See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Mixed aldol condensation between an aldehyde lacking alpha-hydrogens and a ketone with alpha-hydrogens proceeds via enolization of the ketone, nucleophilic addition of the enolate to the aldehyde carbonyl, followed by dehydration to give an alpha,beta-unsaturated carbonyl compound (chalcone/enone type). Step 2 - Identify the components: Benzaldehyde (C6H5CHO) has no alpha-hydrogens (the carbon bearing the CHO is the phenyl ring), so it acts as the electrophilic carbonyl partner. Acetone (CH3COCH3) has alpha-hydrogens on both methyl groups and acts as the nucleophilic enolate donor. Step 3 - Reaction mechanism: The alpha-carbon of acetone (one of the CH3 groups) attacks the carbonyl carbon of benzaldehyde to form a beta-hydroxy ketone intermediate: C6H5CH(OH)CH2COCH3. Dehydration then removes water from the beta-hydroxy ketone to give the conjugated enone. Step 4 - Product structure: After dehydration, the double bond forms between the alpha and beta carbons relative to the ketone, yielding C6H5CH=CHCOCH3, i.e., (E)-4-phenylbut-3-en-2-one (benzalacetone). This corresponds to option (a): C6H5CH=CHC(=O)CH3. Step 5 - Why other options fail: - Option (b) C6H5CH=C(CH3)2 is an alkene with no carbonyl group; it lacks the ketone functionality and would not arise from aldol condensation. - Option (c) C6H5C(=O)CH=CHCH3 places the carbonyl on the phenyl-bearing carbon, which would require benzaldehyde to donate its carbonyl carbon into the chain differently; this is not the product of this reaction. - Option (d) C6H5CH2C(=O)CH=CH2 has the phenyl group on a CH2 and a terminal alkene, inconsistent with the straightforward mixed aldol between benzaldehyde and acetone. Therefore, the correct answer is A.