See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction: Alcoholic KOH promotes E2 elimination (dehydrohalogenation) of alkyl halides, removing HX to form alkenes. Step 2 - Identify the substrate: The compound has a secondary carbon bearing iodine (C–I). This carbon is bonded to: (a) a quaternary carbon (bearing cyclobutyl and two methyls), (b) a CH2 group connected to a methylcyclopentane ring (with defined stereochemistry), and (c) a CH(CH3) group connected to a methylcyclohexane ring (with defined stereochemistry). Step 3 - Identify beta-hydrogens available for E2 elimination: E2 requires an anti-periplanar beta-hydrogen. The beta carbons adjacent to C–I are: (i) The quaternary carbon on one side — this has NO beta-H (it is quaternary), so elimination cannot occur in that direction. (ii) The CH2 group on the cyclopentane side — this has 2 beta-H atoms, giving one type of alkene (toward the cyclopentylmethyl side). (iii) The CH(CH3) group on the cyclohexane side — this has 1 beta-H, giving one type of alkene (toward the cyclohexyl side). Step 4 - Count alkene products and their chirality: Product A (elimination toward CH2-cyclopentane side): Forms a trisubstituted alkene. The cyclopentane ring carbon bearing the methyl (originally a stereocenter) may or may not remain chiral. The cyclopentane ring methyl stereocenter is retained. Additionally, the cyclohexane-bearing carbon (CH-CH3) retains its stereocenters. Each product from this elimination pathway can exist as stereoisomers due to existing stereocenters in the cyclopentane and cyclohexane portions. Two chiral stereoisomers arise from this pathway. Product B (elimination toward CH(CH3)-cyclohexane side): Forms an alkene. The cyclopentane stereocenter is retained, and the cyclohexane stereocenters are retained (1,3-dimethylcyclohexane). Two chiral stereoisomers arise from this pathway. Step 5 - Total count: Each elimination direction gives 2 chiral products (due to retained stereocenters in the ring systems), and there are 2 possible directions of elimination, giving 2 + 2 = 4 chiral products total (counting stereoisomers). Step 6 - Why other options fail: (a) 2 — undercounts; ignores both elimination directions and their stereoisomers. (c) 6 and (d) 8 — overcount; there are only two viable beta-carbon sites and limited stereocenters retained per product. Therefore, the correct answer is B.