See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify compound (A). Benzene reacted with Fenton's reagent (Fe2+ / H2O2) undergoes hydroxylation via a free radical or electrophilic mechanism. Fenton's reagent generates hydroxyl radicals (•OH), which attack benzene to give phenol. Therefore, (A) = phenol. Step 2: React phenol with Br2/H2O (aqueous bromine). Phenol is a highly activated aromatic compound. The –OH group is a strong ortho/para director. In aqueous conditions (no Lewis acid catalyst needed), bromine in water is sufficiently reactive toward phenol. Because phenol is so activated, all three available ortho and para positions are brominated. - Positions ortho to OH: C2 and C6 - Position para to OH: C4 This gives 2,4,6-tribromophenol as the major (essentially quantitative, 100%) product. Step 3: Why other options fail. (a) 2-bromophenol: This would be obtained only under very mild, controlled conditions (e.g., non-aqueous solvent, limited Br2). Aqueous Br2 with phenol gives tribromination, not monobromination. (b) 3-bromophenol: The meta position is deactivated by the OH group, so bromination does not occur at meta. This product would not form. (c) A bromobenzaldehyde: Fenton's reagent gives phenol, not benzaldehyde. This option is incorrect at the first step. (d) 2,4,6-tribromophenol: Consistent with the strong ortho/para directing effect of the OH group and the high reactivity of phenol toward electrophilic aromatic substitution in aqueous bromine (all three activated positions brominated). This matches the 100% yield noted. Therefore, the correct answer is D.