Question

$C{H}_{3}OH\stackrel{HBr}{\to }A\stackrel{KCN}{\to }B\stackrel{H_2/Ni}{\to }C\stackrel{HN{O}_2}{\to }D$

Consider the above reaction sequence and choose the correct option regarding the different products obtained in the above reaction sequence.

• A. 'D' is an alcohol consisting one carbon more than the starting alcohol
• B. Product 'B' is formed via ${S_N}^2-$pathway
• C. 'C' is an amine
• D. All of the above

Answer 1

The correct option is D All of the above

$C{H}_{3}OH\underset{S_{N}2}{\overset{HBr}{\to }}C{H}_{3}Br\underset{S_{n}2}{\overset{KCN}{\to }}\underset{\left(B\right)}{C{H}_{3}CN}\underset{Reduction}{\overset{H_2/Nl}{\to }}\underset{\left(C\right)}{C{H}_{3}C{H}_{2}N{H}_2}\stackrel{HN{O}_2}{\to }\underset{\left(D\right)}{C{H}_{3}C{H}_{2}OH}$

Diazotisation followed by hydrolysis

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., $C{H}_{3}CN$ is a product formed vis $S_{N}2$ pathway on the treatment of $C{H}_{3}Br$ and KCN.

Product 'C' is ethylamine formed by the reduction of methyl cyanide.

Hence, the correct option is $\text{D}$