JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Alcohols Phenols and Ethers

The reaction involves benzyl chloride ($\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_2\mathrm{Cl}$) reacting with aqueous KOH followed by acidification.

Step 1: $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_2\mathrm{Cl}$ + $\mathrm{KOH}(\mathrm{aq})$ → $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_2\mathrm{OH}$ (benzyl alcohol) + $\mathrm{KCl}$

Step 2: Benzyl alcohol is then treated with $\mathrm{HI}$ and red phosphorus, which converts the $-\mathrm{OH}$ group to $-\mathrm{I}$.

$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_2\mathrm{OH}$ + $\mathrm{HI}$ → $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CH}{}_2\mathrm{I}$ (benzyl iodide) + $\mathrm{H}{}_2\mathrm{O}$

However, looking at option (A), it shows p-iodophenol, which suggests aromatic substitution occurred. The correct product B is p-iodophenol formed via electrophilic aromatic substitution.