See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: This reaction involves the use of a nitrile as a masked carbanion equivalent (via the alpha-carbon of phenylacetonitrile), followed by acylation and hydrolysis. Step 1 - Deprotonation with EtONa: Sodium ethoxide (EtONa) is a strong base that deprotonates phenylacetonitrile (Ph–CH2–CN) at the alpha carbon (the CH2 group), because the CN group strongly stabilizes the adjacent carbanion through resonance. This generates the stabilized carbanion Ph–CH(–)–CN. Step 2 - Acylation with acetyl chloride (CH3COCl): The carbanion acts as a nucleophile and attacks the electrophilic carbonyl carbon of acetyl chloride (CH3–C(=O)–Cl), displacing chloride. This yields Ph–CH(CN)–C(=O)–CH3, i.e., the alpha-carbon now bears both the CN group and the acetyl group. Step 3 - Acid hydrolysis with H3O⊕/Δ: Under acidic aqueous conditions with heat, nitriles (–CN) hydrolyze. However, when a nitrile is alpha to a carbonyl (beta-keto nitrile), it undergoes decarboxylative hydrolysis. The CN hydrolyzes to –COOH (forming a beta-keto acid: Ph–CH(COOH)–C(=O)–CH3), and then the beta-keto acid readily decarboxylates upon heating (loss of CO2), giving Ph–CH2–C(=O)–CH3. Result: The product is Ph–CH2–C(=O)–CH3, which is phenylacetone (1-phenylpropan-2-one). Why other options fail: - (a) Ph–CH2–CHO: This would require reduction or a different reaction pathway; no aldehyde-forming step is present. - (c) Ph–CH(CH3)–CHO: This would require introduction of a methyl group on the benzylic carbon and aldehyde formation, which does not match the reagents. - (d) Ph–CH(CH3)–C(=O)–CH3: This would require alkylation at the benzylic carbon with a methyl group, but acetyl chloride provides an acyl group, not a methyl group, and after decarboxylation the CH2 is restored. Therefore, the correct answer is B.