Question

$CH _{3} CONH _{2}, Br _{2} \& KOH$ give $CH _{3} NH _{2}$ as the product. The intermediates of the reaction are
(a)$CH _{3}- \overset{\underset{||}{O}}{C} - NHBr$
(b)$CH _{3}- N = C = O$
(c)$CH _{3} NHBr$
(d)
The correct answer is:

• A. a, b
• B. a, c
• C. b, d
• D. c, d

Answer 1

Answer: (A)

In hoffmann's bromamide degradation reaction.
$CH_3 - \overset{\underset{||}{O}}{C}- NH_2 \ce{->[NaOH/Br_2][-H_2O, - NaBr]} \underset{\text{(N - Bromamide) Intermediate}}{CH_3 - \overset{\underset{||}{O}}{C} - NHBr}$