Question
Acetamide is treated with the following reagents separately. Which one of these
would yield methyl amine?
A.
$$\frac{{NaOH}}{{B{r_2}}}$$
B.
$${\text{Sodalime}}$$
C.
$${\text{Hot}}\,\,conc.\,{H_2}S{O_4}$$
D.
$$PC{l_5}$$
Answer :
$$\frac{{NaOH}}{{B{r_2}}}$$
Solution :
Key Idea The reagent which can convert $$ - CON{H_2}$$ group into $$ - N{H_2}$$ group is used for this reaction.
Among the given reagents only $$\frac{{NaOH}}{{B{r_2}}}$$ converts $$ - CON{H_2}$$ group to $$ - N{H_2}$$ group, thus it is used for converting acetamide to methyl amine. This reaction is called Hoffmann bromamide reaction, in which primary amides on treatment with $$\frac{{B{r_2}}}{{NaOH}}$$ form primary amines.
$$\mathop {C{H_3}CON{H_2}}\limits_{{\text{Acetamide}}} + NaOH + B{r_2}$$ $$ \to \mathop {C{H_3}N{H_2}}\limits_{{\text{Methyl amine}}} + NaBr + N{a_2}C{O_3} + {H_2}O$$