Stereochemistry of acyclic compounds Isomerism Classification Geometrical Isomerism

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Stereochemistry
Stereochemistry of acyclic
compounds
Geometrical and Conformational Isomerism
Dr. Satish Dhirendra Mitragotri
(M.Sc., Ph.D., SET, NET, GATE and M.B.A. (Finance))
Department of Chemistry
Walchand College of Arts and Science, Solapur
Compounds with identical molecular formulas but
distinct arrangements of atoms in space are called
isomers. Isomers
Conformational
Isomers
Geometrical
Isomers
Stereo Isomers Optical Isomers Structural Isomers
Positional Isomers
Functional
Chain
Types of Isomers
Geometrical isomerism
• Geometrical isomerism: Compounds in which atoms or
groups exhibit different spatial arrangements on either
side of a chemical bond or ring structure due to
restricted rotation are called as Geometrical isomers.
• Geometric isomerism is also called configurational
isomerism or cis-trans isomerism.
• Examples
CH3
H
C
H3
H
H
CH3
C
H3
H
Cis Trans
Cl
H
Cl
H
H
Cl
Cl
H
Cis Trans
Examples
Cl
H
H
H
H
Cl
Cl
H
H
H
Cl
H
Trans
Cis
H
H
H
H
H
H
H
H
H
H H
H
Cl
Cl Cl
Cl
Cis Trans
N
OH
C
H3
H
N
OH
C
H3
H
Syn
Anti
Ph
N
OH
C
H3
Ph
N
OH
C
H3
Syn
Anti
phenyl Syn
Anti methyl
methyl
phenyl
Geometrical isomerism
Cause of Geometrical Isomerism
• Restricted rotation due to presence of double bond or ring
system.
• In order to convert one isomer into another we need to break
the pi bond or ring.
• At RT it is not possible to rotate the bond freely.
C
H3
H
CH3
H
C
H3
H
CH3
H
H
CH3
C
H3
H
Cis Trans
pi bond will be broken
Cl
H
H
H
Cl
H
Cl
H
H
H
Cl
H
Cis
Cl
H
H
H
H
Cl
Trans
Cl
H
H
H
Cl
H
Cis
Geometrical isomerism in aldoximes and ketoximes
O
C
H3
H
N
H2 OH
N
OH
C
H3
H
N
OH
C
H3
H
Syn
Hydroxilamine
+
Aldehyde
H2O
Anti
O
C
H3
Ph
N
H2
OH
Ph
N
OH
C
H3
Ph
N
OH
C
H3
Hydroxilamine
+
H2O
Ketone
Syn
Anti
phenyl Syn
Anti methyl
methyl
phenyl

