JEE Main 2025 Apr 2 Shift 2, Chemistry Q9: Preparation of Carboxylic acids

Examining each route for the actual major product:

(A) Nitrile + dilute acid under mild conditions stops at the amide ($\mathrm{R}\text{-}\mathrm{CON}{\mathrm{H}}_2$); vigorous hydrolysis would be needed to reach the carboxylic acid. Not a COOH route.

(B) Grignard + $\mathrm{C}{\mathrm{O}}_2$ followed by aqueous workup is the classical synthesis of $\mathrm{R}\text{-}\mathrm{COOH}$. $✓$

(C) Stephen reduction ($\mathrm{SnC}{\mathrm{l}}_2/HCl$ then hydrolysis) converts a nitrile to an aldehyde ($\mathrm{R}\text{-}\mathrm{CHO}$), not an acid.

(D) PCC is a mild oxidant: primary alcohol $\to$ aldehyde, stopping before the acid.

(E) Rosenmund reduction (${\mathrm{H}}_2/Pd\text{-}\mathrm{BaS}{\mathrm{O}}_4$) gives benzaldehyde; subsequent $\mathrm{B}{\mathrm{r}}_2/{\mathrm{H}}_2\mathrm{O}$ acts as a mild oxidant and converts the aldehyde to benzoic acid. $✓$

$\therefore$ B and E only.