Iso Full
Uploaded by
Mihir DhankarIso Full
Uploaded by
Mihir DhankarM.
PRAKASH INSTITUTE ISOMERISM Std: XI
ISOMERISM
S.N. TOPICS Page
Structural isomerism
1. Chain isomerism 3
2. Position isomerism 4
3. Functional group isomerism 5
4. Ring-Chain isomerism 7
5. Metamerism 7
6. Tautomerism 8
Geometrical isomerism
7. Baeyer-Strain Theory 19
8. Geometrical isomerism 23
9. Cis-trans nomenclature of Geometrical isomerism 25
10. E-Z nomenclature of Geometrical isomerism 25
11. Syn-Anti nomenclature of Geometrical isomerism 29
12. Geometrical isomerism across single bond 31
13. Geometrical isomerism in cumulated system 31
14. Geometrical isomerism in spiro system 32
15. Geometrical isomerism in triphenyl system 32
16. Calculation of Geometrical isomerism 33
17. Properties of Geometrical isomerism 35
Conformational isomerism
18. Conformational isomerism 38
19. Conformational isomerism of alkanes 40
20. Gauche effect 43
21. Conformation of cyclohexane 46
Optical isomerism 53
22. Cause of optical activity element of symmetry 57
23. Chiral & Achiral molecule 62
24. Topomerization 66
25. Molecules with chiral axis 67
26. Hemispiranes molecules 68
27. Spiranes 69
28. Biphenyls (Atropisomerism) 69
29. Prochiral centre & molecules 70
30. Enantiomers 72
31. Diastereomers 72
32. Erythroses & Theroses 73
33. Meso compounds 73
34. Calculation of optical isomerism 75
35. Interconversion of structures 77
36. Fischer projection formula 79
37. D & L configuration 81
1
M. PRAKASH INSTITUTE ISOMERISM Std: XI
38. R & S configuration 82
39. Homomers & Anomers 86
40. Anomerization 88
41. Mutarotation 89
42. Epimers 91
43. Epimerization 91
44. Racemic mixture 92
45. Resolution 93
46. Racemisation 95
47. Pseudo-chiral carbon 97
48. Transannular interaction 97
49. Bredt’s rule 98
50. Problems 99
51. Stereochemical reactions 112
2
M. PRAKASH INSTITUTE ISOMERISM Std: XI
ISOMERISM
Compound having same molecular formula but diff properties are known as isomers ,and
this phenomena is called isomerism
E.g. C2H6O EtOH liq. Me – O – Me gas
(1) A Structural isomerism: Compound having same M.F. but different properties due to
difference in their structure.
(i) Chain isomerism: same M.F. but nunber of carbon atoms in parent c-chain are different
i.e. different. root ward
e.g
(i) Butane 2-Me-propane
(ii)
C6H14
(iii)
3
M. PRAKASH INSTITUTE ISOMERISM Std: XI
COOH
C4H8O2
(iv) COOH (v)
(vi)
(2) Position isomerism: same M.F. but different positions of multiple bonds,
functional groups and substituents i.e word root remain same.
e.g.
Cl OH
OH (iii)
Cl (ii)
(i)
Cl
(v) Cl
(iv) Cl Cl Cl
Cl Cl Cl
CN COOH
(vi)
C.I (vii)
COOH
COOH
(a) (b) (c)
a – b C.I
b – c P.I
a – c P.I
Q. Write the min no. of e-atoms required for chain of position isomerism in
following compounds.
Compound Chain Position Alkane Chain + Position
isomers
Alkane 4 6 C4 2
Alkene 4 4 C5 3
Alkyne 5 4 C6 5
4
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Alcohol 4 3 C7 9
- COOH 4 5 C8 18
C9 35
C10 75
(3) Functional group isomerism: same M.F. but different functional group
(1) Alcohol-ether (2) Aldehyde – Ketons
(3) Caboxylic acid – ester (4) Cyanide – Isocyanide
(5) Nitro – Nitrito (6) 1°-2° 3° Amine
(7) 1°-2° 3° Amide (8) Alcohol – Phenol
(9) Alkyne – Alkadiene
(i) Alcohol – Ether: CnH2n+2O
C2H6O CH3CH2OH CH3 O CH3
OH O
C3H8O
OH
C3H7 OH CH 3 CH OH CH 3 O C2 H 5
| Total = 3
CH 3 1x1 =1
2 CH 3 O C2 H 5
CH 4 CH 3 , C2 H 6 C2 H 5 (1) C3 H 8 C3 H 7 (2)
C4 H10 C4 H 9 (3) C5 H12 C5 H11 (4)
CH 3 O C3 H 7
C3H10 O 1 2 2
3 + 4 alcohols = 7 isomers
C2 H 5 O C2 H 5
1 1 1
4
C5H11 exists in eight different forms of alkyl groups.
(ii) Aldehyde – ketone : CnH2nO > C = O
e.g. C3H6O 1-aldo + 1 ketone ; C4H8O Ald 2, + ketone – 1
O
OH , C
C5H10O Ald – 4 + Ketone – 3 , C4H8O dou = 1
OH O
5
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Other function isomers of aldehyde & ketone.
1. Unsaturated alcohol 2. Unsaturated ether
2. Cyclic alcohol 4. Cyclic ether (Expoxy compound)
Note : Aldehydes, carboxylic acids (& their derivatives) & cyanides do not exhibit
position isomerism.
(iii) Acid- ester ; CnH2nO2
(1) C2 H 4O2 1 1 AcOH HCOOMe
(2) C6 H 6O2 1 2 EtCOOH AcOMe HCOOEt
(3) C4 H 8O2 24
Others C4 H 8O2 C C C C OH can changeto ether
||
O
C C C C
| |
OH OH
1 2 3 1 2 3
C 2 H 7 N 1 1 2 1 , C3 H 9 N 2 1 1 , C4 H10 N 4 3 1
(iv) Phenolic & Alcoholic : (v) Alkyne – Alkadiene
OH OH
O
CC CC C C
C C C C C C C C
C C C C
(4) Ring –Chain isomerism : One compound is cyclic while other is acyclic.
(i) Alkene - Cycloalkane
C3 H 6 C C C
R .C .
, C4 H 8 C C C C
P.I .
C C C C
R .C .
R .C .
Note: If not mentioned; ring chain is included in functional isomerism.
6
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(II) Alkyne-Alkadiene-Bicyclo-spiro. DOU2
C C C C C C C C
C4H6 has 5 cyclic isomers & C5H8 has 17 cyclic isomers.
(5) Metamerism: Same M.F. but different alkyl groups attached to same polyvalent
functional group.
O S O O O
|| | || | || || ||
O, C N , S C , NH , C O , C O C
e.g.
C4 H10 O C C O C C, C O C C C, C OC C
|
C
(a) (b) (c )
a,b b, c & a, c metamer
for ketones metamers also related as positional isomers.
(6) Tautomerism/Cryptomeris/Metotropy/ Desmotropism/Dynomic isomerism: If
structural isomers are in dynamic equilibrium then they are called tautomers.
It arises due to rapid oscillation of an atom usually hydrogen between 2 polyvalent atom
in a molecule .
A B H ,
H A B C N H
H C BN
It exists in a state of dynamic equilibrium. At equilibrium amount of that tautomer will
be more stable.
Tautomers have different functional groups but not called as a functional groups isomers
because they are interchangeable to each other.
If H –atom oscillates between 1st & 2nd atom diad system.
HCN st rd
HNC
If between 1 and 3 then triad system.
Toutomerism is actually intermolecular acic base reaction in which at equilibrium that
tautyomer will be more which is less acidic & less basic.
CH 3 C N C N CH 3
Not tautomerism, but fucntional isomersm
Becausae energy required for - CH3 oscillation is high.
Condition for tautomerism :
Multiple bond should be present (like : C O, C N , C N , N O, N N etc )
Wrt to multiple bond next atom should be SP3 (carbon) hybridized & other atom SP2
hybridised, then tautomerism takes place. At provided at α-carbon or α-atom should have
atleast one α-H atom.
Tautomerism observed in the following systems :
7
M. PRAKASH INSTITUTE ISOMERISM Std: XI
O OH
(1) Keto-Enol : CH3 C CH2 C
(2) Amido-Imido : HN C O N C OH
R R
(3) Amide-Imidol : HN C O N C OH (R = CH3, -Et etc)
(4) Imine-Enamine :CH3 CH NH CH2 CH NH2
(5) Nitroso-Oximino : CH3 N O CH2 N OH
O O
(6) Nitro-Acinitro : CH3 N CH2 N
O OH
(7) Amidine system : H2N CH N HN CH NH
(8) Triazene system : H2N N N HN N NH
(9) Azo-Hydrazone : CH N NH C N NH2
(1) Keto –Enol Tautomerism:
O OH
|| |
CH 2 C CH 3
CH 3 C CH 3
Tautomerism is catalysed by acid & base
Base –cat Mechnism:
8
M. PRAKASH INSTITUTE ISOMERISM Std: XI
O OH OH
|| | |
OH / H 2 O
CH 3 C CH 2 CH 3
CH 3 C CH CH 3 CH 2 C CH 2 CH 3
Minor Major
More acidic less acidic
OH
O O
|| ||
CH 2 C CH 2 CH 3 CH 3 C C H CH 3
More stable Less stable
O O
| |
CH 2 C CH 2 CH 3 CH 3 C CH CH 3
Less stable More stable
H2O H2O
OH OH
| |
CH 2 C CH 2 CH 3 CH 3 C CH CH 3
Note: In base catalysed mechanism, enol from depnes on the stability of carbanion and
enolate ion.
1
In basic mediucm enol form stab. Cabanion. enolate ion.
stb.
Acid catalysed mechanism:
O OH OH
|| | |
H
CH 3 C CH 2 CH 3 CH 2 C CH 2 CH 3 CH 3 C CH CH 3
Major
Minor
In acid catalysed mechanism enol form depens on stability of alkene & H – atoms.
OH OH
OH Cl
OH
O O
OH OH
Cl
H H
9
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Space Tautomerism:
O OH
|| |
OH
CH 3 C CH CH CH 3 CH 3 C CH CH CH 2
OH H2O
O O
|| |
CH 3 C CH CH CH 2
CH 3 C CH CH CH 2
e.g. : Space Taoutomerism
O OH O OH
(1) OH
(2) OH
O OH O OH
(3)
OH (4)
OH
O OH O
(5)
(6) OH
No Tautomerism
N N
|
H
Question : Which of the following show tautomerism :
O
O O O
O
HCHO , CH 3CHO , , , Ph C , Ph , Ph C Ph
O O O H
OH OH OH
10
M. PRAKASH INSTITUTE ISOMERISM Std: XI
O O O O
C1
Becouse after enolisation C1
SP hybridised i.e. linear
Me3CCHO MeNO2 PhNO2 Et3C N O CH 2 CH OH
O
O O O Cl
O
N H
O O
O O
O O
O O
N O
O O O H
OH
O
O O
O N OH
O
CD3 C CD3
O O
O NH2
O
N
H
11
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Stability order of keto-enol form :
i. Generally keto form is more stable than enol form.
ii. Because total BDE of keto form is greater than total BDE of enol form.
iii. But for the following systems enol form is more stable than keto form.
O
(2)
(1)
O
O
(3) O
(4)
O O
(5) O Cl (6) O O
O
(8)
(7) N O
O O H
H
N O
(9)
Enol content : Percentage of enol end in equilibrium is called enol content.
Enol content Acidity of - H
Stablity of enolate ion
Intramolecular H –bonding in enol form
Stablity of enol form
Aromaticiation in enol form
1/ stablitiy of keto form
5 or 6 membered chelation in enol form
Presence of EWG (-NO2, -COOEt etc.) at α-atom
Order of enolisation :
Aldehyde < Ketone < β-Ketoester (AcCH2COOEt) < Keto ester with phenyl
group (PhCOCH2COOEt) < 1,4-Dial (OHC-CH2-CH2-CHO) < Ketoaldehyde
(AcCH2CHO) < Diketone (CH3COCH2COCH3) < 1,3-Dial (OHC-CH2-CHO) <
Diketones with phenyl groups < Phenols.
12
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Compound with % of enol Compound with % of enol
1 AcOH = 6 x 10-7 2 MeCOPh = 1.1 x 10-6
3 MeCHO = 6 x 10-5 4 Me2CO = 2.5 x 10-4
5 EtOCOCH2COOEt = 3.3 x 10-3 6 AcOEt = 0%
7 N≡CCH2COOEt = 0.25 8 Cyclohexanone = 1.2
9 AcCH2COOEt = 7.7 – 8.4 10 Ph2CHCHO = 9.0
11 PhCOCH2COOEt = 21 12 CH3COCH2COCH3 = 80
13 PhCOCH2COMe = 89.2 14 Phenol = 99.99
15 PHCOCH2COPh = 100
Effect of solvent on enol content :
In acetyl acetone since enolic form is less polar, due to intramolecular H-bonding, than
Keto form therefore any polar solvent would decrease the enolisation & favours ketofrom
& vice-versa.
Solvent H2O Liquid state Gaseous state
% of enol 16 76 92
Enol content is more stable in liquid state than gaseous state because of strong
intramolecular attraction. Acetylacetone has more enol content in toluene.
Due to intramolecular H-bonding, a six membered ring is formed. This process is called
chelation and makes the compound very stable.
H-Bonding makes enol form of the following compounds stable, so write the enol form
of the following :
(1)
O O
Ph Ph
(2)
O O
O O
(3)
13
M. PRAKASH INSTITUTE ISOMERISM Std: XI
O
NO2
(4)
(5)
O
COOEt
(6)
Question : Arrange the following in increasing order of % of enol content.
