Consider the following reaction sequence Choose the correct option...

We can see in this reaction sequence that ' $D$ ' is ethyl alcohol which consists of one carbon atom more than methyl alcohol, the starting alcohol. Product ' $B$ ', i.e., $\mathrm{CH}_3 \mathrm{CN}$ is a product formed via $S_{\mathrm{N}} 2$ pathway on the treatment of $\mathrm{CH}_3 \mathrm{Br}$ and KCN . Product ' $C$ ' is ethyl amine formed by the reduction of methyl cyanide.