See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Calculate the overall yield for part A. The reaction sequence has three steps with yields of 58%, 54%, and 68% respectively. Overall yield = 0.58 × 0.54 × 0.68 = 0.58 × 0.3672 = 0.2130 ≈ 21% This matches option (c) 21%. Step 2: Identify reagents A, B, C for part B. Reaction A: Me3C-C(=O)-CH3 → Me3C-C(=O)-CH2-Br This is alpha-bromination of a ketone. Under acidic conditions, Br2/H+ brominate the alpha carbon of the ketone selectively. This rules out NBS (which is used for allylic/benzylic bromination) and Br2/HO- (which would give multiple brominations under basic conditions for methyl ketones via haloform reaction). So A = Br2/H+. Reaction B: Me3C-C(=O)-CH2-Br → Me3C-CH(OH)-CH2-Br This is reduction of the ketone carbonyl to an alcohol without affecting the C-Br bond. NaBH4 is a mild reducing agent that selectively reduces ketones/aldehydes but does not reduce alkyl halides. LiAlH4 would also reduce the ketone but is much more reactive and could cause side reactions; however, for selectivity in this context NaBH4 is the standard choice. So B = NaBH4. Reaction C: Me3C-CH(OH)-CH2-Br → Me3C-CH-CH2 (epoxide) This is intramolecular SN2 cyclization of the bromohydrin to form an epoxide under basic conditions. The hydroxide (HO-) deprotonates the alcohol to form an alkoxide, which then displaces the adjacent bromide intramolecularly to form the epoxide. So C = HO-. Therefore A = Br2/H+, B = NaBH4, C = HO-, which corresponds to option (b). Why other options fail for B: (a) Br2/H+, LiAlH4, H(+): LiAlH4 could work for reduction but H(+) would not form an epoxide from a bromohydrin. (c) NBS, AlCl3, HO-: NBS is for allylic/benzylic bromination, not alpha-carbonyl bromination; AlCl3 is a Lewis acid not a reductant. (d) Br2/HO-, BF3, HO-: Br2/HO- would cause haloform reaction on the methyl ketone; BF3 is a Lewis acid, not a reductant. Therefore, the correct answer is {"A": ["C"], "B": ["B"]}.