Question

An organic compound $\left(A\right)$ of molecular formula $C_2{H}_{5}NO$ reacts with ${Br}_2/NaOH$ to give compound $\left(B\right)$ of molecular formula ${CH}_{5}N$. $\left(A\right)$ is reduced by $LiAl{H}_4$ to give cmpound $\left(C\right)$ of formula $C_2{H}_{7}N$. Identify $\left(A\right),\left(B\right)$ and $\left(C\right)$.

Answer 1

Here $B{r}_2/NaOH$ acts as Hoffmann reagent since it gives the product with one carbon less than starting material. Hence $\left(A\right)$ is an amide and $\left(B\right)$ is an amine.

When amide reacts with $LiAl{H}_4$, it gives amine with the same number of carbon atoms.

${CH}_{3}CO{NH}_2\left(A\right)\underset{}{\overset{{Br}_2/NaOH}{\to }}\underset{methylamine}{{CH}_3{NH}_2}\left(B\right)+{CO}_2$

${CH}_{3}CO{NH}_2\underset{}{\overset{LiAl{H}_4}{\to }}\underset{ethylamine}{{CH}_3{CH}_2{NH}_2}\left(C\right)+H_{2}O$

Hence, A= acetamide

B= methylamine

C=ethylamine