See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Hoffmann Exhaustive Methylation (HEM) involves: (1) methylation of the amine with excess CH3I to form a quaternary ammonium salt, then (2) Hofmann elimination (treatment with Ag2O/H2O then heat) to give the least substituted alkene (anti-Markovnikov product). Two sequential HEM steps are applied to each substrate. Concept: In each HEM step, the nitrogen is first fully methylated to a quaternary ammonium salt, then undergoes E2 elimination preferentially at the least hindered beta-carbon (Hofmann's rule), opening the ring. --- Case (d): Piperidine (unsubstituted) --- Step 1 HEM: Piperidine -> N,N,N-trimethylpiperidinium iodide -> elimination gives pent-4-en-1-yl-dimethylamine (5-N,N-dimethylaminopent-1-ene, i.e., CH2=CH-CH2-CH2-CH2-NMe2). Actually the ring opens to give N,N-dimethylpent-4-en-1-amine: (CH2=CHCH2CH2CH2NMe2). Step 2 HEM: The open-chain amine is methylated again to quaternary salt, then elimination gives penta-1,4-diene: CH2=CH-CH2-CH2-CH=CH2 ... wait, that is hexa-1,5-diene: H2C=CH-CH2-CH2-CH=CH2 (product q). But let me reconsider: piperidine ring has 5 carbons + 1 N. Ring opening gives a 5-carbon chain with terminal alkene and dimethylamino group at the other end. Ring opening of piperidine by HEM: N is between C2 and C6. Elimination occurs at the less hindered end, giving: (CH2=CH-CH2-CH2-CH2)-NMe2 = 5-(dimethylamino)pent-1-ene. Second HEM on this: methylate N to give [CH2=CHCH2CH2CH2NMe3]+, elimination gives CH2=CH-CH2-CH2-CH=CH2 = hexa-1,5-diene = product (q)? No - that is 6 carbons but piperidine only has 5 carbons in the ring... Piperidine has 5 CH2 groups + N. After ring opening: N,N-dimethylpent-4-en-1-amine = Me2N-CH2CH2CH2CH=CH2. Second HEM: [Me3N-CH2CH2CH2CH=CH2]+ -> eliminate to give CH2=CH-CH2-CH2-CH=CH2 = penta-1,4-diene? The chain is 5 carbons: H2C=CH-CH2-CH=CH2 = penta-1,4-diene = product (p). So (d) -> (p). This matches the given answer. --- Case (a): 4-Methylpiperidine --- The methyl is at C4 (para to N). Ring opening by first HEM: elimination at less hindered beta carbon gives an open-chain amine. The ring has C4 bearing a methyl. After two HEM steps, the product should be a diene with a methyl branch at the middle. The product assigned is (s): H2C=CH-CH(CH3)-CH=CH2 = 3-methylpenta-1,4-diene. First HEM on 4-methylpiperidine: opens ring to give N,N-dimethyl-4-methylpent-4-en-1-amine or similar. The methyl at C4 means elimination gives a branched intermediate. After two full HEM cycles, we get H2C=CH-CH(CH3)-CH=CH2 = (s). So (a) -> (s). Matches. --- Case (b): 3-Methylpiperidine --- Methyl at C3. After two HEM steps, Hofmann elimination (least substituted alkene preference) yields a product where the methyl ends up on the carbon bearing a double bond as a substituent on a non-terminal carbon. The product is (r): H2C=CH-CH2-C(CH3)=CH2 = 2-methylpenta-1,4-diene. So (b) -> (r). Matches. --- Case (c): 2-Methylpiperidine --- Methyl at C2 (alpha to N). After two HEM steps, the product is the longer diene (q): H2C=CH-CH2-CH2-CH=CH2 = hexa-1,5-diene. Wait, but piperidine only gives penta-1,4-diene. With a methyl at C2, after ring opening we get a 6-carbon chain (5 ring carbons + 1 methyl = 6 carbons total), giving hexa-1,5-diene = H2C=CH-CH2-CH2-CH=CH2 = product (q). So (c) -> (q). Matches. Summary: (a) 4-methylpiperidine -> double HEM -> (s) H2C=CH-CH(CH3)-CH=CH2 (b) 3-methylpiperidine -> double HEM -> (r) H2C=CH-CH2-C(CH3)=CH2 (c) 2-methylpiperidine -> double HEM -> (q) H2C=CH-CH2-CH2-CH=CH2 (d) piperidine -> double HEM -> (p) H2C=CH-CH2-CH=CH2 Therefore, the correct answer is {"a": ["S"], "b": ["R"], "c": ["Q"], "d": ["P"]}.