An optically active hydrocarbon X has molecular formula $C_{6} H_{12}$ , X on catalytic hydrogenation given optically inactive $C_{6} H_{14}$ , X could be:

3-methyl-1-pentene

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3-methyl-2-pentene

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4-methyl-2-pentene

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4-ethyl-1-butene

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Solution

The correct option is A 3-methyl-1-pentene
The optically active $C_{6} H_{1} 2$ hydrocarbon is 3-Methylpent-1-ene, having one chiral carbon shown in fig A.
On catalytic hydrogenation , the compound obtained is shown in fig B, and the molecule doesnot have chiral carbon.
The reaction is $C_{6} H_{1} 2 \to C_{6} H_{1} 4$ and this reaction takes place inthe presence of $H_{2}$ and Pd.
Answer is option A.

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