2 Chemistry med easy Isomeris
ISOMERS x = no. of ‘c’ in parent
. k chain
P.S.S tric
Compound having same M.F i c no. P .I y = no. of branches of ‘c’
Mag r CI &
but diff physical & chemical prop. f o
Magic no ⇒ diff ⇒ C.I
XY same ⇒ P.I
Types of Isomerism
2 4 C.I 1 2 21
1. Structural isomerism Mf = same, structure diff, IUPAC name 2 Cl P.I
diff 3 Cl
2. Stereoisomerism Mf = same, diff orientation of groups 1 1 Cl Cl
3
in space, IUPAC name same 40 31 20 20
Types of Structural Isomerism 1
CN
3 1
1. Chain Isomerism CN C.I
2
4 2
Compound having same MF. but different parent 3
chain. 40 31
2 4 3. Metamerism
3 2
1 C.I MF Same but difference in arrangement of atoms
n
1 or groups from either side of Bridging func group
3 (Polyvalent)
*
2. Position Isomerism —O— —NH— O O
MF = same, different position of branch like R or —S— —N —C—O—C—
F.G. like OH, NH2, OR, SH —C— O
O
O —C—O—
3 1 3 1 —C—NH—
2 OH 2 PI —C— O —C—
4 4
OH S —C—N— O
1 2 3
4 4 2
C—C—C—O—C C—C—O—C C—C—O—C—C
2 5 3 PI
5
—
1 ’ C
3 1 3 ‘c al 1‘c’ 1‘c’ 2‘c’ 2‘c’
o r m Branch
n
O 1, 2, 3 are metamers
2
3 4. Functional isomerism
2 PI
1 1 MF same, different F.G.
O
O
keto F.I.
OH
ene alc.
,Some examples of F.I. 2 1O 2 OH
O 1, 5-system sp sp
3 2 3
ail
a. Aldehyde & ketone: H 1,3 f a
Jab hog 4 4
l
O 3 5H Pheno
sp
b. Carboxylic acid & ester: CH3COOH, HCOOCH3 2 1 + –
1,2-system H –C≡N H–N≡C
c. Alcohol & ether: OH
O
d. Cyanide & isocyanide: RCN, RNC
e. 1°, 2°, 3° amines:
CALCULATION
OF NUMBER OF
3°
STRUCTURAL ISOMERS
2°
1° NH N Degree of unsaturation (D.U)
2 NH
or, double bond equivalent (D.B.E)
f. 1°, 2°, 3° amides:
H + X- N
DU=(C + 1) -
NH2 NH N 2
O
O O
g. diene, yne: C = no of 'C' atoms DU = 0 ⇒ all single bond
H = no of 'H' atoms DU = 1
5. Ring Chain Isomerism
X = no of 'X' atoms 1π 1 Ring
MF ⇒ same
N = no of 'N' atoms
1st Ring DU = 2
1π + 1 Ring
2nd Chain 2π
2 Ring
6. Tautomerism
Total structural isomers
Keta-enol system 1 3
O H sp O–H
1, 3-system
CH3–C–CH2 10 + 0 - 0
Condn for exhibition
α
2 3 α-H
CH3–C=CH2
1 C5H10 DU = (5 +1) -
keto ene+ol 2
of tauto. +
Intramolecular H enol
1 α–c should be sp
3 transfer = 6 – 5 = 1
1, 3 system
2 Presence of α–H 1π 1 Ring
Examples of Tautomerism:
O Total 10
O
O
O O
O Alkanes No. of structural isomers
O O
C4H10 2
Ph–CH=CH–OH C5H12 3
167
Isomerism
, C6H14 5 Ring: -
C7H16 9 a) Two different group at different positions.
