See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Conversion of a carboxylic acid to an amide via acid chloride intermediate. Step 1 - Reaction with SOCl2: Cinnamic acid (Ph-CH=CH-COOH) reacts with thionyl chloride (SOCl2) to convert the carboxylic acid (-COOH) into an acid chloride (-COCl), giving cinnamoyl chloride: Ph-CH=CH-C(=O)-Cl. The C=C double bond is unaffected under these mild conditions. Step 2 - Reaction with cyclopropylamine (NH2-cyclopropane): The acid chloride reacts with cyclopropylamine (a primary amine) via nucleophilic acyl substitution. The amine nitrogen attacks the carbonyl carbon, HCl is eliminated, and an amide bond is formed: Ph-CH=CH-C(=O)-NH-cyclopropane. Why other options fail: - Option (a) Ph-CH=CH-C(=O)-CH2-NH-cyclopropane: This would require reduction of the carbonyl to a methylene and then amination, which does not occur here. - Option (c) Ph-CH=CH-C(=O)-H: This is an aldehyde; SOCl2 does not reduce carboxylic acids to aldehydes, and the amine is not incorporated. - Option (d) Ph-CH=CH-NH-cyclopropane: This would require loss of the carbonyl oxygen entirely and direct N-substitution on the vinyl carbon, which does not occur under these conditions. The product is the amide Ph-CH=CH-C(=O)-NH-cyclopropane, which corresponds to option (b). Therefore, the correct answer is B.