1. CHEMISTRY
SOLUTIONS
1. a) p-chlorophenol will have high pKa value, as its phenoxide is not much stabilized as
compared to phenoxide ion of m-chlorophenol. The electron withdrawing Cl group stabilizes m-
chlorophenol due to close proximity position to phenoxide ion.
Cl
O
-
O
-
O
-
Cl
stabilizes less stability contributor
canonical form
stabilizes more
[4]
b) i) CH3
CH3
A =
ii)
Br
B = C =
[3 + 3]
c) CH3
CH3
CH2Cl
CH2Cl
CH2CN
CH2CN
CH2CHO
CH2CHO
OH
CHO


 
 
h
/
Cl2

 
KCN


 


OH
.
dil
1. SnCl2/HCl
2. H2O
[6]

2. 2. a)
(CH2)4
H5C2OC
H5C2OC
O
O
ethanol
.
abs
OEt
Na


 



CH2
CH2
CH2
CH
H5C2O2C
C
O
HCl
Hg
/
Zn


 

O

 
DNP
N
NH
NO2
O2N
1.H3O+
2.
(A)
(C)
(B)
C8H12O3
C5H10
[2 + 2 + 2]
Reaction mechanism of Claisen condensation intramolecular
i.e. Dieckmann condensation
(H2C)3
C
CH2
H5C2OC
O
OC2H5
O
EtOH
EtO


 


(H2C)3
CH
-
C OEt
EtO2C
O

 (H2C)3
C
CH
OEt
CO2Et
O
-


(H2C)3
C
CH
CO2Et
O
[2]
b)
NO2 CH2OH
CH3
CO2
-
reflux
OH

 


+
NO2
HO2C CO2H
(B)
KMnO4
(A)
O2N
CH3
CH2OC
O
[3 + 3]
3. a) i)
O
H2NNH
+ 


 
 H
CO
CH 2
3
N
NH

3. ii) O





 
  3
2
3 CH
CHCO
P
)
Ph
(
CH
COOCH3
[2 + 2]
b) Since compound B reacts with HIO4 to give benzaldehyde so it should contain one benzene
ring and two adjacent – OH groups. Moreover since compound (A) gives iodoform test so it must
contain 3
CH
OH
|
CH
 group. The structures of compounds and reactions are as
CH
CH
CH3
OH
Br
positive
Iodoform


 

HC
C
H3
CH
CH3
OH

 
 4
HIO
CHO
CH3CHO
+
1. KOH
2. H+
[2 × 3]
c)
C
CH3
CH2
C
H3
O
H
/
Zn
O
2
3


 HCHO +
H
CCH2
O C
C
O
CH3
O
+
H3C
CCH2
CHO
O
(A) (B) (C)
[2 × 3]
4. a) There are 5 isomeric hexane (C6H14) of which 2,2-diemthylbutane and n-hexane produces
3 isomers (monochloro derivatives).
A
2
Cl
h


 
B + C + D
C6H14
C6H12 
 
HCl
F
Alkene
(E)
G Zn/CH3CO2H
Isomeric with B, C, D
Isomers
Isomers
Hydrogenation –HCl

4. C
H3 C CH2
CH3
CH3
CH3

 
h
C
H3 C CH2
CH3
CH3
H2C
Cl
+ C
H3 C HC
CH3
Cl
CH3
CH3
C
H3 C CH2
CH3
CH3
CH2 Cl
+
C
H3
HC CH
CH3
CH3
C
H3




 
 COOH
CH
/
Zn 3
C
H3 C HC
CH3
CH3
Cl
CH3

 
HCl
(A) (B) (C) (D)
(E)
(F)
(G)
Isomers
-HCl
-HCl
Me
C C
Me
Me Me
[8]
b) Tollen’s test given by three types of compounds
i) Aldehydes (CnH2nO)
ii) Formic acid (HCO2H)
iii) α-hydroxyketone. Molecular formula of compound C3H6O2 indicates that it must be a α-
hydroxyketone.
C
H3
C
CH2
OH
O
C O
H2C
C
H3
OH
+ HCN C
C
H3
OH
CH2
OH
O
H

 


O
H3 C
C
H3
OH
CH2OH
CO2H
(B)
optically active
(C)
optically active
monobasic
A =