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Geometrical isomerism in aldoximes and ketoximes
In case of aldoximes the syn and anti configuration can be
determined by sequence of reactions
Syn Aldoxime - It on acetylation followed by hydrolysis gives
original aldoxime back.
N
OH
C
H3
H
N
OCOCH3
C
H3
H
Hydrolysis
N
OH
C
H3
H
H+
Syn
+
Geometrical isomerism in aldoximes
N
OH
C
H3
H
N
OCOCH3
C
H3
H
C
H3
CN
(CH3CO)2O
Anti
+
H+
Anti Group Migrates
Nitrile
In case of anti aldoxime the acetylation reaction followed by
hydrolysis give nitrile or caynide
It involves migration of anti atom i.e. H
Geometrical isomerism in ketoximes
Beckmann transformation and configuration of ketoximes.
The acid-catalyzed conversion of an oxime into an amide is
known as Beckmann rearrangement
The Beckmann rearrangement, named after the German chemist
Ernst Otto Beckmann (1853–1923), is a rearrangement of an
oxime functional group to substituted amides
N
OH
R1
R2
H
+
N
H
R1
R2
O
Syn
Beckmann
Transformation
R2
Oxime
Amide
N
OH
R1
R2
H
+
N
H
R2
R1
O
R1
Syn
Beckmann
Transformation
Amide
Oxime
In case of ketoximes the anti group to OH migrates and
amide is formed at the end of Beckmann transformation.
From the hydrolysis products of amide i.e. Acid and Amine
the original amide and hence oxime can be identified
Geometrical isomerism in ketoximes
N
OH
C
H3
Ph
H
+
N
H
C
H3
O
Ph
Methyl
Syn
Beckmann
Transformation
Oxime
Amide
N
H
C
H3
O
Ph
N
H
C
H3
O
Ph
OH
H
N
H2
Ph
C
H3
O
OH
Amide
Hydrolysis
+
Acetic acid Phenyl amine
Oxime
Amide
Geometrical isomerism in ketoximes
N
OH
C
H3
Ph
H
+
N
H
O
CH3
Ph
Oxime
Amide
Syn
Beckmann
Transformation
Phenyl
N
H
O
CH3
Ph
OH
H
N
H2
CH3
N
H
O
CH3
Ph
O
Ph OH
Amide
Hydrolysis
+
Benzoic acid Methyl amine
In this case the oxime has methyl group in anti position to OH
group . After Beckmann Transformation the methyl (CH3) group
migrates and gets attached to Nitrogen . After hydrolysis of amide
the Nitrogen containing part of amide forms the amine
Here methyl amine and benzoic acid are obtained and hence
methyl group must have been attached to nitrogen i.e. it was
migrated during B.T. and must in anti position in case of ketoxime.
Presentation of Molecules
•Structural Formula
•Condensed Structural Formula
•Line-Angle Formula
•Fischer projection formula
•Saw-horse formula
•Flying wedge formula
•Newman projection formula

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Structural Formula
• A structural formula displays the atoms of the molecule
in the order they are bonded. It also depicts how the
atoms are bonded to one another, for example single,
double, and triple covalent bond. Covalent bonds are
shown using lines. The number of dashes indicate
whether the bond is a single, double, or triple covalent
bond.
• Ex. Structural Formula for Ethanol:
Presentation of Molecules
C C O H
H
H
H
H
H
Condensed Structural Formula
• Condensed structural formulas show the order of
atoms like a structural formula but are written in a
single line to faster save space and make it more
convenient and to write out. Condensed structural
formulas are also helpful when showing that a group
of atoms is connected to a single atom in a
compound.
• Ex. Condensed Structural Formula for Ethanol:
CH3CH2OH
• (Molecular Formula for Ethanol C2H6O).
Presentation of Molecules
Line-Angle Formula
• Because organic compounds can be complex at
times, line-angle formulas are used to write carbon
and hydrogen atoms more efficiently by replacing the
letters with lines. A carbon atom is present wherever
a line intersects another line. Hydrogen atoms are
then assumed to complete each of carbon's four
bonds. All other atoms that are connected to carbon
atoms are written out.
• Ex. Line-Angle Formula for Ethanol:
Presentation of Molecules
OH
Fischer projection formula - The Fischer projection,
devised by Emil Fischer in 1891,is a two-dimensional
representation of a three-dimensional organic molecule by
projection.
• All non-terminal bonds are depicted as horizontal or
vertical lines. The carbon chain
is depicted vertically, with carbon
atoms sometimes not shown
and represented by the center
of crossing lines (see figure below).
Presentation of Molecules
C
C
O H
H
H
O
H
O
3 H
H
CO H
O
CH
3
H
O
Cl
• The orientation of the carbon chain is so that the first
carbon with most oxidized state (C1) is at the top.
• Horizontal groups are in front of plane
• Vertical groups are behind plane
Presentation of Molecules
H
H
CO H
O
CH3
H
O
Cl
CO H
O
CH3
H
O
H
H
Cl
C
C
O H
H
H
O
H
O
3
C
O H
H
H
O
H
O
3
C
H H
O
O H
O
C
CH
3
Saw-horse formula
• They are called sawhorse projections because
the eclipsed conformation looks like a
carpenter's sawhorse.
• The molecule is viewed side on
Presentation of Molecules
H
CH
3
H
O C H
OO
H
Cl
H
CH
3
HO
C H
OO
H
Cl
H
CH
3
HO
C H
OO
H Cl

