See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Alpha-halogenation of carbonyl compounds. When a ketone reacts with Br2 in the presence of a Lewis acid catalyst (AlCl3) or under acidic conditions, bromination occurs preferentially at the alpha carbon — the carbon directly adjacent to the carbonyl group. This is because the enol tautomer forms under acid catalysis, and the nucleophilic alpha carbon attacks Br2, placing the bromine at the alpha position. Step 1: Identify the starting material. The starting material is butyrophenone: Ph-C(=O)-CH2-CH2-CH3. The carbonyl is at C1 (attached to phenyl), the alpha carbon is C2 (the -CH2- directly next to C=O), the beta carbon is C3, and the gamma/terminal carbon is C4 (CH3). Step 2: Determine the site of bromination. Under acidic conditions (AlCl3 as Lewis acid), Br2 effects alpha-bromination. The alpha carbon (C2, i.e., -CH2- adjacent to C=O) is the most activated position. The enol forms preferentially at the alpha carbon, and Br2 reacts there. Step 3: Identify the product. Alpha-bromination of Ph-C(=O)-CH2-CH2-CH3 gives Ph-C(=O)-CHBr-CH2-CH3. This is the product in option (b): a benzene ring attached to a carbonyl, with the carbon alpha to the carbonyl bearing a Br, and the remaining chain being -CH2-CH3 (ethyl group beyond the alpha carbon). Step 4: Why other options fail: - Option (a): Br at the beta carbon (C3). This would require non-standard regioselectivity; acid-catalyzed bromination favors the alpha position, not beta. - Option (c): Br at the terminal gamma carbon (C4). This is characteristic of radical (NBS/light) or nucleophilic substitution, not acid-catalyzed alpha-bromination of a ketone. - Option (d): Although it appears structurally similar to (b), the chain drawn is -CHBr-CH2-CH2-CH3 (the chain is a butyl/propyl group beyond alpha), which would correspond to a different starting material or misidentification; option (b) correctly shows Ph-C(=O)-CHBr-CH2-CH3 consistent with alpha-bromination of butyrophenone. The 0.75 mole% of AlCl3 is catalytic, consistent with acid-catalyzed enolization and alpha-bromination, giving 100% selectivity for the alpha-bromo product. Therefore, the correct answer is B.