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Module 8 Stereochemistry
Lecture 20 Stereochemistry I
Stereochemistry is the study of the relative arrangement of atoms or groups in a molecule
in three dimensional space. Stereochemical isomers are molecules, which have the same
chemical formula and bond connectivity but different relative arrangement in three-
dimensional space. In contrast, constitutional isomers have same molecular formula but
different bond connectivity. Thus, n-butane and isobutane are structural isomers while the
isomers of limonene, the compound which gives different taste to lemon and orange are
examples of stereochemical isomers (Figure 1).
H
3
CCH
3
H
3
CCH
3
CH
3
n-butaneisobutane
structural isomers
CH
3
H
3
C
(+)-limonene
CH
3
H
3
C
(-)- limonene
Stereochemical isomers
"orangy taste""lemony taste"
Figure 1
To understand the difference between the two isomers of limonene, introduction to some
new terms and concepts are required. The most important being the concept of chirality.
A chiral object is one that cannot be superposed on its mirror image. The term originates
from the greek term for “hand”. As it is with human hands, the left hand cannot be
superimposed on the right hand. It is the same with chiral molecules. They are non-
superimposable mirror images of each other. Achiral objects, on the other hand, are easily
superimposable on each other. A tennis racquet and a spoon are examples of achiral
objects.
The next question that comes to the mind is how to determine whether a molecule is
chiral or achiral. At times, it becomes extremely difficult to determine with increasing
molecular complexity, to determine the non superimposibity of a compound with its
mirror image. Thus, a mathemathical concept known as group theory can be applied to
determine the symmetry elements in a molecule. There are four symmetry elements
which needs to be considered for this purpose:
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pff
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pf1b
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Module 8 Stereochemistry

Lecture 20 Stereochemistry I

Stereochemistry is the study of the relative arrangement of atoms or groups in a molecule

in three dimensional space. Stereochemical isomers are molecules, which have the same

chemical formula and bond connectivity but different relative arrangement in three-

dimensional space. In contrast, constitutional isomers have same molecular formula but

different bond connectivity. Thus, n-butane and isobutane are structural isomers while the

isomers of limonene, the compound which gives different taste to lemon and orange are

examples of stereochemical isomers (Figure 1).

H 3

C

CH 3 H 3

C CH 3

CH 3

n-butane isobutane

structural isomers

CH 3

H 3

C

(

)

-limonene

CH 3

H 3

C

(

  • ) - limonene

Stereochemical isomers

"orangy taste" "lemony taste"

Figure 1

To understand the difference between the two isomers of limonene, introduction to some

new terms and concepts are required. The most important being the concept of chirality.

A chiral object is one that cannot be superposed on its mirror image. The term originates

from the greek term for “hand”. As it is with human hands, the left hand cannot be

superimposed on the right hand. It is the same with chiral molecules. They are non-

superimposable mirror images of each other. Achiral objects, on the other hand, are easily

superimposable on each other. A tennis racquet and a spoon are examples of achiral

objects.

The next question that comes to the mind is how to determine whether a molecule is

chiral or achiral. At times, it becomes extremely difficult to determine with increasing

molecular complexity, to determine the non superimposibity of a compound with its

mirror image. Thus, a mathemathical concept known as group theory can be applied to

determine the symmetry elements in a molecule. There are four symmetry elements

which needs to be considered for this purpose:

  • Centre of symmetry( i ): The center of symmetry i is a point in space such that if a

line is drawn from any part (atom) of the molecule to that point and extended an

equal distance beyond it, an analogous part (atom) will be encountered. Thus the

molecule 3,6-dimethylpiperazine-2,5-dione has centre of symmetry (sometimes

referred to as centre of inversion) running through the centre of the molecule.

N

H

H

N

O

O CH

3

H

3

C

  • Plane of symmetry (σ): A plane of symmetry is a reflection plane which brings

into coincidence one point of the molecule with another one through the mirror

reflection. Thus, glyoxal has a plane of symmetry running through the molecular

plane.

