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Polymerization Isomerism

An optically ...

Question

An optically active hydrocarbon(X) on catalytic hydrogenation gives an optically inactive compound (Y), C $_{6}$ H $_{14}$ . The hydrocarbon (X) is-

A

3-methyl-1-pentene

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B

3-methyl-2-pentene

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C

2-ethyl-1-butene

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D

3-methylcyclopentene

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Solution

The correct option is A 3-methyl-1-pentene
The optically active hydrocarbon X is 3-methyl-1-pentene $C H_{2} = C H - C H (C H_{3}) C H_{2} C H_{3}$ . On catalytic hydrogenation, it forms, 3-methyl pentane $C H_{3} C H_{2} C H (C H_{3}) C H_{2} C H_{3}$ , which is optically inactive.

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