See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The haloform reaction occurs when a methyl ketone (or a compound oxidizable to a methyl ketone) is treated with a halogen (X2) and excess base (OH-). The key feature is that excess base drives complete trihalogenation of the more reactive methyl group. Step 1 - Identify the substrate: Ethyl methyl ketone is CH3CH2COCH3. It has two alpha positions: the methyl group (CH3CO-, 3 alpha-H) on one side and the methylene group (CH3CH2CO-, 2 alpha-H) on the other side. Step 2 - Reactivity under basic conditions: Under basic (excess OH-) conditions, once one alpha hydrogen on a carbon is replaced by Cl, the inductive effect of Cl makes the remaining alpha-H on that same carbon more acidic, accelerating further halogenation on the SAME carbon. This leads to exhaustive halogenation at the most reactive position. Step 3 - Which alpha position is more reactive? The methyl group directly attached to the carbonyl (CH3CO-) undergoes the haloform reaction. Under excess base, all three hydrogens of the CH3 group are replaced by Cl, giving CCl3 on that carbon. Step 4 - Haloform mechanism: The base-catalyzed halogenation proceeds: CH3CH2COCH3 -> CH3CH2COCHCl2 -> CH3CH2COCCl3 (trihalogenation of the methyl group). With excess OH-, the CCl3 group is then susceptible to nucleophilic attack by OH-, but the question asks for the major product formed, which is CH3CH2COCCl3 before the haloform cleavage step completes, OR the question is simply identifying that the methyl side undergoes complete chlorination to give CH3CH2COCCl3. Step 5 - Why not the other options? (a) ClCH2CH2COCH3: This would require monohalogenation at the ethyl (methylene) side, which is not favored under excess base conditions. (c) ClCH2CH2COCH2Cl: Requires halogenation at both sides equally, not the major product. (d) CH3CCl2COCH2Cl: Requires halogenation at ethyl carbon and partial halogenation at methyl, not the haloform product. The haloform reaction selectively and exhaustively halogenates the terminal methyl group of ethyl methyl ketone under excess OH-, giving CH3CH2COCCl3 as the major product. Therefore, the correct answer is B.