See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Hofmann Elimination (Exhaustive Methylation/Elimination). When a quaternary ammonium salt is treated with a strong base (HO-) and heat, it undergoes E2 elimination via the Hofmann rule. The Hofmann rule states that elimination preferentially occurs to give the LESS substituted (less stable) alkene, because the bulky trimethylamino leaving group causes the base to abstract the more accessible (less hindered) proton. Step 1: Identify the substrate. The substrate is N,N,N-trimethylcyclohexylammonium ion. The NMe3+ group is attached to the cyclohexane ring (at C1). The adjacent carbons (C2 and C6) each bear hydrogens that can be abstracted. Step 2: Apply Hofmann's rule. In Hofmann elimination, the less substituted alkene is the major product. For a cyclohexyl substrate with NMe3+ at C1, elimination of a proton from the methyl group of NMe3 (if it were an open-chain system) or from the ring carbon would give either an endocyclic or exocyclic alkene. Since NMe3+ is directly on the ring and the only beta-hydrogens available are on the ring carbons adjacent to C1, the product would be cyclohexene with the double bond between C1 and C2 (an endocyclic alkene). However, if there is a methyl group on the nitrogen's carbon or if the structure is 1-(trimethylammonio)methylcyclohexane, then exocyclic elimination to give methylenecyclohexane is possible. Step 3: Re-examine the structure. The image shows NMe3+ on top of the arrow, suggesting the full substrate is (trimethylammoniomethyl)cyclohexane or N,N,N-trimethyl-cyclohexylammonium. Given that the correct answer is methylenecyclohexane (option B, exocyclic =CH2), the substrate must be cyclohexyltrimethylammonium where the NMe3+ is on a CH2 attached to the ring (i.e., the substrate is (N,N,N-trimethylaminomethyl)cyclohexane: cyclohexane-CH2-N+Me3). Hofmann elimination removes a beta-H from the cyclohexane ring carbon adjacent to CH2-N+Me3, giving methylenecyclohexane (exocyclic double bond) as the Hofmann (less substituted) product. Step 4: Why other options fail. - Option (a) and (c): 1-methylcyclohexene is a more substituted (trisubstituted) endocyclic alkene — this would be the Zaitsev product, not the Hofmann product. - Option (d): 1-methylenecyclohex-2-ene has two degrees of unsaturation and is not expected from simple elimination. - Option (b): Methylenecyclohexane is the less substituted exocyclic alkene, consistent with Hofmann elimination giving the least substituted alkene as the major product. Therefore, the correct answer is B.