Identify the major product of the following reaction sequence: CH ₃ CO ₂ H [ (ii) 3-Methylaniline ] (i) SOCl ₂ (iii) LiAlH ₄ , then H ₃ O ^+ ?

Step-by-step solution

Acetic acid ( CH ₃ COOH ) reacts with thionyl chloride ( SOCl ₂ ) to form acetyl chloride ( CH ₃ COCl ). Acetyl chloride reacts with 3-methylaniline to form an amide. The nucleophilic - NH ₂ group of 3-methylaniline attacks the carbonyl carbon of acetyl chloride, eliminating HCl to yield N-(3-methylphenyl)acetamide. The amide is then reduced by lithium aluminium hydride ( LiAlH ₄ ). LiAlH ₄ reduces the carbonyl group ( - C (= O )- ) of the amide to a methylene group ( - CH ₂- ), converting the amide into an amine.