See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When an aryl alkyl ether is treated with excess HBr (or HI) under heat, the C-O bond that cleaves is the alkyl C-O bond (not the aryl C-O bond), because the aryl C-O bond has partial double bond character (resonance stabilization) and is much stronger and harder to break under normal acidic conditions. The oxygen is protonated first, then the weaker alkyl C-O bond undergoes nucleophilic attack by Br-. Step 1: The substrate is phenoxyethanol: Ph-O-CH2CH2OH. With excess HBr and heat, the ether oxygen is protonated to give Ph-O+(H)-CH2CH2OH. Step 2: The alkyl C-O bond (the bond between oxygen and the -CH2CH2OH group) is cleaved by Br- acting as a nucleophile in an SN2 fashion (or via protonation and displacement), giving phenol (Ph-OH) and BrCH2CH2OH initially. Step 3: Since HBr is in excess, the -OH group of BrCH2CH2OH is also converted: BrCH2CH2OH + HBr → BrCH2CH2Br + H2O. Step 4: The aryl C-O bond is NOT cleaved under these conditions (no bromobenzene forms), so the phenol (Ph-OH) remains intact. Final products: Phenol (benzene ring-OH) + BrCH2CH2Br (1,2-dibromoethane). Why other options fail: - (a) requires electrophilic aromatic substitution at para position, which does not occur here under HBr/heat conditions with no catalyst. - (b) requires cleavage of the aryl C-O bond, which is not favored; also the -CH2CH2OH alcohol would be further brominated with excess HBr. - (d) requires cleavage of aryl C-O bond to give bromobenzene, which does not occur under these mild acid conditions. Therefore, the correct answer is C.