See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Saytzeff's rule states that in elimination reactions, the more substituted (more stable) alkene is the major product. Hofmann's rule applies when bulky bases or specific leaving groups (like NR3+) force formation of the less substituted alkene. Step 1 - Analyze each option for whether Saytzeff or Hofmann product predominates: Option (a): This is an E2 elimination of a quaternary ammonium salt (Hofmann elimination). Ammonium salts are poor leaving groups and follow Hofmann's rule, giving the LESS substituted alkene as major product. So Saytzeff product is NOT major here. Option (b): CH3CH2CH2-CH(F)-CH3 with EtO- (moderately sized base). Fluorine is a poor leaving group in E2 due to the strong C-F bond; also, F tends to give Hofmann-type selectivity. The less substituted alkene tends to be favored. Saytzeff product is NOT reliably major here. Option (c): CH3-CH2-C(CH3)(Br)-CH3 with t-BuOK (bulky base). A bulky base like t-BuOK favors the less hindered (less substituted) proton abstraction, giving the LESS substituted (Hofmann) alkene as the major product. So Saytzeff product is NOT major here. Option (d): CH3CH2CH2-C(Br)(CH3)-CH3 with CH3OK (small, non-bulky base). CH3O- is a small base, and with a simple alkyl bromide leaving group, Saytzeff's rule applies. The substrate has a tertiary bromide. Possible alkenes: removing H from the CH2CH2 side gives a more substituted (trisubstituted) alkene, while removing H from one of the CH3 groups gives a less substituted (disubstituted) alkene. With a small base (CH3OK), the more substituted Saytzeff alkene is the major product. Step 2 - Confirm: Only option (d) uses a small base (CH3O-) with a normal alkyl bromide leaving group under conditions that favor Saytzeff elimination. Therefore, the correct answer is D.