O

H

H

O

σ

  • Axis of symmetry (C n

): Symmetry axis C

n

, also called n-fold axis, is an axis

which rotates the object (molecule) around by 360°/n, such that the new position

of an object is superimposable with the original one. For example, (1Z,4E,8E)-

3,7,11-trimethylcyclo-dodeca-1,4,8-triene has 3-fold rotation axis.

CH

3

H

3

C

CH

3

C

3

If a molecule has only either centre of symmetry or plane of symmetry then it is achiral.

However, in most cases, molecules have more than one element of symmetry. In such

cases, it becomes important to know the point group to which the molecule belongs. A

point group reflects the combination of symmetry elements present in the structure. The

point group of a molecule can be determined by following the algorithm given below.

The point groups of high symmetry are usually not important in simple organic

molecules. The point groups C

1,

C

n

and D

n

are chiral groups and they contain chiral

molecules while all other groups are achiral (Figure 2).

Group of Low Symmetry

Yes

No

C

1

,C

s

,C

i

Group of High

Symmetry

T

d

, O

h

, I

h

, C

∞ v

, D

∞ h

Yes

No

Highest Order

Rotational Axis

C

n

Perpendicular

C

2

Axes

Yes

No

C or S groups D groups

σ

h

Yes

No

σ

d

No

D

n

D

nd

D

nh

σ

h

No

Yes

C

nh

σ v

No

S

2n

C

nv

Yes Yes

C

nv

No

C

n

Figure 2

Enantiomers

If a molecule is non-superimposable on its mirror image, then the molecule is said to

have enantiomeric relationship with its mirror image molecule. For example, in 2-

chloropropane, the molecule is superimposable with its mirror image, so they are

identical molecules, but in 2-chlobutane, the molecule is not superimposable with its

mirror image and the two molecules are called enantiomers (Figure 3). Thus, enantiomers

are stereoisomers since they differ only in the relative arrangement of the different groups

in space but not in bond connectivity. Enantiomers are identical in all physical properties

(except optical rotation) and chemical properties and reactivity compared to an achiral

reagent in reactivity

Cl

Me Me

H

Cl

Me Me

H

mirror

Cl

Me Et

H

Cl

Et Me

H

Cl

Et Me

H

Cl

Me Et

H

2-Chloropropane

2-Chlorobutane

all groups align

therefore, molecules are

superimposable

two groups fail to align

therefore, molecules are

nonsuperimposable

Cl

Me Me

H

Cl

Me Me

H

Figure 3

Obviously, the next issue is how to detect and analyse the enantiomers physically. In this

respect, in 1801, Haüy, a French mineralogist observed that some quartz crystals rotate

polarized light clockwise, while other crystals rotate polarized light to the left. Haüy also

noticed that quartz crystals exhibit the phenomenon of hemihedrism (externally, some

crystals are non-identical mirror images of other crystals). This is referred to as optical

activity. Followed by this, J. B. Biot observed the optical activity in certain organic

compounds and was able to conclude that it is a molecular property. In 1884, Louis

Pasteur in an ingenius experiment crystallized and physically separated two types of

crystals of tartaric acid –one of which was hemihedral to the left while the other was

mixture of enantiomers and the mixture contains more of the enantiomer with the S

configuration than the enantiomer with the R configuration. From the observed specific

rotation, we can calculate the optical purity of the mixture.

optical purity =

observed specific rotation

specific rotation of the pure enantiomer

x 100

For example, if a sample of 2-bromobutane has an observed specific rotation of +9.2°, its

optical purity is 0.40. In other words, it is 40% optically pure-40% of the mixture consists

of an excess of a single enantiomer.

optical purity =

x 100

Because the observed specific rotation is positive, we know that the solution contains

excess ( S )-(+)-2-bromobutane. The enantiomeric excess ( ee ) tells us how much excess

( S )-(+)-2-bromobutane is in the mixture. As long as the compound is chemically pure,

enantiomeric excess and optical purity will be the same.

enantiomeric excess =

excess of a single enantiomer

entire mixture

x

x

If the mixture has a 40% enantiomeric excess, 40% of the mixture is excess S enantiomer

and 60% is a racemic mixture. Half of the racemic mixture plus the amount of excess S

enantiomer equals the amount of the S enantiomer present in the mixture. Thus, 70% of

the mixture is the S enantiomer and 30% is the R enantiomer.

