Main products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are :

Solution

Chemistry ma’am · 4 steps

With an aryl methyl ether, iodide always attacks the methyl carbon, never the ring carbon.

1. Protonate the oxygen. HI first protonates the ether oxygen. This turns the -OCH₃ group into a good leaving group: the C–O bond can now break and leave a neutral -OH.
2. Pick the bond. The ring C–O bond has partial double-bond character from resonance, and its carbon is sp², so I⁻ cannot do backside attack there. An aryl cation is also far too unstable to form. The methyl carbon is small and unhindered, so I⁻ attacks it by S_N2.

   C₁₀H₇-O-CH₃ + HI → C₁₀H₇-OH + CH₃I

3. Naphthol stays. Even with excess HI, 1-naphthol does not turn into 1-iodonaphthalene, for the same reason: its C–O bond is on an sp² ring carbon and does not break.
4. So. The main products are 1-naphthol and CH₃I, which is option (d).

The idea to keep. Cleaving an aryl alkyl ether with HI always gives a phenol and an alkyl iodide. The aryl–O bond never breaks, because the ring carbon is sp² and has partial double-bond character.

Answer from the NTA final answer key. Solution by PrepWiser.