See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The reaction of acetophenone with benzaldehyde under acidic conditions (HCl/heat) is a Claisen-Schmidt condensation. In this reaction, the alpha-carbon of the ketone (acetophenone has two alpha-H atoms on the methyl group) is activated, and the aldehyde (benzaldehyde, which has no alpha-H) acts as the electrophilic carbonyl partner. Step 1 – Identify the nucleophile and electrophile: Acetophenone (Ph-CO-CH3) provides the alpha-carbon (the CH3 group adjacent to the carbonyl). Benzaldehyde (Ph-CHO) is the electrophile (no alpha-H, so it cannot self-condense). Step 2 – Mechanism: Under acidic conditions (HCl), the enol form of acetophenone attacks the carbonyl carbon of benzaldehyde in an aldol-type addition, giving initially a beta-hydroxy ketone: Ph-CO-CH2-CH(OH)-Ph. Step 3 – Dehydration: With heat, the beta-hydroxy ketone undergoes dehydration (elimination of water) to give the alpha,beta-unsaturated ketone (enone). The product is Ph-CO-CH=CH-Ph, which is chalcone (trans-1,3-diphenyl-2-propen-1-one). Step 4 – Match to options: Option (a) shows exactly this structure: a benzene ring connected to C=O, then CH=CH (double bond), then another benzene ring — this is chalcone. Why other options fail: - Option (b) Ph-CO-CH2-CH2-Ph would require reduction of the double bond; it is not the direct condensation product. - Option (c) Ph-CH=CH-CH(OH)-Ph is the beta-hydroxy intermediate after dehydration of a different type; it does not match the Claisen-Schmidt product and has the wrong connectivity. - Option (d) Ph-CH=C=CH-Ph is a cumulene (allene-type), which is not formed in this reaction. Therefore, the correct answer is A.