Another class of stereoisomers is the so called diastereomers which have different

chemical and physical properties. Such compounds may include geometrical isomers- the

cis and trans isomers (Figutre 4).

H

H

3

C

H

CH

3

CH

3

H

3

C

H

H

cis -but-2-ene trans -but-2-ene

Figure 4

A different type of isomerism may exist in disubstituted cyclic compounds. Thus, in 4-

tert-butylcyclohexanol, two isomers- cis and trans exist (Fig 5).

H

3

C

CH

3

H

3

C

OH

OH

H

3

C

CH

3

H

3

C

cis -4- tert -butylcyclohexanol

trans

tert -butylcyclohexanol

OH

CH

3

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

OH

Figure 5

Diastereomeric compounds may or may not be chiral. The above two examples are both

achiral, each having a plane running through them. However, when cis and trans

epoxides are compared, it can be easily seen that they may be chiral compounds. As an

example, the comparison of cis and trans isomers of 2,3-dimethyloxirane, the cis -isomer

is achiral having a plane of symmetry in the molecule. However, the trans isomer does

not have any plane of symmetry through the molecule and as such it is a chiral molecule

(Figure 6).

O

CH

3

H

3

C

O

CH

3

H

3

C

cis -2,3-dimethyloxirane

O

CH

3

H

3

C

trans -2,3-dimethyloxirane

Achiral compound Chiral compound

σ

Figure 6

Though, here, the enantiomers are represented in the flying wedge form where two of the

groups around the chiral centre are depicted in the plane of the paper and groups towards

us in bold bonds and groups away from us in broken bonds, there are other forms of

depictions of chiral compounds. These are discussed below.

  • Fischer projection formula. It is a representation of a 3D molecule as a flat

structure where a tetrahedral carbon is represented as two crossed lines. The two

vertical bonds about the stereocentre are above the plane of paper (towards the

viewer) while the horizontal bonds are below the plane of the paper (away from

the viewer) (Figure 8).

s s t t

e e

r r

e e

o o gg

e e

n n i i

c c

c c

e e

n n t t

r r

e e

F

F

i i

s s

c c h h

e e

r r pp

r r

o o j j

e e

c c t t i i

o o

n n

B

B

o o

n n d d

s s

a a b b

o o

v v

e e

t t

h h

e e p p

l l

a a

n n

e e

o o

f f p p

a a p p

e e

r r

B

B

o o

n n d d

s s b b

e e l l

o o

w w

t t h h

e e

p p

l l

a a

n n

e e

o o

f f p p

a a p p

e e

r r

Figure 8

A few examples of depiction of molecules in Fischer projection formula is given

below. It must be noted that when bonds are rotated by 180°, they result in the

same identical molecule (Figure 9).

Ph COOH

HO

H

COOH

OH

Ph H

Ph

CH

3

OH

NHCH

3

H

H

3

C NHCH

3

OH

H Ph

Figure 9

  • Sawhorse projection formula. Sawhorse projection formulas are used to denote

two principal stereocentres. It is a view of a molecule down a particular carbon-

carbon bond, with the groups connected to both the front and back carbons are

drawn using sticks at 120° angles. Sawhorse Projections can also be drawn so

that the groups on the front carbon are staggered (60 ° apart) or eclipsed (directly

overlapping) with the groups on the back carbon. The overall representation is

given below (Figure 10).

