See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Gilman reagents (lithium dialkylcuprates, R2CuLi) are well known for performing conjugate (1,4-) addition to alpha,beta-unsaturated carbonyl compounds rather than direct (1,2-) addition to the carbonyl carbon. Step 1: Identify the substrate. Methyl vinyl ketone (MVK) is CH2=CH-C(=O)-CH3, an alpha,beta-unsaturated ketone (enone). Step 2: Identify the reagent. LiCuMe2 is lithium dimethylcuprate (Gilman reagent). It delivers a methyl nucleophile in a 1,4-conjugate fashion. Step 3: Mechanism of conjugate addition. The methyl group from LiCuMe2 adds to the beta-carbon (C1, the terminal alkene carbon) of MVK. This generates an enolate intermediate at the alpha-carbon (adjacent to C=O). Upon aqueous workup, the enolate is protonated to give the saturated ketone. Step 4: Determine the product. Adding CH3 to the beta-carbon of CH2=CH-C(=O)-CH3 gives: CH3-CH2-CH2-C(=O)-CH3, which is pentan-2-one (methyl propyl ketone). This corresponds to option (a): Me-CH2-CH2-C(=O)-Me. Step 5: Why other options fail. - Option (b): This would result from 1,2-addition of two methyl groups to the carbonyl and elimination, which is not the pathway for Gilman reagents. - Option (c): This would require 1,2-addition to the carbonyl carbon giving a tertiary alcohol with two added methyls, which is characteristic of organolithiums (RLi), not Gilman reagents. Gilman reagents strongly prefer 1,4-addition. - Option (d): Cyclopropyl methyl ketone would require an intramolecular cyclization, which does not occur here. Therefore, the correct answer is A.