(1) CH3CHO ClCH2CHO NC-CH2CHO
O O O O
(2)
CH3 Br NO2
(3)
O O O
O2N Et
O O O O
(4)
CD3 CD3
O
(5)
O O OMe
14
M. PRAKASH INSTITUTE ISOMERISM Std: XI
O O
(6) CH3 CHO
OEt
O O O O O O O O
(7)
H H Ph Ph
O O O
EtO OEt
O O O
NO2
(8)
(9)
O O
O O O
(10)
O O
(11)
O O
O O
(12)
N
H
15
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(13)
O O O O
O O
(14)
O O O
O O
(15)
O
O O
(16)
H H
N O N O
(17)
O
O
(18) CH3 CHO
O O
O
NO2 O2N NO2
(19)
16
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(20) CH CHO C CH OH
Duteium Exchange reaction:
O O
(i) ||
||
CH 3 C H
O D / D2O
CD3 C H
On keeping for a low time all hydrogen atoms which are present on those carbon atom
on wh ich – ve charge ddisperxse by resonance.
Mechnism:
O O
||
||
CH 3 C H O D / D2 O
CD3 C H
HOD OD
O O OD O O
|| | |
OD
| ||
D2 O
CH 2 CH
CH CH
CH CH
CH CH
CH 2 C H 2 2 2
D 2O
O
||
D CH 2 CH
Question : Write the deuterium exchange products for the following.
O
(1)
O
(2)
O
(3)
17
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(4)
(5)
(6)
18
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Difference between Tautomerism & Resonance:
S.N. Tautomerism Resonance
1. Involves a change in position of atom Involves a change in position of the
(generally hydrogen) unshared of pi-electrons.
2. They are definite compounds and Resonating structures are only
may be separated and isolated. imaginary and can’t be isolated.
3. Two tautomeric forms have different Various resonanting structures have
structures (functional group) the same functional group.
4. They are in dynamic equilibrium with They are not in dynamic equilibrium.
each other.
5. It has no effect on bond length. It affects the bond length.
6. It does not lower the energy of the It decreases the energy and hence
molecule and hence does not play increases the stability of the molecule.
any role in stabilizing the molecule.
7. It can occur in planar as well as non- It occur only in planar molecules
planar molecules.
8. It is indicated by " " . It is indicated by " " .
Baeyer Strain Theory :
Van’t Hoff and Lebel proposed tetrahedral geometry of carbon.
Baeyer postulated that any deviation of bond angles from the normal tetrahedral value
would impose a condition of internal strain on the ring.
The bond angle is of 109028’ (or 109.50) for carbon atom in tetrahedral geometry
(methane molecule)
Baeyer observed different bond angles for different cycloalkanes and also observed some
different properties and stabiliry.
On the basis, he proposed angle strain strain theory.
The theory explains reactivity and stability of cycloalkanes.
Baeyer proposed that the optimum overlap of atomic orbitals is achieved for bond angle
of 109.50. In short, it is ideal bond angle for alkane compounds.
19
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Effective and optimum overlap of atomic orbitals produces maximum bond strength and
stable molecule.
Higher the strain higher the instability.
Higher strain produce increased reactivity and increases heat of combustion.
Baeyer proposed “Any deviation of bond angle value (109.50) will produce a strain in
molecule. Higher the deviation lesser the instability.”
Baeyer’s theory is based upon some assumptions as following :
1. All ring systems are planar. Deviation from normal tetrahedral angles redults in to
instable cycloalkanes.
2. The large ring systems involve negative strain hence do not exists.
3. The bond angles in cyclohexane and higher cycloalkanes (cycloheptane, cyclooctane,
cyclononane etc.) are not larger than 109.50 because the carbon rings of those compounds
are not planar (flat) but they are puckered (Wrinkled).
4. All carbon atoms are SP3 hybridised. Expected bond angle = 109.50.
These assumptions are helpful to understand instability of cyclohexane ring system.
The ring of cyclopropane is triangle. All the three angles are of 600 in place of 109.50
(normal bond angle for carbon atom) to adjust them into triangle ring system.
In same manner, cyclobutane is square and bond angle are of 900 in place of 109.50
(normal bond angle for carbon atom) to adjust them into sruare ring system.
The deviation of cyclopropane and cyclobutane ring systems then normal tetrahedral
angle will produce strain in ring. The strain will make them unstable as compare to
molecules having tetrahedral bond angle.
So, cyclopropane and cyclobutane will easily undergo ring opening reactions to form
more stable open chain compounds.
Now compare the stability of cyclopropane and cyclobutane.
1. The bond angle in cyclopropane is 600.
20
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The normal tetrahedral bond angle is 109.50.
Therefore, deviation = (normal tetrahedral bond angle) – (actual bond angle)
Deviation = 109.50 - 600 = 49.50.
2. The bond angle in cyclobutane is 600.
The normal tetrahedral bond angle is 109.50.
Therefore, deviation = (normal tetrahedral bond angle) – (actual bond angle)
Deviation = 109.50 - 900 = 19.50.
The deviation is higher for cyclopropane (49.50) than cyclobutane (19.50) therefore
cyclopropane is more prone to ring opening reactions.
As a result of this, the strain is more in cyclopropane as compare to cyclobutane. It will
make cyclopropane less stable than cyclobutane. So, syslopropane easily undergoes ring
opening reaction as compare to cyclobutane.
According to Baeyer, the relative order of stability for some common cycloalkanes is as
under.
Cyclopentane > Cyclobutane > Cyclopropane
Angle strain in Cyclopropane :
Limitations of Baeyer’s Strain Theory :
1. Successfulness of Baeyer’s angle strain theory:
Baeyer rightly proposed that cyclopropane and cyclobutane are flat molecule
and having angle of 600 and 900 those are much deviated from the ideal
tetrahedral value of 109.50 hence these ring systems are unstable and easily
undergo ring opening reactions. There is much angle strain in cyclopropane
and cyclobutane.
21
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Baeyer also proposed that Cyclopentane is not need to be plalnar but it is
plalnar as in that condition the angle (1080) is much near to ideal tetrahedral
angle.
2. Unsuccessfulness of Baeyer’s angle strain theory :
Baeyer was not able to explain the effect of angle strain in larger ring systems.
According to Baeyer Cyclopentane should be much stable than cyclohexane
but practically it is reversed.
Larger ring system are not possible according to Baeyer as they have negative
strain but they exist and much stable.
Larger ring systems are not planar but puckered to eliminate angle strain.
Sachse Mohr’s Theory (Theory of strainless rings) :
Baeyer suggested that both large and small polymethylene rings should be
strained, but Hermann Sachse, soon pointed out that large rings need not be
strained, because the carbons need not be coplanar.
Sachse Mohr’s theory proposed that higher member ring can become free from
strain if all the ring carbons are not forced into one plane. They exhibit in two
non-planar puckerd conformations both of which are completely free from strain.
These are called Chair form and Boat form.
22
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Stereoisomerism
Isomers having the same constitution but different spatial (3-Dimentional)
arrangement of their atoms or groups are known as stereoisomers and the phenomenon is
called stereoisomerism.
Types of stereoisomers :
A. Configurational isomerism :
i. Geometrical isomerism
ii. Optical isomerism
B. Conformational isomerism
A. Conformational isomerism :
i. Geometrical isomerism : Compounds having same structural formula but different
arrangements of groups in 3-D space due to restricted rotation about C-C bond.
ii. Rotation is not possible due to :
a. Presence of pi-bond.
b. Presence of cyclic ring
c. Presence of cyclic ring
1. Geometrical isomerism in alkene: Each sp2 hybridised C-atom must be attached to
two different groups.
Me Me x a a a Cl H Me Me
H H y b b a H H H Me
But-1-ene 1,1-Dichloroethane 2-Butyne
2,3-Dichloropropane 2-Methylbut-2-ene 3,4-Dimethylpent-2-ene
2. Geometrical isomerism in cycloalkens : In cyclic system one cyclosystem is equal to
one double bond.
It possible when any two carbon atonms of the ring having individually two different
groups attached with carbon.
Or Atleast two atoms of cyclic ring must have two different groups.
Cyclopropane, cyclobutane, cyclopentane, cyclohexane etc. never show
geometrical isomerism.
Monosubstituted cycloalkane never show geometrical isomerism.
G G
23
M. PRAKASH INSTITUTE ISOMERISM Std: XI
1,1-Disubstituted cycloalkane never show geometrical isomerism.
x G a
y G b
Benzene ring ring never show geometrical isomerism.
1,2; 1,3; 1,4-type cycloalkanes always show geometrical isomerism.
Br
COOH Ph Ph
Me
COOH Ph I
Cyclobutene, cyclopentene do not show geometrical isomerism, because trans
form not existing.
Although the trans form is quite unstable when n 8 but it exist.
H
H
cis
trans
Decalin or Bicyclo [4,4,0] decane or Decahydronaphthalene also show cis-trans
isomerism.
H H
H Cis-Decalin H
24
M. PRAKASH INSTITUTE ISOMERISM Std: XI
H H
H Trans-Decalin H
Question: Which of the following gives geometrical isomerism:
H
(1) (2) (3) (4) (5)
H H H
Cl
H
(6) (7) (8) (9) Cl (10) C
H H H
Cl
Nomanclature system in Geometrical Isomerism:
Geometrical isomers may be considered diastereomers since they have
different physical properties.
(I) Cis-trans nomenclature of geometrical isomers :
Cis isomer : The isomer in which similar groups or atoms lie on same side of the
double bond is called cis isomer.
Trans isomer : The isomer in which similar groups or atoms lie on opposite side
of the double bond is called trans isomer.
CH3 H Cl
C C
H H Cl H
CH3
cis trans trans
(II) E-Z system nomenclature of geometrical isomers:
German word : E- Entgegan; meaning against, Z = Zusammen; meaning together
Higher priority is given to atoms with greater atomic number.
If higher priority groups are on same side, then it will be Z.
25
M. PRAKASH INSTITUTE ISOMERISM Std: XI
If higher priority groups are on opposite side, then it will be E.
2 Cl I1
C C
Br F2
1
Cohn – Ingold – Prelog Sequence Rule: (C.I.P)
(1) If different groups are attached to sp2- C, then priority order given on the basis of
their atomic number. If isotopes are present, then the atomic mass is the criteria.
Cl I
HO CH3
C C
CC
Br F
F CH2 CH3
E Z
(2) If 1st atoms are same then priority is given on the basis of next atom.
HO CH3
C C
H2N CH2 CH3
(3) If multiple bond containing groups are present then every double and triple bonded
atoms are considred as 2 or 3 such atoms.
O C
* *
F CH = CH2 C, C, H | |
CC CH = CH2 C O
CH3 CH2 CH3 C, H, H , |
OH
Z
Examples :
C C O C
| | | |
C CH C CH , CH O CH O , C O C O
| | | | | |
C C O C OH OH
26
M. PRAKASH INSTITUTE ISOMERISM Std: XI
C C * *
| | CH3 CH2 CH CH2 H2 N CH CH 2
C N C N , CC , C C
CH= CH2 CH 3 HN CH CH 2 CH 3
| | F CH2
N C * * |
E CH 3
E
(4) if there is a ‘E’ and ‘Z’ configuration groups at asymmetric centre than the carbon
with ‘Z’ should be given more priority over ‘E’.
Cl E
C C
Br Z
* *
E
(5) If asymmetric centre attached ‘R’ and ‘S’ configuration groups, then the carbon
having ‘R’ configuration should be given priority over ‘S’.
Cl S
C C
Br R
* *
E
(6) Priority sequence of some substituent :
i I Br Cl F O N C H
ii CMe3 CHMe2 CH 2CH 3 CH 3
iii Cl SH OH H
iv CH 2 Br CH 2Cl CH 2OH CH 3
v OH CHO CH3 H
vi OCH 3 NMe2 CH 3 H
vii C CH CMe3 CH CH 2 CHMe2
27
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Examples :
* I
I Cl * I F F 1 & 2 Geo.
(2) (3)
(1)
Cl Br Cl 1 & 3 Posi.
Br F Br
E 2 & 3 Posi.
Z E
Total = 6 isomers
* * Br Br
CH3O CH2NH2 Me F
(4) (5) (6) (7)
CH3 CH3 Cl CH3
Z E
CH3
CH3
* 16 F CH = CH2
NC C Cl H OH
(9) (10)
(8) Cl
CN C C OH CH CH2 CH3
D
17
Z C
F
Cl NC C C C
CH = CH2 Me OK
(12) (13)
(11) CN C C C
Br C=CH2 Et OMe
Cl
CH3
NC CHO Me2N COOH C=C CH2Cl
(15) (16)
(14)
HO COOH COOH CHCl2
HO C=C C
CH3 H
(17) (18) (19)
H CH3
H
CH3 H
O O O
(20) (21) O
O
O
28
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(III) Syn-Anti nomenclature of geometrical isomers : Aromatic aldoximes and
aromatic ketoximes also show geometrical isomerism.
Formation of aldoximes and ketoximes :
R R OH
1
C=N 1
C=N
C O H 2 N OH C N OH R OH R
Oxime
G.I. Where R R 1
i. Syn-anti isomerism in aldoximes :
Syn isomers :When H & OH groups are on same side the isomer is known as syn
(analogous to cis).
Anti isomers : When H & OH groups are on opposite side the isomer is known as anti
(analogous to trans).
ii. Syn-anti isomerism in ketoximes :
Syn isomers : When OH & alkyl groups (a/c to alphabetical) are on same side is known
as syn isomers.
Anti isomers : When OH & alkyl groups (a/c to alphabetical) are on opposite side is
known as anti isomers.
eg.
Me Me OH Me
C N C N C N OH
Et OH Et Et
Anti syn & anti
syn
Eg. “E” of ethyl comes 1st ‘m’ of methyl.
Me .. Me OH
C N C .N.
H OH H
syn Anti
29
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(IV) Syn-anti in hydrazone, phenylhydrazone, and azo compounds :
a. Geometrical isomerism due to N N Bond
Syn isomer : If Lone pairs are same side
Anti isomer: if Lone pairs are on opposite side.
RN RN
R N N R
syn Anti
e.g.
Ph N Ph N Ph N Ph N
NaO N N ONa NC N N CN
syn Anti syn Anti
Ph N N Ph Me N N Me
O O O
b. Geometrical isomerism due to hydrazone & phenyl hydrazone :
Syn isomers : When NH2 & alkyl (aryl) groups or NHPh & alkyl (aryl) groups (a/c to
alphabetical) are on same side is known as syn isomers.
