JEE Main 2026 — Carboxylic Acid Derivatives Question with Solution

Step 1: Hydrolysis of the nitrile.
Propanenitrile on basic hydrolysis followed by acidic workup gives propanoic acid:
$C{H}_3-C{H}_2-C\equiv N\underset{(ii)H_3{O}^{+}}{\overset{(i)O{H}^{-}/H_{2}O/\Delta }{\to }}C{H}_3-C{H}_2-COOH$

Step 2: Hell$-$Volhard$-$Zelinsky (HVZ) reaction.
Propanoic acid has an $s{p}^3$ $\alpha$-carbon bearing two $\alpha$-hydrogens. In the HVZ reaction, $C{l}_2$ in the presence of red phosphorus selectively replaces one $\alpha$-hydrogen by chlorine, giving the $\alpha$-chloro acid after aqueous workup:
$C{H}_3-C{H}_2-COOH\underset{(iv)H_{2}O}{\overset{(iii)C{l}_2/\text{Red P}}{\to }}C{H}_3-CH(Cl)-COOH$

Hence, the major product $P$ is $2$-chloropropanoic acid, which corresponds to option $(1)$.