See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When an alkyne is treated with H2 (2 moles) over Pt catalyst, the triple bond is fully hydrogenated to give an alkane. The question asks which alkyne gives an optically INACTIVE alkane product. For a compound to be optically inactive, it must either: 1. Have no chiral center, OR 2. Have a chiral center but be a meso compound (internal plane of symmetry), OR 3. Be a racemic mixture (but since hydrogenation of an achiral alkyne gives a single achiral product in this context, we focus on symmetry). Step 1: Identify the alkane products after adding 2 moles of H2. (a) 3-Methyl-1-pentyne: CH≡C-CH(CH3)-CH2-CH3 After adding 2 H2: CH3-CH2-CH(CH3)-CH2-CH3 = 3-Methylpentane In 3-methylpentane, C3 has: CH3, CH2CH3, CH2CH3, H The two CH2CH3 groups are identical, so C3 is NOT a chiral center. Therefore, 3-methylpentane has no chiral center and is optically inactive. (b) 4-Methyl-1-hexyne: HC≡C-CH2-CH(CH3)-CH2-CH3 After adding 2 H2: CH3-CH2-CH2-CH(CH3)-CH2-CH3 = 4-Methylhexane C4 has: CH3, CH2CH3, CH2CH2CH3, H — all four groups are different, so C4 IS a chiral center. This compound is optically active. (c) 3-Methyl-1-heptyne: HC≡C-CH(CH3)-CH2-CH2-CH2-CH3 After adding 2 H2: CH3-CH2-CH(CH3)-CH2-CH2-CH2-CH3 = 3-Methylheptane C3 has: CH3, CH2CH3, CH2CH2CH2CH3, H — all four groups are different, so C3 IS a chiral center. This compound is optically active. Step 2: Why options (b) and (c) fail: Both 4-methylhexane and 3-methylheptane possess a true chiral carbon with four different substituents, making them optically active. Step 3: Option (a) gives 3-methylpentane, where C3 bears two identical ethyl groups, so it is not a chiral center. The molecule is symmetric and optically inactive. Therefore, the correct answer is A.