See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional groups: The starting material is 1-(2-(2-aminopropyl)phenyl)ethan-1-one. It has an ortho relationship on the benzene ring between an acetyl group (-C(=O)CH3) and a -CH2CH(NH2)CH3 side chain. Step 2 - Identify the intramolecular reaction type: The primary amine (-NH2) can attack the ketone carbonyl (C=O) intramolecularly to form a hemiaminal, which then dehydrates to give an imine (C=N). This is a classic intramolecular condensation (Schiff base formation). Step 3 - Determine ring size and regiochemistry: The amine nitrogen attacks the carbonyl carbon of the acetyl group. Counting atoms in the ring formed: N-CH(CH3)-CH2-(benzene C)-C(=N): this forms a 6-membered ring fused to the benzene, giving a 3,4-dihydroisoquinoline skeleton. The nitrogen ends up in the ring with a C=N double bond (imine). Step 4 - Assign substituents: The CH3 from the acetyl group ends up on the carbon bearing the C=N (C1 position of the dihydroisoquinoline), and the CH3 from the aminopropyl side chain ends up on C3 (the carbon that bore the amine). This gives 1,3-dimethyl-3,4-dihydroisoquinoline. Step 5 - Match to answer choices: Option (d) shows a 3,4-dihydroisoquinoline ring system with CH3 at C1 (adjacent to N via C=N) and CH3 at C3 (on the saturated carbon bearing N), which perfectly matches the product of intramolecular imine formation. Step 6 - Eliminate other options: (a) retains the ketone and has no nitrogen in the ring - this would require C-C bond formation, not amine condensation; (b) has a different connectivity/regiochemistry of methyl groups; (c) retains the free NH2 and does not show imine formation, meaning no cyclization/dehydration occurred. Therefore, the correct answer is D.