JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Alcohols Phenols and Ethers

Hydroiodic acid protonates the ether oxygen, so the fragment that leaves in the substitution is a neutral molecule and not an alkoxide. Iodide then attacks the carbon that can take a backside displacement. At the ring carbon the oxygen lone pair is delocalised into the aromatic system, which shortens and strengthens that $\mathrm{C}-\mathrm{O}$ bond, and an $s{p}^2$ carbon cannot accept backside attack in any case, so substitution occurs at the methyl carbon. The oxygen stays on the ring and keeps the proton. The products are 1-naphthol and methyl iodide, $\mathrm{C}{\mathrm{H}}_3\mathrm{I}$.