See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Hofmann exhaustive methylation (Hofmann elimination) involves: (1) methylation of the amine with excess CH3I to form a quaternary ammonium salt, (2) treatment with AgOH/heat to give Hofmann elimination. For each nitrogen atom, one cycle of methylation + elimination is required to cleave a C-N bond. The number of steps equals the number of C-N bonds that must be broken to liberate all nitrogen as a non-amine product. Key rule: The number of Hofmann exhaustive methylation-elimination sequences needed equals the number of rings containing nitrogen plus the number of C-N bonds in an acyclic amine. Specifically: - An acyclic secondary amine requires 1 step (odd). - A monocyclic amine (one ring with N) requires 2 steps (even). - A bicyclic amine (two rings sharing N) requires 3 steps (odd). Analysis of each option: (a) Piperidine: monocyclic secondary amine with N in a 6-membered ring. It has 2 C-N bonds in the ring, requiring 2 Hofmann elimination steps to open the ring and then eliminate. Number of steps = 2 (even). ✓ (b) Octahydroquinoline (decahydroquinoline): this is a bicyclic compound where N is at the ring junction of two fused 6-membered rings (one ring is a piperidine-type, fused to a cyclohexane). However, the nitrogen is only in ONE ring (the heterocyclic ring); the other ring is carbocyclic. Therefore it behaves as a monocyclic amine with respect to Hofmann elimination — N participates in one ring. Steps required = 2 (even). ✓ (c) The bicyclic structure shown is a bridged bicyclic amine (like 9-azabicyclo[3.3.1]nonane or similar) where N is at the bridgehead connecting two rings — N is part of TWO rings. Such a compound requires 3 Hofmann elimination steps (odd). ✗ (d) Diethylamine (acyclic secondary amine, NH with two ethyl groups): requires only 1 Hofmann elimination step (odd). ✗ Therefore, the correct answer is A,B.