Question

In the following reaction
Br₂ / P excessNH₃
RCH₂COOH □(→┴(Br₂/P) ) X □(→┴(excess NH₃) )
The major compounds X and Y are
(a) RCHBrCONH₂ ; RCH(NH₂)COOH
(b) RCHBrCOOH ; RCH(NH₂)COOH
(c) RCH₂COBr ; RCH₂COONH₄
(d) RCHBrCOOH ; RCH₂CONH₂

Answer 1

I think the correct answer is (d)

Answer 2

The major compounds X and Y are :

Reaction : RCH₂COOH □(→┴(Br₂/P) ) X □(→┴(excess NH₃) ) Y

1. RCH₂COOH □(→┴(Br₂/P) ) X

• Above reaction is Hell Volhard Zelinsky

halogenation

• Br- ion will replace alpha hydrogen of acid molecule

X = RCHBrCOOH

2. RCHBrCOOH (→┴(excess NH₃) ) Y

• Hoffman's bromamide reaction

• NH2 will replace OH- ion and forms amide. Also Br- ion gets removed while forming HBr.

Y = RCH₂CONH₂

Correct option : (d) RCHBrCOOH ; RCH₂CONH₂