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• Plane line represents the bonds which are in plane
• Wedge represents the bonds which are in front of
plane
• Dotted line represents bonds which are behind the
plane
Presentation of Molecules
H
CH3
HO
C H
OO
H
Cl
CH3
HO
H
C H
OO
H
Cl
H
CH3
H
O C H
OO
H
Cl
CH
3
H
O
H
C H
OO
H
Cl
• Fisher to saw horse conversion
• The bottom chiral center is written at the front
• Top chiral center is written at rear, Substituents
placed in front are Up, Substituents placed behind
are Below Rotate front or behind chiral center
through 1800
Fischer formula represents molecule in most unstable
conformation i.e. eclipsed conformation
Presentation of Molecules
H
H
CO H
O
CH
3
H
O
Cl
H
CH
3
HO
C H
OO
H Cl
H
CH
3
HO
C H
OO
H
Cl
H H
O
O H
O
C
CH
3
Flying wedge formula
• It uses a wedge to show a bond coming out of
the paper (or up) and a dash to show a bond
going into the paper (or down).
• These are very often used to depict molecules
with single asymmetric (chiral) center
Presentation of Molecules
C
C
CH3
H
HO
CO H
O
H
Cl
• Newman projection formula :we are looking
straight down the bond between two carbon
atoms. Therefore, the first carbon ends up
completely blocking the second carbon from
view
Presentation of Molecules
H
H
CH
CO H
O
H
O
Cl
H
H
CO H
O
CH
H
O
Cl
3
Interconversion of Molecular Formula
H
H
CO H
O
CH
3
H
O
Cl H
CH
3
H
O C H
OO
H
Cl
H
CH
3
HO
C H
OO
H
Cl
C
C
CH
3
H
HO
CO H
O
H
Cl
Representation of ethane and n-butane
H
H
H
H
H
H
H H
H
H H
H
C
H3
CH3
H
H
H
H
H
H

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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
C
C
H H
H
H
H
H C
C
H
H
H
H
H
H
Rotate
through
1800
1 2
3
4
5
6
7
H
H
H
H
H
H
H H
H
H H
H
C
H3 CH3
H
H
H
H
H
H
H
H
H
H
H
H
Rotate back
C through
1800
Representation of ethane and n-butane
CH3
H
H
CH3
H
H
H2C CH2
CH3
CH3
H H
CH3
H H
CH3
CH3
H
H
H
H
CH3
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
C
C
H
H
H
H
H
H
C
C
H
H
H
H
H
H
Rotate
through
1800
Conformational Isomerism
• Conformational isomers – They are
stereoisomers that can be converted into one
another by rotation around a single bond or
by flipping in case of ring system.
• Conformational analysis - The study of the
energetics between different conformations is
referred to as conformational analysis.
• It is useful for understanding the stability of
different isomers
•
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
Staggered
Staggered
Skew Eclipsed
Skew
Conformational Isomerism in ethane
• Conformational analysis of ethane with the help of
energy profile diagram
H
H
H
H
H H
H
H
H
H
H
H
P.E.
12.0KJ/Mol
Staggered
Eclipsed
Skew
60 120 180 240 300 360

