See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Hofmann Exhaustive Methylation involves (i) exhaustive methylation of the amine to form a quaternary ammonium salt, followed by (ii) treatment with moist silver oxide (Ag2O/H2O) and then heating to cause elimination (Hofmann elimination). The elimination follows E2 mechanism and the base (hydroxide) abstracts a beta-hydrogen. In Hofmann elimination, the less substituted (less hindered) beta-carbon is preferentially deprotonated, giving the less substituted (Hofmann) alkene. Step 2 - Structure analysis: The starting amine is N-ethyl-1-ethylcyclohexylamine (a secondary amine). The nitrogen bears one ethyl group and one H. C1 of cyclohexane also bears one ethyl group. Step 3 - Exhaustive methylation: The secondary amine (-NH(Et)) is methylated with excess CH3I to give the quaternary ammonium salt: the nitrogen becomes N+(CH3)3 with an ethyl group... Actually, let me re-examine. The amine shown is a secondary amine: the nitrogen has one H, one ethyl group, and is attached to the C1 of cyclohexane (which also has an ethyl group). Exhaustive methylation converts -NH(Et) to -N+(CH3)2(Et) quaternary salt (adding methyl groups until quaternary). With excess CH3I: N goes from secondary (-NH-Et) to quaternary N+(CH3)2(Et) attached to C1-cyclohexyl. Step 4 - Hofmann elimination: The quaternary ammonium hydroxide is heated. Beta-hydrogens available are: (a) on the ethyl group attached to C1 of cyclohexane (the -CH2CH3 on C1), (b) on the ring carbons adjacent to C1, (c) on the N-ethyl group (-CH2CH3 on N), (d) on N-methyl groups (only one H... actually CH3 has 3H but gives ethylene equivalent). Hofmann's rule favors elimination toward the least substituted position. The least hindered beta-hydrogens are on the N-methyl or N-ethyl groups rather than on the ring or the C1-ethyl group. The N-ethyl group provides a simple primary beta-carbon (-N-CH2-CH3), abstraction of H from CH2 gives ethylene (H2C=CH2) and releases the amine. Hofmann elimination preferentially cleaves toward the least substituted carbon, and the N-ethyl group's terminal CH3... the beta carbon of the N-ethyl is -CH2-, giving ethylene. Since Hofmann elimination preferentially occurs at the least substituted beta-carbon, and the ethyl group on nitrogen gives ethylene (CH2=CH2) as the simplest least-substituted alkene, this is the major product. Step 5 - Why other options fail: Options (a) and (b) would result from elimination involving the ring or the C-ethyl group, which are more hindered (Hofmann elimination disfavors more substituted alkenes). Option (c) cyclohexene would require ring opening or endocyclic elimination, which is less favored. Option (d) H2C=CH2 (ethylene) results from the least hindered Hofmann elimination from the N-ethyl group. Therefore, the correct answer is D.