See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The transformation converts a methyl ester (CO2CH3) on a para-methylbenzene ring into an amide (C(=O)-NH-CH3). We need to identify the reagent sequence that achieves this. Step 1: Identify the starting material and product. Starting material: methyl 4-methylbenzoate (para-toluic acid methyl ester, H3C-C6H4-CO2CH3) Product: N-methyl-4-methylbenzamide (H3C-C6H4-C(=O)-NH-CH3) Step 2: Determine the most direct route. To convert a methyl ester to an amide (specifically N-methylamide), one efficient route is: 1. Hydrolyze the ester to the carboxylic acid (using H3O+, i.e., aqueous acid hydrolysis) 2. Convert the carboxylic acid to the acid chloride (using SOCl2) 3. React the acid chloride with methylamine (CH3NH2) to give the N-methylamide This corresponds exactly to option (a): H3O+; SOCl2; CH3NH2 Step 3: Evaluate each option. (a) H3O+; SOCl2; CH3NH2 — H3O+ hydrolyzes the methyl ester to the carboxylic acid (4-methylbenzoic acid). SOCl2 converts the acid to the acid chloride. CH3NH2 reacts with the acid chloride to give N-methyl-4-methylbenzamide. This is correct and efficient. (b) HO-/H2O; PBr3; Mg; CO2; H3O-; SOCl2; CH3NH2 — This sequence involves a Grignard synthesis step that is unnecessarily long and would not give the correct product directly. The use of PBr3 on a carboxylic acid to form an aryl bromide, then Grignard with CO2 would regenerate a carboxylic acid, making the sequence redundant and circuitous. (c) LiAlH4; H2O; HBr; Mg; CO2; H3O+; SOCl2; CH3NH2 — LiAlH4 would reduce the ester to a benzyl alcohol, then HBr would convert it to benzyl bromide, Mg makes a Grignard reagent, CO2 gives a homologated acid — this changes the carbon skeleton and would not give the same product. (d) Incorrect because option (a) does yield the desired product. Therefore, the correct answer is A.