations are higher in energy than chain conformations.
The more stable conformation of a substituted cyclohexane is the one that have
substituents at equatorial position that minimizes diaxial interactions.
Unit -4
Stereochemistry
Additional Problemsproblems
1. Write all possible constitutional isomers for the following compounds.
a) Alkanes with formula C7 H16
b) Alcohols with formula C6H14O
c) Ketones with formula C6H12O
d) Amines with formula C6 H15 N
2. Draw all structures for the geometrical isomers of each of the following compounds.
Label each double bond as Cis- or Trans (Z or E) and write the name of each isomer.
a) 3- methyl -3- hexane
b) 2,3- Dimethyl -2- pentene
c) 1,2-Dibromoethane
d) 2-chloro -2- butane
e) 4- Methyl -3- hexen-1- ol.
3. Which of the following compounds are chiral and thus capable of existing in optically
active forms? Identify the chiral center or centers.
a)
b)
CH3CH2CHCH2CH3
Cl
OH
c)
1 - Chlorobutane
e)
Br
g)
C6H5CHCH3
d) 3 - Chloro- 2- butanol
Cl
OH
CH2 = CH CH CH3
f) 4 -Bromo-1chlorocycdohexene
h) 1,3 dimethylbenzene
Unit -4
Stereochemistry
4. Assign an ( R ) or (S) absolute configuration to each of the following molecules
Cl
a)
C
H
CH2CH3
b)
HO
COOH
CH3
H3C
c)
CH2Cl
C
C
NH2
COCH3
O2N
CH=CH2
CH(CH3)2
d)
C
CH2OCH3
COOH
e)
H
O
CH2CH3
Cl
5. Using R and S descriptors, write all possible combinations (pairs) of enantiomers,
diastereomers and meso forms.
a) CH3 CH CH CH3
b) CH3 CH CH CH2Br
c) HO CH2 CH CH CH2OH
Br OH
Br Br
OH OH
6. Draw Newman projections for staggered and eclipsed conformations formed by
rotation about C2 – C3 bond for the following compounds.
a) Br CH2CH2CH2Br
b ) CH3 CH CH CH3
OH OH
7. Draw the two chair conformations for each of the following compounds and indicate
which conformer is more stable.
a) Trans -1- ethyl -2- methylcyclohexane
b) Cis -1, 3- Dimethylcyclohexane.
c) Trans -1,4- Diethyl cyclohexane .