Question

(ii) Complete the following reactions; C6H5CH = CH2
Br2 (1) NaNH2
(a) C H CH = CH2
А
. В
(ii) CH3I
CH3
Br
C2H5ONa
-
C2H5OH
A+B
CH3
· Hot
(C) CH3CH2CH=C- CH3 -
Kmn04
A+B​

Answer 1

Explanation:

Correct option is

B

H

2

C=CH−CH=

CH

3

∣

C

−

CH

3

H

2

C=CH−CH=

CH

3

∣

C

−

CH

3

First dibromide forms which reacts with Mg and forms grinard reagent, which then reacts with ketone and in presence of acid it looses one molecule of water and forms final product.

Answer 2

$C_5H_5-CONH_2 \to C_5H_5NH_2 (aniline) \to \phi-NC \to \phi - NHCH^+$

Explanation:

• It's termed Hoffmann bromamide reaction when amide combines with Br$_2$ in the presence of KOH to generate an amine with one carbon less than the parent substrate,
• and it's named carbylamine reaction when aniline (a primary amine) reacts with KOH/CHCl$_3$. Sn/HCl is used to reduce a variety of carbon compounds.
• $C_5H_5-CONH_2 \to C_5H_5NH_2 (aniline) in \ the \ presence \ of \ Br_2/KOH$

• $C_5H_5NH_2 (aniline) \to \phi-NC in \ the \ presence \ of \ C-OH/CHCl_3 \\(Carbylamine \ reaction)$

• $\phi-NC \to \phi - NHCH^+ \in \ the \ presence \ of \ Sn/HCl$