Anti isomers : When NH2 & alkyl (aryl) groups or NHPh & alkyl (aryl) groups (a/c to
alphabetical) are on opposite side is known as anti isomers.
30
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Me NH2 Me
C N C N
Et Et NH2
Syn or E
A n ti o r Z
Ph NHPh Ph
C N C N
M e Syn or E Me N H Ph
A n ti o r Z
NO2
Ph
Ph C N
C N Me
NO2
Me Syn or E NO2 A n ti o r Z
NO2
(V) Geometrical isomerism across a single bond :
s-cis isomer : In molecules of the type A CH CH B , the fully eclipsed
conformation with a dihedral angle of 00 is termed as s-cis isomer.
s-trans isomer : In molecules of the type A CH CH B , the conformation
with a dihedral angle of 1800 is termed as s-trans isomer.
"single"-trans "single"-cis
s-trans s-cis
(VI) Geometrical isomerism of Cumulated bonding system : Cumulated bonding
system with an odd number of bonds, in which the substituted groups at the two
ends of the cumulated chain lie in the same plane and geometrical (E-Z)
isomerism is shown.
Simplest cumulene : CH2 = C= CH2
H H
C=C=C
H H
Px Pz Px Py
C – H bonds are in perpendicular plane.
Examples :
CH3 H
H H Cl Cl
C=C=C C=C=C C=C=C
Cl H Br Br H CH3
31
M. PRAKASH INSTITUTE ISOMERISM Std: XI
CH3
H
C =C C = C= C = C= C = C
H H
x
x=1
(VII) Geometrical isomerism in spiro compounds : The replacement of double bonds in
cumulated bonding system by ring systems gives a spiran, in which the
substituted ringss at the two ends of the spiran chain lie in the same plane and
geometrical (E-Z) isomerism is shown.
a a a
a
b b b
b
Hz
Hz Vert
Br
CH3
Cl Br CH3 CH3
Br Cl Cl Cl
Br
Cl
Cl
Cl O
CH3 CH3 F
F O
Cl Br Br
Cl
Cl
(VIII) Geometrical isomerism in triphenyl system: Geometrical isomerism are also
possible in terphenyl derivatives, where restricted rotation may arise aroud pivotal
(centric or central) bonds, consequently the two terminal phenyl groups are co-
32
M. PRAKASH INSTITUTE ISOMERISM Std: XI
axial as well as coplanar. The terphenyl compounds exist in three stereoisomeric
forms: a chiral cis isomer (exist as enantiomeric pair) while the trans isomer with
a center of symmetry is as optically inactive meso form.
Me Br OH Me Me Br OH
HO Br HO Br Me
Cis: Chiral molecule Trans : Meso molecule
COOH NO2 COOH NO2 Br Br
Cl Cl
Free rotation NO2 COOH Br NO2 Br COOH
NO2 Br OH COOH Br COOH Me Me COOH
Et Et
COOH Br OH NO2 COOH Me Me CO
OH Br
Calculation of Geometrical isomers in polyenes :
a. When compound has ‘n’ double bonds and ending groups of a polyene are different,
the number of geometrical isomers = 2n.
Example-1 : C6 H 5 CH CH CH CH CH CH CH CH Cl
Since, the number of double bonds is four and two ends are different, one is C6H5 and
other is Cl. Therefore, Number of geometrical isomers = 2n = 24 = 16.
b. When the ending groups of polyene are same
Case-I : When number of double bonds is even then the number of geometrical
n
1
isomers 2n1 2 2 .
Example-2 : Cl CH CH CH CH CH CH CH CH Cl
n=4 (Even)
33
M. PRAKASH INSTITUTE ISOMERISM Std: XI
n
1
n 1
Number of geometrical isomers 2 2 2
23 21 10
Case-II : When number of double bond is odd then number of geometrical isomer
n 1 1
n 1
2 2 2
Example-2 : C6 H 5 CH CH CH CH CH CH C6 H 5
n = 3 (odd)
n 1
1
number of geometrical isomers 2n 1 2 2
22 22 1 6
Examples :
(1) CH 3 CH CH CH CH CH 2CH 3
n = 2, 22 = 4 , i.e. 0
(2) CH 3 CH CH CH CH CH 3
N = 2, G.I. = 221 211 3 EE/EZ/ZE/ZZ
(3) CH 3 CH CH CH C C CH CH CH 2
22 = 4
(4) (5)
22 4 n3 G.I . 231 221 6
(6) CH 2 CH CH CH CH C CH CH CH CH 3 G.I. = 2 2 = 4.
(7) C C C C C C C C C C C C C
||
C
34
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Nomanclature:
1
E
6 Z
(2E - 4Z - 6Z) non -2-,4,6-triene
2 Z
9
4
Properties of Geometrical Isomerism:
1. Stability: Stability of cis isomer is less than corresponding trans isomer, as
molecules of trans isomer are more tightly held in the crystal lattice than the
corresponding cis-isomer.
COOH COOH
< <
cis COOH
trans HOOC
COOH C
O H2O
COOH C
O
O
< 290°C
COOH
O
> 290°C
HOOC
- H2O
Proof: O
2. Dipole moment :
Cl
Cl Cl
> , -I -I
Cl
0 0 0 0
35
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cl
Cl
+I +I
> , <
0 0 Cl
Dichloro benzene O > m > P,
H H
O O
>
O O
H
H
0 0
Dipole moment of some organic compounds :
Compounds Ortho Meta Para
Chlorotoluene 1.35 1.78 1.9
nitrotoluene 3.66 4.17 4.4
Chloro-nitro 4.6 3.69 2.7
bezene
Nitro aniline 4.26 4.85 5.21
Chloro 4.75 3.4 2.5
benzonitrile
dichlorobenzene 2.5 1.72 00
Dinitrobenzene 6.58 4.23 00
Acetic acid 1.74 Formic acid 1.5
Methanol/Ethanol 1.7 Bromobenzene 1.5
Phenol 1.4 Benzoic acid 0.9
Ethyl chloride 2.04 Chloro benzene 1.73
Ethanamine 1.3 Ethylcyanide 3.57
Toluene 0.4 Aniline 1.56
Benzaldehyde 2.8 Nitro benzene 3.9
36
M. PRAKASH INSTITUTE ISOMERISM Std: XI
3. Boiling Point : Boiling point of cis > trans (Because of steric hindrance in cis
compound and high polarity of cis compound).
(i) For ionic compound: B. P
(ii) For non –polar compound : B.P. vander Wall force surface area.
1
For isomers, B.P. .m.m H .bond
Branching
>
eg. ,
C
|
C C C C C , C C C C C C C C C C C C
| |
C C
C C C C C O C C C C OH
0 0 H-bonding
4. Melting point: melting point of trans > cis (Because of symmetrical packing of
trans form in its crystal lattice)
close packing sysmmetry
Cl Cl
Cl
< , < , >
Cl
5. Solubility : Solubility of cis isomer is greater than trans isomer, as molecules of
trans isomer are more tightly held in the crystal lattice than the corresponding cis-
isomer.
37
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Conformational isomerism
Conformational isomerism : An important aspect of organic compounds is that the
compound is not static, but rather has conformational freedom by rotating, stretching and
bending about bonds. Each different arrangement in space of the atoms is called
“Conformer” and this phenomenon is called conformational isomerism.
Free rotation between C-C bond is responsible for different conformers:
For free rotation, energy barier = 0.6 Kcal/mol.
For restricted rotation energy barier is in between 0.6 < energy barier < 16 Kcal/mol.
Frozen energy barier ≥ 16 Kcal/mol (Stable in any condition)
Different conformers can have great different energies and the relative proportion of each
conformer is related to the difference between them.
Conformers will be of different energy due to strain. Sources of strain are generally
categorized in one of the three types :
Torsional strain : It is caused by the repulsion of the bonding electrons of one
substituent with bonding electrons of a nearby substituent. It is maximum in case of
eclipsed conformation & i.e. why this conformation is less stable. OR
Torsional strain is due to interactions as groups change relative position with a
change in torsional bond angle (Dihedral angle).
Steric strain or Van-der Waals strain : It is caused by atoms or groups of a atoms
approaching each other too closely. OR
Destabilization due to the repulsion between the electron cloud of atoms or groups.
Groups compete to occupy common space. Strain due to close contact of atoms
reparated by four or more bonds. OR
It is observed when groups are placed in position closer than the sum of their Van-der
Waals radii.
Angle strain : Molecules that are forces to have a bond angle far from ideal (109.50
for SP3).
A large source of strain for ring compounds :
Ring size Cycloalkane Total ring strain Ring strain per
Kcal/mol CH2
Kcal/mol
3 Cyclopropane 27.4 9.1
4 Cyclobutane 26.4 6.6
5 Cyclopentane 5.8 1.2
6 Cyclohexane 0.1 0
7 Cycloheptane 6.0 0.9
8 Cyclooctane 9.5 1.2
10 Cyclodecane 12.1 1.2
12 Cyclododecane 3.8 0.3
38
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Strain is large for small rings, but reaches minimum at 6-membered ring. This angle
strain is due to forcing the electron density in bonds at angles that are not ideal.
The angle strain becomes lowest with a 6 membered ring due to the ability to
form a conformer that has nearly perfect torsional angles and a lack of ring strain.
Planar cyclohexane has 1200 bond angle for <C-C-C, where all the hydrogens are in
eclipsed position. Chair cyclohexane has 111.40 bond angle (nearly tetrahedral) for
<C-C-C, where no hydrogens are in eclipsed position.
Dihedral angle : Angle between two bonds on adjacent atoms.
The angle defined by X-C-C and C-C-Y plane is termed as dihedral angle.
Torsional angle is an alternative term used for dihedral angle. While dihedral
angle is measured from 0-3600, torsional angle is measured from 0-1800, as a positive
value in the clockwise direction and a negative value in the anti-clockwise direction.
Bond angle : Angle between two bonds on same C-atoms.
In different conformers, various bond length, bond angle, basic structure remains
ssame. Only & only dihedral angle is changed. Any molecule exists in infinite
numbers of conformers, out of them only two forms are defined, as most stable and
least stable. The most stable form is Staggered & least stable firm is called Eclipsed
form. Various isomers between 00 to 3600 are called Skew forms & stability order is :
Staggered > Skew > Eclipsed.
Staggered : A low energy conformation where the bonds on adjacent atoms bisect
each other (600 dihedral angle), maximizing the separation.
The staggered conformer has a better orbital match between bonding and antibonding
states.
The staggered conformar can form more delocalized molecular orbitals.
Eclipsed : A high energy conformation where the bonds on adjacent atoms are
aligned with each other (00 dihedral angle).
39
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Gauche : Description given to two substituents attached to adjacent atoms when their
bonds are at 600 with respect to each other.
a a
a
b b b ba b b
a a a a
b ab b
Staggered Eclipsed Gauche
1. Conformation of ethane : The staggered conformation is more stable than the
eclipsed conformation by 12 kJ/mol or 2.9 kcal/mol. There are infinite number of
conformations in between both these conformers.
2. Conformation of propane : The staggered conformation is more stable than the
eclipsed conformation by 14.3 kJ/mol or 3.4 kcal/mol. There are infinite number of
conformations in between
three conformers.
40
M. PRAKASH INSTITUTE ISOMERISM Std: XI
3. Conformation of butane : “Totally eclipsed” conformation (which has largest
groups eclipsing each other) is higher in energy than other eclipsed conformations.
“Gauche” conformation is higher in energy than anti (both are “staggered”
conformations).
Structure conversions :
41
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question-1 : Write the most & least stable conformer in the following :
i. Normal pentane (about C2-C3 bond)
ii. Normal hexane (about C2-C3 bond & C3-C4 bond)
iii. 3-Methylpentane (about C2-C3 bond)
iv. 2,3-Dimethylhexane (about C3-C4 bond)
v. 2,2,3,4,5,5-Hexamethylhexane (about C3-C4 bond)
vi. 3,4-Dimethyl hexane (about C3-C4 bond)
vii. 2-Methyl hexane (about C3-C4 bond)
viii. 1,2-Diphenyl ethane
Question-2 : Compare the stability in the following conformers
Case-I : If benzene ring is coplanar to paper plane.
Me Me Me
H Me H H Me H
H H H H H H
H Me H
A B C
Case-II : If benzene is perpendicular to paper plane.
Me Me Me
H Me H H Me H
H H H H H H
H Me H
A B C
42
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Effect of temperature on dipole moment :
On increasing temperature dipole moment of the compound increases. Consider the
following process :
X X
H X H H
Decreasing T
H H Increasing T H H
H X
Dipole moment 0 Dipole moment = 0
Halogens Increasing T Dipole moment
F Increasing Increases
Cl Increasing Increases
Br Increasing Increases
I Increasing Increases
But the dipole moment guache form of 1,2-dichloro ethane is less than the dipole moment
of gauche form of 1,2-bromo ethane.
Gauche effect :
The gauche effect denotes the existence of of a gauche conformer, which is more stable
than the anti conformer. This effect is present in 1,2-difluoroethane (CH2FCH2F) for
which the gauche conformation is more stable by 3.4 kJ/mol in the gas phase. Another
example is 1,2-methoxyethane.
F F
H H H F
H H H H
F H
FF HF F F
H F H H
H H H H H H
H HF H
HH H F
Gauche Anti
A B C D
Order of stability : C > D > B > A
ClCl HCl Cl Cl
H Cl H H
H H H H H H
H H H
HH Cl H Cl
Gauche Anti
A B C D
43
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Order of stability : D > C > B > A
The increased stability of the gauche conformer is explained based on the possibility of
hyperconjugation between C-F and C-h bonds. This is initiated by the very high
electronegativity of fluorine.
In the hyperconjugation model, the donation of electron density from the C-H σ bonding
orbital to the C-F σ* antibonding orbital is considered the source of stabilization in the
gauche isomer. Due to the greater electronegativity of fluorine, the C-H σ orbital is a
better electron donor than C-F σ orbital, while the C-F σ* orbital is a better electron
acceptor than the C-H σ* orbital orbital. Only the gauche conformation allows good
overlap between the better donor and the better acceptor.
For effective Intramolecular H-bonding to occur, the donor and the acceptor groups must
be close to each other, which is possible only in the eclipsed or gauche conformation.
