See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Reaction of a carboxylic acid with organolithium reagents. When a carboxylic acid reacts with an alkyllithium (RLi), the first equivalent deprotonates the acidic OH of the carboxylic acid (pKa ~5) to form a carboxylate salt (RCO2^- Li^+). The second equivalent of RLi adds to the carboxylate to form a tetrahedral dianion intermediate (a dilithio alkoxide-alkoxide species), which is stable and does not collapse further under the reaction conditions because the resulting alkoxide is a poor leaving group and the dianion is stabilized. Upon aqueous acidic workup (H+, H2O), the tetrahedral intermediate collapses: loss of the lithium alkoxide gives a ketone intermediate, but since we have an alpha-hydroxy group already present, the net result is formation of a ketone at the former carboxyl carbon. Step-by-step reasoning: 1) The starting material is 2-cyclohexyl-2-hydroxyacetic acid [cyclohexyl-CH(OH)-COOH]. 2) First equivalent of EtLi deprotonates the carboxylic acid OH to give the carboxylate: cyclohexyl-CH(OH)-COO^- Li^+. Actually the alpha-OH may also be deprotonated by a second equivalent, but the key reaction is: first EtLi deprotonates COOH, second EtLi adds to the carboxylate carbon to give a tetrahedral dianion, third EtLi (excess) ensures complete reaction or deprotonates the alpha-OH. 3) Upon workup, the tetrahedral intermediate at the carboxylate carbon collapses to give a ketone: cyclohexyl-CH(OH)-C(=O)-Et. 4) This is an alpha-hydroxy ketone: the carbon bearing OH (from the original alpha-hydroxy acid) is retained, and the carboxyl carbon becomes a ketone with one ethyl group. Only ONE equivalent of EtLi adds to the carboxylate (the second equivalent), giving a methyl ketone analog—here an ethyl ketone. The product is 1-hydroxy-1-cyclohexyl-2-butanone, i.e., cyclohexyl-CH(OH)-CO-CH2CH3, which matches option (b): cyclohexyl attached to CH(OH) adjacent to C(=O)-Et. Why other options fail: (a) would require addition of two ethyl groups to the carbonyl carbon (like a Grignard giving tertiary alcohol directly from ester), which does not happen cleanly from a carboxylate with this protocol at room temperature yielding the ketone stage; (c) and (d) involve OEt groups which would require esterification, not present here. The reaction of RCO2H + 2 RLi → ketone (after workup) is a well-known synthetic method: the carboxylate formed after first deprotonation reacts with one more equivalent of RLi to give, after hydrolysis, a ketone. Therefore, the correct answer is B.