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Answer to Assignments (Subjective)
LEVEL – I

1
...
Furthermore H and I are monosubstituted
*

benzene derivatives
...
C HC6 H 5COOH with one
asymmetric carbon atom
...
COOH, giving C6H5CH2CH2COOH with no
asymmetric carbon
...
COOH ⎯⎯
Pd

CH2CH2COOH

(I)

2
...


3
...
Since the alcohol is not
oxidized by acidified KMnO4 and gives cloudiness at once with Lucas reagent, hence it is
a t-alcohol
...


C

HOH
→
CH3 ⎯⎯⎯ H3C

C

OH + H3C

(A)

CH3

C

OH

CH3
(B)

This is because C2H5OH undergoes dehydration to form C2H4 at 170°C in presence of
excess of conc
...

H 2SO4 (conc
...


In the presence of strong acids, the H⊕ is captured by the carboxylic acid and the
following equilibrium is established:
OH
O
–
+ H 2SO 4
R C
+ H 2SO 4
R C
OH
OH
O

6
...


It is because the carboxylate group (—COO–) of the branched acid is more shielded from
the solvent molecules, therefore, it cannot be stabilized effectively by salvation
...


It is because carboxylic group does not have a true carbonyl group due to resonance
...


9
...

Aldehyde

O
H

C

OH
Hence it has reducing character of aldehydes
...


Both these unsaturated acids have two ionisable hydogens
...

H

C

O⎫
⎬
C
O⎭

H

C

C

OH

⎧O
⎨
⎩
O

H-bond

H
C
C

OH

H

O

O

(Maleiate ion)

(Fumarate ion)
(H-bond not possible)

Due to the formation of H-bond in maleiate ion more enegy is required to remove H ⊕
from it than from fumarate ion, in which H⊕ release is easy comparatively
...

LEVEL - II

1
...
The remaining oxygen may be present as —OH or —
OR
...
Only (D) is reduced to an achiral product
...
(E) gives a positive iodoform test
...


CH3
HCl
C ⎯⎯⎯→ H3C
⎯
Peroxide

C

CH2

H3C

CH3

CH3

CH3

Mg
Cl ⎯⎯⎯ H3C
→
Ether

CH3
CO2
MgCl ⎯⎯⎯ H3C
→

C
CH3

C

COOMgCl

CH3
CH3
+

H 2O / H
⎯⎯⎯⎯→ H3C
⎯
– Mg(OH)Cl

C

COOH

CH3

3
...
COOMgB r
→
→
– CH 4
(G )

+

HgSO4 / H 2SO4
H
⎯⎯ H — C ≡ C — COOH ⎯⎯⎯⎯⎯ H
→
→
H 2O
(I)

(H)

C

C

Tautomerism
COOH ⎯⎯⎯⎯→
⎯

OH H
COOH
H

C

CH2

O

KMnO4
COOH ⎯⎯⎯→ H2C
(O)

(J)

COOH

O

4
...


NaN 3
SOCl2
D
HYdrolysis
RCO 2 H ⎯⎯⎯ RCOCl ⎯⎯⎯ RCON 3 ⎯⎯ RNCO ⎯⎯⎯⎯ RNH 2
→
→
→
→

6
...
Formly chloride is not stable above –60°C
...


An extremely mild but selective oxidizing agent for aldehydes is silver oxide suspended
in aqueous base
...

CHO

H
C

⎯⎯
→

C
CH3

H2CH3C

8
...


(A) (CH3CO)2O
(C) CH3COOC2H5
(E) CH3COCH3

(B) CH3COOH
(D) C2H5OH

10
...


i)

–

CN
CH3CH2CH2CH2Br ⎯⎯⎯ CH3CH2CH2CH2CN
→
– Br –
+

2HOH / H
⎯⎯⎯⎯ CH3CH 2CH 2 CH 2COOH
→
n − Pen tan oic acid

CH3

ii)
H3C

C

CH3

C

CH3

SOCl2
OH ⎯⎯⎯ H C
→
3

CH3

Mg
→
Cl ⎯⎯⎯
Ether

C

H 3C

C
CH3

CH3
H

CHBr2
⎯⎯⎯⎯
→
t − BuOK

C

Br
Br

KMnO4
⎯⎯⎯→

H
H

H
HOOCCH2

C

HOOCCH2
H

E
Meso compound

Br
Br

2H 2
⎯⎯⎯
→
Ni

CH3

CH3

CH3

2
...


CH3

CH3

CH3

H

COOH

and

C=C

C=C
H

COOH

CH3

(A) Geometrical isomers
CH3

CH3

C2H5—C—COOH

or

HOOC—C—C2H5

H

H
B

4
...


is a saturated monoester with M
...


CH3

H3C

Cl
H3C
O

F=

CH3
N

H3C
O

CH3

Br

7
...


A

C = CH—C—OCH2CH2CH2CH3
O

CH3
B

CH3
H3C

C = CHCO2H ;

or

CH3CHCH2CH3
OH

C

H3C

C = CH—C—O—CH—CH2CH3 ;
O
CH3
CH3CH2CH2CH2OH ;

Dehydration
2 - Butene

CH3

D

C—CH—CO2H ;

H3C

OH OH

E

O=CH—CO2H
MF : C2H2O3

M
...


The given reactions are as follows
...
Therefore, its
structure would be CH3COCH3 (acetone)
...

Since B is obtained by oxidation of D with KMnO4, the compound D must be an alcohol
with molecular formula CH3CH2OH (ethanol)
...

Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treating A with
ethanol, the compound A must be an anhydride (CH3CO)2O (acetic anhydride)
...


Acetoacetic ester shows tautomerism and the two forms are called as keto and enol forms
...
When Br2 is added,
of the enol form
...
As keto and enol forms are in equilibrium, when enol form
is used, the equilibrium shifts to right hand side to give more enol form which discharges
the colour of excess of Br2 and gives blue violet colour with excess of FeCl3 present in
the reaction mixture
...


(A) (CH3CO)2O (Acetic anhydride)
(B) CH3COOH (Ethanoic acid)
(C) CH3COOC2H5 (Ethyl ethanoate)
(D) C3H5 OH (Ethanol)
(E) CH3COCH3

2
...
CH2
...
CH
...
CH2
...
CH
...


(A) CH3
...
CH2
...
CH3 or CH3COOCH2CH2CH2CH3
(B) C2H5OH
(C) CH3
...
CH2
...
CHO
(F) CH3COOH

4
...
CH2
...


(A)
HCOOH
(C) COOH
...


A
B
C
D
E

7
...


OH

A=

OH

B=
CH3

H2C

O

H2C

O
CH3

H2C

N

Cl

C=

D=

CH3

O

O

H3C

OH

H3C

E=

F=
O

O

G=
9
...


H3C

O
CH3

H3C

X=

Y=

O
propyl propionate

O
HO
propionic acid

H3C

H 3C

Z=

CH2

D=

OH
propan-1-ol

H3C

Objective
LEVEL - I

1
...

5
...

9
...

4
...

8
...


c
d
b
d
a, d

11
...

15
...

19
...

14
...

18
...


a
c
a
d
c



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