See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material structure: The starting material is a dilithium dienolate derived from ethyl acetoacetate (ethyl 3-oxobutanoate). Treatment of ethyl acetoacetate with 2 equivalents of a strong base (like LDA or n-BuLi) generates a dianion (dilithio enolate). The structure shown is the dilithio dianion of ethyl acetoacetate: the two OLi groups are on C1 (the terminal methylene, giving CH2=C(OLi)-) and on C3 (the alpha carbon to the ester, giving =C(OLi)-OEt). This is the dianion of ethyl acetoacetate where both the methyl ketone terminus (as a vinylogous enolate) and the ester alpha position are deprotonated. Step 2 - Reactivity of the dianion with alkyl halide: In the dianion of ethyl acetoacetate, the more reactive (harder, more carbanion-like) nucleophilic site is the terminal carbon (the gamma carbon, C1, which bears the exo-methylene). The less reactive site is the alpha carbon between the two carbonyls. Under kinetic conditions with a primary alkyl halide (CH3CH2I), alkylation occurs preferentially at the terminal (gamma) position, the more reactive carbanion. Step 3 - Alkylation at terminal carbon: The ethyl iodide alkylates the terminal carbon (the gamma carbon of the dianion, i.e., the carbon bearing the exo =CH2). This converts CH2=C(OLi)- to CH3CH2-CH2-C(OLi)= (after alkylation at the terminal carbon of the vinylogous enolate, the terminal CH2 picks up an ethyl group). Step 4 - Aqueous workup (H3O+): Protonation of both OLi groups regenerates the carbonyl functions. The terminal OLi becomes a ketone C=O, and the internal OLi on the vinyl position becomes the ester C=O (restored). This gives: CH3CH2-CH2-C(=O)-CH2-C(=O)-OEt, which is ethyl 3-oxohexanoate (ethyl 3-oxopentanoate with propyl group), consistent with option (a). Step 5 - Why other options fail: - Option (b): Would require alkylation at the oxygen rather than carbon, giving an O-alkylated product (vinyl ether), which is not the major product with a primary alkyl iodide under these conditions. - Option (c): Would require alkylation at the internal alpha carbon (between the two carbonyls), which is the less reactive site in the dianion; gamma-alkylation is preferred. - Option (d): Would require double O-alkylation or alkylation at the ester oxygen, which is not favorable. Step 6 - Conclusion: The major product is the gamma-alkylated compound after hydrolysis: CH3CH2CH2-C(=O)-CH2-C(=O)-OEt, matching option (a). Therefore, the correct answer is A.