Previous Year NEET Question- Alcohols L-9

Question: Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?#

A) $\mathrm{NaOH} / \mathrm{Br}_2$

B) Sodalime

C) Hot conc. $\mathrm{H}_2 \mathrm{SO}_4$

D) $\mathrm{PCl}_5$

Answer: $\mathrm{NaOH} / \mathrm{Br}_2$#

Solution:#

Among the given reagents only $\mathrm{NaOH} / \mathrm{Br}_2$ converts $-\mathrm{CONH}_2$ group to $-\mathrm{NH}_2$ group. This is known as Hoffmann bromamide reaction. $$ \mathrm{CH}_3-\mathrm{CONH}_2+\mathrm{NaOH}+\mathrm{Br}_2 $$ $$ \rightarrow \mathrm{CH}_3 \mathrm{NH}_2+\mathrm{NaBr}+\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O} $$