This is on stereochemistry.

1. 1
CH 103: STEREO CHEMISTRY
AND KINETIC MOLECULAR
THEORY
By
Dr. M. Sithambaresan
B.Sc.special (Jaffna), M.Phil. (Peradeniya), Ph.D. (Kerala)

2. Stereo chemistry:
Relationship between the structure and properties is
called stereochemistry. Stereochemistry deals with the
spatial arrangements of atoms in group of molecule.
Isomers
Structural isomers /
constitutional isomers
Stereoisomers
Positional
isomers
Functional group
isomers
Conformational
isomers
Configurational
isomers
Geomatrical
isomers
Optical
isomers
Enantiomer
Diastereo isomer
Eg: Cis and Trans
Types of isomerism and Nomenclature of isomers

3. 3
Structural isomers/ Constitutional isomer:
Isomers whose atoms have a different connectivity is called as constitutional
Isomers. Eg: Butane, Isobutane.
CH3CH2CH2CH3 H C
CH3
CH3
CH3
1) Positional isomers
Compounds having the same number and same kinds of atom, but having
different arrangements between the atoms are called positional isomers
Eg: C3H6O2
HOCH2CH2CHO CH3CCHO
OH
H

4. 4
2) Functional group isomers
Compounds having the same number and same kind of atoms,
but having different functional groups.
Eg: C3H6O
HOCH2CH2CHO / CH3CH2COOH
Stereo isomers:
Isomers having the same structures but different spatial arrangements
1) Conformational isomers/ Rotational isomers/ Conformers
Different conformations corresponding to energy minima
Conformation
Different arrangements of atoms that can be converted into one another by rotation
about single bonds are called conformation.
Eg: Butane
1
0
120
C
H3C
CH3
2
3
4
C
H
#
*
H
H
H
1
C
CH3
2
3
4
CH
#
*
H
H
H
CH3
(1) (11)

5. 5
• If the second carbon atom of structure I is rotated by
120o
around the C2-C3 carbon-carbon single bond you
get the structure II
• These conformational isomers differ in energy and
exist in different proportions in any sample of the
compound. In such situations, the lowest energy of the
isomer is the main isomer. Conformational isomers
can be interconverted by rotation about single bond.
• Normally we use two types of representation to denote
the conformation of the molecules.
– Newman projection
– Sawhorse representation

6. 6
• We view the molecule directly down the C-C
bond axis. So the ‘C’ in front hides the ‘C’
behind.
CC
H
H
H
HH H
H
H H
HH
H
H
0
60
H H
H
H
H
eclipsed conformation Staggered
conformation(less stable)
(more stable)

7. 7
ii) Sawhorse representation
• Here we are viewing the molecule slightly
from above and from the right
H
C
C
H
H
H H
H
60
0
C
C
H
H H
H
HH
eclipsed
conformation
staggered
conformation
• In these two conformations staggered is more stable, because
there of the less repulsion between the H atom in C1 –C2. This
is called non-bonded interaction or steric interaction

8. 8
C
H
H
HC
H1
H2H3
600
C
H
H
HC
H1H2
H3
60 0
C
H
H
H
H1
H2
H3
C
H1
H2 H3
C
C
H
H
H
60
0
60
0
C
H
H
HC
H1
H2
H3

9. 9
Potential
energy
dihedral angle
E
60
0
120
0
180
0
240
0
300
0
360
0
eclipsed
staggered
CH3
H
CH3
H
H H
60
0
H
H
CH3
CH3
HH
If we take butane it is more complex molecule. We can draw several conformations

10. 10
60
0
H
CH3
CH3
H
HH H
60
0
CH3
CH3
H
H
H
H
60
0
CH3
CH3
H
HH H
CH3
CH3
H
H
H
H
60
0
CH3
H
HH
H
CH3
60
0
CH3
CH3 H
HH
fully eclipsed
less stable
high potential enegry
gauche
eclipsed anti conformation
more stable form
less potential energy
gauche
eclipsed

11. 11
Potential
energy
dihedral angle
60
0
120
0
180
0
240
0
300
0
360
0
fully eclipsed
eclipsed
gauche
anti conformation
Fully eclipsed conformation is less stable than the eclipsed form because
large methyl groups are very close to each other and steric interaction
between these groups raises the energy of this conformation.
At room temperature butane exists as a mixture of about 70% anti form
and 30% gauche conformations.

12. 12
Configurational isomers
• Two or more molecules that have same
constitution but different configuration are
called configurational isomer.
• Configurational isomers can be
interconverted only by breaking and making
the bonds.
C
Br
H
Cl
C
1
11
H
H
Cl
Br
H
To change configuration I in to II we have to break and remake the
bonds. It has to be done chemically

13. 13
Geometrical isomers
• Consider alkenes molecule
HH
H H
sp2sp2
Due to this double bond, rotation about one of the CH2 groups is difficult

14. 14
Chirality
• Chirality is a property characterizing three dimensional forms which are
not superimposable on their mirror images. Chiral molecules or chiral
objects can’t be superimposed on its mirror image.
• Achiral molecules can be superimposed on its mirror image.
• Chiral molecules are optically active. These optically active compounds
can rotate the plane polarized light.
• If you want to find out whether a molecule is chiral or achiral we have to
look the symmetry elements. There are 4 symmetry elements present in a
molecule.
• Axis of symmetry (Cn).
• Center of symmetry
• Plane of symmetry
• Alternative axis of symmetry.

