See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: EtLi is a strong organolithium reagent. When a carboxylic acid reacts with organolithium reagents, the reaction proceeds stepwise. Step 1 - Acid-base reaction: The first equivalent of EtLi deprotonates the carboxylic acid OH (pKa ~5) to form the carboxylate salt (RCO2- Li+). This consumes 1 equivalent of EtLi (producing ethane gas). Step 2 - Addition to carboxylate: A second equivalent of EtLi adds to the carboxylate to form a dilithiated tetrahedral intermediate (dianion). This requires 1 more equivalent of EtLi. Step 3 - The substrate also has an alcohol (OH) group. The third equivalent of EtLi deprotonates the alcohol OH, protecting it as the lithium alkoxide and preventing side reactions. So with 3 equivalents EtLi: 1 eq deprotonates COOH, 1 eq deprotonates OH, 1 eq adds to the carboxylate carbon nucleophilically. The note says 3.5 eq were used (the extra 0.5 eq ensures complete addition to the carboxylate). Upon aqueous workup (H+, H2O), the dilithiated tetrahedral intermediate collapses to give a ketone (the standard carboxylate + RLi → ketone pathway stops at ketone because the carboxylate dianion intermediate is less reactive toward further addition under these conditions). The reaction of RCO2H with 2 eq RLi (after deprotonation) gives a ketone R-CO-R' after workup. Here the carboxylic acid portion is cyclohexyl-CH(OH)-CO2H, and EtLi adds to the carbonyl of the carboxylate. After workup: the -CO2H becomes -CO-Et (a ketone), and the -OH is restored from the lithium alkoxide. Product: cyclohexyl-CH(OH)-CO-Et, which is an alpha-hydroxy ketone: the cyclohexane-bearing carbon retains its OH group, and the carboxylic acid is converted to an ethyl ketone. This matches option (b): cyclohexane ring with CH(OH)-C(=O)-Et. Why other options fail: - Option (a) shows a 1,3-diketone, which would require two separate ketone-forming reactions or a different mechanism - not consistent with this substrate and reagent. - Options (c) and (d) show alpha,beta-unsaturated ketones (enones), which would require elimination - there is no beta-leaving group to form an alkene under these conditions; the OH is on the alpha carbon to the acid, not beta. The extra 0.5 eq mentioned ensures complete conversion of the carboxylate intermediate to product, confirming the ketone (b) is formed cleanly. Therefore, the correct answer is B.