Identify the final product (C) in the reaction given below;

$CH_3CH_2I+ NaCN →A → \text{(Partial hydrolysis}/OH^-) → B + NaOH + Br_2 → C$?

$CH_3CH_2NH_2$

The correct answer is Option (2) → $CH_3CH_2NH_2$

Step 1: Alkyl halide + NaCN

The reaction of an ethyl halide with sodium cyanide follows an $S_N2$ mechanism, resulting in a nitrile.

$\text{CH}_3\text{CH}_2\text{I} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN}$

Product A: Propionitrile (ethyl cyanide)

Observation: The carbon chain length increases by one carbon.

Step 2: Partial hydrolysis of nitrile

Nitrile, on partial hydrolysis (usually using cold concentrated $\text{HCl}$ or alkaline $\text{H}_2\text{O}_2$), yields an amide.

$\text{CH}_3\text{CH}_2\text{CN} \rightarrow \text{CH}_3\text{CH}_2\text{CONH}_2$

Product B: Propanamide

Step 3: Reaction with $\text{Br}_2 + \text{NaOH}$

This is known as the Hofmann Bromamide degradation reaction.

$\text{RCONH}_2 \overset{\text{Br}_2/\text{NaOH}}{\longrightarrow} \text{RNH}_2$

In this step, the carbonyl group is removed as a carbonate/$\text{CO}_2$ equivalent, resulting in an amine with one less carbon than the starting amide.

$\text{CH}_3\text{CH}_2\text{CONH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2$

Final Product (C)

The final product obtained after the degradation is Ethylamine.

$\text{CH}_3\text{CH}_2\text{NH}_2$