Acetamide is treated with the following reagents separately. Which...

Key Idea The reagent which can convert $-\mathrm{CONH}_2$ group into $-\mathrm{NH}_2$ group is used for this reaction. Among the given reagents only $\mathrm{NaOH} / \mathrm{Br}_2$ converts $-\mathrm{CONH}_2$ group to $-\mathrm{NH}_2$ group, thus it is used for converting acetamide to methyl amine. This reaction is called Hofmann bromamide reaction. $$\begin{aligned}& \underset{\text{ acetamide }}{\mathrm{CH}_{3}\mathrm{CONH}_{2}}+\mathrm{NaOH}+\mathrm{Br}_{2} \longrightarrow \begin{array}{c} \mathrm{CH}_{3}\mathrm{NH}_{2} \\ \text{ methyl amine } \end{array}\\ &+\mathrm{NaBr}+\mathrm{Na}_{2}\mathrm{CO}_{3}+ \mathrm{H}_{2}\mathrm{O} \end{aligned}$$