Ph

CH

3

OH

NHCH

3

OH

H Ph

NHCH

3

H

CH

3

NHCH

3

H

3

C H

OH

H Ph

Eclipsed Staggered

60° rotation

1

2

2

1 1

2

Figure 10

  • Newmann projection formula. In this notion, the molecule is again viewed by

looking down a particular carbon-carbon bond. The front carbon of this bond is

represented by a dot, and the back carbon is represented by a large circle. The

three remaining bonds are drawn as sticks coming off the dot (or circle),

separated by one another by 120°. Just like Sawhorse projection formula,

Newman Projection can be drawn such that the groups on the front carbon are

staggered (60 ° apart) or eclipsed (directly overlapping) with the groups on the

back carbon (Fig 11).

H

H 3

3

C

C

O

O

H

H

O

O

H

H

C

C

O

O

O

O

H

H

H

H

H

H

O

O

O

O

C

C

O

O

H

H

O

O

H

H

C

C

H

H

3 3

H

H

H

H

H

H

O

O

O

O

C

C

O

O

H

H

O

O

H

H

C

C

H

H

3 3

H

H

r r

o o t t

a a t t i i

o o

n n

e e

c c l l i i pp

s s

e e d d

s s

t t

a a gg gg

e e

r r

e e

d d

Figure 11

Step1: Put the groups at the vertical bonds on the tails of the “Y” of sawhorse projection.

Step 2: Put the other groups on the head of “Y”.

In other words, imagine that C

1

-C

2

bond is in plane of the paper and look across the C

1

C

2

bond, the groups which are coming out of the plane of paper form the tails of “Y”

while the groups below the plane of the paper form the heads of “Y”.

In order to convert a molecule from Fischer projection to sawhorse projection, the

molecule has to be first depicted in eclipsed form. Then the steps outlined above are to

be carried out in a reverse manner (Figure 3).

Fi scher pro j

ection

Br

C

2

H

5

CH

3

Br

CH

3

C

2

H

5

180° rotation of C 2

2

1

Br

C

2

H

5

CH

3

Br

H

3

C C

2

H

5

Br

H

3

C C

2

H

5

Br

H

3

C C

2

H

5

Sawhorse pro j

ection

Figure 3

In other words, again look along the C

1

-C

2

bond as being the plane of paper, then the

groups lying above the paper form the vertical line while the groups below the paper

form the horizontal bonds in Fischer projection.

To convert a Sawhorse projection to Newmann projection look along the C

1

-C

2

bond

through C

1

such that C

2

is not visible. Now the groups on C

1

are same for both sawhorse

and Newmann projection. The C

2

carbon is replaced by a circle and the bonds emanating

from C

2

being retained in their actual spatial location (Figure 3).

H

1

H H

COOH

2

HO H

COOH

H

OH

H

H H

COOH

H OH

H

H H

H

H

H

COOH

H H

Line of sight

from C 1

to C 2

through C 1

-C

2

Figure 3

Now coming back to the issue to assign the relative configurations of a stereocentre,

several approaches were used. The first one being the stereochemical descriptor D/L

which relies on the chemical correlation of the configuration of the chiral center to D-

glyceraldehyde. The compounds which can be correlated without inverting the chiral

center are named D, those correlated to its enantiomer are designated as L. It is worth a

while to note that though D-glyceraldehyde is dextrorotatory (rotates the plane of

polarized light to the right), the compounds correlated to D-glyceraldehyde do not have to

be dextrorotatory and vice versa (Figure 4).

CHO

CH

2

OH

H

OH

D-glyceraldehyde

CHO

CH

2

OH

HO

H

L-glyceraldehyde

Figure 4

This system is not much in use today. However, traditionally it is still used for sugars and

amino acids. In cases of carbohydrates, only the carbon at the end of carbon chain is

considered for assigning D/L after writing the molecule in Fischer projection with the

anomeric carbon on the top. Now if the carbon at the end of chain (farthest from

anomeric carbon) has a hydroxyl group to the left, it is denoted as an L-sugar and it has a

hydroxyl group to the right, it is called a D-sugar (Figure 19).

