See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The Hofmann rearrangement (OH⁻/Br₂) converts a primary amide (RCONH₂) to a primary amine (RNH₂) with loss of CO and retention of configuration at the migrating carbon (intramolecular migration, inversion at the carbon bearing the amide group is not occurring; rather the carbon migrates with retention to the nitrene/isocyanate nitrogen). The key stereochemical feature is that the alkyl group migrates with RETENTION of configuration at the carbon that migrates. Step 2 - Starting materials: Compound 1 is a cyclohexane ring bearing a CH3 group and a CONH2 group at the same carbon (C1), with CH3 on wedge (coming forward). Compound 2 is the same framework but with an ethyl (C2H5) group on wedge and a 15N-labeled amide. Step 3 - Hofmann rearrangement mechanism: The amide nitrogen is brominated, then base removes H to give N-bromoamide anion, which rearranges via intramolecular migration of the C1 carbon (with its substituents) to nitrogen with RETENTION of configuration at C1 (the migrating group), forming an isocyanate intermediate, which hydrolyzes to give the amine. The nitrogen ends up where the carbonyl carbon was, and the configuration at C1 is retained. Step 4 - Stereochemical outcome: In compound 1, if CH3 is on a wedge at C1, the amine (NH2) produced will be on a wedge (same face) at C1, with CH3 also on wedge — wait, more precisely: after Hofmann rearrangement, the C1 carbon that bore CONH2 and CH3 now bears NH2 and CH3. The migration proceeds with retention at the carbon, so the relative configuration of substituents at C1 is retained. The NH2 appears on the same face as CH3 (both wedge) in the product for compound 1, and similarly 15NH2 on same face as Et for compound 2. But looking at the answer choices, option (b) shows NH2 on wedge and CH3 on dash (trans relationship on the ring) for compound 1, and 15NH2 on wedge with Et on dash for compound 2 — these represent the 1,3-disubstituted cyclohexane products where both groups are trans (one wedge, one dash), consistent with retention of configuration during intramolecular migration in a cyclic system where the amide and methyl are trans to each other on the ring. Step 5 - Why option (b) is correct: The starting amide has the CONH2 and CH3 at C1 of a methylcyclohexane (they are on the same carbon — quaternary), but the structures shown in the question appear to be 1-methyl-3-(amide)cyclohexane or similar. Looking more carefully, the starting materials show the amide and the methyl/ethyl on different carbons of the ring with a specific relative configuration. With OH⁻/Br₂ and intramolecular migration, the Hofmann product places NH2 where CONH2 was, with retention, giving a trans-1-amino-3-methylcyclohexane (NH2 on wedge, CH3 on dash) for compound 1, and trans-1-(15NH2)-3-ethylcyclohexane (15NH2 on wedge, Et on dash) for compound 2. This matches option (b). Step 6 - Why other options fail: (a) shows cis products (both substituents on same face/both wedge or both dash) — wrong stereochemistry for Hofmann with retention. (c) and (d) either swap the 15N label placement or show wrong facial selectivity. Therefore, the correct answer is B.