C8H18 18 b) Hybn sp3
C9H20 35 b
a a
C10H22 75
Cis-
TYPES OF STERO ISOMERISM a sp 3
a
H H
Configurational Isomerism b
a H
Trans
IUPAC name same H a b
TYPES OF CONFIGURATIONAL TYPES OF G.I
ISOMERISM Case 1: Cis -Trans
1. Geometrical Isomerism similar groups same side cis
otherwise trans
Arises due to restricted rotation around
C=C, C=N, N=N, Ring Me Me Me H
C = C C = C
Condn :- Restricted site must have diff groups
H H H Me
Cis
a X a Y Trans
diff. C = C diff. C = C
b Y b X Case 2:
Z/E
Example of GI zusammen Entgegen
. same side opp. side
No P.S.S
H
Cis/trans fail hoga toh Z/E Sochna hai
H I
1) Same C = C 5) C 1
H3C Br
H NH2 Cl
(No) C = C
1
HO E NH2
Br ××
diff H
2) diff C=N 6) C
C.I.P rules
Cl Yes OH Yes Cl
1. Atomic no , priority
H 2. Atomic no of 1st atom same
3) ×× ××
7) C
N=N
azo
Cl go for next atom apply rule 1
Yes
R
Always
R Comp 3. In Isotopes mass no , priority
GI O C
C = O > C – O'
H H
O
4) C 8) C Same N C C
Cl No H
No C N < C = O
N C
168
Organic Chemistry
, SPECIAL CASES n +1
P = (n = odd)
Case 1: Odd no. of cumulative double bond 2
Py–Py P –P P –P X CH3–CH=CH–CH=CH–CH=CH–CH3
a z z y y
C C C C n = 3
b
Y 3+1 4
P = = = 2
Br Me 2 2
C=C=C=C
Total G.I. = 23–1 + 22–1
Cl G.I. H
= 22 + 21
Case 2: Odd no. of Rings (spiro) = 4 + 2 = 6
X
a Properties
of Geometrical
Isomerism
b Y
a. BP: Cis > Trans (BP ∝ DM)
Case 3: Cumulative double bond + Ring = odd Me
Me Me H
a DM
X C=C C=C m=0
H
DM ≠ 0 H Me
b Y m≠0
Case 4: Cycloalkene
8 memb b. MP: Cis < Trans
c. Stability: Cis < Trans
d. Solubility: Cis > Trans
2. Optical Isomerism
Chiral carbon
sp3 'c' attached with 4 diff groups
CALCULATION OF G.I. A
OH
X *C B * H CH3
Case 1: If molecule is Unsymm H *
*
n
Total G.I. = 2 , n = sites where G.I. exhibits OH H
Y
CH3–CH=CH–CH=CH–CH=CH–CH2Br
dextrorotatory
n = 3, Total G.I. = 23 = 8. If compound CW 'd' or +
rotates PPL
ACW 'ℓ' or –
Case 2: If molecule is symm. O.A
laevorotatory
Total G.I. = 2n–1 + 2P–1 If compound does not
rotate PPL
n O.I.A
P = (n = even)
2
169
Isomerism
ISOMERS x = no. of ‘c’ in parent
. k chain
P.S.S tric
Compound having same M.F i c no. P .I y = no. of branches of ‘c’
Mag r CI &
but diff physical & chemical prop. f o
Magic no ⇒ diff ⇒ C.I
XY same ⇒ P.I
Types of Isomerism
2 4 C.I 1 2 21
1. Structural isomerism Mf = same, structure diff, IUPAC name 2 Cl P.I
diff 3 Cl
2. Stereoisomerism Mf = same, diff orientation of groups 1 1 Cl Cl
3
in space, IUPAC name same 40 31 20 20
Types of Structural Isomerism 1
CN
3 1
1. Chain Isomerism CN C.I
2
4 2
Compound having same MF. but different parent 3
chain. 40 31
2 4 3. Metamerism
3 2
1 C.I MF Same but difference in arrangement of atoms
n
1 or groups from either side of Bridging func group
3 (Polyvalent)
*
2. Position Isomerism —O— —NH— O O
MF = same, different position of branch like R or —S— —N —C—O—C—
F.G. like OH, NH2, OR, SH —C— O
O
O —C—O—
3 1 3 1 —C—NH—
2 OH 2 PI —C— O —C—
4 4
OH S —C—N— O
1 2 3
4 4 2
C—C—C—O—C C—C—O—C C—C—O—C—C
2 5 3 PI
5
—
1 ’ C
3 1 3 ‘c al 1‘c’ 1‘c’ 2‘c’ 2‘c’
o r m Branch
n
O 1, 2, 3 are metamers
2
3 4. Functional isomerism
2 PI
1 1 MF same, different F.G.
O
O
keto F.I.
OH
ene alc.
