See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In E2 elimination reactions, the ratio of Saytzeff (more substituted/internal alkene) to Hoffmann (less substituted/terminal alkene) products depends on the nature of the leaving group and the base used. The key principle here involves the C–X bond length and the E2 transition state geometry. Step 1 – Saytzeff vs Hoffmann selectivity in E2: The Saytzeff product (more substituted alkene) is thermodynamically more stable and is favored when the leaving group departs more readily (i.e., is a better leaving group) and when the transition state is more product-like or when steric effects are minimal. Step 2 – Effect of leaving group on product distribution: For E2 reactions with a given base, the leaving group ability order is: I > Br > Cl > F. When the leaving group is a poor leaving group (like F), the C–X bond is very strong and has a short bond length. In the E2 transition state with a poor leaving group, the transition state is more 'carbanion-like' or has more E1cb character, meaning the base abstracts the proton to a greater extent before the leaving group departs. This leads to a more planar, less hindered transition state that favors removal of the more accessible (less hindered) proton, giving the Hoffmann (less substituted) product. Conversely, when the leaving group is excellent (like I), the C–X bond breaks more readily, the transition state is earlier (more reactant-like), and the more stable (Saytzeff) product is favored because the reaction follows the thermodynamic preference. Step 3 – Answer to Part A (maximum Saytzeff product): Maximum Saytzeff product is obtained when the leaving group is the best (most readily departing), which corresponds to X = -I (iodide is the best leaving group among the options). With iodide, the E2 reaction proceeds with the least E1cb character, and the more substituted (Saytzeff) alkene is the major product. Answer A: option (a) X = -I. Step 4 – Answer to Part B (Hoffmann product is major): Hoffmann product is favored when the leaving group is the poorest, i.e., when there is the most E1cb character in the transition state. Among the options, F is the worst leaving group (C–F bond is strongest, shortest). With X = -F, the transition state resembles E1cb (carbanion-like), and the base preferentially removes the less hindered (more accessible) proton, giving the less substituted (Hoffmann) alkene as the major product. Answer B: option (d) X = -F. Step 5 – Why other options fail for Part B: -I, -Br, -Cl are all better leaving groups than -F and would give increasing amounts of Saytzeff product relative to Hoffmann product. Only -F gives predominantly Hoffmann product due to its poor leaving group ability and the resulting E1cb-like transition state. Therefore, the correct answer is {"A": ["A"], "B": ["D"]}.