NEET 2020 Chemistry Question 135 Solution

In the presence of hydrogen iodide ($$\text{HI}$$), anisole (methoxybenzene) is first protonated at the ether oxygen to form a methylphenyl oxonium ion: $$\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HI} \rightleftharpoons \text{C}_6\text{H}_5-\text{O}^+(\text{H})-\text{CH}_3 + \text{I}^-$$

The oxonium ion has two different $$\text{C}-\text{O}$$ bonds: 1. The $$\text{C}_{\text{phenyl}}-\text{O}$$ bond: Due to resonance with the benzene ring, this bond has partial double bond character ($$sp^2$$ hybridized carbon) and a shorter bond length ($$136\text{ pm}$$), making it exceptionally strong and resistant to cleavage. 2. The $$\text{C}_{\text{methyl}}-\text{O}$$ bond: This is a weaker, single $$\sigma$$-bond ($$sp^3$$ hybridized carbon).

The nucleophilic iodide ion ($$\text{I}^-$$) selectively attacks the less sterically hindered methyl carbon via an $$S_N2$$ pathway, displacing phenol as the neutral leaving group: $$\text{I}^- + \text{CH}_3-\text{O}^+(\text{H})-\text{C}_6\text{H}_5 \xrightarrow{S_N2} \text{CH}_3\text{I} + \text{C}_6\text{H}_5\text{OH}$$ Phenol does not react further with HI because the hydroxyl group cannot be displaced by iodide from the $$\text{sp}^2$$ phenyl ring.

The cleavage products are unequivocally Phenol ($$\text{C}_6\text{H}_5\text{OH}$$) and Methyl iodide ($$\text{CH}_3\text{I}$$), corresponding to Option (1).