The gauche conformation with a torsion angle of 60-700 between the interacting groups
are ideally suited for intramolecular H-Bonding.
Intramolecular H-Bonding is less effective in eclipsed position due to torsional & Van
der Waal’s repulsion.
OHOH OHH OH OH
H OH H H
H H H H H H H H
H H H H
H OH
A B C D
Order of stability : C > D > B > A. The energy difference between D and C conformer is
about 50 kcal/mol.
In some compounds, Gauche form is more stable than anti form due to intramolecular H-
bonding or electrostatic force of attraction.
In general 2-substituted ethanols of the type, X CH 2 CH 2 OH , where
X OH , NH 2 , F , Cl , Br , OCH 3 , NH 2 , NHMe, NMe2 CHO, COOH , NO2 etc. have
preferred gauche conformations with OH and X forming intramolecular H-bonds.
Usually the dihedral angle is slightly greater than 600.
In the following cases gauche form is more stable than anti form due to electrostatic force
of attraction.
44
M. PRAKASH INSTITUTE ISOMERISM Std: XI
NMe3 NMe3
H COO H H
CH2 NMe3
Gauche is more stable
CH2 COO H H H H
H COO
Gauche Anti
H H
H Cl H H
CH3CH2Cl Both are same
H H H H
H Cl
Gauche Anti
CH3 CH3
H Cl H H
CH3CH2CH2Cl Gauche is more stable
H H H H
H Cl
Gauche Anti
In case of lone pair containing compounds bigger group should be placed at gauche to
lone pair. In such cases lone pair containing orbital is partial negative due to high
electronegativity and carbon of methyl group has partial positive charge due to its + I
effect.
Example :
CH3 CH2 NH2
CH3 H
H CH3 H H
N N
H N N
H H H H H CH3 H H
H H H H CH3
Order of stability : B > D > A > C
45
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question : Me2CH - OH
CH3 CH3 CH3 H
CH3 CH3
d>b>a>c
H
H H H H CH3 H CH CH3 H CH3
3
Order of stability : D > B > A > C
Most stable form of 1,2-dibromo-1,1-difluoro-2,2-dichloroethane
Br
F F
Cl Cl
Br
Some other examples of Gauche effect :
2-Chloroethane, 1,2-difluoroethane, 2-fluroethyltriacetate gauche form is more
stable than their anti form.
2-Chloroethanol, 2-bromoethanol also prefer gauche conformers. There is as yet
no explanation for this behavior.
1,1,2,2-Tetrachloroethane & 1,1,2,2-tetrabromoethane both prefer gauche but
1,1,2,2 tetrafluoroethane prefer anti form.
Both 2,3-dimethylpentane & 3,4-dimethylhexane prefer the gauche form.
2,3-Dinitro-2,3-dimethylbutane exists entirely in gauche form in solid state, but in
benzene (as solvent) gauche-anti ratio is 79 : 21 while in CCl4 as solvent gauche-
anti ratio is 42 : 58.
Conformation in cyclohexane :
Cyclohexane is the most stable (least reactive, least strained) cycloalkane which
exist in different numbers of conformers mainly chair, boat, twisted-boat, half-chair,
inverted-chair which keeps on interconverting into eachother by a process called
“Flipping”.
Chair Boat Half chair Twisted boat inverted chair
(1) (2) (3) (4) (5)
46
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Potential energy : Half-chair > Boat > Twist-boat > Chair.
Stability order : Chair > Twist-boat > Boat > Half-chair.
Chair and boat form both have minimum angle strain but boat form is less stable due to
“Flagpole-Flagpole” repulsion.
47
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Chair form is most stable conformer in which two types of bonds are present :
Axial bonds are more hindered than equatorial bonds due to 1-3,1-3-diaxial repulsion.
a a a a a a
e a e
e e
e e
e e
e e
a
e e a
a a a
a
Axial bonds Equatarial bonds Combined
At room temperature, one chair form converts into inverted chair form by “Flipping”.
During flipping all axial bonds are converted into equatorial bonds & vice-versa.
“A Values”
It’s nice to have some shorthand. For a mono-substituted cyclohexane, the energy
difference between axial and equatorial conformers with a given substituent is known as
its A-value.
For example, the A value of methyl is 1.70 , ethyl is 1.75, OH is 0.87, Br is 0.43, i-Pr is
2.15, and t-Bu is 4.9 .
A-values are useful because they are additive. We can use them to figure out the energy
differences between di- and trisubstituted cyclohexanes.
Using the equation G RT ln K , we can calculate the energy difference.
48
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cyclohexane :
%e
In Cyclohexane cyclic ring K eq 1 ratio. i.e. 50% equatorial & 50% axial form.
%a
And both are equal in energies.
Keq H Keq = 1
H
H
Any group bigger than hydrogen have tendency to stay at equatorial position to avoid 1-
3,1-3-diaxial repulstion.
Methylcyclohexane :
%e 95
If Mehtyl group (Methylcyclohexane) is attached to cyclic ring then K eq 19
%a 5
ratio. i.e. 95% equatorial & 5% axial form. Equatorial form is 1.70 kcal/mol more stable
than its axial form.
H
H H
CH3
H H
Keq1 > 1
CH3
H
More stable
Less stable
Ethylcyclohexane :
%e 95.84
If ehtyl group (Ethylcyclohexane) is attached to cyclic ring then K eq
ratio
%a 4.16
K eq 23 . i.e. 95.84% equatorial & 4.16% axial form. In case of ethyl group
(Ethylcyclohexane) equatorial form is 1.75 kJ/mol more stable than its axial form.
Et
Keq2 > 1 Et
49
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Isopropylcyclohexane :
If isopropyl group (Isopropylcyclohexane) is attached to cyclic ring then
%e 97
K eq K eq 32.33 ratio. i.e. 97% equatorial & 3% axial form. In case of
%a 3
isopropyl group (Isopropylcyclohexane) equatorial form is 2.15 kcal/mol more stable
than its axial form.
Me Me
Me
CH
Keq3 > 1 CH
Me
Tert-butylcyclohexane :
If tert-butyl group (tert-butylcyclohexane) is attached to cyclic ring then
%e 99.97
K eq K eq 3332.33 ratio. i.e. 99.97% equatorial & 0.03% axial form.In
%a 0.03
case of tert-butyl group (tert-butylcyclohexane) equatorial form is 4.9 kcal/mol more
stable than its axial form.
Me
Me
Me C Me
Keq4 > 1 C Me
Me
Keq4 >> Keq3 > Keq2 > Keq1 > Keq
3332 32 23 19 1
Cyclohexanol :
Given that oxygen has a larger atomic number than carbon, it’s not unreasonable to think
that the OH group might be “bulkier” than carbon. When you think about the source of
strain in CH3, however, you realize that it’s not necessarily the size of the carbon atom
itself but the hydrogens of CH3 interacting with the axial hydrogens on the ring that lead
to strain. Oxygen, having only one hydrogen, can always rotate such that the H is
pointing away from the cyclohexane, thereby leading to very little in the way of diaxial
interactions with the ring.
50
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The value for OCH3 is even less (0.6 kcal/mol).
Bromo cyclohexane :
Along similar lines one could be forgiven for thinking that Br, being such a heavy and
large atom, might exert a large destabilizing influence when in the axial position.
However, the difference is only 0.43 kcal/mol, less than that for OH. Why might this be?
The answer here is bond length. The average C-Br bond is about 193 picometers in
length (1.93 Angstroms) – compare this to 1.50 for the bond between C and CH3 in
cyclohexane. The Br, being farther away, will thus have less interaction with the axial
hydrogens. [Note – this A value of 0.43 is the average of two experimentally determined
values [0.38 and 0.48].
Interestingly, despite their great difference in size, the A values for Cl, Br, and I are all
roughly similar (about 0.43 or so). This is because the increased size is balanced by the
increased bond length – the halogens might be increasing in size along Cl <Br < I – but
they are also getting farther away.
51
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question: Draw the most stable confirmer of the following:
CH3 CH 3
CH 3
CH 3 CH 3 CH3
(1) (2)
CH 3
CH 3 CH3
CH3
Me
(3) Et
Me
Geometrical isomerism in cyclohexane:
(i) G.I in 1,2-dimethylcyclohexane.
CH3 CH3
CH3 CH3
CH3 CH3
CH3 CH3 (e,e)
(a,a) (a,e) (e,a)
trans
trans cis cis
(A) (B) (C) (D)
D >B = C > A
(ii) G.I. in 1-Ehtyl-2-methylcyclohexane:
Tip: Put – Et in place of Me at top position only in (i) above case then D > C>B> A
(iii) G.I. in 1,3-dimehtyl chyclohexane:
CH3 CH3 CH3 Me
CH3 CH3
CH3
(a,a) (a,e) (e,e)
cis trans (e,a)
cis
trans
(A) (B) (C) (D)
D >B = C > A
52
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Important Concept
e
a a
Position Cis Trans
e ae
a ea (1,2) (1,4) (1,6) (a,a) or (e,e)
(a,e) or (e,a)
e e (1,3) (1,5) (a, a) or (e,e) (a, e) or (e,a)
a
Exceptions:
(1) cis-1,3-cyclohexan-diol:
OH OH
> OH OH
(a,a) more stable (e,e) less stable
(2) cis-1,4 cyclohexandiol :
OH
OH OH
OH
<
(a,e)
(3) cis-1,4-ditert.butyl cylohexane:
Twisted boat most stable
H
|
N CH3
(4) cis & trans -2-mehtyl pepyridine:
N
H N
H
CH3 > H
H
CH3
cis (on the basis of H-atom) trans
53
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Optical Isomerism
Optical isomers have same MF, EF, MW, EW, G.F. same structure. Vapour density atleast
one, physical or chemical properties different & different orientation in space.
Definition of Specific rotation :
1 gm/ml solution of a compound rotate ppl in 10 cm polarimeter tube than the obs. rotation is
known as SP rotation.
A monochromatic light consist of E.M. waves which oscillates /vibrates in all direction
parallel to the of light. When monochromatic light is passed through Nicol prism of calcite or
CaCO3. Only plane light oscillates which is known as polarized light.
Biot is the first scientist who used PPL on O.C. and explained some oscillation show rotation
towards PPL are called OAC. Those O.C. does not show rotation towards PPL are called
O.I.C.
The rotation of polarized light can be in the clockwise or counterclockwise direction :
If the rotation is clockwise (to the right from from the noon position), the compound is
called dextrorotatory. The rotation is labeled d or (+).
If the rotation is counterclockwise (to the left from the noon position), the compound is
called laevorotatory. The rotation is labeled l or (-).
O.A.C d/l/uneq.mix
O.I.C (i) symmetric (ii) meso (iii) R.M.
obs .
Factors affecting obs. Rotation : sp rot t
lc
(i) wavelength of light (ii) temperature (iii) solvent (iv) conc. (gm l1).
(iv) length of tube (in dm) (v) Nature of compound.
54
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Enantionmericallly Excess/Optical purity : A sample of an O.A.C. that consist of a single
enatiomer is said to be E.E. or O.P.
mol of one enatiomer mol of other enatiomer
% of E.E./O.P =
Total no. of mol of both enatiomer
OR
obs. rotation (mix) RS d l
% e.e. /O.P. = 100 100 100
sp. rotation ( pure ) R S d l
Mass per ml of pure
%O.P. x100
Mass per ml of mixture
obs. rotation mass per ml of pure form
O.P. =
sp. rotation mass per ml of mixture
Question-1: The sp. Rotation of a pure enantiomer is + 10°. What will be its obs. rotation if
it is isolated from a reaction with (i) 30% racemisation & 70% retention.
(ii) 70% recamisation & 30 % retention.
Sol: (i) obs - 0.70 10 = + 7.0° (ii) obs = 0.30 (- 10) = - 3.0°
Question-2: What is % composition of mix two enantiomers whose rotation is -10°?
The sp. Rotation of pure enantiomers is – 20°.
obs 10
Sol: % O.P. 100 100 50%
sp 20
% of l = 50 + 25 = 75% & % d = 25%
Question-3: What is the O.P. of a sample having an obs = +9° & sp = + 12°
90
% O.P. = 100 75%
12
% of d form = 75+ 12.5 = 87.5% , % of l form = 12.5%
55
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question-4: An aq. solution containing 10 mg of optically pure fructose was diluted to 500
ml with water and placed in a polarimeter tube 20 cm. long. The measured rotation was – 5°.
To this solution 500ml of a solution containing 10 gm. of recaminc fructose is added . What
will be change in sp. rotation?
Solution:
0 5
sp = 125
l c 2dm 10 / 500
Total vol = 500 + 500 = 1000 ml 10/1000 0.01 g ml 1 of pure fructose.
Mass per ml of mix = (10 + 10 = 20) 20/1000 = 0.02 g ml 1
obs mass ml 1 of pure 0.01
% O.P = 100 1
100 100 50% l form.
sp mass ml of mix 0.02
new sp = 125 0.5 62.5 and change in sp rotation = 62.5 125 62.5 .
%d %l obs RM E.E
100% + 100%
90% 10 + 20 80%
75 25 + 50 50%
50 50 0 100
(OIC)
40 60 80 20
20 80 40 60
2 98 4 96
100 100
56
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cause of Optical Activity
The compound must be unsymmetrical. Two know the molecule unsymmetrical, we have to
know about the symmetry of element.
Elements of Symmetry:
The type or elements of symmetry are as follows,
(1) P.O.S () internal mirror plane (2) C.O.S. (i) (3) A.O.S. (C) (4) A.A.O.S.
(1) P.O.S.: Plane of symmetry (σ) : An imaginary plane which bisects the molecule into
two equal half and they are the mirror image of each other & known as plane of symmetry or
internal mirror plane.
Question: Find the number of atom bisected using plane of symmetry.
