See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Hoffmann exhaustive methylation (Hofmann elimination) involves: (1) quaternizing the amine with excess CH3I to form a quaternary ammonium salt, (2) treating with Ag2O/H2O to convert the iodide to hydroxide, and (3) heating (pyrolysis) to cause elimination, producing an alkene via an E2-like mechanism that favors the less substituted (Hofmann) alkene. Step 1 – Identify the starting amine: The compound is Me2CHCH2NHCH2CH2Me, a secondary amine. The nitrogen is connected to two carbon chains: - Chain A: -CH2CH(Me)2 (isobutyl group, i.e., -CH2-CHMe2) - Chain B: -CH2CH2Me (n-propyl group) Step 2 – Quaternization: Excess CH3I methylates the nitrogen fully to give the quaternary ammonium iodide: [Me2CHCH2-N+(Me)(CH2CH2Me)(CH3)] I- ... actually N gets two methyl groups added to reach quaternary: [Me2CHCH2-N+(Me2)-CH2CH2Me] I- . Wait, it's a secondary amine so needs 2 CH3I to become quaternary: the ammonium salt is [(Me2CHCH2)(CH2CH2Me)(Me)2N+] I-. Step 3 – Conversion to hydroxide with Ag2O. Step 4 – Hofmann elimination (pyrolysis): Hofmann elimination preferentially removes a proton from the beta-carbon that is least substituted (least hindered), giving the less substituted alkene. The nitrogen in the quaternary salt has four groups: 1. -CH2CH(Me)2 (isobutyl): the beta-carbon to N is -CH(Me)2, which bears 2 methyl substituents (tertiary H, more substituted) 2. -CH2CH2Me (n-propyl): the beta-carbon to N is -CH2Me, which bears 1 methyl (secondary H, less substituted) 3. Two -CH3 groups: no beta-H available for elimination (methyl groups have no beta-carbon relative to N in the elimination sense... actually CH3 attached to N has no beta-H). Hofmann's rule: elimination occurs toward the less substituted beta-carbon. Between chain A (beta-carbon is CHMe2, tertiary) and chain B (beta-carbon is CH2Me, secondary), Hofmann elimination favors chain B because it gives the less substituted alkene. Elimination from chain B (-CH2CH2Me): removes H from the -CH2Me (beta carbon) to give CH2=CHMe, i.e., Me-CH=CH2 (propene / 1-propene), with loss of the amine fragment. This matches option (a): Me-CH=CH2. Why other options fail: - (b) H2C=CH2 (ethylene) would require elimination from an ethyl chain, but there is no ethyl group directly; the n-propyl group gives propene, not ethylene. - (c) 2-methylpropene (Me2C=CH2) would be the Zaitsev product from the isobutyl chain, but Hofmann elimination avoids the more substituted alkene. - (d) CH3CH(Me)CH=CH2 (3-methyl-1-butene) would come from elimination at the terminal position of the isobutyl chain, but the isobutyl group's structure Me2CHCH2- has no beta-H on the terminal side that would give this product under Hofmann conditions. Therefore, the correct answer is A.