Consider the following reaction sequence
${CH}_{3} OH \overset{HBr}{\to} A \overset{KCN}{\to} B \overset{H_{2} / Ni}{\to} C \overset{{HNO}_{2}}{\to} D$
Choose the correct option regarding the different products obtained in the above reactions sequence.

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‘D’ is an alcohol consisting one carbon more than the startng alcohol

Product ‘B’ is formed via S_N2-pathway

‘C’ is an amine

All of the above

answer is D.

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Detailed Solution

${CH}_{3} OH \to_{{SN}_{2}}^{HBr} {CH}_{3} Br \to_{{SN}_{2}}^{KCN} \underset{B}{{CH}_{3} CN} \to_{Reduction}^{H_{2} / Ni} \underset{C}{{CH}_{3} {CH}_{2} {NH}_{2}} \to_{Diazotization followed by hydrolysis}^{{HNO}_{2}} \underset{D}{{CH}_{3} {CH}_{2} OH}$ 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., CH₃CN is a product formed via SN₂ pathway on the treatment of CH₃Br and KCN.
Product 'C' is ethylamine formed by the reduction of methyl cyanide.