15. 15
1) Axis of symmetry (Cn)
• A molecule has an axis of symmetry if rotating
the molecule about the axis by an angle of
360/n produces a new structure
indistinguishable from the original molecule
0
120
H
C
H
H
H
H
C
H
H
H
360/n = 120
n=3 So methane has C3
axis
C6 axis (principle rotation axis)
C C
H
HH
H
C2 axis

16. 16
2) Center of symmetry (i)
• A molecule is said to have a center of
symmetry if all straight lines that can be
drawn through the center of the molecule
meet identical atoms at same distance
from the center.
C
CO
NH
NH
CO
C
CH3
H H
CH3
no center ofsymmetry
cis compound
C
CO
NH
NH
CO
C
CH3
CH3
H
H
center ofsymmetry
trans compound

17. 17
3) Plane of symmetry (σ)
• A plane passing through the molecules such
that all the atoms on one side of the plane
are reflect through the plane. So, this plane
functions as a mirror
COOH
H OHC
C
COOH
OHH
B
Cl
Cl
Cl
COOH
H ph
H H
HHOOC
ph

18. 18
4) Alternating axis of symmetry (sn) or
rotation reflection axis of symmetry or
improper axis of symmetry
• A molecular has an alternating axis of
symmetry of order (n) if rotation about the
axis by 360/n degree following by reflection
in a plane perpendicular to this axis
produces an equivalent structure.
COOHCOOH Ph Ph
H
H H
H
COOH Ph
COOH
H
Ph
H H H
0
180
rotation
Ph
H
COOH
HH
Ph
H
COOH
reflection

19. 19
• According to the symmetry elements molecules
are classified in to two types.
– Dissymmetric molecules => does not have i, σ or Sn .
– Asymmetric molecules => does not have Cn, i, σ or Sn
• Both asymmetric and dissymmetric molecules will
be chiral molecules.
• Chiral molecules do not have center of symmetry,
plane of symmetry alternative axis of symmetry.
• Achiral molecules will have above this.
• When a molecule contains n chiral centers there
are 2n
chiral molecules

20. 20
Optical isomers
• Enantiomers
Two molecules that are related as objects and non super
imposable mirror image are called enantiomers.
• Enantiomers are identical in chemical and physical properties.
• Eg. Same boiling point and melting point.
They differ in their reaction to plane polarized light and also
differ in their reactions to chiral reagents
OHH
CH3
CHO
CH3
HO H
CHO

21. 21
Diastereo isomers
• Configurational isomers which are not mirror
images and are not identical but which can be
converted one in to the other only by
changing the configuration at one or more
chiral centers are called diastereo isomers/
diasteromers
CHO
CH OH2
OH
H
H
HO
CHO
CH OH
2
OH
H
H
OH

22. 22
Keep fit
• Verify the following isomers which are
enantiomers or diastereo isomers.
CHO
CH2OH
OH
HO H
H
CHO
CH2OH
HO H
H OH
CHO
CH2OH
HO
HO
H
H H
CHO
CH2OH
OH
OH
H
1 11 111 1V

23. 23
Meso compound
• Meso compound is one whose molecules are
super imposable on their mirror images even
though they contains chiral centers. These
molecules are optically inactive
COOH
H OHC
C
COOH
OHH

24. 24
Optical activity
• A substance which changes the direction of plane polarized
light is called the optically active substances.
• If a compound rotates the plane of polarization in a clockwise
direction the compound is said to have positive rotation or to
be dexorotatory (D).
• If it rotates the plane in anticlockwise direction it is said to
have a negative rotation or to be laevorotatory (L).
• Plane polarized light
When a light is pass through a Ni prism the light is said to be
plane polarized. I.e. all vibrations are in one plane.
Ordinary light has vibrations in all directions.

25. 25
Specific rotation
• Rotations can be measured using a polarimeter, in
which monochromatic light (i.e. light of a single
wave length) is passed through a polarizer to
convert it into plane polarized light (Usually sodium
light is used to produce the plane polarized light).
• Then the plane polarized light is directed through a
tube containing the sample dissolved in achiral
solvent.
• Then the angle of rotation (α) can be measured by
analyzer

26. 26
• The observed angle of rotation is proportional to the
concentration of the solution and the length of the
sample tube
25
D
25
D
α ∞ C l
α = [α] C l
C – concentration in g/cm3
L – path length in dm
D – Wave length of the light (wave length of sodium light – 589 nm)
If the substance is liquid,
[α] = α / ρ. L
Specific rotation is constant for one compound and it is a characteristic property

27. 27
Racemic mixture or Racemic
modification ( )
• A mixture of equal parts of enantiomers is called a
racemic modification.
• A racemic modification is optically inactive.
Separation of the two enantiomers in a racemic mixture.
 This process is called as resolution.
 The enantiomers making up a racemic mixture have
identical physical properties and hence cannot be
separated by the usual methods of fractional distillation
or fractional crystallization.
 There are several ways to resolve the racemic mixture.
±

28. 28
1. Mechanical separation
This method is applicable only for racemic mixtures
where the crystal form of the enantiomers looks
quite different. From the shape of the crystal we can
separate the two enantiomers by hand. But normally
it is difficult. Therefore it is not a useful method.
2. Inoculation
We make a supersaturated solution of racemic
mixture and then introduce one crystal of pure
enantiomer of the solution.
3. Bio-chemical separation
Certain bacteria and moulds when they grow in a
dilute solution of a racemic mixture, it destroys one
enantiomer more rapidly than the other.
Eg. Penicillium glaucum (a mould) , when grow in a
solution of racemic ammonium tatrate, attracks the
(+) form and leaves (-) form.