F Cl
Cl
Cl
(2) (3)
(1)
Cl
Cl Cl
Cl
Br
(5) (6)
(4)
Cl
Cl
Cl
H
H
(7) (8) (9) H
H
H
H
F
CH3
(11) (12)
(10)
57
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cl Cl
(13) (14) (15)
Cl
Cl
Cl
Cl Cl
(16) (17) (18)
Cl Cl
Cl
Cl
H Cl
H H
(20) (21) (22) H2
(19)
Cl H
Cl Cl
H H
(23) O= C= O (24) >C= C=C < (25)
Cl
Cl
O O
(26) O (27) (28)
O
Cl Cl Cl
Cl Cl
(3 0 ) (3 1 )
(2 9 )
Me Me
H Me Me
CH3
Fan/Cycle/Hockey Stick/Right hand/Human being No P.O.S.
POS in Alphabet:
A B C D E F G H I J
1 1 1 1 1 0 0 2 2 0
K L M N O P Q R S T
1 0 1 0 0 0 0 0 1
U V W X Y Z
1 1 1 2 0 0
(2) Centre of Symmetry COS (i): An imaginary point in a molecule through which draw a
line in opposite direction at 180° and we find that similar atom exists at similar distance are
known as centre of symmetry (C.O.S). a centre of stmmetry is usually present only in an
even membered ring.
58
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(trans-2,4-dimethylcyclobutane trans-1,3-dicarboxylic acid)
Question: Find the number of atom bisected using centre of symmetry.
O
(1) (2) (3)
H H
(4) H B
< H
H
(5)
Cl Cl
(6) (7) (8) (9) (10)
Cl Cl Cl
Cl
Cl
(11) (12) (13)
Cl
Cl
Cl
COS in alphabets:
A B C D E F G H I J
K L M N O P Q R S T
U V W X Y Z
(3) Axis of symmetry (A.O.S.) (Cn) :
Subscript (n) denotes the number of times that object should be rotated to regain its original
form.
59
M. PRAKASH INSTITUTE ISOMERISM Std: XI
“An imaginary axis around which rotate a molecule in such a way that again reappear of a
molecule is known as axis of symmetry ”
360
General Formula : n
O O
(1) H H H H
180° reappear
Axis of symmetry for H2O = C2 .
subsidiary
C
(2)
main - C C3 Passing through centre parallel to plane.
Main axis of symmetry and subsidiary axis of symmetry should be noted.
Cl
(3) (4) (5) (6)
Cl
Cl
Cl Cl Cl
(7) (8) (9) (10)
Cl Cl Cl
Cl
Cl
(11) (12) (13) (14)
Cl
60
M. PRAKASH INSTITUTE ISOMERISM Std: XI
OH
N
(15) (16)
HO OH
(4) A.A.O.S. Alternating axis of symmetry Sn: In case of alternating axis of symmetry,
rotate the molecule by 2π/n degrees and then along the axis, place a mirror which is
perpendicular to the axis. The obtained mirror image should be identical with the original
one. Identical means, appearance should be identical with the original.
Example-1 : Following molecule has S2 AAOS.
Example-2 : Following molecule has S4 AAOS.
Example-3 : Following molecule has S2 AAOS.
Cl Cl Cl
x-axis S2
180°
Cl Cl Cl
61
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Condition for optical activity:
(i) Compound must be unsymmetrical
Original Compound
Unsymmetrical
Symmetrical
POS/COS/AAOS
Any one are present
PPL X, O. Inactive Assymetrical Dissymetrical
POS , COS POS x, COS x
AAOS , AOS AAOS x, AOS
(ii) Compound having dipole moment :
(iii) Compound having non-superimpossable mirror image on its compound (Enantiomer)
(iv) Compound having chiral atom (not necessary)
1. Chiral & Achiral molecules
A molecule (or an object) is said to be chiral or dissymmetric, if it is not superimposable
on its mirror image and the property of non-superimposability is called chirality.
On the other hand, a molecule (or an object) which is superimposable on its mirror image
is called achiral (non-dissymmetric or unsymmetrical).
Chiral carbon atom (Chiral centre/Stereo centre): Carbon atom bonded to four different
atoms or groups is called as asymmetric carbon atom or a chiral atom. A chiral atom is
indicated by an asterisk (*). OR Those SP3 atom having 4 different valencies.
Note : Isotopes of a particular atom behave as defferent groups in stereoisomerism.
D H OH
35 37
H T Cl Cl CH 3 Cl
Br D H
Chiral Chiral Chiral
If a molecule contains only one chiral centre/atom, then the molecule has to be optically
active (i.e. non superimposable on its mirror image) as it will not contain any element of
symmetry. Molecule containing two or more chiral centres may or may not be chiral
(optically active).
The property of nonsuperimposability of an object on its mirror image is called chirality.
62
M. PRAKASH INSTITUTE ISOMERISM Std: XI
It is necessary to distinguish chiral and chiral centre. The word chiral is used for molecule
as a whole which is optically active, whereas chiral cnetre is for an atom which is attached to
form different atoms/groups.
Following is the structure of cholesterol, it has total eight chiral centres.
Relationships between chiral centers and chiral molecules :
i. The term chiral center refers to an atom in the molecular structure. The term chiral
molecule refers to the entire molecule.
ii. The presence of on chiral center renders the entire molecule chiral. The presence of two
or more chiral centers may or may not result in the molecule being chiral. In the examples
given below the chiral centers are indicated with an asterisk. The vertical broken line
represents a plane of symmetry.
Ibuprofen : one chiral Cis-1,2- Trans-1,2-
center renders the dimethylcyclohexane is dimethylcyclohexane is
molecule chiral. an achiral molecule. a chiral molecule.
Examples :
(1) CH4 6POS, COS , 4C3 (2) CH3Cl 3 POS, COS , 1 C3
(3) CH2Cl2 2POS, COS , 1 C2 (4) CH2ClBr 1POS, COS , 1C1
(5) *CHClBrF POS , COS , AOS
63
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Chirality in compounds lacking a stereogenic C-atom :
i. tetrahedral stereocenters other than carbon :
R' R' R O O16
R Si R'' R Ge R'' N R P R'' Me S Me
R' R'''
R''' R''' R' R' O18
ii. amines, phosphines, sulphoxides :
1. 2. Cl
N
N
Ph Me
H
3. O 4. O
|| ||
CH3 S Ph Me S Me
||
O
5. O 6. As
|| Me
Me S Me D
|| H
18 O
7. O 8. Cl
Br
P P
MeO Cl Me OMe
OM e OMe
9. 10. Me
O
x
H N
N
H Me
CH3 Ph D
11. 12.
N
N
4 chiral atom
13. O 14. ..
|| O S OMe
R S OR |
.. Ort
64
M. PRAKASH INSTITUTE ISOMERISM Std: XI
|
15. N
16.
17. 18. OH OH
OH OH
19. 20. H H
Me COOH
21. COOH 22. COOH
H
H
23. N 24. N
Br Br
25. F 26.
O O
27. Cl O 28. N
N
Me N
Me Cl
29. Me 30. OH
Me
OH
65
M. PRAKASH INSTITUTE ISOMERISM Std: XI
31. OH 32. OH Br
Br OH
OH
33. H H 34.
H H
Topomerization : Generally, enantiomers are non-interconvertible into each other, but in the
following cases where enetiomers are interconvertible which is known as Topomerization.
Case-I : Cyclohexane
Me
Me a
a Keq = 1 Me
Me e
50% e 50%
Optically inactive racemic mixture
Case-II : Amine inversion
Some facts about nitrogen flipping:
Ammonia, oxygen, F-, CH3 show flipping (amine inversion)
Due to inversion basicity increases.
Due to inversion , optically inactive due to racemic mixture production.
Nitrogen flipping not possible in sp2 hybrid nitrogen.
As, P, S do not show flipping at room temp. due to large size, as well as due to d-orbital.
Bridge head nitrogen doe not show flipping (Bredt’s rule)
66
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Flipping is not possible in high acedic medium.
Flipping will be less if alkyl groups size is more or in small cyclic system.
Examples :
H H
| |
N H N
N N
(1)
N N N
(2)
Et
H
C N N
D D N CD 3
Me Me Me
H H H
Hybrid sp3 sp3 sp3 sp3
Flipping
Chiral
Optical Activity O.A O.I. O.I(RM) O.A.
(RM)
Resolution
iii. Molecules with the Chiral axis (Stereoaxis) : Allene itself is achiral since it has two
planes of symmetry. In order to generate chirality, the two planes of symmetry must be
eliminated. When unlike substituents are added at each end of the C=C=C unit, the
substituted allene becomes chiral. This is so in 2,3-pentadiene substituent at one end (H and
CH3; H ≠ CH3) and the other end (H and CH3; H ≠ CH3).
The cumulated bonding systems with an even number of double bonds do not have a plane of
symmetry of a center of symmetry and therefore, must show optical isomerism and must be
resolvable into enantiomers.
67
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Example :
H Me H Cl H Cl
C C C C C C C C C
Me H Cl H H H
H Cl H COOH R R
C C C C C C C C C
Me Cl Cl H R' R'
R R Cl Cl Me H
C C C C C C C C C C C
R' R' F F Me3C Cl
iv. Hemispiranes : The replacement of one double bond in an allene by a ring gives
alkylidenecycloalkanes (Hemispiranes) does not alter the basic geometry of the system of
allenes and suitably substituted compounds, therefore, exist in optically active form. Related
compounds in which SP2-carbon is replaced by nitrogen, compound has also been obtained
as enantiomers. Alkyl cycloalkyl ketone (Achiral) is more stable than its enol form (chiral).
H Me H COOH Me Cl
Me H Me H H Me
H H Me
N N N
Me OH H OH Me OH
H O H OH
C C
Me Me Me Me
More stable achiral Leass stable chiral
O OH
C C
Me Me
Me Me
H H
More stable achiral Less stable chiral
68
M. PRAKASH INSTITUTE ISOMERISM Std: XI
v. Spiranes : The replacement of both double bonds in an allene by ring system gives spiran
; appropriately substituted compounds have been obtained in optically active forms.
H Me H H
Me H Me Me
Me COOH H COOH
H H Me H
H Me H H
Me H Me Me
Note : Cummulele with odd number of double bonds, hemispiranes with two rings one
double bond or one ring two double bonds or spirane with odd numbers of ring system show
geometrical isomerism.
vi. Biphenyl (Atropisomerism): Resolvable (Chiral) biphenyls must contain two different
bulky ortho substituents on each ring and these make each ring unsymmetrical and two rings
are held in perpendicular planes. Rotation about the bond linking the two phenyl rings does
not occur due to steric hindrance between the bilky ortho substituents.
Atropisomerism is also called axial chirality and the chirality is not simply a centre or a plane
but an axis.
NO2 COOH Cl COOH
HOOC O 2N O 2N
O 2N O 2N O 2N
COOH COOH COOH
F HOOC F HOOC F HOOC
COOH F COOH F COOH F
Chiral Achiral (i) Chiral
Ac
O H
H HOOC NH HOOC C N
COOH H COOH HN N C
H O
C h ira l Ac
A c h ira l (i)
Polunuclear aromatic system such as Binol also exist as enantiomers.
69
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question : Identify the chiral molecule & number of chiral atoms in the following:
(1) (2) (3) (4) (5)
CHO
(6) (7) (8) (9) (10) H OH
H CH2OH
OH Cl Br
(11) (12) (13)
| |
(14) CH CH CH CH
3 3
* *
Cl
Cl
HO Cholestrol
Question : Write them Minimum number of carbon atoms required for the chirality:
(i) Alkane (ii) Alkene (iii) Alkyne (iv) R
– OH
(v) RCOOH (vi) RCHO (vii) RCOCH3 (viii)
RNH2
(ix) Enyne (x) Diene (xi) R- Cl
2. Prochiral center & Prochiral molecule :
Those achiral carbon atoms converting into chiral carbon atom after the chemical reaction.
Example-1: A tetrahedrally bonded atom (of an achiral compound) of the general formula
Cabc2 (as in propanoic acid) which becomes a stereocenter Cabcd and the compound becomes
chiral on replacement of one of the identical groups with a different group ‘d’ is called a
prochiral center and the molecule as prochiral.the two identical ligands in propanoic acid i.e.
H atoms of the methylene group are called “homomorphic” from greek homos meaning same
and morphe meaning form, these are indistinguishable when considered in isolation.
70
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Example-2: when the carbonyl group in pyruvic acid is reduced by the addition of hydride
from e.g. NaBH4 the enantiomers of lactic acid are again obtained by the equally feasible
addition to either the front or the rear face. The SP2 hybrid carbon of the carbonyl group in
pyruvic acid is called prochiral and is said to have two heterotropic faces which are
enantiotopic.
Note : In formaldehyde there is no way to distinguish between the two faces for
formaldehyde- addition of CH3MgBr to either face gives the same compound ethanol. This
shows that the two faces in formaldehyde are also homotopic.
Example : Identify the prochiral molecule and number of prochiral centers.
(1) C
l2
+ 2- Prochiral
Cl
Sp 3
Cl
H OK
Cl
(2)
Cl KC
N H (3)
C l2 Cl
+
Sp 2 CN Cl
Cl
2- Prochiral
Cl
Cl
x Cl
(4) Cl
2
+ + + +
Cl
4- Prochiral
Cl
Question : Write the numbers of chiral products on monochlorination : (MCP)
Cl
Cl
(1) CH4 0 (2) CH3 CH3 0 (3) 0 (4)
Cl
Cl Cl
(5) (6) Cl
+ +
+
Cl Cl
(7)
Cl +
+
Cl Cl
CH 3 Cl
(8)
Cl H Cl MCP
1 (10)
(9)
H H Br
1
CH 3
71
M. PRAKASH INSTITUTE ISOMERISM Std: XI
(11) (12)
3. Enantiomers :
Real image not superimposable on its mirror image.
Molecule must contain atleast one chiral centers.
Enantiomers having same molecular formula & structural formula but different in
stereochemical arrangement.
Enantiomers having same physical properties like boiling point, melting point, vapour
density but differ in plane of polarized light.
Enantiomers having same chemical properties & rate of reaction except towards optically
active compounds.
Racemic mixture contains isomers of the same compound, which are mirro image of each
other. It is known as racemisation.
Racemic mixture may be optically inactive (contain equal proportion of both the isomers)
or optically active (contain unequal proportion of both the isomers).
Separation of enantiomers is very tough, which is known as resolution.
