See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: Ethyl acetoacetate is CH3-CO-CH2-CO-OEt, which contains a ketone (at C1) and an ester (at C4). Step 2 - Formation of (A): Treatment with ethylene glycol (HO-CH2-CH2-OH) under acid (H+) selectively protects the more reactive ketone carbonyl as a cyclic 1,3-dioxolane acetal. Esters do not react with alcohols under simple acid catalysis in the same way under these mild conditions. So (A) is the compound where the ketone is protected as a 1,3-dioxolane and the ester (CO-OEt) remains intact: [1,3-dioxolan-2-yl]-CH2-CO-OEt with methyl on the dioxolane ring (i.e., 2-methyl-1,3-dioxolane protecting the acetyl group), giving structure: 2-methyl-1,3-dioxolan-2-yl-CH2-CO-OEt. Step 3 - Reaction of (A) with 2 PhMgBr: Grignard reagents react with esters in a 2:1 ratio. The first equivalent of PhMgBr adds to the ester to give a ketone intermediate (after loss of OEt-), and the second equivalent adds to the resulting ketone to give a tertiary alkoxide. So the ester -CO-OEt reacts with 2 PhMgBr to give -C(Ph)2-OMgBr at that carbon. This gives (B) as the magnesium alkoxide of a tertiary alcohol with two phenyl groups, while the acetal-protected ketone remains intact. Step 4 - Hydrolysis with H3O+: Aqueous acid workup (H3O+) does two things: (i) protonates the alkoxide to give the tertiary alcohol -C(Ph)2-OH, and (ii) hydrolyzes the acetal back to the ketone (CH3-CO-). Step 5 - Structure of (C): The product is CH3-CO-CH2-C(Ph)2-OH, which is 4-hydroxy-4,4-diphenylbutan-2-one. This matches option (b): a ketone (CH3CO-) at one end, a methylene (-CH2-), and a quaternary carbon bearing two phenyl groups and a hydroxyl group. Why other options fail: - (a) shows a single Ph added and incorrect connectivity (would require only 1 PhMgBr and addition at different site). - (c) retains the ester group, meaning the Grignard reaction with the ester did not proceed to completion, which is incorrect since 2 equivalents are used. - (d) shows a completely different skeleton with a terminal phenyl ketone and alcohol, inconsistent with the starting material and reagents. Therefore, the correct answer is B.