Question

When acetamide is treated with $Br_2$ and caustic soda, the product formed is

• A. $N$ -bromamide
• B. bromoacetic acid
• C. methenamine
• D. ethanamine

Answer 1

Answer: (C)

methenamine

Answer 2

The correct option is(C): methenamine.

Amides react with bromine and caustic soda to give their corresponding primary amines. Thus, acetamide gives methanamine. This reaction is known as Hofmann's bromamide degradation reaction.

$\underset{\text{acetamide}}{H_3C-\overset{\underset{||}{O}}{C} - NH_2} + Br_2 + 4KOH \ce{->[343K]}$
$2 KBr + K _{2} CO _{3}+ \underset{\text{methanamine}}{H _{3} C - NH _{2}}+2 H _{2} O$