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H H
H
H
CH3
CH3
Syn-periplanar
CH3 - CH3 angle
00
H H
H
H
CH3
CH3
Skew
Gauche
CH3 - CH3 angle
600
syn-clinal
CH3
H H
H
H
CH3
anti-periplanar
CH3 - CH3 angle
1800
H H
H
H
CH3
CH3
Skew
Gauche
CH3 - CH3 angle
3000
anti-clinal
H
H
H
H
CH3
CH3
anti-clinal
CH3 - CH3 angle
1200
H
H
H
H
CH3
CH3
anti-clinal
CH3 - CH3 angle
2400
600 600
600
600
600
600
Conformational Isomerism in n-butane
• Conformational analysis of n-butane with the
help of energy profile diagram
60 120 180 240 300 360
P.E.
18.25KJ/Mol
Syn-periplanar
anti-periplanar
anti-clinal
gauche
14.70KJ/Mol
4.00KJ/Mol
D and L is (used for absolute configuration) totally different than
d and l [or (+) or (-) used for optical rotation]
• System introduced by Fischer
• Configuration is compared to glyceraldehyde and assigned
•For D configuration the OH group is on right hand side
•For L configuration the OH group is on left hand aside
D & L Nomenclature
CHO
OH
H
CH2
OH
C
CHO
H
CH2
OH
C
HO
D-glyceraldehyde L-glyceraldehyde
D & L Nomenclature
COOH
OH
H
CH2OH
C
COOH
OH
H
CH3
C
CHO
OH
H
CH2
CH3
C
All have OH on right hand side so all are D
CHO
H
CH2OH
C
H2N
COOH
H
CH2OH
C
H2N
COOH
H
CH2
Ph
C
H2N
All have NH2 on right hand side so all are L
R & S Nomenclature
Sequence rules or Standard rules [R = RECTUS, S = SINISTER]
1.Higher atomic number precedes lower S>F>O>N>H
2. Higher atomic mass number precedes lower T > D > H
3.Cis precedes trans, R precedes S and Z precedes E
How to apply rules
1. Consider the atoms directly attached to Carbon
2. Assign them priorities as per sequence rules
3. If decision is not reached at first connected atom then move
on to next atom
4. In case atoms with multiple bonds are present then atoms to
which they are attached they are duplicated or triplicate
depending on double or triple bond
5. In case of tri substituted N the lone pair has zero value i.e.
lowest priority
R & S Nomenclature
Sequence rules or Standard rules
1.Higher atomic number precedes lower e.g. S>F>O>N>H
H < C < N < O < F < P < S < Cl < Br < I < ..
2. Higher atomic mass number precedes lower T > D > H
3.Cis precedes trans, R precedes S and Z precedes E
How to apply rules
1. Consider the atoms directly attached to Carbon
C CH2
OH
I
OH
NH2
Consider C not O Four atoms directly
attached to “C” are
I, O, N and C
Priority will be like
I(53) > O(8) > N(7) > C(6)

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R & S Nomenclature
2. Assign them priorities as per sequence rules
3. If decision is not reached at first connected atom then move
on to next atom
C CH2OH
I
CH2CH3
NH2
Both are C
O > C
C < O
1
2
4
3
R & S Nomenclature
4. In case atoms with multiple bonds are present then atoms to
which they are attached they are duplicated or triplicate
depending on double or triple bond
5.In case of tri substituted N the lone pair has zero value i.e.
lowest priority
C
O
H
O
C COOH
I
CHO
NH2
1
2
4
3
C
O
H
C
O
OH
C
O
H
O
X
O 2 = 8 X 2 = 16
X
O 3 = 8 X 3 = 24
C
N
N X 3, 7 X 3 = 21
R & S Nomenclature
VERY good Rule
Vertical is good or correct
1. Assign the priorities to all four groups using sequence rules
2. Move from 1 to 2 , 2 to 3 if it is clockwise the configuration
is R and if anti-clockwise then it is S
3.Lowest priority group should be in vertical position (at top or
bottom ) then follow normal assignment.
4.If lowest priority group is in horizontal position the REVERSE
THE ASSIGNMENTi.e. R to S and to R
H
C C
F
I
Cl
Br
1
2
3
4
4th priority group at
VERTICAL position
1
2
3
4
H
C
1
2
3
4
C CN
I
OH
NH2
1
2
3
4
4th priority group at
HORIZONTAL position
R & S Nomenclature
VERY good Rule
Central C is attached to four groups or atoms as O, N, C
and H
Atomic numbers are O = 8, N = 7, C = 6 and H = 1
OH will have priority 1,
H2N will have priority 2,
CH2OH will have priority 3
H will have priority 4
It is clockwise so R
NH2
H
CH2
OH
C
HO
1
2
3
4
1
2
3
Clockwise so R
R & S Nomenclature
VERY good Rule
CHO
H
CH2OH
C
HO
1
2
3
4
1
2
3
Clockwise so R
Central C is attached to four groups or atoms as O, C, C
and H
Atomic numbers are O = 8, C = 6, H = 1
OH will have priority 1, H will have priority 4
CHO can be taken as H-C-O X 2 [1+8+8 = 17]
CH2OH can be taken as O-C-H X 2 [8+1+1 = 10]
Thus CHO will have higher priority and will be 2
CH2OH will have priority 3 It is clockwise so R
R & S Nomenclature
VERY good Rule
Central C is attached to four groups or atoms as O, C, C
and H
Atomic numbers are O = 8, C = 6, H = 1
OH will have priority 1, H will have priority 4
CHO can be taken as H-C-O X 2 [1+8+8 = 17]
CH2OH can be taken as O-C-H X 2 [8+1+1 = 10]
Thus CHO will have higher priority and will be 2
CH2OH will have priority 3 It is anti-clockwise so S
CHO
OH
H
HOH2C C
1
2
3
Anti-clockwise so S
1
2
3
4

