See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Intramolecular nucleophilic substitution (cyclization) of a dihalide with a primary amine. Step 1: Identify the reactants. The dihalide is Br-CH2-(CH2)2-CH2-Br, which is 1,4-dibromobutane (4-carbon chain with bromine at both ends). The amine is CH3NH2 (methylamine), a primary amine. Step 2: First substitution. Methylamine acts as a nucleophile and displaces one bromide via SN2, forming Br-CH2-(CH2)2-CH2-NH-CH3 (with HBr also formed or the free amine attacking). The nitrogen now bears one methyl group, one H, and is connected to the 4-carbon chain with a terminal Br. Step 3: Intramolecular cyclization. The nitrogen lone pair performs an intramolecular SN2 attack on the terminal carbon bearing the remaining bromine, displacing Br- and forming a 5-membered ring. The ring contains 4 carbons and 1 nitrogen. Step 4: Identify the product. The nitrogen carries the methyl group (from CH3NH2) and is part of the 5-membered ring with no N-H bond remaining after cyclization (since the two bonds from N go to the ring carbons and the methyl group, with the H lost as HBr or by deprotonation). This gives 1-methylpyrrolidine: a pyrrolidine ring with a methyl substituent on nitrogen (N-methyl group), no N-H. Step 5: Match to options. Option (b) shows a 5-membered ring with N bearing a methyl group (N-CH3) and no N-H, which is 1-methylpyrrolidine. This matches the product. Why other options fail: - Option (a): Shows pyrrolidine with N-H. This would be the product if NH3 (ammonia) were used instead of CH3NH2, since the methyl group would be absent. - Options (c) and (d): Show 2-methylpyrrolidine (methyl on carbon-2 of ring, not on nitrogen). The methyl group comes from CH3NH2 and ends up on nitrogen, not on the ring carbon. Therefore, the correct answer is B.