29. 29
Disadvantages of this method (Bio-chemical separation)
 Dilute solution must be used, and so the amounts obtained will
be small.
 One form is always destroyed and the other form is not always
obtained 50% yield since some of this may also be destroyed.
 It is necessary to find a micro organism which will attack only
one of the enantiomer.

30. 30
Conversion of enantiomers in to
diastereo isomers
• Seperation of racemic mixture of carboxylic acids (we react
the enantiomer with chiral reagents)
.
.
(-)
dil.HCl
(+)RCOOH RCOOH
dil.HCl
RCOOH+( )_ react with
optically active
alkaloid
(-)2 brucine
+ R-COO-N
(-) (+)
(-)(+)
Salts ofdiastereo
isomer
R-COO-N
(-) (+)
(-) (-)
(+)
R-COO-N
(-) (+)
(-)
R-COO-N
(-) (+)
(-) (-)
fractional
crystallization

31. 31
• These two salts have different solubility.
Therefore we can separate them by fractional
crystallization.
• After the separation of two diastereoisomers,
the salt was allowed to react with dil. HCl.
• Three steps involved in this process.
• Chemical reaction.
• Physical reaction => crystallization.
• Chemical reaction.
• The alkaloids commonly used for the
resolution are
(-) Brucine
(-) Quinine
(-) Strychine
(+) Cinchonine.

32. 32
Separation of racemic mixture of
the bases
(+)_
Salts ofdiastereo
isomer
amines
(+) acid
(+) (+) salt and (-) (+) salt
(base)
fractional
crystallization
(+) (+) Salt (-) (+) salt
NaOH(base hydrolysis)
(+) amine + acid (-) amine + acid

33. 33
Examples of chiral acids
COOH
CH - OH
COOH
CH2 O
CH3H3C
HO3SH2C
Campher – 10 – sulphonic acidMalic acid

34. 34
Separation of racemic mixture of an
alcohol
• Since alcohols are neither basic nor acidic, they cannot be
resolved by direct formation of salts.
• First they were attached to an acidic ‘handle’ which permits
the formation of salts and then can be resolved
ROH + dicarboxylic acid or
O
O
O
anhydride
pyridine
O
( )+-
( )+-
O
OR
OH
half est ers
+
(-) Brucine
OH.Brucine
O
O
OR
OH.Brucine
O
O
OR
(-) (-) salt (+) (-) salt
diast ereomeric salt
(-) (-) salt (+) (-) salt
fractional crystallization
(-) ROH (+)ROH
NaOH NaOH

35. 35
Recemisation
• Conversion of one enantiomer in to racemic mixture is
called recemisation

36. 36
Base
R1
R2
R3
HC
R1
R2
R3
C(-)
R1
R2
R3
C(-)
H
+
R1
R2
R3
H C+-

37. 37
Nomenclature of optically active
compounds
1) Erythro / Threo system of nomenclature
• If the same groups are attached in the same direction then the
compound is erythro compound.
• If the same groups are attached in the opposite direction then the
compound is threo
H
CHO
CH2OH
OH H
HO
CHO
H
HHO
HO
CH2OH
OH
CHO
H
HHO
CH2OH

38. 38
R / S system of nomenclature
• According to this system it specifies the configurations of each chiral
centre in a molecule. Here we use a set of rules called sequence rules.
These rules are used to determine the order priority
• Sequence rule.
• 1) Atoms of higher atomic number get highest priority than the atoms of
lower atomic number
• 2) Isotopes of higher atomic weight gets highest priority than the isotope
of lower atomic weight.
• 3) Lone pair electrons get lowest priority.
I Br Cl F
1 2 3 4
, , ,
H H
1
1 1
2
,
(2) (1)
P
C2H5H3C
H
(1)(2)
(3)
(4)

39. 39
4) The atom directly bonded to the chiral carbon determines the priority of
the groups. If two atoms attached to the chiral centre are the same, we
consider the next atom attached to the first atom
5) Where there is a double or triple bond, both atoms considered to be
duplicated or triplicated.
(4)
C
HOH2C
C2H5
CH3
H
(1)
(2)
(3)
1) CH3
CH3CH2
CH(CH3)2
C
H (1)
(2)
(3)
(4)
2)
O O
O
CH2NH2
C
N
C1)
C2)
Priority
N
N
N
C
N
N
N
C

40. 40
• The molecule is then oriented so that the atom
with the lowest priority is directed away from
the observer.
• The other three substituents will then face the
observer radiating outwards from the central
carbon atom.
• If the order of priorities of these three groups is
in a clockwise manner the molecule is said to
have the R – configuration while if the order
follows an anti-clockwise manner it is said to
have the S-configuration.