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8
R & S Nomenclature
VERY good Rule
Central C is attached to four groups or atoms as O, N, C
and H
Atomic numbers are O = 8, N = 7, C = 6 and H = 1
OH will have priority 1, H2N will have priority 2,
CH2OH will have priority 3 and H will have priority 4
It is clockwise so R but lowest priority group is
horizontal so R will Change to S
NH2
H
CN
C
HO 1
2
3
Clockwise so R
1
2
3
4
but 4th priority group
is horizontal so
R will be changed to S
R & S Nomenclature
VERY good Rule
Central C is attached to four groups or atoms as O, N, C
and H
Atomic numbers are I = 53, O = 8, N = 7 and C = 6
I will have priority 1, OH will have priority 2
H2N will have priority 3, and CN will have priority 4
It is clockwise so R but lowest priority group is
horizontal so R will Change to S
C CN
I
OH
NH2
1
2
3
4
1
2
3
Clockwise so R
but 4th priority group
is horizontal so
R will be changed to S
R & S Nomenclature
CN
H
CH2OH
C
HO
1
2
3
4
COOH
H
CH3
C
HO
1
2
3
4
1
2
3
Clockwise so R
I
CN
C
HO
H2
N
1
2
3 4
NH2
H CN
C
HO 1
2
3
4
1
2
3
Anti-clockwise so S
but 4th priority group
is horizontal so
S will be changed to R
E & Z Nomenclature
Z = ZUSAMMEN indicating together
E = ENTGEGN indicating on opposite side
This nomenclature is more clear than cis –trance as in some
cases the cis –trans system fails e.g.
CH3
H
C
H3
H
H
CH3
C
H3
H
Cis Trans
CH3
O
H
H Ph
Cl
OH
C
H3
H
E & Z Nomenclature
How to apply?
1.Use same sequence rules as that of R and S
2. Assign priorities to substituent present on both side of double
bond
3. If same priority substituents are on same side ( both 1,1 and
2,2 on above or below side) the assign Z
4. If same priority group are not on same side and combination is
like (1,2 and 1,2) then assign E
CH3
O
H
H
Ph
1
2
1
2
TOP
BOTTOM
1 and 1 on same side
so it is Z
CH3
O
H
H Ph
1
2
1
2 TOP
BOTTOM
1 and 1 on same side
so it is E
E & Z Nomenclature
Ph
N
OH
C
H3
2
1 1
2
N
OH
C
H3
H
1
1
2
2
E
O = 8 and Electon = 0
so OH will have 1st priority
Z
E
N
OH
H
Ph
1
1
2
2 H
N
OH
C
H3
Z
1 1
2
2

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E & Z Nomenclature
Cl
H
Cl
H
H
Cl
Cl
H
E
Z
1 2
1 1
1
2
2
2
CH3
O
H
C
H3
I
1
2
1
2
TOP
BOTTOM
Z
CH3
O
H
H
Cl
1
2
1
2
TOP
BOTTOM
Z
E & Z Nomenclature
CH3
O
H
C
H3
CH2CH3
E
CH3
O
H
Br
CH2
CH3
1
2
1 2
TOP
BOTTOM
1
2
1
2
TOP
BOTTOM
Z
CH2Br
O
H
C
H3
CH2CH3
1
2
1
2 TOP
BOTTOM
Z E
CH2CH2Cl
O
H
Br CH2CH3
1
2
1 2
TOP
BOTTOM