CHO

H OH

HO

H

H

OH

H OH

CH

2

OH

D-glucose

CHO

H OH

HO H

CH

2

OH

HO

H

L-arabinose

CHO

HO H

HO H

H

OH

H OH

CH

2

OH

D-Mannose

CHO

H OH

HO

H

CH

2

OH

H OH

D-Xylose

Figure 5

  • In case of ties, use the next atoms along the chain of each group as tiebreak-

ers. For example, we assign a higher priority to isopropyl -CH(CH

3

2

than to

ethyl -CH

2

CH

3

or bromoethyl -CH

2

CH

2

Br. The first carbon in the isopropyl

group is bonded to two carbons, while the first carbon in the ethyl group (or

the bromoethyl group) is bonded to only one carbon. An ethyl group and a -

CH

2

CH

2

Br have identical first atoms and second atoms, but the bromine atom

in the third position gives -CH

2

CH

2

Br a higher priority than -CH

2

CH

3

. One

high-priority atom takes priority over any number of lower-priority atoms.

CH

2

CH

2

Br

H

3

CH

2

C

H

CH(CH

3

2

  • Treat double and triple bonds as if each were a bond to a separate atom. For

this method, imagine that each pi bond is broken and the atoms at bothends

duplicated. Note that when you break a bond, you always add two imaginary

atoms. (Imaginary atoms are circled below.)

R C C

H

H

H

R

C C

H

H

C

H

C

R C N

H

N

H

C

R C C H

C

C

C

C

R

C O

OH

O C

R

C

N

H

H

R C C H

R C

O

OH

break and duplicate

becomes

becomes

becomes

becomes

break and duplicate

break and duplicate

break and duplicate

  • Using a three-dimensional drawing or a model, put the fourth-priority group

away from you and view the molecule along the bond from the asymmetric

carbon to the fourth-priority group. Draw an arrow from the first-priority

group, through the second, to the third. If the arrow points clockwise, the

asymmetric carbon atom is called ( R ). If the arrow points counter-clockwise,

the chiral carbon carbon atom is called ( S ).

Here, the order of priority is assigned as follows:

Following rule 1, bromo group gets first priority.

Following rule 3, carbonyl group gets second priority.

O

becomes

O

(C) (O)

Following rule 2,

H

H

group gets third priority over methyl.

Br

2

CH

3

1

CH

3

OH

H

R

1

CH

3

OH

R =

Br

CH

3

Here, for C

1

, the priority order is set to be as follows.

Following rule 1, hydroxyl group gets first priority.

Following rule 3, ethenyl group gets second priority ahead of R and methyl groups.

(C)

becomes (C)

Following rule 2, R group gets priority over methyl.

R =

Br

CH

3

R

CH

3

OH

2 S

R CH

3

OH

Since, in this case, the lowest priority group is on the plane and not below it, so, it is

converted to an identical form by keeping one of the groups on plane constant and

rotating the other 3 groups clockwise. Now, it can be seen that the configuration is S.

Br

2

CH

3

CH

3

OH

H

Br

2

R

CH

3

H

Here, the priority order is decided as follows:

Following rule 1 bromo gets first priority

Following rule 2, R gets second priority

Following rule 2, methyl gets third priority.

Br R

CH

3

H

2

S

R =

CH

3

OH

Hence the molecule is

Br

2

CH

3

CH

3

OH

H

1

S ,7 S

-7-bromo-3-methyloct-1-en-3-ol

Sometimes, the determination of configuration of a compound drawn as a Fischer

projection is necessary. The rules are given below.

  • Rank the groups (or atoms) that are bonded to the asymmetric carbon in order

of priority.

  • Draw an arrow from the group (or atom) with the highest priority (1) to the

group (or atom) with the next highest priority (2). If the arrow points

clockwise, the enantiomer has the R configuration; if it points anti-clockwise,

the enantiomer has the S configuration, provided that the group with the

lowest priority (4) is on a vertical bond.

H

Cl

CH

2

CH

2

CH

3

H

3

CH

2

C

H

Cl

H

3

CH

2

CH

2

C CH

2

CH

3

R

-3-chlorohexane (

S

-3-chlorohexane