See image — Amines Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Beckmann rearrangement converts a ketoxime (formed from a ketone + NH2OH) into an amide via PCl5 (or other dehydrating agents) with heat. The group anti to the hydroxyl on the oxime migrates to nitrogen, giving an amide. For an unsymmetrical ketone, two oxime geometric isomers (E and Z) form, and each undergoes migration of the group anti to -OH, yielding two different amide regioisomers. Step 2 - Oxime formation: The ketone ArCH2-C(=O)-Me reacts with NH2OH/HCl to give an oxime ArCH2-C(=NOH)-Me. Because the ketone is unsymmetrical (one side is ArCH2-, the other is Me), two geometric isomers of the oxime form: the E-oxime and the Z-oxime. Step 3 - Beckmann rearrangement of each oxime isomer: - In one isomer, the ArCH2 group is anti to OH: migration of ArCH2 gives ArCH2-NH-C(=O)-Me (option a, an N-H amide where ArCH2 is on N and acetyl is the carbonyl part). - In the other isomer, the Me group is anti to OH: migration of Me gives Ar-CH2-C(=O)-NH-Me (option b, an N-methylamide). Thus both products (a) and (b) are formed as the two regioisomeric amides from the major Beckmann pathway. Step 4 - Minor product via Beckmann fragmentation/cyclization: Under PCl5/heat conditions, especially with an arylmethyl group adjacent to nitrogen in an intermediate nitrilium ion, intramolecular cyclization can occur. The nitrilium ion intermediate (ArCH2-N≡C-Me or related species) can undergo intramolecular electrophilic cyclization onto the electron-rich aromatic ring of the benzodioxole, forming a cyclic imine (option c, the fused ring =N-Me lactam/cyclic imine product). Step 5 - Why all three form: The question explicitly states 'consider minor product also.' Products (a) and (b) are the two major Beckmann amide products from the two oxime geometries. Product (c) is a minor product from intramolecular cyclization of the nitrilium ion intermediate onto the aromatic ring. Step 6 - Why answer is D: Since all three structures — (a) the N-H amide regioisomer, (b) the N-methyl amide regioisomer, and (c) the cyclic imine minor product — are obtained from this Beckmann-type rearrangement when minor products are considered, all of these are correct. Therefore, the correct answer is D.