41. 41
C
OH
CO2H
H
CH3
1
2
3
4 1
HO
4
3
2
CH3
H
C
CO2H
1200
anti clockwise=> S
If the lowest priority compound is in the vertical line (top/bottom) the sequence give
correct descriptor. If it is in the horizontal line the descriptor should be reversed

42. 42
E/Z system of nomenclature
• This naming is used for double bonds. If the groups of higher
priority are opposite sides it is named as E.
• If the groups of higher priority are in same sides it is named
as Z.
C C
Cl
Cl
H
Br
Z compound
1 1
22
Cl
HBr
Cl
C C
E compound
1
1
2
2

43. 43
Conformation in cyclic system
• A stable chemical bond is formed when the orbitals
of two bonding atoms overlap.
• For a given pair of atoms, the greater the overlap of
atomic orbital, the stronger the bond.
• When carbon is bonded to four other atoms, it’s
bonding orbitals (sp3
) are directed to the corners of
a tetrahedron; the angle between any pair of
orbitals is 109.5o
.
• Formation of a bond with another carbon atom
involves overlap of one of these sp3
orbital with a
similar sp3
orbital of another carbon atom. C-C-C
bond angle should be 109.5o
.

44. 44
Cyclopropane
• Cyclopropane is a planar molecule. Bond angle (C-C-
C) is 60o
.
• According to this, cyclopropane has a bond angle
compression of 109-60 = 49o
.
• It has angle strain or baeyer strain and therefore it is
highly reactive.
• Here the H atoms above the plane of the ring are
eclipsed with one another and those below the ring
are also eclipsed with one another.
• The highly unstable character of cyclopropane is due
to both ring strain and the eclipsing of the ring
substitution.

45. 45
• Cyclopropane is more reactive than normal
propane. Why?
C CC
C
C C
Poor or partial overlap takes place. Due to this poor overlap cyclopropane
bonds are weaker and more reactive (angle strain). In addition to angle strain
there is eclipsing strain /torsional strain.

46. 46
Cyclobutane
• Same as cyclopropane. But here angle strain is lesser
than that of cyclopropane.
• Bond angle compression = (109o
-90o
) =19o
.
• Cyclobutane has more eclipsing strain than cyclopropane
by its larger number of ring H’s. To decrease the eclipsing
strain cyclobutane take the bend conformation.
H
H H
H
H
H
H
H

47. 47
Cyclopentane
• Has practically no angle strain. But it has
considerable eclipsing strain.
• To decrease the eclipsing strain, it take envelop
confirmation.
• In this conformation four carbon atoms are in a
plane and one is above the plane.
109 108
00

48. 48
Cyclohexane
• No angle strain practically. But the planar molecule has eclipsing /
torsional strain. To decrease the torsional strain, it take chair conformation
or boat conformation.
Chair conformation
109 120
00
equitorial
axial
flipping
H
H
H
H
H
H
H
H
H
H
H
HH H
H
H
H
H
H
H
H
H
H
H 1
2
3
4
5
6
1
2
3
4
5
6
newman
projection
1a, 2a  Trans
1e, 2a  Cis
1a, 3a  Cis
1a, 3e  Trans
Chair conformations are strain free.
They have neither angle strain nor
eclipsing strain.

49. 49
Boat conformation
• Interaction between this flagpole hydrogen is called flagpole
interaction / steric interaction.
H
H
H HH
H
H
H
H
H
H H
flagpole/ steric interaction
newman
projection
H
H
H
H
H
H
H
1
2
3
4
5
6
H

50. 50
The order of stability
• Chair conformation 〉 twist boat conformation〉 boat conformation
chair
potential
energy
half chair
boat
twist
boat
Chair half chair twist boat/
skew boat
boat

51. 51
Mono substituted cyclohexane
H
me
two 1,3 diaxial interaction/
two gauche butane
high energy
less stable
5%
95%
less energy
stability high
H
me
me
H
ring flipping

52. 52
Disubstituted cyclohexane
a) 1,2-disubsituted cyclohexane
cis
me
me
50% 50%
ring flipping
me
me
H
H
me
me
H
H
2 gauche-butane
int eract ion &
1 gauche int eract ion.
2 gauche-butane
interact ion &
1 gauche int eract ion.

53. 53
Trans
ring flipping
me
me
me
me
me
me
4 gauche-butane
interaction
energy high
stability less
>1%
<99%
1 gauche interaction
less energy
highly stable

54. 54
1,3 disubstituted cyclohexane
cis
ring flipping
me
me
me
me
me
me
4 gauche-butane
interaction
energy high
stability less
>1%
<99%
no gauche
-
butane
interaction

55. 55
Trans
ring flipping
me
me
me
me
me
me
2 gauche-butane
interaction
50% 50%
2 gauche-butane
interaction

56. 56
ExampleCl
Cme3
ring flipping
Cl
Cl
2-Cl - H (1,3 diaxial
interaction) = 2*0.25
= 0.5 Kcal/mol
Stable conformation
2
1,3 diaxial interaction
2*2.7=5.4 Kcal/mol
(CH3)2C H-

57. 57
1,4 disubstituted cyclohexane
Cis
ring flipping
me
me
me
me
me
2 gauche-butane
interaction
50% 50%
me
2 gauche-butane
interaction

58. 58
Trans
ring flipping
me
me
me
me
me
me
4 gauche-but ane
int eract ion

59. 59
Optical activity of biphenyls and
allenes
• It was subsequently established by dipole moment and x-ray
diffraction data, that the benzene rings in biphenys are co-axial
• i.e. 2 benzene rings are perpendicular to each other.
• The steric interaction of the H atoms is not however large enough to
prevent free rotation completely.
• Compounds in which at least three of the four ortho-positions in
biphenyls are occupied by certain groups could be resolved =>
optically active.

