Question

Acetamide is treated separately with the following reagents. Which one of these would give methyl amine?

• A. $P{Cl}_5$
• B. $NaOH+{Br}_2$
• C. soda lime
• D. Hot conc. $H_2{SO}_4$

Answer 1

The correct option is B $NaOH+{Br}_2$

Acetamide is a primary amide and methylamine also a primary amine and methylamine contain one less carbon atom from acetamide.

The Conversion of a primary amide to a primary amine containing one carbon atom less than the original amide can be carried out by Hofmann bromamide rearrangement reaction which involves the mixture of $B{r}_2$ in the presence of $NaOHorKOH$ as a reagent.

Hence, option $B$ is correct.