[2 × 3]
5. a) Neopentyl chloride (CH3)3CCH2Cl, a 1° RCl, does not participate in typical SN2 reactions
because the bulky (CH3)3C – group sterically hinders backside attack by a
nucleophile. [3]
b) The H’s on the C attached to the ring (the benzylic H’s), although they are in this case 2°, are
nevertheless more reactive toward Brx
than are ordinary 3° H’s like a C = C group in the allylic system,
the Ph group can stabilize the free radical by electron donation through extended p orbital overlap.
C
CH3
CH3
C
CH3
CH3
x
x
[3]
6. a) The compound A must be a ketone as it forms oxime but does not reduce Tollen’s reagent.
The compound D must be an aldehyde. Its structure does not include the fragment CH3 –
O
||
C – as it
does not respond idoform test. The compound E must be a ketone containing CH3CO – fragment. Let
the compounds D and E be RCH2CHO and R′COR″ respectively. Where R, R″, R′ are all alkyl groups.
From these we get.

5. RCH2CHO + O C
R'
R"


 3
O RCH2
CH
C
R'
R"
R
CH2
CH
OH
CH
R'
R"


 ]
H
[
R
CH2
C
O
CH
R'
R"
-H2O
Since the M.F. of A is C6H12O, it follows that R = R′ = R″ = CH3. Hence, the structure of
molecules (A) to (E) and the reactions are as follows.


 ]
H
[
CH2CH3
CH
CH
CH3
CH3
O
H

 
 O
H2
H2CH3C
CH
C
CH3
CH3
CH3CH2CHO + CH3COCH3
(D) (E)
(B)
(A)
(C)
CH
CH3
CH3
CO
CH2
C
H3
[2 × 5]
b) C
H3
CH
CO2H
NH2
)
D
(
e
lim
soda


 
 H3CH2C NH2
CH3CH2OH as one of
the products
(N)
(M)
C
H3
CH
CO2H
OH
HNO2
a hydroxy
acid (P)
HNO2
oaptically active amino acid
[2 + 2]
‘α’ hydroxy acids from lactides, these are 6 membered ring compounds formed by reaction
between two molecules of the hydroxy acid.
CH
OH
C
OH
O
C
H3
C
O
C
CH3
O
H
O
H


 
 O
H
2 2
C
H3
CH
C
O
O HC
CH3
C O
O
3,6-dimethyl-1,4-dio
xane-2,5-dione
[2]
7. a) i) Due to delocalisation of lone pair electron of nitrogen in benzene ring.

6. ii) In the SN1 reaction mechanism carbocation are formed as intermediates. The carbonium
ion is sp2
hybridized, the structure of which permits the attack of nucleophile from above and below
the plane, consequently, racemisation occurs.
C
C
H3 CH3
Nu





 
 below
from
attack
C Nu
C
H3
C
H3
+





 
 above
from
attack
C
C
H3
C
H3
Nu
CH3
iii) This is in accordance with Saytzeff rule.
iv) Because Cl–
is good leaving group in comparison to –
OCOCH3
v) Due to rearrangement of intermediate carbonium ion.
C
H3 CH3
CH3
CH3
Cl
C
H3
CH3

 
 


Cl
+
+
vi)
2
CONH
NH



group lone pair electrons is involved in delocalisation with
O
||
C
 of
2
CONH
NH



NH
C
NH2
O
NH
C
NH2
O
-
[2 × 6]
b)
C5H4O2 




 
 reaction
Cannizaro
B + C
C8H8O2
CH3COCH3
dil. OH–
C4H4O )
G
(
)
F
( HBr
Pt
]
H
[ 2

 


 

(E) dibromoalkane
adipic acid
HO2C(CH2)4CO2H
1. KCN
2. H2O
200-225°C

[O]
O CHO O CO2
-
O CH2OH

 

_
OH
+
dil. OH- CO2
200-225°C
CH3COCH3
Claisen
condensation
Eq. wt.
112.08
O CH=CH-COCH3
O O

 
HBr
Pt
/
Ni
or
H2



C4H4O
Br
Br

 
KCN
NC
CN

 
 O
H2
HO2C
CO2H
Adipic acid Dibromoalkane
[O]
[F]
O
H
Br
HBr
[G]
[10]
6