,Some examples of F.I. 2 1O 2 OH
O 1, 5-system sp sp
3 2 3
ail
a. Aldehyde & ketone: H 1,3 f a
Jab hog 4 4
l
O 3 5H Pheno
sp
b. Carboxylic acid & ester: CH3COOH, HCOOCH3 2 1 + –
1,2-system H –C≡N H–N≡C
c. Alcohol & ether: OH
O
d. Cyanide & isocyanide: RCN, RNC
e. 1°, 2°, 3° amines:
CALCULATION
OF NUMBER OF
3°
STRUCTURAL ISOMERS
2°
1° NH N Degree of unsaturation (D.U)
2 NH
or, double bond equivalent (D.B.E)
f. 1°, 2°, 3° amides:
H + X- N
DU=(C + 1) -
NH2 NH N 2
O
O O
g. diene, yne: C = no of 'C' atoms DU = 0 ⇒ all single bond
H = no of 'H' atoms DU = 1
5. Ring Chain Isomerism
X = no of 'X' atoms 1π 1 Ring
MF ⇒ same
N = no of 'N' atoms
1st Ring DU = 2
1π + 1 Ring
2nd Chain 2π
2 Ring
6. Tautomerism
Total structural isomers
Keta-enol system 1 3
O H sp O–H
1, 3-system
CH3–C–CH2 10 + 0 - 0
Condn for exhibition
α
2 3 α-H
CH3–C=CH2
1 C5H10 DU = (5 +1) -
keto ene+ol 2
of tauto. +
Intramolecular H enol
1 α–c should be sp
3 transfer = 6 – 5 = 1
1, 3 system
2 Presence of α–H 1π 1 Ring
Examples of Tautomerism:
O Total 10
O
O
O O
O Alkanes No. of structural isomers
O O
C4H10 2
Ph–CH=CH–OH C5H12 3
167
Isomerism
, C6H14 5 Ring: -
C7H16 9 a) Two different group at different positions.
C8H18 18 b) Hybn sp3
C9H20 35 b
a a
C10H22 75
Cis-
TYPES OF STERO ISOMERISM a sp 3
a
H H
Configurational Isomerism b
a H
Trans
IUPAC name same H a b
TYPES OF CONFIGURATIONAL TYPES OF G.I
ISOMERISM Case 1: Cis -Trans
1. Geometrical Isomerism similar groups same side cis
otherwise trans
Arises due to restricted rotation around
C=C, C=N, N=N, Ring Me Me Me H
C = C C = C
Condn :- Restricted site must have diff groups
H H H Me
Cis
a X a Y Trans
diff. C = C diff. C = C
b Y b X Case 2:
Z/E
Example of GI zusammen Entgegen
. same side opp. side
No P.S.S
H
Cis/trans fail hoga toh Z/E Sochna hai
H I
1) Same C = C 5) C 1
H3C Br
H NH2 Cl
(No) C = C
1
HO E NH2
Br ××
diff H
2) diff C=N 6) C
C.I.P rules
Cl Yes OH Yes Cl
1. Atomic no , priority
H 2. Atomic no of 1st atom same
3) ×× ××
7) C
N=N
azo
Cl go for next atom apply rule 1
Yes
R
Always
R Comp 3. In Isotopes mass no , priority
GI O C
C = O > C – O'
H H
O
4) C 8) C Same N C C
Cl No H
No C N < C = O
N C
168
Organic Chemistry
, SPECIAL CASES n +1
P = (n = odd)
Case 1: Odd no. of cumulative double bond 2
Py–Py P –P P –P X CH3–CH=CH–CH=CH–CH=CH–CH3
a z z y y
C C C C n = 3
b
Y 3+1 4
P = = = 2
Br Me 2 2
C=C=C=C
Total G.I. = 23–1 + 22–1
Cl G.I. H
= 22 + 21
Case 2: Odd no. of Rings (spiro) = 4 + 2 = 6
X
a Properties
of Geometrical
Isomerism
b Y
a. BP: Cis > Trans (BP ∝ DM)
Case 3: Cumulative double bond + Ring = odd Me
Me Me H
a DM
X C=C C=C m=0
H
DM ≠ 0 H Me
b Y m≠0
Case 4: Cycloalkene
8 memb b. MP: Cis < Trans
c. Stability: Cis < Trans
d. Solubility: Cis > Trans
2. Optical Isomerism
Chiral carbon
sp3 'c' attached with 4 diff groups
CALCULATION OF G.I. A
OH
X *C B * H CH3
Case 1: If molecule is Unsymm H *
*
n
Total G.I. = 2 , n = sites where G.I. exhibits OH H
Y
CH3–CH=CH–CH=CH–CH=CH–CH2Br
dextrorotatory
n = 3, Total G.I. = 23 = 8. If compound CW 'd' or +
rotates PPL
ACW 'ℓ' or –
Case 2: If molecule is symm. O.A
laevorotatory
Total G.I. = 2n–1 + 2P–1 If compound does not
rotate PPL
n O.I.A
P = (n = even)
2
169
Isomerism
