See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Alcoholic KOH (alc. KOH) promotes E2 elimination reactions. With 3 moles of alc. KOH acting on a hexachlorocyclohexane (benzene hexachloride isomer), three successive E2 eliminations occur, removing three HCl molecules (or three Cl2 equivalents in pairs) to ultimately give benzene (1,3,5-cyclohexatriene → benzene) as the final aromatic product. Step-by-step reasoning: 1. Each of the three reactions starts with a hexachlorocyclohexane (C6H6Cl6) - a cyclohexane ring bearing one Cl on each carbon (six Cl total). These are different stereoisomers of benzene hexachloride (BHC/HCH). 2. Alc. KOH promotes E2 elimination. With 3 moles of alc. KOH, three E2 eliminations occur sequentially, each removing one molecule of HCl from the ring. 3. First elimination: removes HCl → trichlorocyclohexene or proceeds further. Second elimination: removes another HCl → dichlorocyclohexadiene. Third elimination: removes another HCl → trichlorobenzene or continues to benzene. 4. However, the key insight is that regardless of the stereoisomer of hexachlorocyclohexane used, when 3 moles of alc. KOH are used, the final product after three eliminations is benzene (C6H6). This is because the driving force of aromatization is so strong that all three double bonds form to give the fully aromatic benzene ring, regardless of the initial stereochemistry of the Cl substituents. 5. All three isomers - whether all-axial (alpha-BHC), mixed configurations (beta, gamma, delta isomers) - upon treatment with 3 moles alc. KOH, ultimately yield benzene as the thermodynamic/aromatic product through three sequential dehydrochlorinations. 6. Since all three reactions (1), (2), and (3) produce benzene, A = B = C = benzene. Why other options fail: - Options (a), (b), (c) suggest that one of the products is different, which would only be true if the stereochemistry prevented some eliminations - but with sufficient base (3 moles) and the strong driving force of aromatization, all isomers proceed to benzene. - The stereochemistry of the starting material affects the rate but not the final product when enough base is provided and the aromatic product is accessible. Therefore, the correct answer is D.