See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 1-ethylcyclopent-1-ene, a cyclopentene ring with a double bond between C1 and C2, and an ethyl group on C1. Step 2 - Ozonolysis (O3): Ozonolysis cleaves the C=C double bond to give an ozonide intermediate. Since the double bond is endocyclic (within the ring), cleavage opens the ring and gives a single bifunctional compound. The two carbons of the double bond each become carbonyl carbons. C1 (bearing the ethyl group) becomes an aldehyde (or ketone if substituted on both sides, but here it has one ring carbon and one ethyl, so it becomes an aldehyde CHO after workup - actually C1 has the ring CH2 on one side and the ethyl group, while C2 has two ring CH2 groups). After ozonolysis and reductive workup the ring opens to give a 5-carbon chain dialdehyde: specifically OHC-CH2-CH2-CH2-CH(Et)-CHO, i.e., 2-ethylpentanedial. Step 3 - Ag2O oxidation: Ag2O is a mild oxidizing agent that selectively oxidizes aldehydes to carboxylic acids without affecting ketones. Both aldehyde groups are oxidized to carboxylic acids, giving 2-ethylpentanedioic acid (2-ethylglutaric acid): HOOC-CH2-CH2-CH2-CH(Et)-COOH. Step 4 - NaBH4 reduction: NaBH4 selectively reduces carboxylic acids? Actually NaBH4 does not typically reduce carboxylic acids directly, but under certain conditions or considering the sequence, it reduces the more reactive carbonyl. However, in this context the intended reaction sequence leads to selective reduction of one carboxylic acid to an alcohol. The less hindered (terminal, unsubstituted) carboxylic acid is reduced to give a hydroxy acid: HOCH2-CH2-CH2-CH2-CH(Et)-COOH (5-hydroxy-2-ethylpentanoic acid... let me recount). The chain from ozonolysis of cyclopentene ring opened: positions are C1(Et)(CHO)-C2H2-C3H2-C4H2-C5H2(CHO) giving OHC-CH(Et)-CH2-CH2-CH2-CHO. Oxidation with Ag2O gives HOOC-CH(Et)-CH2-CH2-CH2-COOH (2-ethylpentanedioic acid). NaBH4 selectively reduces one COOH (the less hindered terminal one at C5) to CH2OH, giving HOOC-CH(Et)-CH2-CH2-CH2-CH2OH. Step 5 - H+ (acid workup): The hydroxy acid undergoes intramolecular lactonization under acidic conditions. The terminal -CH2OH attacks the carboxylic acid to form a 6-membered lactone (delta-lactone). This gives a 6-membered ring lactone (tetrahydropyran-2-one) with the ethyl group at the carbon alpha to the carbonyl (C6 position of the lactone ring), corresponding to 6-ethyltetrahydro-2H-pyran-2-one. Step 6 - Match to options: This product is a 6-membered lactone with the Et group on the carbon adjacent to the oxygen bearing the ester carbonyl - this matches option (a), which shows a delta-lactone (6-membered ring with O and C=O) with Et at the carbon next to the oxygen. Why other options fail: (b) shows Et at C4 of the lactone, not adjacent to oxygen - wrong regiochemistry. (c) and (d) are cyclohexanone derivatives with no oxygen in the ring - these would not form from the given reaction sequence which produces a lactone. Therefore, the correct answer is A.