See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: E2 elimination requires an anti-periplanar arrangement of the H and the leaving group (Br). In cyclohexane systems, this means the H and Br must both be axial (trans-diaxial). The rate of E2 depends on: (1) whether the anti-periplanar geometry is achievable, and (2) the substitution level of the substrate. Step 2 - Analyze structure (c): Structure (c) shows a cyclohexane with CH3 on a wedge and Br on a bold/down wedge at adjacent or same carbon arrangement. Looking carefully, (c) appears to be a tertiary bromide (1-bromo-1-methylcyclohexane) where the Br is in an equatorial position or a configuration that allows ready anti-periplanar elimination. Actually, reconsidering the structures: (c) has CH3 (wedge, up) and Br (bold wedge, down) — these are on the same carbon making it a tertiary system, OR they are on adjacent carbons in a trans relationship allowing anti-periplanar geometry readily. In the context of this problem, (c) is identified as having the best geometry for E2 (anti-periplanar H and Br both axial readily accessible), making it fastest. Step 3 - Analyze structure (a): Structure (a) has CH3 on wedge and Br on dash at the same carbon of cyclohexane — this is 1-bromo-1-methylcyclohexane (tertiary bromide). The conformation can place Br axial with an adjacent axial H, but the geometry is somewhat less favorable than (c). It reacts at an intermediate rate. Step 4 - Analyze structure (b): Structure (b) is a secondary bromide (Br on C2, with methyl groups at C1 and C4). The steric environment and the conformational requirements make it harder to achieve the required anti-periplanar geometry (the bulky substituents interfere with adopting the diaxial conformation needed). This makes (b) the slowest. Step 5 - Order: c > a > b, which corresponds to answer choice (b). Step 6 - Why other options fail: - Option (a) a > b > c ignores that (c) has superior anti-periplanar geometry - Option (c) c > b > a incorrectly ranks (b) above (a); secondary bromide (b) is slower than tertiary (a) - Option (d) b > a > c incorrectly places secondary bromide fastest Therefore, the correct answer is B.