See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Concept: E2 (bimolecular elimination) reactions proceed through an anti-periplanar transition state in which C–H and C–X bonds break simultaneously. The nature of the leaving group and the extent of C=C double bond formation in the transition state (Hammond postulate / E2 spectrum) both influence regioselectivity. Step 2 – Evaluate statement (a): Leaving group ability correlates inversely with basicity of the leaving group (or directly with the acidity of its conjugate acid). The conjugate acid pKa values are: HI = –10, HBr = –9, HCl = –7, HF = 3.2. Lower pKa means stronger acid, meaning the conjugate base (halide) is a weaker base and therefore a better leaving group. Hence I- (pKa –10) is the best leaving group and F- (pKa 3.2) is the poorest. Statement (a) is TRUE. Step 3 – Evaluate statement (b): Zaitsev's rule predicts the more substituted (more stable) alkene is the major product. 2-hexene is more substituted than 1-hexene. For X = I (81% 2-hexene), X = Br (72% 2-hexene), and X = Cl (67% 2-hexene), the more substituted alkene predominates, so Zaitsev's rule is followed. Statement (b) is TRUE. Step 4 – Evaluate statement (c): F- has the highest pKa conjugate acid (3.2), meaning F- is the strongest base among the halogens listed and the poorest leaving group. Because F- is a poor leaving group, the C–F bond breaks less in the transition state (the transition state is more reactant-like / has less double bond character), consistent with the Hammond postulate applied to an E2 reaction where C–X bond breaking is less advanced. The data show that with F, Hofmann product (1-hexene, less substituted, 70%) predominates, indicating the TS geometry/electronics differ. Statement (c) is TRUE. Step 5 – Conclusion: Since statements (a), (b), and (c) are all true, the correct answer is (d). Therefore, the correct answer is D.