See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In elimination reactions, the nature of the base and the substrate determine which alkene is the major product. Hofmann elimination (bulky or poor base context) favors the less substituted alkene (anti-Zaitsev), while Zaitsev elimination (strong, small base) favors the more substituted alkene. Reaction 1: CH3CH2CHCH3 with +N(CH3)3 -OH (a quaternary ammonium hydroxide) undergoes Hofmann elimination upon heating. The leaving group is the trimethylamine, and the hydroxide acts as base. Because the nitrogen bears three methyl groups making approach to the more hindered beta-carbon difficult, and because Hofmann elimination rules apply to quaternary ammonium salts, the major product is the LESS substituted alkene. The beta-carbon with more hydrogens (the terminal CH3CH2- end) is preferentially deprotonated, giving CH3CH2CH=CH2 (1-butene), which is product Y. Reaction 2: CH3CH2CHCH3 with Br at C2 reacts with CH3CH2ONa (sodium ethoxide), a strong but relatively small base, undergoing E2 elimination following Zaitsev's rule. The base removes a proton from the more substituted beta-carbon, giving the more substituted and more stable alkene CH3CH=CHCH3 (2-butene), which is product X. Summary: - Reaction 1 (Hofmann elimination with quaternary ammonium salt): major product is Y (less substituted alkene, 1-butene) - Reaction 2 (Zaitsev elimination with sodium ethoxide): major product is X (more substituted alkene, 2-butene) This matches option (b): 1-Y, 2-X. Why other options fail: - (a) 1-X, 2-X: Incorrectly assigns Zaitsev product to reaction 1. - (c) 1-X, 2-Y: Both are reversed from the correct reasoning. - (d) 1-Y, 2-Y: Incorrectly assigns Hofmann product to reaction 2. Therefore, the correct answer is B.