See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: An optically active compound has a chiral center (stereocenter). When treated with one mole of H2, the double bond is hydrogenated. The product (B) is optically inactive, meaning hydrogenation destroys the chirality — either by removing the chiral center or creating a meso compound or a compound with no stereocenters. Step 1: Identify the molecular formula. C5H9Cl has degree of unsaturation = (2×5 + 2 - 9)/2 = (10 + 2 - 9)/2 = 3/2... recalculating: (2×5 + 2 - 9 - 1)/2 = (10 + 2 - 10)/2 = 2/2 = 1. So one degree of unsaturation, meaning one double bond (with Cl counted, formula C5H9Cl: DBE = (2×5 + 2 - 9 - 1)/2 = (12-10)/2 = 1). Confirmed: one C=C double bond. Step 2: For compound (A) to be optically active, it must have a chiral center. After hydrogenation of the double bond (adding H2 across C=C), compound (B) must be optically inactive. Step 3: Analyze option (d): CH3 — CH2 — CH(Cl) — CH = CH2. The chiral center is C3 (bearing Cl, CH2CH3, CH=CH2, and H). This compound is optically active. On hydrogenation of CH=CH2, it becomes CH3 — CH2 — CH(Cl) — CH2 — CH3, which is 3-chloropentane. In 3-chloropentane, C3 bears Cl, H, and two identical ethyl groups (CH2CH3 on both sides), so C3 is no longer a stereocenter. Thus compound (B) is optically inactive. This satisfies both conditions. Step 4: Check why other options fail. - Option (a): CH3—CH(CH2Cl)—CH=CH2. The chiral center is C2. After hydrogenation: CH3—CH(CH2Cl)—CH2—CH3, which is 1-chloro-2-methylbutane. C2 bears CH3, CH2CH3, CH2Cl, H — all four groups are different, so C2 remains chiral. Product (B) would still be optically active. Fails. - Option (b): Cl—CH(CH3)—CH=CH—CH3. After hydrogenation: Cl—CH(CH3)—CH2—CH2—CH3, which is 2-chloro-2-methylpentane... actually it is 1-chloro-1-methylbutane = Cl—CH(CH3)—CH2CH2CH3. C1 bears Cl, CH3, CH2CH2CH3, H — all different, still chiral. Fails. - Option (c): CH3—CH(Cl)—CH2—CH=CH2. After hydrogenation: CH3—CH(Cl)—CH2—CH2—CH3, which is 2-chloropentane. C2 bears Cl, CH3, CH2CH2CH3, H — all different, still chiral. Product remains optically active. Fails. Therefore, the correct answer is D.