60. 60
Two conditions were necessary for biphenyl
compounds to exhibit optical activity
1) Neither ring must have a vertical plane of symmetry.
Eg=>
=> Plane of symmetry. Optically inactive (not resolvable)
=> Absence of plane of symmetry. Optically active
A
A B
B
A
A B
BA B

61. 61
2) The substitutent in the ortho positions must have a large size
Eg. i) 6- nitro diphenic acid
ii) 6,6’ dinitro diphenic acid
• Here ortho H’s in biphenyl are replaced by bulky groups. These
groups will sterically interact strongly with each other and
completely preventing the free rotation.
• So the molecule cannot exist in the plannar conformation.
COOHHOOC
NO2
COOHHOOC
NO22ON

62. 62
Allenes
• General structure of allenes
• Some allenes can also show chirality in the
absence of asymmetric C atoms. Both
terminal C atoms in allenes must be sp2
hybridized and since the central atom has to
form 2π bonds, it should be sp hybridized.
• The overlap of P orbital in allene
C C C
H
H
H
H
CCC
H
H
H
H

63. 63
• The p orbitals of the terminal C atoms must be
perpendicular to each other.
• If an allene is substituted it will exist in enantiomeric
forms because the substituent at the two ends is in
planes at right angle to each other
• Note: The biphenyls and the allenes provide
examples of compounds which are chiral although
they don’t possess asymmetric carbon atoms
CCC
H
H
Cl
Cl
C C C
H
H
Cl
Cl

64. 64
Stereo selective reaction
• A stereo selective reaction is a reaction that yields
predominantly one enantiomer of a possible pair
or one diastereo isomer of several
diastereoisomers.
• i.e. There are many possibilities to give products.
But one reaction will takesplace selectively.
• Eg. E2 elimination (highly stereo selective).
• Some elimination reaction undergo cis elimination
and some under go trans elimination. After the
elimination reaction the product will have specific
structures according to the type of reaction taking
place

65. 65
Elimination reaction
1) Cis or syn Elimination reaction.
2) Trans or anti elimination reaction.
Cis elimination
• If the groups are lost from the same face it
is known as syn or cis elimination
-(ZY)
Z
Y
C C

66. 66
Anti elimination
• If the eliminated groups are from opposite faces of
the developing double bond it is known as anti or
trans elimination
• E2 elimination typically involves anti-elimination.
• In the transition state the H and the leaving groups
are located in the anti-relationship (Trans coplanar
elimination)
-(ZY)Z
Y
C C
X
H
B
HB
+
+C C

67. 67
In newman projection
Trans elimination is easy than cis elimination because of the trans coplanar
elimination. Vacant p orbitals of C atoms are found on one plane and thus it is
easy to overlap and form the new double bond
Rate of elimination of HCl from chlorofumaric acid is very much faster
than the elimination of HCl from chloromalic acid.
Because in the chlorofumaric acid trans elimination is possible
H
X
B
HB
+
+
containing = bond between C1 and C2
1
2
C C
ClHOOC
COOHH
H
COOHHOOC
Cl
CC
Chlorofumaric
acid
Cholromalic
acid

68. 68

69. 69

70. 70
Examples
C C
ClHOOC
COOHH
H
COOHHOOC
Cl
CC
Trans
Cis
elimination
elimination
COOHHOOC
HOOC COOH
In the chloromalic acid cis elimination takes place. It is difficult than the trans elimination.
C
N
Ph H
CO2me
Ph NC
C
N
Ph H
CO2me
syn elimination
difficult

71. 71
PhCH CHPh Ph Ph
BrBr
H H
C C
Br2
free
rotation
trans elimination
Br
Ph
Br
H
H
Ph
Ph
H
Br
H
Br
Ph
C C
Br
H Ph
Ph
trans elimination
free
rotationBr
Ph
Br
H
H
Ph
Ph
H
Br
H
Br Ph

72. 72
• The elimination of D and L forms give trans olifine product. Elimation of
meso compounds give only cis olifine product.
• The final product of the elimination depends on the conformation of
starting material and type of elimination taking place.
• Here the elimination is stereo specific
free
rotation
trans elimination
Br
Ph
Br
H
H
Ph
Ph
H
Br
H
Br
Ph
C C
BrH
PhPh
cis compound

