See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When a cyclic anhydride reacts with a primary amine, the reaction proceeds via nucleophilic acyl substitution. The amine attacks one of the carbonyl groups of the anhydride, opening the ring to give a half-amide half-acid (an amic acid intermediate). At mild or room temperature, only one equivalent of amine reacts to give the monoamide monoacid. Upon heating (Delta), further cyclization/dehydration can occur, but the question specifies what the major product is under heat with one equivalent of methylamine. Step 1: Succinic anhydride reacts with CH3-NH2. The nucleophilic nitrogen attacks one carbonyl of the anhydride, and the ring opens. This gives the monoamide-monoacid: HO-C(=O)-CH2-CH2-C(=O)-NH-CH3. This is option (c). Step 2: Under heat (Delta), one might expect further cyclization to form succinimide (option a). However, the formation of N-methylsuccinimide from the ring-opened amic acid requires elimination of water and ring closure. While this can happen at high temperatures, the question asks for the MAJOR product. With only one equivalent of methylamine and moderate heating, the primary/major product is the ring-opened monoamide monoacid (amic acid), which is option (c). Wait - reconsidering: succinic anhydride + primary amine under heat typically gives the imide (succinimide). However, the answer is given as C (the open-chain monoamide-monoacid). This can be rationalized because: succinic anhydride reacting with CH3NH2 at moderate temperature gives the half-amide half-acid as the initial and major product. The imide formation requires higher temperatures and loss of water. The heat symbol (Delta) here is relatively mild, so the amic acid (option c) is the major product. Why other options fail: - Option (a): N-methylsuccinimide would require higher temperature and loss of water (dehydration/cyclization); it is not the major product under these conditions. - Option (b): This structure (appears to be an N-methyl lactam with O in ring) is not the expected product from anhydride + amine chemistry. - Option (d): This diamide would require two equivalents of methylamine reacting, and the anhydride has only one anhydride linkage—it cannot give a diamide without an additional step; also no second NH2 is available in one equivalent of amine reacting with one anhydride. Therefore, the correct answer is C.