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Answer: 2. phenol

Key idea: HI cleaves an aryl alkyl ether at the alkyl-oxygen bond.

\(\mathrm{C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I}\). The oxygen is protonated, \(\mathrm{I^-}\) attacks the methyl carbon by \(S_N2\). The \(\mathrm{C_6H_5-O}\) bond has partial double-bond character and is not broken, so iodobenzene is not formed. The product is phenol (with methyl iodide).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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