See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is a cyclohexane with a methyl group (wedge, equatorial or axial) at C1, a chloro group (wedge) at C3, and an isopropyl group at C4. The stereochemistry shows Cl and the adjacent H groups in specific orientations. Step 2 - E2 requirement: E2 elimination requires an anti-periplanar arrangement between the H being removed and the leaving group (Cl). In a cyclohexane ring, anti-periplanar means both must be axial (diaxial relationship). Step 3 - Conformational analysis: For E2 to occur, the Cl and the beta-H must be diaxial. The molecule must adopt a chair conformation where Cl is axial. When Cl is axial at C3, we examine which adjacent carbons (C2 and C4) have axial hydrogens available for elimination. Step 4 - Zaitsev's rule and substitution: E2 follows Zaitsev's rule preferentially, giving the more substituted (more stable) alkene. Elimination toward C4 (which bears the isopropyl group) would give a trisubstituted alkene (more substituted, endocyclic double bond at C3-C4), which is more stable than elimination toward C2 giving a disubstituted endocyclic alkene. However, the key constraint is the anti-periplanar geometry. Step 5 - Identify major product: When the conformation is arranged so that Cl at C3 is axial and the H at C4 (on the isopropyl-bearing carbon) is also axial (diaxial), elimination proceeds to give the endocyclic double bond between C3 and C4. With the methyl group at C1 (wedge) and isopropyl at C4, this produces a trisubstituted cyclohexene - matching structure (b), which shows the double bond conjugated/positioned between C3-C4 giving the more substituted alkene consistent with Zaitsev's rule and the anti-periplanar diaxial requirement. Step 6 - Elimination toward C2 would give a less substituted alkene (option a), and exocyclic products (c, d) would require breaking the C4-C(isopropyl) bond which is not the beta-carbon bearing H in context. Options (c) and (d) show exocyclic double bonds which would not arise from simple E2 on this substrate under these conditions. Option (a) gives the less substituted product. Option (b) gives the more substituted endocyclic alkene satisfying both diaxial anti-periplanar geometry and Zaitsev preference. Therefore, the correct answer is B.