For resolution chiral reagent or enzymes are required.
4. Diastereomers :
Those stereoisomers which are not mirror image of each other.
Molecule must contain more than one chiral centers.
They have same chemical properties but different physical properties.
Their separation is very easy due to difference in physical properties like melting and
boiling points.
Erythro and Threo are diastereomers of each other.
Cis-trans isomers sre also diatereomers of each other.
72
M. PRAKASH INSTITUTE ISOMERISM Std: XI
5. Erythroses :
In a fischer projection if two adjacent stereocenters having similar groups are present on
same side.
It is observed in unsymmetrical optical active compounds.
6. Theroses :
In a fischer projection if two adjacent stereocenters having similar groups are present on
opposite side.
It is observed in unsymmetrical optical active compounds.
Note :
If two molecules are mirror images, then their configurations are exactly opposite and
they are enantiomers. For example D & L erythroses are mirror image, so they are
enantiomers, same way D and L threoses are enantiomers.
If two molecules are not mirror images, but still they are stereoisomers they are
diastereomers, D-erythrose and D-threose are not mirror images so they are siastereomers,
same way D-erythrose and L-threose; L-erythrose and D-threose and L-erythrose and L-
threose are diastereomeric pairs.
Diastereomers are different geometrical entities. But enantiomers are identical entities.
For the diastereomeric relationship chirality is not essentiality. For example cis- and
trans-isomers are diastereomers.
Single asymmetric centre cannot show daistereomerism.
Epimers are also diastereomers.
7. Meso compounds :
An optically inactive compound whose molecule is superimposable on its mirror image
inspite of the presence of chiral atom is called a meso compound.
If a molecule has two or more chiral centers, it is usually chiral. The exceptions are meso-
molecules, which are not chiral. These are molecules that due to symmetry have chiral
centers that “cancel” each other out. It is known as internal companssation.
Meso compounds must have two or more chiral centers.
73
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Meso compounds have optical inactive nature due to the presence of plane of symmetry
or centre of symmetry.
Meso compounds are considered as optical isomers.
Two different meso forms of the same optical active compound are diastereomers, it is
observed when the molecules contains more than two chiral atom.
A person is meso (not chiral) even though they have
chiral elements (hands and feet). There is a plane of
symmetry down the middle of a person, which makes a
person the same as their mirror image.
NH 2
This molecule is meso (not chiral). It has two chiral
centers,and a plane of symmetry.
N H2
NH2 NH2 This molecule is not meso (chiral). It has two chiral
centers but no plane of symmetry.
NH2 This molecule is not meso (not chiral). It has a plane of
symmetry but no chiral centers. The carbons attached to
H2N the NH2 groups may look like chiral centers but they are
not.
In Tarteric acid : HOOC CH CH COOH
OH OH
The molecule contains two chiral carbons and the number of optical isomers should be 2n =
22 = 4 but number of optical isomer is reduced to 3 because one molecule has a plane of
symmetry. The steroisomers of tartaric acid are,
I and II are enantiomers (non-superimposable); III and IV are meso form (superimposable).
74
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Calculation of number of optical isomers in compounds (containing ‘n’ chiral atoms) :
S.N. Compounds No. of O.A. No. of No. of Total no.
forms (a) meso Racemic of optical
forms (m) mixture isomers
(a/2)
1. The molecule 2n 0 2n1 am
has no
symmetry
2a. The molecule
has symmetry :
Case-I : When 2n1 n
1 n
1
am
compound has 2 2
2 2
even no. of
chiral Carbon
atoms.
2b. The molecule
has symmetry : n 1 n1 a/2 2n1
n 1
Case-I : When 2 2 2
2 2
compound has
odd no. of chiral
Carbon atoms.
Examples :
1. C6H5 CH CH CH CH3
OH OH OH
Number of optically active forms a = 2n = 23 = 8
Number of racemic mixture : (a/2) = 2n1 = 23-1 = 4
Total number of optical active forms : a + m = 8 + 0 = 8
2. HOOC CH CH COOH
OH OH
Number of optically active forms a = 2n1 = 22-1 = 2
n
1
Number of meso forms m = 2 2
= 22/2 – 1 = 20 =1
Total numbers of optical isomers : (a + m) = 2+1 = 3
75
M. PRAKASH INSTITUTE ISOMERISM Std: XI
3. HOCH2 CH CH CH CH2OH
OH OH OH
n 1 31
Number of optically active forms a = 2n 1 2 2
= 231 2 2
42 2
n1 31
Number of meso forms m = 2 2
=2 2
2
Number of racemic mixture : (a/2) = 2/2 =1
Total numbers of optical isomers : (a + m) = 2n1 231 4
Question : Identify the following in the given molecules
POS / COS / AOS / Asym / Dissym / sym / Chiral carbon
A B C D E F G
Cl Cl Cl Cl Cl
(1) (2) (3)
Cl (4) (5)
Cl Cl
Cl
CH3
Cl CH3 H H Cl H
(6) (7) (8) (9) (10)
Cl Cl H Cl
CH3 CH3
Cl
Cl Cl Cl
(11) (12) (13) (14)
Br
76
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Interconversion of structures
(i) Inter conversion of wedge – Dash formula to Fischer formula :
a d a a
(i) d a c d c d
(ii)
b b c
c b b
a a a a
d
(iii) d c (iv) d c d
b b c
c b b
Tip: bond will move bond opposite side
(2) Sawhorse projection formula
c' c'
b' b' a'
a' a' b'
c a' c
b' c b' c'
b
b
c b
a a
a
b
a
(1) (2)
(3) (4)
180°
Structure (1) (4)
(3) Newmann projection formula :
a a
f d
f d
b c c
b
e
e
Front C - atom
Back C - atom
(4) Convert sawhorse & Newmann projection formulae in Fischer and vice-versa.
H C OO H
CO O H OH
H OH HO H
(i) C OO H 180°
H
H OH OH
H OH
C O OH C OO H
2-O H sam e side C OO H
2- OH opposite side
77
M. PRAKASH INSTITUTE ISOMERISM Std: XI
H COOH
COO H OH
H OH HO H
(ii) HO COOH 180°
HO H H
HO H
COOH COOH
2- O H opposite side COOH
2-O H sam e side
H COOH
OH COOH
CH 3 COOH
H OH HO
(iii) Me COOH 180° H HO
180° H HO
H Me 180° H
HO H H H
COOH OH HO
HO Me
HO
Me
H
2 - OH same side
HO COOH
COOH
Me H
HO H H
(iv) Me COOH 18 0° OH H
18 0° OH
H Me
HO H H H
COOH Me
HO
HO
H
COOH
COOH
H OH HO
H
(v )
H H
HO OH
COOH
COOH
Br
H
Me
Br COOH
Me
H 180°
(vi) Me Br H Me H Me
Me 180°
H H
Me H Me H
COOH COOH COOH COOH
COOH Me
H CO O H
H
Me 180 °
(vii) Me CO O H Me H
Me
H H
Me H
Br Br Br
78
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Fischer projection formula :
Fischer projections are abbreviated structural forms that allow one to convey valuable
stereochemical information.
Fisher is always eclipsed confirmation.
OH CHO
H
H Cl
H OH
CH3 Cl H
CHO
CH3
N ew m an n
F i sh e r
The definition is that every carbon is specified completely by a cross designining the
carbon (at the center) and the four bonds to that carbon. The stereochemistry of the bonds is
defined (now) as the horizontal bonds are in front of the plane (coming towards the viewer);
the vertical bonds are behind the plane (going away from you).
When relating one Fischer projection to another, it’s important to realised that it may
only be manipulated within the 2-D plane.
A 900 rotation is equivalent to breaking bonds and exchanging two groups, which would
result in the formation of the other enantiomer.
79
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Fischer projections can also be used to represent molecules with more than one chirality
center.
A Fischer projection or Fischer projection formula is a convention used to depict a stereo-
formula in two dimensions without destroying the stereochemical information. i.e. absolute
configuration, at chiral centers.
To convert this stereoformula into a Fischer projection use the following procedure
Step-1: Hold the molecule so that
i. The chiral center is on the plane of paper,
ii. Two bonds are coming out of the palne of the paper and are on a horizontal plane.
iii. The two remaining bonds are going into the plane of the paper and are on a vertical plane.
Step-2: Push the two bonds coming out of the plane of the paper onto the plane of the paper.
80
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Step-3: Pull the two bonds going into the plane of the paper onto the plane of the paper.
Step-4: Omit the chiral atom symbol for convenience.
Conventions for configuration D-L and R-S systems :
A. D-L Configuration : It is relative configuration/nomenclature:
It is applicable only for carbohydrates & amino acids in fFischer projection formulae.
D : Dexus : The (+) enantiomer of glyceraldehydes has its –OH group on the right of the
Fischer projection.
Thus the D-series of sugars are those with the –OH group on the highest numbered
stereocenter (i.e., the bottom stereocenter) on the right in the Fischer projection.
L : Laevus : The (-) enantiomer of glyceraldehydes has its –OH group on the left of the
Fischer projection.
Thus the L-series of sugars are those with the –OH group on the highest numbered
stereocenter (i.e., the bottom stereocenter) on the left in the Fischer projection.
D & L forms of a molecule are enantiomers of each other.
Rules for writing D & L forms :
For D-L nomenclature, Fischer projection must be correct. i.e. main functional group or
high oxidation number group always should be at the top & follow the IUPAC rule.
If amino acids or α-Hydroxy acids, then consider the –NH2 group or –OH group
respectively.
If two or more chiral carbon, then consider the highest numbered stereocenter (i.e., the
bottom stereocenter).
D-L configurations are not related to the optical rotations of the sugars to which they are
applied i.e. some sugars are D-(+)-while others are D-(-).
81
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The direction of rotation is independent of the stereochemistry of the sugar, so it may be
designated D (-), D (+), L (-) or L (+). For example naturally occurring fructose is the D-(-)-
isomer.
There is no relation between D-L & d-l, R-S & d-l, D-L & R-S.
Examples :
OH CHO CH 2OH CHO
(1) H CH 2O H H OH (2) H OH (3)
OH
C HO CH 2O H CHO C H 2 OH H
CHO
CH O
COOH COOH H OH
H OH HO H
(4) H NH2 (5) HO H (6) (7)
HO H OH
H
CH 3 CH3 H OH
C H 2O H
C H 2O H
B. Absolute configuration: R,S-System:
An absolute configuration refers to the spatial arrangement of the atoms of a chiral
molecular entity (or group) and its Stereochemical description e.g. R (Rectus) or S (Sinister).
The arrangement of atoms in an optically active molecule, based on chemical
interconversion from or to a known compound, is relative configuration. Relative because it
is no way to knowing just looking at a structure whether the assignment of (+) or (-) is
correlated to a particular isomer, R or S.
R and S notations are used only to describe asymmetric molecule following Cahn-Ingold-
Prelog (CIP) sequence rules.
To assign absolute configuration as R or S the 3-D drawing of the molecule has to be
viewed with the lowest priority group at the back.
After giving priority order for the groups at asymmetric centre, if priority direction is
clockwise the configuration is specified ‘R’ (Latin: rectus, right); if anticlockwise is specified
‘S’ (Latin sinister, left).
If the lower priority group is not away from the viewer, then to bring the bring the group
at the position where it is away from the viewer do even matual exchenges of the groups so
that the least priority group come to away from the viewer.
82
M. PRAKASH INSTITUTE ISOMERISM Std: XI
2
Cl Cl
2
2
Cl Cl
3 18 0
1
F
3
1
o ut o f plan e 3 1
Br F Br H Br
H H Br F
4
4 F H
4
n eg le ct o n re a c tio n c o n fig u ra tio n d o e s n o t c h a n g e
R S
F
Br Br
I
A
I Br
I
Cl Cl
Cl
F F
S
B C D E
Cl I Br
Br
F
I Cl F Cl Br Cl I
Br
F I F
S
S
Question : Draw all the possible stereoisomer of 2-butanol in wedge-Dash & fischer
projection.
Case- I : If one chiral carbon.
Rule: In Wedge-dash:
Real mirror image Rotation (conformation) 3 groups
R S R R
Odd interchange Even interchange
83
M. PRAKASH INSTITUTE ISOMERISM Std: XI
In Fischer:
90 / 90
Rotation Rotation (3 groups)
R S Out of plane RR
Real mirror image
Odd interchange Even interchange
Example:
Br
in ter
cha ng e Cl Me
Me
H S
Cl Br Br
H 2 - in ter Me
H
cha ng e
R R
Cl
Br
a nticlo ck Cl
Me H
R ota tio n
Cl Br Me R
H
R
Case- II: 2- unsysmmetrical chiral carbon .
Que: Draw all the possible stereoisomer of 2-Bromo-3-butane.
Me
H Br Br H H Br
Br H
H Cl Cl H Cl H
H Cl
Same side Me
Opposite
B Side C
Erythreo
Threo D
A
If mirror image are superimposable on the its mirror image. It means identical. Those stereo
isomer which are identical but having different representation are known as homomer.
Those stereoisomers & its mirror image are non-superimposable it means enantiomers.
Those stereo isomers which are not mirror images of each other are called diastereomers.
A – B Ena, A – C Dia, A – D dia, B– C dia, B – D dia, C – D Ena.
84
M. PRAKASH INSTITUTE ISOMERISM Std: XI
RS
RR Identical (no change)
Identical (no change)
RR
RS
Dia (1 change) RS Dia (1 change)
RR SR SS
SS SR
Ena. (Both change) Ena. (Both change)
Ques: Draw all the possible streoisomer of 2,3-dichlorobutane in fischer as well as wedg-
dash formual also count the RS configuration.
H Cl Cl H Cl H H Cl
H Cl Cl H
Total - 3
H Cl Cl H
Cl Cl Cl Cl
PO S
R & S nomenclature for the compounds of N, P & S :
In such cases lone pair has lowest priority.
Tetrahedral compounds like amines, phosphine, phosphineoxides, amineoxides,
sulphoxides display chirality provides the ligands are non-equivalent.
