Identify the substrate

The compound is 1-methoxy naphthalene, i.e. a naphthalene ring attached to a methoxy group: $\text{Ar-O-CH}_3$ where $\text{Ar}$ represents the 1-naphthyl group.

Reagent involved

Hydroiodic acid, $\text{HI}$, cleaves ethers after protonation of the oxygen atom.

General rule for cleavage of aryl-alkyl ethers

For an ether of the form: $\text{Ar-O-R}$ treatment with $\text{HI}$ gives cleavage at the alkyl-O bond, not at the aryl-O bond, because:

• the $\text{C}_{\text{aryl}}-O$ bond has partial double bond character due to resonance,
• cleavage at aryl carbon by $S_N1$ or $S_N2$ is not feasible,
• iodide attacks the alkyl group.

Mechanism

First protonation of ether oxygen: \text{Ar-O-CH}_3 + H^+ \rightarrow \text{Ar-O(H)^+-CH}_3

Then iodide ion attacks the methyl group via $S_N2$: \text{Ar-O(H)^+-CH}_3 + I^- \rightarrow \text{Ar-OH} + CH_3I

Products formed

Therefore, the main products are:

• 1-naphthol
• methyl iodide

Reaction: $\text{1-methoxy naphthalene} + HI \rightarrow \text{1-naphthol} + CH_3I$

Conclusion

So the correct option must be the one corresponding to: $\boxed{\text{1-naphthol and methyl iodide}}$

Comparison with stored answer

The stored correct answer is D. This matches the expected products for cleavage of an aryl methyl ether with HI, so I agree with the stored answer.