stereo chemistry and kinetic molecular theory
TRANSCRIPT
1
CH 103: STEREO CHEMISTRY AND KINETIC MOLECULAR
THEORY
By
Dr. M. SithambaresanB.Sc.special (Jaffna), M.Phil. (Peradeniya), Ph.D. (Kerala)
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
OpticalisomersEnantiomerDiastereo isomer
Eg: Cis and Trans
Types of isomerism and Nomenclature of isomers
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
CH31) 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
2) Functional group isomersCompounds 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
ConformationDifferent arrangements of atoms that can be converted into one another by rotation about single bonds are called conformation.Eg: Butane
1
0120
C
H3C
CH3
2
3
4
C
H#
*HH
H
1
C
CH3
2
3
4
CH#
*H
H
H
CH3
(1) (11)
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
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• We view the molecule directly down the C-C bond axis. So the ‘C’ in front hides the ‘C’ behind.
CCH
H
H
HH H
H
H H
HH
HH
060
H H
HH
H
eclipsed conformation Staggered conformation(less stable)(more stable)
7
ii) Sawhorse representation• Here we are viewing the molecule slightly
from above and from the right
H
CC
H
H
H H
H60 0
CC
H
H HH
HH
eclipsedconformation
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
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CH
H
HC
H1
H2H3
600
CH
H
HCH1H2
H3
60 0
CH
H
H
H1
H2
H3
C
H1
H2 H3
CC
H
H
H
60 0
60 0
CH
H
HC
H1
H2
H3
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Potentialenergy
dihedral angle
E
60 0 120 01800 2400
30003600
eclipsed
staggered
CH3
H
CH3
HH H
60 0H
H
CH3
CH3
HH
If we take butane it is more complex molecule. We can draw several conformations
10
60 0
H
CH3CH3
HHH H
60 0
CH3
CH3
HH
H
H600
CH3
CH3
H
HH H
CH3
CH3
H
H
H
H
600
CH3H
HHH
CH3
60 0
CH3
CH3 H
HH
fully eclipsedless stablehigh potential enegry
gauche eclipsed anti conformationmore stable formless potential energy
gauche
eclipsed
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Potentialenergy
dihedral angle
60 0120 0 180 0
240 0 300 0360 0
fully eclipsedeclipsed
gauche
anti conformationFully 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.
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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
BrH
ClC
111
H
HCl
Br
H
To change configuration I in to II we have to break and remake the bonds. It has to be done chemically
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Geometrical isomers• Consider alkenes molecule
HH
H H
sp 2sp 2
Due to this double bond, rotation about one of the CH2 groups is difficult
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.
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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
0120
H
C
H
HH
H
C
H
HH
360/n = 120n=3 So methane has C3 axis
C6 axis (principle rotation axis)
C CH
HH
H
C2 axis
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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.
CCO
NH
NH
COC
CH3
H H
CH3
no center of symmetrycis compound
CCO
NH
NH
COC
CH3
CH3
H
H
center of symmetrytrans compound
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3) Plane of symmetry ()• 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
ClCl
COOH
H ph
H HHHOOC
ph
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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
0180
rotation
Ph
H
COOH
HH
Ph
H
COOH
reflection
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• According to the symmetry elements molecules are classified in to two types.
– Dissymmetric molecules = does not have i, or Sn .
– Asymmetric molecules = does not have Cn, i, 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
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
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 OH2
OH
H
H
OH
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Keep fit• Verify the following isomers which are
enantiomers or diastereo isomers.
CHO
CH2OH
OHHO H
H
CHO
CH2OH
HO HH OH
CHO
CH2OH
HOHO
H
H H
CHO
CH2OH
OH
OHH
1 11 111 1V
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
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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 lightWhen 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
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 () can be measured by analyzer
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• The observed angle of rotation is proportional to the concentration of the solution and the length of the sample tube
25D
25
D
C l = C lC – concentration in g/cm3
L – path length in dmD – Wave length of the light (wave length of sodium light – 589 nm) If the substance is liquid,
= / . L Specific rotation is constant for one compound and it is a characteristic property
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.
±
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1. Mechanical separationThis 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. InoculationWe make a supersaturated solution of racemic mixture and then introduce one crystal of pure enantiomer of the solution.
