See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Determine x — number of HEM steps to remove nitrogen from quinolizidine. Quinolizidine is a bicyclic amine where nitrogen sits at the junction of two fused six-membered rings (a [4.4.0] bicyclic system with N at the bridgehead). To remove nitrogen via Hofmann Exhaustive Methylation, we must open all rings containing nitrogen. Concept: Each HEM cycle involves (a) methylation of N to quaternary ammonium salt, then (b) Hofmann elimination to give an alkene and a tertiary amine (or open-chain amine). For a bicyclic amine with N at the ring junction, two rings must be opened. - First HEM (methylation + elimination): Opens one ring. The nitrogen is still part of a ring (now monocyclic amine with a pendant alkene chain). This gives a monocyclic tertiary amine with an alkene. - Second HEM (methylation + elimination): Opens the second ring. Nitrogen is now in an open-chain trialkylamine (still attached). - Third HEM (methylation + elimination): Eliminates the nitrogen entirely as trimethylamine, giving a diene or open-chain product. For quinolizidine (bicyclic, N at junction of two 6-membered rings): it requires 3 HEM steps to completely remove nitrogen. Therefore x = 3. Step 2: Determine y — number of possible E2 products (including stereoisomers) from 2-bromobutane with alc. KOH. 2-Bromobutane is CH3-CHBr-CH2-CH3. E2 elimination can remove H from either the adjacent carbon on the left (C1, giving 1-butene) or the right (C3, giving 2-butene). - From C1 (removing H from CH3): gives 1-butene (CH2=CH-CH2-CH3). This is one product, no E/Z isomerism possible (terminal alkene). → 1 product. - From C3 (removing H from CH2): gives 2-butene (CH3-CH=CH-CH3). This can be E or Z (cis/trans). → 2 products (E-2-butene and Z-2-butene). Total y = 1 + 2 = 3. Step 3: Sum x + y = 3 + 3 = 6. Therefore, the correct answer is 6.