Example :
Me
P Switch
Me 2 groups
Ph C Me Ph C6F5 Ph
6 F5 C6F5
S = Correct R = Incorrect
Question: Assign R or S configuration
H CN O-
Br CH2NH2 EtO
Me Me
CHO Me H Me2N OMe H Et
1 2 3 4
85
M. PRAKASH INSTITUTE ISOMERISM Std: XI
NSO2Ph Me Me O Me O
n-Bu S
S - S Me -
Ph MeO -
O Ph O
O Me O Me
5 6 7 8 9
H
O O
Me N O
Me NH Me Me NH H
&
O N H NH H H NH Me
Me
H O O
10
H
Me H N H
Me
Cl
Me COOH
11 12 NH2 13
Ph
HOCH2 NH2 COOH O
Me
Me H CH2OH HO CH2Br MeO
CBr3 Br PhO SMe
14 15 16 17
8. Homomers : Homomers are the identical representation of the some compound i.e. the
molecular models of these representations are superimposable. In other words when two
molecules, at different times occupy the same position in space, they are identical &
homomers. e. g. the four homomers of S-2-Bromobutane.
H H Br Br H
OR OR
OR Br Me OR Et Me Br Me
Br Et Me
Me Et H Et H Et
86
M. PRAKASH INSTITUTE ISOMERISM Std: XI
9. Alpha (α) and Beta (β) Anomers:
An anomer is a type of geometric variation found at certain atoms in carbohydrate
molecules.
An amoner is an epimer at the hemiacetal/acetal carbon in a cyclic saccharide, an atom
called the anomeric carbon.
Anomerisation is the process of conversion of one anomer to the other.
Alcohols react readily with aldehydes to form hamiacetal:
Similarly alcohols react with ketones to produce hemiketals:
Since aldoses and ketoses contain alcohol groups, in addition to their aldehyde or ketone
groups, they have the potential to react to form cyclic forms.
Cyclization creats an anomeric carbon (the former carbonyl carbon), generating the α and
β configurations of the sugar, for example, α-D-glucopyranose and β-D-glucopyranose.
These two sugars are both glucose but are anomers of each other.
In the α-configuration, the –OH on the anomeric C projects to the same side as the ring in
a modified Fischer projection formula. In a Haworth projection formula of the α-
configuration, this –OH is trans to the –CH2OH group.
Since the α and β forms are not mirror images, they are referred to as diastereomers.
87
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Enzymes are able to distinguish between these two structures and use one or the other
preferentially. For example, glycogen is synthesized from α-D-glucopyranose, whereas
cellulose is syntsihezed from β-D-glucopyranose.
The cyclic α and β anomers of a sugar in solution are in equilibrium with each other and
can be spontaneously interconverted which is known as mutarotation.
10. Anomerization :
It is the process of conversion of one anomer to the anomer. For reducing sugars,
anomerization is referred to as mutarotation and occurs readily in solution and is catalyzed by
acid and base.
This reversible process typically leads to an anomeric mixture in which eventually an
equilibrium is reached between the two single anomers.
The ratio of the two anomers is specific for the regarding sugar. For example, regardless
of the configuration of the starting D-glucose, a solution will gradually move towards being a
mixture of approximately 64% β-D-glucopyranose and 36% of α-D-glucopyranose. As the
ratio changes, the optical rotation of the mixture changes, this phenomenon is called
Mutarotation.
88
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The alpha and beta anomers are each in equilibrium with the “linear” form and therefore
with each other. At equilibrium the mixture consists of 36% α-D-Glucose and 64% β-D-
Glucose.
11. Mutarotation : “Muta” = Change, so “change of rotation”
Mutarotation is the change in the optical rotation because of the change in the equilibrium
between two anomers, when the corresponding stereocentre interconvert.
Cyclic sugars show Mutarotation as α and β anomeric forms interconvert.
The optical rotation of the solution depends on the optical rotation of each anomer and
their ratio in the solution.
Example :
i. In the “alpha” (α) anomer, the –OH group on C-1 is on the opposite side of the ring as the
chain on C-5.
ii. In the “beta” (β) anomer, the –OH group on C-1 is on the same side of the ring as the C-
5.
89
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The alpha (α) anomer of D-glucose has a specific rotation of +1120 in water.
The beta (β) anomer of D-glucose has a specific rotation of +190 in water. (18.70 actually,
but rounding up to 19).
When 100% form of either anomer is dissolved in water, the value of the specific rotation
changes over time, eventually reaching tha same value of 52.50.
The specific rotation of α-D-glucopyranose decreases from 1120 to 52.50.
The specific rotation of β-D-glucopyranose increases from 190 to 52.50.
This behavior is called mutarotation.
90
M. PRAKASH INSTITUTE ISOMERISM Std: XI
12. Epimers :
Those stereoisomers which are differing in its configuration at only one chiral carbon
atom are called as Epimers.
Epimers are siastereomers that contain more than one chiral center but differ from each
other in the absolute configuration at only one chiral center.
In epimers the chiral carbon atoms whose absolute configuration makes the two
compounds different are called the epimeric carbons.
13. Epimerization :
The chemical conversion of one epimer to another is called epimerization.
If the interconversion is catalyzed by an enzyme, the enzyme is an epimerase.
i. Enolate Mechanism :
91
M. PRAKASH INSTITUTE ISOMERISM Std: XI
ii. Enediol Mechanism :
14. Isomerization :
Isomerization is the process by which one molecule is transformed into another molecule
which has exactly the same atoms, but the atoms have a different arrangement e.g. A-B-
C→B-A-C (these related molecule are known as isomers).
Endiol rearrangement : The position of carbonyl group may shift via enediol intermediate
under basic condition. For example, rearrangement of D-glucose gives D-fructose.
15. Racemic Mixture :
A racemic mixture is equimolar mixture of two enantiomers (Each of a pair of
molecules that are mirror images of each other)
No matter how many molecules are in a mixture, it is racemic if there are equal
numbers of the two enantiomers.
The racemic mixture produces a net optional rotation – of plane of polarized light – of
zero degree. This is because the mixture contains equal amounts- of both enantiomers
that have opposite rotation. Known as external compensation.
A racemic mixture is a solution containing equal amount of a pair of enantiomers.
A solution containing an excess of either the (R)-enantiomer or the (S)-enantiomer
would be Enantioenriched.
92
M. PRAKASH INSTITUTE ISOMERISM Std: XI
A solution containing only the (R)-enantiomer or the (S)-enantiomer will be
Enantiomerically pure.
16. Resolution of racemic mixtures :
The separation of racemic mixture into the individual enantiomerically pure
enantiomes is called resolution.
Since enantiomers have identical physical properties, such as solubility, boiling point
and melting piint, they cannot be resolved by common physical techniques such as
direct crystallization, distillation or basic chromatography. However, the following
methods can be used for this purpose.
i. Mechanical separation :
If d or (+) and l or (-) forms of a substance exists in well-defined crystalline forms,
the separation can be done by hand picking with the help of magnifying lens and a
pair of tweezers.
For example, the d and l forms of sodium ammonium tartarate can be separated by
this methods.
The method has very limited application and applies to only few crystalline
constituents having different shape.
ii. Biochemical separation :
In this method, the resolution is done by the use of microorganism.
When certain bacteria or moulds are added to a solution of a racemic mixture, they
decompose one of the optically active forms more rapidly than the other.
For example, when the mould, racemic ammonium tartarate, the mouls completely
decomposes the d form while l form is left practically unaffected. The drawback of
the method is that half of the material is destroyed during resolution. The process is
very slow and only small amount of the materials can be separated.
iii. Chemical separation :
This is probably the best method of resolution. The racemic mixture is made to
combine with another optically active compound and resulting solubility in various
solvents.
By fractional crystallization from a suitable solvent, they can be separated.
For example, the racemic mixture of lactic acid is allowed to combine with the
optically active base (-) strachnine or (+) brucine.
Example of resolution of racemic mixtures
93
M. PRAKASH INSTITUTE ISOMERISM Std: XI
i. (S)-1-phenylethylamine combines with a racemic mixture of lactic acid to
form diastereomeric salts. The diastereomers are separated by fractional
crystallization.
After the separation process, each of the diastereomers is subsequently treated with a
strong acid such as hydrochloric acid to regenerate the corresponding of lactic acid.
Note that the lactic acid would be soluble in organic layer, while the ammonium salt
would be in the water layer.
Since enantiomerically pure compounds are very expensive, it is usually necessary to
recover and reuse the chiral amine. This is achieved by treating the (S)-1-
phenylammonium chloride salt with a base such as sodium hydroxide to regenerate and
recover the chiral amine.
94
M. PRAKASH INSTITUTE ISOMERISM Std: XI
17. Racemization :
Racemization is the conversion of an enantiomerically pure mixtue (one where only
one enantiomer is present) into a mixture where more than one of the enantiomers are
present. (or) Conversion of an optically active substance to a raceme.
Opticallt active carbonyl compounds of the –CHC=O, in which the alpha carbon is
asymmetric, are racemized by both acids and bases.
The racemization of an optically active secondary halide with the chiral carbon
carrying the halogen (e.g., 2-chlorobutane) may occur in the solution and, usually, the
more polar and better ionizing the solvent is, the more readily the substance is
racemized.
1. O OK
|| |
CH 3 C H
KCN
CH3 C H
|
sp 2 OR CN
Me OK
2. C O KCN
Me
C
Et CN
Et
RM
H O H OK H CN
| || | | | |
3. CH 3 C C H
CH 3 C C H
KCN
CH 3 C C H
| | | | |
D D CN D OK Diasteromers
O O O O
4. H Me
H Me
OH
CH3
Me H
sp 2
R .M .
95
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Note: where negative charge (anion) there H+ change the position from up to
down.
O
5.
Me OH
Identical
H
D H
| |
6. CH3 C C CHO OH
Diastereomer
| |
H H
CH3
7.
CH3 OH
D iastereom er
H
O
O
8. CH 3
OH Dia. as cis-trans
configurations a/c to G.I.
O
CH 3
Racemisation:
9.
( i ) LAH
H
( ii ) H 2O OH
O
OH
H
Enantiomers
O H OH
( i ) LAH
( ii ) H 2O OH H
Diastereomerisation (Epimerisation)
Note : When there is chiral carbon along with prochiral, it results in
diasteromerisation.
96
M. PRAKASH INSTITUTE ISOMERISM Std: XI
18. Pseudo-asymmetric carbon (Pseudo chiral carbon) :
The traditional name for a tetrahedrally coordinated carbon atom bonded to four
different entities, two and only two of which have the same constitution but opposite
chirality sense. The r/s descriptors of pseudo-asymmetric carbon atoms are invariant
on reflection in a mirror (i.e. r remains r, and s remains s), but are reversed by the
exchange of any two entities (i.e. r becomes s, and s becomes r). An example is C-3
of ribaric (C-3 is r) or xylaric acid (C-3 is s) or hyoscyamine (C-3 is r). The hyphen
in pseudo-asymmetric may be omitted.
H E H I
CH 3 H
C C C C
H Me
H C CH Me H C CH Me
| |
OH OH
Me Me
C C C C
H H
H Z H Z
P seu d o c h ira l
19. Transannular interaction :
Trans cyclo-octene can be resolved into its two enanatiomers due to restricted
rotation about single bond.
Its planar form is achiral but it is less stable than its non-planar form that is chiral,
due to stric hindrance.
Methylene chain tries to flip through the planar conformation to the other face of
the double bond. This steric interference makes it difficult for the molecule to
achieve the planar conformation.
Because of this destabilizing trans-annular interaction. Trans cycloalkenes with
upto 11 carbons do not exist as planar molecules they exist in a twisted
conformation.
97
M. PRAKASH INSTITUTE ISOMERISM Std: XI
In trans cyclo octane to pull the stresh the methylene chain in order to convert it
into its mirror image. This rotational movement is highly restricted due to internal
hydrogen atom. It is this barrier to rotation, about single bond that allows trans
cyclo-octene to be resolved into its two mirror images forms at room temperature.
The barrier to racemise (mixture of two optical active forms) (i.e. passing through
planar conformation) is 36 kcal/mol.
The barrier to interconvert the two enantiomers of trans cyclo-nonene is only 20
kcal/mol and trans cyclodecene 10 kcal/mol.
20. Bredt’s rule :
Bredt’s Rule Definition: The rule that states that it is not possible for double bonds to
be formed on the bridgehead carbon atoms of a bicyclic system if it involves a trans π
bond being incorporated in a ring comprised of fewer than eight atoms.
Bredt’s Rule Explanation:
In principle, the Bredt’s rule is simply a consequence of the strain induced by a planar
bridgehead carbon.
A few key concepts to keep in mind when understanding the Bredt’s rule are –
A double bond comprises one sigma bond and a one π bond.
The pi bond is formed by the overlap of p-orbitals. Therefore these p-orbitals
must be on the same planar. If they are not, the chances of an overlap are
negligible and the bond formed will be unstable.
A double bond on a bridgehead would be equivalent to having a trans pi double
bond on a ring, which is not stable for small rings (fewer than eight atoms) due to
a combination of ring strain, and angle strain (nonplanar alkene).
Bridged bicyclic compounds can only exhibit a double bond at a bridgehead
position if one of the rings has at least eight carbon atoms
And therefore, bridgehead double bonds are unstable due to poor orbital overlap.
Examples –
98
M. PRAKASH INSTITUTE ISOMERISM Std: XI
The compound given in box number 1 cannot maintain a parallel overlap of the p
orbitals, has less than 8 carbon atoms in the ring forming the trans pi double bond
and as a result, it is not stable at room temperature. Note – the p -orbital of the
bridgehead carbon and the p-orbital of the adjacent carbon are not on the same
plane and therefore exhibit poor orbital overlap.
The compounds under “Other Examples’ can maintain a parallel overlap of the p
orbitals, has more than 8 carbon atoms in the ring forming the trans pi double
bond, and as a result, it is stable at room temperature
The ring for trans pi double bond is shown in last structure is possible.
Question-1: Identify the meso in the following :
COOH
(i) (ii) (iii) (iv)
COOH
Cl Cl
COOH Cl
COOH COOH
H HO H
H Cl (vii)
(v) (vi) H OH (viii)
Cl H OH
OH H
COOH
Cl
99
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cl
Cl Cl Cl Cl
R R
S S
(ix) (x)
E Z
H Cl Cl H
Cl
In guard E-Z or cis-trans are diaastereomers of each other but in some cases it is
enantiomers.’
