# 13.5     How will you convert:                 (i) Ethanoic acid into methanamine

Ethanoic acid on reacting with $SOCl_{2}$ replaces the OH by $Cl$ and then reacts it with an excess of ammonia molecule to produce methanamide ($CH_{3}CONH_{2}$) which by Hoffmann bromamide degradation reaction gives us desired product (methenamine)
$\\CH_{3}COOH+SOCl_{2}\rightarrow CH_{3}COCl\overset{NH_{3}(excess)}{\rightarrow}CH_{3}CONH_{2}$ $\overset{Br_{2}+KOH}{\rightarrow}CH_{3}NH_{2}$