See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When anisole (methoxybenzene, C6H5-O-CH3) is treated with excess concentrated HI under reflux, the C-O bond of the ether is cleaved by HI. Step 1 - Identify the ether cleavage reaction: Anisole is an aryl alkyl ether. When treated with HI, the bond that cleaves is the weaker alkyl C-O bond (the Ar-O bond is stronger due to resonance with the ring). The iodide ion acts as a nucleophile and attacks the methyl carbon (SN2), not the aryl carbon (aryl C-O bonds are not cleaved by SN2). Step 2 - Determine the products: The nucleophilic iodide (I-) attacks the methyl group via SN2, breaking the O-CH3 bond. This produces phenol (C6H5-OH) and methyl iodide (CH3I). Step 3 - Role of excess HI: The proton from HI first protonates the oxygen of anisole (forming the oxonium ion), making the methyl carbon more electrophilic. Then I- attacks the methyl group to give phenol and CH3I. Step 4 - Why other options fail: - Option (a): iodobenzene + CH3I would require cleavage of the Ar-O bond, which does not occur under these conditions. - Option (b): iodobenzene + CH3OH would require Ar-O bond cleavage and no iodide incorporation on carbon, which is incorrect. - Option (d): phenol + CH3CH2I is incorrect because there is no ethyl group present in anisole. Option (c) phenol + CH3I is the correct product of HI cleavage of anisole at the alkyl C-O bond. Therefore, the correct answer is C.