73. 73
Consider dehydrogenation of Sec-butylchloride
alcoholic
KOH but-1-ene
but-2-ene
+
CH3CH2CHCH3
Cl
CH3CH CHCH3
CH3CH2CH CH2
cis trans
1 6:
Because the trans-2-butane is more stable than the cis.
The difference in stability is attributed to a difference in Vander Waals strain.
In the cis-isomer => 2 butyl -me groups are crowded together on the same side of the
molecule.
In trans isomer situation is not like that => less Vander Waals strain
C
me me
H H
CC
H
H me
me
C

74. 74
Elimination in cyclohexane
compounds
• This doesn’t undergo trans-diaxial elimination.
• In the axial orientation the bonds are perpendicular and in equatorial
orientation the bonds are in the plane. Due to the ring strain, there is
no C-C free rotation. Therefore, the conformation of the compound
is not changed.
• The equatorial substituted conformation is more stable than the
axial substituted conformation, because of the absence of 1,3
diaxial interaction of Cl and H atoms.
• The preference for anti-elimination from halides can be very strong.
In cyclohexane rings 1,2-substituents can take up anti-conformation
only by occupying axial position. This is possible only if they are
trans to each other
H
Cl
Cl
H

75. 75
• Eg. E2 elimination converts neomenthyl chloride into a
mixture of 75% 3 menthene and 25% 2 menthene.
me
H
(CH3)2HC
Cl
H
(CH3)2HC
me
me
(CH3)2HC
H 25%
75%

76. 76
• In menthyl chloride on the other hand only one ‘H is trans to the Cl and it
is the only one that is eliminated despite the fact that this yields the less
stable ene
Cl
CH3
CH(CH3)2
Cl
(CH3)2CH
CH3
(CH3)2CH
CH3
me
H
(CH3)2HC
Cl
me
(CH3)2HC
H
H

77. 77
KINETIC MOLECULAR THEORY
Properties of gases
1) According to the Boyle’s law
V ∞ 1/P
PV= constant [T] [m]
2) According to Charles law
V∞ T [P] [m]
3) According to Avagadro’s hypothesis
V∞ n [P] [T]
By combining Boyle’s law, Charle’s law and avagadro’s hypothesis
V∞ 1/P.n.T
V = RnT/P
PV = nRT
R – Proportionality constant/ Gas constant

78. 78
The Kinetic Molecular Theory of Gases
• This theory is applicable only to a perfect or ideal gas. The main
postulates of the kinetic theory are,
1) A gas is made up of a large number of particles or molecules that are
small in comparison with both the distance between them and the size
of the container.
2) The molecules are in continuous, randomly directed motion.
3) Newtonian mechanics can be used to describe the interaction of the
molecules with the walls of the vessel containing the gas.
4) The molecules are independent of each other and interact only during
brief collisions. Their collisions are perfectly elastic. So, none of the
translational energy is lost by conversion in to internal energy of the
molecules.
5) The kinetic energy due to the translational motion of a mole of a gas
molecules is equal to 3/2 RT

79. 79
Derivation of kinetic gas equation
• Let us consider a certain mass of gas enclosed in a cubic
box at a fixed temperature.
x axis
y axis
z axis
A
B Vx
-Vx
The length of each side of the box = l cm
Total number of gas molecules = N
The mass of one molecule = m
The velocity of a molecule = C

80. 80
Step I: Resolution of velocity C of a single
molecule along X, Y, Z axis
• According to the kinetic theory, a molecule of a gas
can move with velocity C in any direction.
• Velocity is a vector quantity and can be resolved in to
the components Vx,Vy,Vz along the X,Y,Z axis.
• These components are related to the velocity C by the
following expression.
• C2
= Vx2
+ Vy2
+ Vz2

81. 81
Step II: The number of collisions per second
on face A due to one molecule
• Consider a molecule moving in X direction between
opposite faces A and B. It will strike the face A with
velocity Vx and rebound with velocity –Vx.
• To hit the same face again the molecule must travel l cm
to collide with the opposite face B and then again l cm to
return to face A.
• Time between two collisions of face A = 2l/Vx seconds
• Number of collisions per second on the face A
perpendicular to the X direction = Vx/2l

82. 82
Step III: The total change of momentum on
all faces of the box due to one molecule only
• Each impact of the molecule on the face A causes a change of momentum (mass
X velocity)
• Momentum before the impact = mVx
• Momentum after the impact = m(-Vx)
• The change of momentum = mVx – (-mVx) = 2mVx
• But the no. of collisions per second on face A due to one molecule = Vx/2l
• Total change of momentum per second on face A due to one molecule
= 2mVx (Vx/2l)
= mVx2
/l
• The change of momentum on both the opposite faces A and B along X axis =
2mVx2
/l
• Similarly the change of momentum along Y axis and Z axis will be 2mVy2
/l and
2mVz2/l respectively.
• Total change of momentum per second on all faces of the box by one molecule
= 2mVx2
/l + 2mVy2
/l + 2mVz2
/l
= 2m/l (Vx2
+Vy2
+Vz2
).
= 2mC2
/l.

