Q75 / 200 · NEET 2010 (Screening) · Chemistry
Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine ?
- Correct answerA$\mathrm{NaOH}-\mathrm{Br}_2$
- BSodalime
- CHot conc $\mathrm{H}_2 \mathrm{SO}_4$
- D$\mathrm{PCl}_5$
Solution
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.
$$
\begin{aligned}
& \underset{\text { acetamide }}{\mathrm{CH}_3 \mathrm{CONH}_2}+\mathrm{NaOH}+\mathrm{Br}_2 \longrightarrow \begin{array}{c}
\mathrm{CH}_3 \mathrm{NH}_2 \\
\text { methyl amine }
\end{array} \\
&+\mathrm{NaBr}+\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.