See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Hofmann Exhaustive Methylation (Hofmann Elimination) is used to open cyclic amines by repeated exhaustive methylation followed by elimination. Each Hofmann elimination round opens or shortens the ring by eliminating the nitrogen as trimethylamine and producing a terminal alkene. Step 1: Identify the final product. The final alkene is H2C=CH-CH(CH3)-CH=CH2, which is 3-methyl-1,4-pentadiene (penta-1,4-dien-3-ylmethane), i.e., a five-carbon chain with double bonds at C1-C2 and C4-C5, and a methyl group at C3. Step 2: Work backward. Two rounds of Hofmann elimination are performed (A->B->final product). Each elimination removes NMe3 and forms a C=C bond at the site of the C-N bond. Step 3: The second Hofmann elimination (B -> final product) produces H2C=CH-CH(CH3)-CH=CH2. This diene has two terminal vinyl groups attached to a central CH(CH3). This means B is an open-chain amine where nitrogen is attached at one end of a chain that, upon elimination, gives both double bonds. Since the second step also uses excess CH3I (making a quaternary ammonium salt), B must be a secondary or tertiary amine. After the second exhaustive methylation and elimination of NMe3, we get the diene. The carbon skeleton of the diene (5 carbons + methyl branch at C3) tells us B must be: (CH3)2N-CH2-CH=CH-CH(CH3)-CH2... Actually, let us think more carefully. Step 4: The product has carbons: C1=C2-C3(CH3)-C4=C5. Working back from the second elimination: the nitrogen was attached to either C1 or C5 (terminal carbon) giving one double bond, and the other double bond came from the ring opening in step 1. Since two eliminations are done sequentially on a cyclic amine, after the first Hofmann elimination the ring opens to give an open-chain amino alkene (B). In the second elimination, B loses NMe3 to give the diene. Step 5: If B is an open-chain amine like (CH3)2N-CH2-CH2-CH(CH3)-CH=CH2 (from ring opening of 4-methylpiperidine), then the second Hofmann elimination would give H2C=CH-CH(CH3)-CH=CH2 by eliminating from the N-CH2 end to form the second double bond. This matches the final product perfectly. Step 6: Starting material A = 4-methylpiperidine. First Hofmann elimination: 4-methylpiperidine is exhaustively methylated to give N,N,N-trimethyl-4-methylpiperidinium salt, then Ag2O/H2O converts to hydroxide, and heating causes elimination to open the ring, giving an open-chain dimethylaminoalkene (B) with the structure: (CH3)2N-CH2CH2-CH(CH3)-CH=CH2 (or similar open-chain amino-alkene). Second Hofmann elimination of B gives H2C=CH-CH(CH3)-CH=CH2. Step 7: Why other options fail: - (a) Piperidine (no methyl substituent): Would give penta-1,4-diene (H2C=CH-CH2-CH=CH2) without the methyl branch, not matching the product. - (b) 3-methylpiperidine: Ring opening would place the methyl branch at C2 or C3 of the open chain in an asymmetric position, giving a different substitution pattern. - (d) 2-methylpiperidine: The methyl group at C2 (adjacent to N) would give a different diene after two eliminations, not the symmetric 3-methyl-1,4-pentadiene. - Only 4-methylpiperidine (option c) has the methyl group at the central carbon (C4 of ring = C3 of the open chain after ring opening), which after two Hofmann eliminations gives the symmetric product H2C=CH-CH(CH3)-CH=CH2. Therefore, the correct answer is C.