(II) C6H6Cl6 666 Lindane/Gammaxene:
Cl Cl Cl Cl Cl
Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl
6 POS 1 POS 1 POS 1 POS POS
Cl Cl Cl Cl Cl Cl Cl Cl Cl Cl
Cl Cl Cl Cl Cl
Cl Cl Cl
Cl Cl Cl Cl Cl Cl
POS POS Number of POS/COS/AAOS
COS COS Optically Active
Cl Cl Cl Cl Cl Cl
Cl Cl Cl
Question-2 : Draw the possible stereoisomers:
HOOC Ph
POS
(12) POS POS COS POS
Ph COOH
Trusealic acid
COOH Ph
(13)
Ph COOH
100
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cl Cl
| |
(14) CH 3 CH CH CH CH CH 3
Cl Cl Cl Cl
Cl Cl
Cl Cl
| |
(15) CH 3 CH C C CH CH 3
| | Total = 7
Cl CH CH Cl
| |
CH 3 CH 3
Cl Cl
Question-3 : Calculate number of isomers of 2,3,4 trichloropentane.
Ans: O.A = 2 meso 2 Total = 4
Question-4 : Identify the pseudo chiral carbon in the meso or in O.A. isomer of the above :
Me Me
H Cl Cl H
H Cl H Cl
Pseudo chiral H Cl Achiral carbon
H Cl
Me Me
Meso O.A.
Question-5 : Draw the stereoisomers of the following:
Cl
Cl
(1) Cl
CO S AAOS
Cl (2) (2) (1) (1)
Total= 6
101
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Cl Cl Cl Cl
Cl Cl Cl Cl
POS
(2) POS POS COS
Total= 4
(1) (2)
(1)
COOH COOH COOH
COOH
Cl Cl
(3 )
H COOH H OO C H
H COOH H COOH
Cl Cl F F
N
OH
(4) (5) (7) (8)
(6)
C Cl
Cl Cl Br
OH
N N
H CO O H
OH
Cl
(9) (4)
(1) (2) (1)
Br Br
(10) (10)
Cl Cl
(2) (2 ) (2 ) (2)
(1) (1 )
Br Cl
(11) (8)
Cl Br
(2) (2) (2) (1) (1)
(12) 9 POS, COS COS C4 AAOS S1, S2, & S4
Cube
102
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question-6 : At – 80° which conformation of n- butane is chiral?
Ans:
Me
Me H
H H
H
Question-7 :
Cl Cl
Cl Cl
Cl H
P H
Q R S
Cl Cl
Cl
Pai Relatio High. Chira PO CO AO Equimola
r n B.P. l S S S r mix.
Rotate
PPL
P – Position R P C2
R
Q– Dia Q.Dipol Q
R e
R– Ena Same R C2
S
Representation of compounds in different forms :
OH
OH
OH H H
H OH H
R Cl Cl
R
OHC OHC R
Cl H CHO CH3
Cl CHO CH3
Zig-zag Eclipsed
Staggered Shawhorse
103
M. PRAKASH INSTITUTE ISOMERISM Std: XI
OH CHO
H
H Cl
H OH
Cl H
CHO
CH3
Newmann
Fisher
Wedg-Dash
Cl
Cl
H Cl H Cl Cl
CH3 Cl H
R CH3 H
COOH COOH
Cl H COOH CH3
Cl COOH CH3
Zig-zag Eclipsed Shawhorse
Staggered
Cl H COOH COOH
Cl H R
Cl H H Cl
CH3 Cl H H Cl
COOH
CH3 CH3
Newmann
Fisher Fisher
Make 2 cases for fisher and then check it confirmation.
CHO
CH3 CH3
H OH OH H H
OH OH
H
H Cl HO H CHO
(1) (2)
H Cl HO H H OH CH2OH
CH3 H OH H OH
CH3 OH H
Newmann
CH2OH
Staggered
104
M. PRAKASH INSTITUTE ISOMERISM Std: XI
CH3 COOH
OH
OH OH
H OH H OH
HO H (2) OH H OH
(1)
H OH H CH3
OH COOH
CH3 CH3
fisher fisher
Question-8 : Differentiate between conformation and configuration.
Ans:
Stereo
Conformational Configurational
G.I. O.I
Acyclic Cyclic
CH3 CH3
H H POS H CH3
POS
COS 3 POS
COS
H H H H
CH3 H
chiral
CH3
1 POS
CH3 CH3
H
H
H
H H 2 POS
H H
H
CH3
POS
POS x COS x
No POS
chiral
chiral
105
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Question-9 : Find the number of stereoisomes:
1. CH 3 CH CH CH CH CH CH 3
|
Br
2. O
||
Ph S CH CH CH CH CH 3
3. Ph CH C CH CH 3
| |
OH CH 3
4. Ph CH C CH CH CH CH CH 3
2 O.I . |
Br
5.
OR
6. H
Cl
7.
N
OH
8.
26= 64
Cl
Note: If at the same centre G.I. & O.I. both are present then O.I. are considered
9.
106
M. PRAKASH INSTITUTE ISOMERISM Std: XI
10. CHO
H OH
HO H
H OH
H CH = CH CH3
CH3
11. CH3
Cl
CH CH3
CH3
CH CH CH CH 3
2. O. I.
Br
Question-10 : write the relation between the given pairs of molecules :
1. OH OH
&
OH OH
2. OH OH
&
OH OH
O O
3.
&
H Me H
Me
not chiral
Identical
4.
&
COOCH3 COOCH3
107
M. PRAKASH INSTITUTE ISOMERISM Std: XI
OH OH
5. OH OH
&
COOH Dia. COOH
Note: Two different meso are diastereomes.
6. Me Et
H OH HO H
&
H OH HO H
Et Me
7. H Me Me H
&
Me H Me H
8. O O
S S
&
9. Me Me
Me
&
Me
10.
&
OH OH
11.
12.
&
13.
&
Cl
Cl
108
M. PRAKASH INSTITUTE ISOMERISM Std: XI
14. Me OH
H OH Me H
&
H OH Me H
Me OH
15. CH3 Me
D
D OH &
H OH
H
16.
&
17. O H H
||
CH 3 S OH & CH 3 S OH
Me
||
O Me
18. O H H
||
CH 3 C OH & CH 3 C OH
||
Me O Me
19. OH
OH OH OH
& OH
OH Meso
Meso
Diff.
20.
&
21. Cl Cl Cl Cl
&
E Z
H F F H
109
M. PRAKASH INSTITUTE ISOMERISM Std: XI
22.
N & N
23.
&
24.
&
25. CHO CHO
HO H OH
H
& H OH
HO H
CH2OH CH2OH
26.
&
27. F H
Cl Br & Cl Br
H F
28. Cl Cl
29. Cl Cl
Cl Cl
110
M. PRAKASH INSTITUTE ISOMERISM Std: XI
30. OH OH
OH OH
OH OH
O O
31. CHO
OH H
H OH
CH2OH
HO H
&
H OH CHO
H OH OH
OH
CH2OH
32. COOH H COOH
H
&
HOOC H H
HOOC
111
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Types of reactions :
1. Stereospecific reactions :
A stereospecific reaction is one which, when carried out with stereoisomeric
starting materials, gives a product from one reactant that is a stereoisomer of the
product from the other.
‘Stereospecific’ relates to the mechanism of a reaction, the best-known example
being the SN-2 reactio, which always proceeds with inversion of stereochemistry
at the reacting centre.
Streospecificity in substitution reaction :
112
M. PRAKASH INSTITUTE ISOMERISM Std: XI
2. Stereoselective reactions:
A stereoselective process is one in which one stereoisomer predominates over
another when two or more may be formed.
If more than one reaction could occur between a set of reactants under the same
conditions giving products that are stereoisomers and if one product forms in
greater amounts than the others, the overall reaction is said to be stereoselective.
A stereoselective reaction in which the possible products are enantiomers is said
to be Enantioselective.
A stereoselective reaction in which the possible products are diastereomers is said
to be Diastereoselective.
113
M. PRAKASH INSTITUTE ISOMERISM Std: XI
3. Regioselective reaction :
If more than one reaction could occur between a set of reactants under the same
conditions giving products that are constitutional isomers and if one product
forms in greater amounts than the others, the overall reaction is said to be
regeoselective.
4. Chemioselective reaction :
If an organic compound contains more than one different functional groups or
more than one like functional groups that are not equivalent and, if a reagent
reacts exclusively or predominantely with one of them, the reaction is said to be
Chemioselective reaction. Nature of product depends upon the nature of reagent
used.
114
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Difference between stereospecific and stereoselective reactions :
Stereospecific reactions Stereoselective reactions
Definition A stereospecific reaction is a reaction in A stereoselective reaction is a reaction
which the sterochemistry of the in which there is a choice of pathway,
reactant completely determines the but the product stereoisomer is formed
stereochemistry ot the product without due to its reaction pathway being more
any other option. favourable than the others available.
Number A stereospecific reaction gives a specific A stereoselective reaction can result in
of product from a certain reactant. multiple products.
products
Effects The final product of a stereospecific The selectivity of the reaction pathway
reaction depends on the depends on differences in steric effects
stereochemistry of the reactant. (presence of bulky groups cause steric
hindrance) and electronic effects.
115
M. PRAKASH INSTITUTE ISOMERISM Std: XI
Bredt’s Rule :
The rule that states that it is not possible for double bonds to be formed on the bridgehead carbon
atoms of a bicyclic system if it involves a trans pi bond being incorporated in a ring comprised of
fewer than eight atoms.
Bredt’s Rule Explanation :
In priciple, the Bredt’s rule is ismply a consequence of the strain induced by a planar bridgehead
carbon.
A few key concept to mind when understanding the Bredt’s rule are :
1. A double bond has one sigma bond and a one pi bond.
2. The pi bond is formed by the overlap of p-orbitals. Therefore these p-orbitals must be on
the same planar. If they are not, the chances of an overlap are negligible and the bond
formed will be unstable.
3. A double bond on a bridgehead would be equivalent to having a trans pi double bond on
a ring, which is not stable for small ringss (fewer than eight atoms) due to a combination
of ring strain and angle strain (nonplanar alkene).
4. Bridgehead bicyclic compounds can only exhibit a double bond at a bridgehead position
if one of the ring has at least eight carbon atoms.
And therefore, bridgehead double bonds are unstable due to poor orbital overlap.
Example :
1. The following compound can not maintain a parallel overlap of the p-orbitals, has less
than 8 carbon atoms in the ring forming the trans pi double bond and as a result, it is
not stable at room temperature. Note : The p-orbitals of the brigehead carbon and the
p-orbitals of the adjacent caron are not on the same plane and therefore exhibit poor
orbital overlap.
Not possible according to Bredt’s rule: trans
pi double bonded carbon is in a ring with less
than eight carbon atoms.
Possible because trans pibonded carbon is
not the brigehead carbon.
116
M. PRAKASH INSTITUTE ISOMERISM Std: XI
2. The following compounds can maintain a parallel overlap of the p-orbitals, has more
than 8 carbon atons in the ring forming the trans pi double bond, and as a result, it is
stable at room temperature.
Possible because trans pi-double bonded carbon is
in a ring with eight carbon atoms.
Possible because in cis form 6 carbon atoms in the
ring and in trans form 10 carbon atoms in the ring.
117
M. PRAKASH INSTITUTE ISOMERISM Std: XI
*****************
118
You might also like
- Isomerism Complete Chapter Notes For Iit-Jee100% (3)Isomerism Complete Chapter Notes For Iit-Jee54 pages
- 15 Isomerism Formula Sheets Getmarks AppNo ratings yet15 Isomerism Formula Sheets Getmarks App6 pages
- Isomerism Theory Solved and Unsolved With Anwers NKB100% (1)Isomerism Theory Solved and Unsolved With Anwers NKB42 pages
- CH - 11 Some Basic Principles and Techniques IsomerismNo ratings yetCH - 11 Some Basic Principles and Techniques Isomerism47 pages
- Edt L.No. 2 Isomerism Final 15.06.2021 - DoneNo ratings yetEdt L.No. 2 Isomerism Final 15.06.2021 - Done57 pages
- Matriculation Chemistry (Introduction To Organic Compound) Part 3No ratings yetMatriculation Chemistry (Introduction To Organic Compound) Part 325 pages
- 9-230305 - 5mar23 Exam QP With SolutionsNo ratings yet9-230305 - 5mar23 Exam QP With Solutions6 pages
- PEDIASURE Case For Class Discussion Spring 2024No ratings yetPEDIASURE Case For Class Discussion Spring 202420 pages
- Article150 - Disobedience To Summonscriminallaw (Revisedpenalcode)No ratings yetArticle150 - Disobedience To Summonscriminallaw (Revisedpenalcode)2 pages
- 1) Justification For The Story Chosen Based On The Selection Criteria For Young Learners100% (1)1) Justification For The Story Chosen Based On The Selection Criteria For Young Learners2 pages
- Configure Disney VPN F5 On Linux: Credentials and Needed FilesNo ratings yetConfigure Disney VPN F5 On Linux: Credentials and Needed Files4 pages
- The Hatching A Novel Ezekiel Boone PDF Download100% (3)The Hatching A Novel Ezekiel Boone PDF Download34 pages
- Chapter 1 Partnership Basic Consideration and FormationNo ratings yetChapter 1 Partnership Basic Consideration and Formation14 pages
- Why Banya Is A Beacon of Hope For The People of Koboko CountyNo ratings yetWhy Banya Is A Beacon of Hope For The People of Koboko County4 pages
- Accounting Level II - Third Term Assignement I0% (1)Accounting Level II - Third Term Assignement I3 pages
- Electrical Estimator's Manual - Electrical KnowhowNo ratings yetElectrical Estimator's Manual - Electrical Knowhow3 pages
- Do We Need A New Species Concept-Coyne, Orr & Futuyma 1988No ratings yetDo We Need A New Species Concept-Coyne, Orr & Futuyma 198812 pages