See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is 1-methyl-3-chloro-3-isopropylcyclohexane. The Cl is at C3 (tertiary carbon, bearing both the Cl and the isopropyl group). NaOEt is a strong, bulky base that promotes E2 elimination. Step 2 - Identify possible beta-hydrogens for E2: From the tertiary C3 bearing Cl, the adjacent carbons (C2 and C4) carry beta-hydrogens. Elimination toward C2 gives a double bond between C2 and C3 (endocyclic), and elimination toward C4 gives a double bond between C3 and C4 (endocyclic). There is no gamma-hydrogen pathway for exocyclic alkene from these positions under normal E2. Step 3 - Zaitsev's rule and alkene stability: E2 with NaOEt (moderately bulky) follows Zaitsev's rule, favoring the more substituted (more stable) alkene. The double bond between C3 and C4 in the ring, where C3 also bears the isopropyl group, gives a trisubstituted endocyclic alkene. The double bond between C2 and C3 gives a less substituted endocyclic alkene. Step 4 - Stereochemical requirement of E2: E2 requires anti-periplanar geometry between the H and the leaving group (Cl). The conformation of the cyclohexane ring and the stereochemistry (both CH3 at C1 and Cl at C3 on wedge, isopropyl on dash at C3) determines which beta-H is anti to Cl. The anti-periplanar H at C4 relative to the Cl at C3 satisfies this requirement, giving elimination toward C4. Step 5 - Product identification: Elimination of HCl using the H at C4 gives a cyclohexene with the double bond between C3 and C4, retaining the CH3 (wedge, up) at C1 and the isopropyl group adjacent to the double bond. This corresponds to option (a): 1-methyl-4-isopropylcyclohex-3-ene (p-menth-3-ene type) with CH3 up at C1. Step 6 - Why other options fail: Option (b) shows a different regiochemistry of the double bond inconsistent with the preferred anti-periplanar elimination. Options (c) and (d) show exocyclic double bonds (=CMe or =CH2), which would require elimination of a methyl or isopropyl C-H, not a ring C-H; these are not favored under standard E2 conditions with NaOEt. Therefore, the correct answer is A.