83. 83
Step IV: Total change of momentum due to
impacts of all molecules on all faces of the box
• There are ‘N’ molecules in the box each of which is moving with a
different velocity V1, V2, V3 etc.
The total change of momentum due to impacts of all the molecules on all faces
of the box = 2m (C1
2
+C2
2
+C3
2
+…..) / l
Multiply and divide by ‘N’ =
Mean square velocity
Total momentum transferred per second =

84. 84
Step V: Calculation of pressure from change
of momentum
According to Newton’s law of motion
The rate of change of momentum = Force
C
2
2mN
l
F=
Pressure = Total force
Total area
= 1x
6l2
C
2
2mN
l
6l2
= Area of the
six face cubical
vessel
C
2
= mN
3l3
C
2
= mN
3V
l3
Volume ofcube=P
P V = C
2
mN1/3

85. 85
Kinetic gas equation in terms of kinetic energy

86. 86
Molecular velocity
for one mole
P V = C
2
mN1/3
P V = C
2
1/3mN0
P V
=C
2
mN0
3
(1)
For one mole of gas
PV=RT
=C
2
mN0
3 RT
M - molar mass
C root mean square velocity
C
2 mean square velocity
=C
2 3 RT
M
= 3 RT
C
M

87. 87
Distribution of molecular velocities
• While deriving kinetic gas equation, it was assumed that
all molecules in a gas have the same velocity. But it is not
so.
• When any two molecules collide one molecule transfers
kinetic energy to the other.The velocity of the molecule
which gains energy increases and that of the other
decreases.
• Millions of such molecular collisions are taking place per
second. Therefore, the velocities of molecules are
changing constantly.
• Since the no. of molecules is very large, a fraction of
molecules will have the same particular velocity.
• James Clark Maxwell calculated the distribution of
velocities from the laws of probability.
• He derived the following equation for the distribution of
molecular velocities.

88. 88
dnc
n 4 ( )M
2 RT
3/2 e
2RT
2
mc-
C 2 dc.
n - total no. of molecules
M - molecular mass
T - Temperature on absolute scale(K)
fraction ofthe total no.ofmolecules having velicities between C and (C-
d nc
n
d nc )dc+
no.of molecules having velocities between C and (C+ )dc
velocity
fraction of
molecules
distribution of molecular velocities
most probable velocity
mean(average) T1<T2<T3
T1
T2
T3
Root mean square velocity

89. 89
The important features of the curve are,
1) A very small fraction of molecules has either very
low (close to zero) or very high velocities.
2) There is a certain velocity for which the fraction of
molecules is maxima.
Most probable velocity: is the velocity possessed by
maximum number of molecules of the gas at a given
temperature. This particular velocity corresponds to
the peak of the curve.
Effect of temperature on distribution of molecular
velocities: The entire distribution curve shifts to the
right with rise in temperature. The rise in temperature,
increases the fraction of the molecules having high
velocities.

90. 90
Relation between Average velocity, Root
mean square velocity and most probable
velocity
• Suppose C1, C2, C3….Cn are the velocities of individual
molecules in a gas and N is the total number of molecules
present in the gas.
• Average velocity (Vav)
N
Vav C1 C2 C3 ...... Cn= ++ + +
From Maxwell equation Vav is given by Vav 8RT
m
=
Root mean square velocity (Vrms) Vrms
Vav
=
2 22 2
N
C1 C2 C3 ...... Cn++ + +
We already have seen 3RT
M
=Vrms

91. 91
• Most probable velocity (Vm)
• According to the calculations made by Maxwell
M
Vm = 2RT
The ratio of the three velocities is,
Vm: Vav: Vrms = 1: 1.128: 1.224.

92. 92
Behaviour of real gases
The general gas equation
• PV=nRT
• Derived from the postulates of the kinetic theory is
valid for an ideal gas only. Real gases obey this
equation only approximately and that also under
conditions of low pressure and high temperature.
The higher the pressure and the lower the
temperature the greater are the deviations from the
ideal behavior.
• In general, the most easily liquefiable and highly
soluble gases show larger deviations. Thus gases
like CO2, SO2, and NH3 show much larger deviations
than H2, O2, N2 etc.

93. 93
Deviations from Boyel’s law
P(atm)
ideal gas
CO
H2
He
CH4
PV(litre atm)
ideal gas
PV(litre atm)
P(atm)
N2
H2
0
C0
C
0
40
0
C0
CO2
If Boyle’s law is obeyed, the value of PV for a given quantity of a gas should be
constant at all pressure.
But real gases deviate from this ideal behavior.
CO and CH4 at low pressures are more compressible than Boyle’s law requires.
This continues with increase in pressure till PV passes through a minimum at a
certain stage with further increases in pressure the compressibility is less than
expected and this continues throughout.

94. 94
Effect of temperature on deviations from
ideal behaviour
PV
P(atm)
-25
0
C
20
0
C
50
0
C
Variation of pressure-volume of N2
The PV-P plots of N2 at different temperature varying between -25 o
C and 50o
C.
It is seen that as the temperature is raised the depth in the curve becomes smaller and
smaller.
At 50o
C the curve seems to remain almost horizontal for an appreciable range of
pressure varying between 0 and 100 atm.
General nature of the deviations from ideal behaviour doesn’t depend on the gas, but
rather on the temperature.
Actually the determining factor is the temperature relative to the critical temperature of
the particular gas.