3. Bio-chemical separationCertain 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.
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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
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 activealkaloid
(-)2 brucine+ R-COO-N
(-) (+)
(-)(+)
Salts of diastereo isomer
R-COO-N(-) (+)
(-) (-)
(+)R-COO-N
(-) (+)
(-)R-COO-N
(-) (+)
(-) (-)
fractionalcrystallization
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
Separation of racemic mixture of the bases
(+)_
Salts of diastereo isomer
amines (+) acid(+) (+) salt and (-) (+) salt
(base)
fractionalcrystallization
(+) (+) Salt (-) (+) salt
NaOH(base hydrolysis)
(+) amine + acid (-) amine + acid
33
Examples of chiral acids COOH
CH - OH
COOHCH2 O
CH3H3CHO3SH2C
Campher – 10 – sulphonic acidMalic acid
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
( )+-
( )+-
OOR
OH
half esters+
(-) Brucine
OH.BrucineO
OOR
OH.BrucineO
OOR
(-) (-) salt (+) (-) salt
diastereomeric salt
(-) (-) salt (+) (-) salt
fractional crystallization
(-) ROH (+)ROH
NaOH NaOH
35
Recemisation• Conversion of one enantiomer in to racemic mixture is
called recemisation
36
Base
R1
R2
R3
HC
R1
R2
R3
C(-)
R1
R2
R3
C(-)
H+
R1
R2
R3
H C+-
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
CHOCH2OH
OH HHO
CHO
HHHO
HO
CH2OH
OH
CHO
HHHO
CH2OH
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 F1 2 3 4
, , ,
H H1
1 1
2,
(2) (1)
PC2H5H3C
H
(1)(2)
(3)
(4)
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)CHOH2C
C2H5
CH3
H
(1)(2)
(3)1) CH3
CH3CH2CH(CH3)2
C
H (1)(2)
(3)
(4)
2)
O O
O
CH2NH2
C
N
C1)
C2)
PriorityN
N
N
C
N
N
N
C
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
COH
CO2HH
CH3
1
2
3
4 1HO
4
3
2
CH3
H
CCO2H
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
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 CCl
Cl
H
BrZ compound
1 1
22
Cl
HBr
ClC C
E compound
1
12
2
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
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
• Cyclopropane is more reactive than normal propane. Why?
C CCC
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
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 HH
HH
H
H
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 10800
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
HH
H
H
HHH H
H
H
H
H
H
H
H
H
H
H 1
2
3
4
5
6
12
3
45
6
newmanprojection
1a, 2a Trans1e, 2a Cis1a, 3a Cis1a, 3e TransChair conformations are strain free. They have neither angle strain nor eclipsing strain.
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
newmanprojection
HHH
HH
H
H1
2
3
4
56
H
50
The order of stability• Chair conformation twist boat conformation boat conformation
chair
potentialenergy
half chair
boat
twistboat
Chair half chair twist boat/skew boat
boat
51
Mono substituted cyclohexane
H
me
two 1,3 diaxial interact ion/two gauche butane
high energyless stable
5%
95%
less energystability high
H
me
me
H
ring flipping
52
Disubstituted cyclohexanea) 1,2-disubsituted cyclohexanecis
meme
50% 50%
ring flippingme
me
H
H
me
me
H
H
2 gauche-butane interact ion &1 gauche interact ion.
2 gauche-butane interact ion &1 gauche interact ion.
53
Trans
ring flipping
meme
meme
me
me
4 gauche-butane interact ionenergy highstability less
>1% <99%1 gauche interact ionless energyhighly stable
54
1,3 disubstituted cyclohexanecis
ring flipping
me
me
me
me
me
me
4 gauche-butane interact ionenergy highstability less
>1% <99%
no gauche-butane interaction
55
Trans
ring flipping
me
me
meme
me
me
2 gauche-butane interact ion
50% 50%
2 gauche-butane interact ion
56
ExampleCl
Cme3
ring flipping
Cl
Cl
2-Cl - H (1,3 diaxialinteract ion) = 2*0.25 = 0.5 Kcal/mol
Stable conformation
2 1,3 diaxial interact ion2*2.7=5.4 Kcal/mol
(CH3)2C H-
57
1,4 disubstituted cyclohexaneCis
ring flipping
me
me
me
me
me
2 gauche-butane interact ion
50% 50%
me
2 gauche-butane interact ion
58
Trans
ring flipping
me
me
me
me
me
me
4 gauche-butaneinteract ion
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
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
2) The substitutent in the ortho positions must have a large sizeEg. 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
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 2 bonds, it should be sp hybridized.