95. 95
Explanation of deviations
Postulate I
The volume occupied by the gaseous molecules themselves is negligibly
small when compared to the total volume occupied by the gas.
Under conditions of high pressure, the volume occupied by the gaseous
molecules will no longer be negligible in comparison with the total
volume of the gas.
The postulate I is not valid at high pressures and low temperatures.
Postulate II
The forces of attraction between gaseous molecules are negligible. This
assumption is valid at low pressures or at high temperature because
under these conditions the molecules lie far apart from one another.
At high pressure or at low temperature, the volume is small and
molecules lie closer to one another. The intermolecular forces of
attraction cannot be ignored.
Hence, the postulate II doesn’t hold under conditions of high pressure
and low temperature.

96. 96
Van der waal’s equation
Van der waals introduce two correction terms in the ideal gas equation
due to two incorrect postulates of the kinetic theory. They are,
1) The molecules in a gas are point masses and posses no volume.
2) There are no intermolecular attractions in a gas
Volume correction
• PV=nRT
• Here volume V of an ideal gas is the same as the volume of the container.
The dot molecules of ideal gas have zero-volume and the entire space in
the container is available for their movement.
• However, Van der waals assumed that molecules of a real gas are rigid
spherical particles which possess a definite volume.
• The volume of a real gas is therefore, ideal gas volume minus the volume
occupied by gas molecules.
• If the volume excluded by one mole of a gas is represented by b, then the
volume correction is (V-b)
• For n moles of gas the volume correction is (V-nb)
• Here b- Excluded volume which is a constant and characteristic for each
gas.

97. 97
Pressure correction
• A molecule in the interior of a gas is attracted by other molecules on all
sides. These attraction forces cancel out.
• But a molecule that strikes the wall of the vessel is attracted by molecules
on one side only. Hence it experiences an inward pull.
• Therefore, it strikes the wall with reduced velocity and the actual pressure
of the gas, P will be less than the ideal pressure.
• P=Pideal – p
• Pideal = P+p
molecular attractions
balanced
inward pull
B
B
B
A

98. 98
• p is determined by the force of attraction between
molecules striking the wall container (A) and the molecules
pulling them inward(B)
• p ∞ CB.CA
• p ∞ n/V. n/V
• a is proportionality constant characteristic of the gas.
• p = an2
/V2
• Van der waals’ equation
• (P + an2
/V2
) (V-nb) = nRT
• Units of a and b
• For a
• p = an2
/V2
• a= PV2
/n2
= (atm) (litre)2
mol-2
• For b
• nb = excluded volume
• b = volume/n = litre mol-1

99. 99
Critical phenomenon and Liquefaction of Gases
• A gas can be liquefied by lowering the temperature and increasing
the pressure.
• At lower temperature, the gas molecules lose kinetic energy. The
slow moving molecules then aggregate due to attractions between
them and are converted in to liquid.
• The same effect is produced by the increases of pressure
Isotherm
• The P-V curves of a gas at constant temperature are called isotherms
or isothermals.
• For an ideal gas PV= nRT and the product PV is constant if
temperature is fixed.
• Andrew plotted the isotherm of carbon dioxide for a series of
temperatures. There are 3 types of isotherms
1) Isotherm above 31o
C
2) Isotherm below 31o
C
3) Isotherm at 31o
C.

100. 100
1) Isotherm above 31o
C
• This isotherm approximates to the isotherm of ideal gas. Thus in the
region above the isotherm at 31o
C CO2 always exists in the gaseous
state.
2) Isotherm below 31o
C
• The isotherm below 31o
C is discontinuous. For example the
isotherm of 21o
C consists of 3 parts.
• i) Curve ‘ab’
• It is a PV curve for gaseous carbon dioxide. Along ab, the volume
decreases gradually with the increase of pressure. At b the volume
decreases suddenly due to the formation of liquid carbon dioxide
having higher density.
pressure
volume
gas &
liquid
Pc
d
c
b 50
0
C
21
0
C
31.1
0
C
0
0
C
f
ea
g

101. 101
ii) The horizontal portion bc
• Along the horizontal part bc of the isotherm, the liquefaction continues
while the pressure is held constant. At c all the gas is converted to
liquid.
iii) The vertical curve cd
• This part of the isotherm is the PV curve of liquid carbon dioxide. This
is almost vertical since the liquid is not very compressible.
iv) Isotherm at 31o
C.
• Above 31o
C there was no possibility of liquefaction of CO2. The critical
temperature of CO2 is therefore, 31o
C. The isotherm efg at this
temperature is called the critical isotherm.
• The ef portion of the critical isotherm represents the PV curve of CO2
gas. At the point f, the curve records a twist which is coincident with the
appearance of liquid CO2. Here the gas and liquid have the same
density and are indistinguishable. The point is called the critical point
and the corresponding pressure is called the critical pressure. Beyond
f the the isotherm becomes nearly parallel to the vertical axis and
marks the bonding between the gases CO2 on the right and the liquid
CO2 on the left

102. 102
Definitions
• Critical temperature (Tc): of a gas maybe
defined as the temperature above which it
cannot be liquefied no matter how great the
pressure applied.
• Critical Pressure (Pc): is the minimum pressure
required to liquefy the gas at its critical
temperature
• Critical volume (Vc): is the volume occupied by
one mole of the gas at the critical temperature
and critical pressure.

103. 103
The End