• The overlap of P orbital in allene
C C CH
HH
H
CCCH
H
H
H
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
CCCH
H
Cl Cl
C C CH
H
ClCl
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
Elimination reaction1) 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
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
HB
HB++C C
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 CClHOOC
COOHH
H
COOHHOOC
ClCC
Chlorofumaric acid
Cholromalic acid
68
69
70
Examples
C CClHOOC
COOHH
H
COOHHOOC
ClCC
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 eliminationdifficult
71
PhCH CHPh Ph Ph
BrBr
H H
C CBr2
freerotation
trans elimination
Br
Ph
Br
H
H
Ph Ph
H
Br
H
Br
Ph
C CBr
H Ph
Ph
trans elimination
freerotationBr
Ph
Br
H
H
Ph PhH
Br
H
Br Ph
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
freerotation
trans elimination
Br
Ph
Br
H
H
PhPh
H
Br
H
Br
Ph
C CBrH
PhPh
cis compound
73
Consider dehydrogenation of Sec-butylchloride
alcoholicKOH but-1-ene
but-2-ene
+
CH3CH2CHCH3Cl
CH3CH CHCH3
CH3CH2CH CH2
cis t rans1 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
Cme me
H HCC
H
H me
meC
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
• Eg. E2 elimination converts neomenthyl chloride into a mixture of 75% 3 menthene and 25% 2 menthene.
meH
(CH3)2HC
Cl
H
(CH3)2HC
me
me
(CH3)2HC
H 25%
75%
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
HH
77
KINETIC MOLECULAR THEORYProperties of gases1) According to the Boyle’s law V 1/P PV= constant T m
2) According to Charles lawV T P m
3) According to Avagadro’s hypothesisV n P T
By combining Boyle’s law, Charle’s law and avagadro’s hypothesis V 1/P.n.TV = RnT/PPV = nRTR – Proportionality constant/ Gas constant
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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
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Derivation of kinetic gas equation• Let us consider a certain mass of gas enclosed in a cubic
box at a fixed temperature.
x axisy axis
z axis
AB 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
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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
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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
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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.
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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+C22+C3
2+…..) / l
Multiply and divide by ‘N’ =
Mean square velocity
Total momentum transferred per second =
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Step V: Calculation of pressure from change of momentum
According to Newton’s law of motionThe rate of change of momentum = Force
C22mN
lF=
Pressure = Total forceTotal area
= 1x6l2
C22mN
l
6l2 = Area of the six face cubicalvessel
C2
= mN3l3
C2
= mN3V
l3 Volume of cube=P
P V = C2mN1/3
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Kinetic gas equation in terms of kinetic energy
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Molecular velocity
for one mole P V = C
2mN1/3
P V = C21/3mN0
P V=C2
mN0
3(1)
For one mole of gas
PV=RT
=C2
mN0
3 RT
M - molar mass
C root mean square velocity
C2 mean square velocity
=C2 3 RT
M
= 3 RTCM
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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.
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dncn 4 ( )M
2 RT
3/2 e2RT
2mc-C 2 d c.
n - total no. of moleculesM - molecular mass
T - T emperature on absolute scale(K)
fraction of the total no.of molecules having velicities between C and (C-
d nc
nd nc )dc+
no.of molecules having velocit ies between C and (C+ )dc
velocity
fract ion ofmolecules
dist ribut ion of molecular velocit ies
most probable velocitymean(average) T1<T2<T3
T1T2
T3
Root mean square velocity
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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.
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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)
NVav C1 C2 C3 ...... Cn= ++ + +
From Maxwell equation Vav is given by Vav 8RTm
=
Root mean square velocity (Vrms) VrmsVav
=2 22 2
NC1 C2 C3 ...... Cn++ + +
We already have seen 3RTM
=Vrms
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• Most probable velocity (Vm)• According to the calculations made by Maxwell
MVm = 2RT
The ratio of the three velocities is,Vm: Vav: Vrms = 1: 1.128: 1.224.
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Behaviour of real gasesThe 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.
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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
0C0
C040
0C0
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.
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Effect of temperature on deviations from ideal behaviour
PV
P(atm)
-250C200
C
500C
Variation of pressure-volume of N2
The PV-P plots of N2 at different temperature varying between -25 oC and 50oC.
It is seen that as the temperature is raised the depth in the curve becomes smaller and smaller. At 50oC 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.
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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 IIThe 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.
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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 gasVolume 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.
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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 attractionsbalanced
inward pull
B
B
B
A
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• 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
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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 pressureIsotherm
• 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 isotherms1) Isotherm above 31oC2) Isotherm below 31oC3) Isotherm at 31oC.
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1) Isotherm above 31oC• This isotherm approximates to the isotherm of ideal gas. Thus in the
region above the isotherm at 31oC CO2 always exists in the gaseous state.
2) Isotherm below 31oC• The isotherm below 31oC is discontinuous. For example the
isotherm of 21oC 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 500C
210C
31.10C0 0C
f
ea
g
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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 31oC.• Above 31oC there was no possibility of liquefaction of CO2. The critical
temperature of CO2 is therefore, 31oC. 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
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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.
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The End