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Question

(A) The molecular formula of $(Y)$ is $C_6H_{14}$,which corresponds to an alkane. Since $(X)$ is a hydrocarbon that undergoes hydrogenation to form $C

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$3$-methyl-$1$-pentene

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(A) The molecular formula of $(Y)$ is $C_6H_{14}$,which corresponds to an alkane. Since $(X)$ is a hydrocarbon that undergoes hydrogenation to form $C_6H_{14}$,$(X)$ must be an alkene with the formula $C_6H_{12}$.$(X)$ is optically active,meaning it must contain a chiral center. In $3$-methyl-$1$-pentene $(CH_2=CH-CH(CH_3)-CH_2-CH_3)$,the carbon at position $3$ is chiral because it is bonded to a hydrogen atom,a methyl group,an ethyl group,and a vinyl group.Upon catalytic hydrogenation,$3$-methyl-$1$-pentene forms $3$-methylpentane. In $3$-methylpentane,the carbon at position $3$ is bonded to two identical ethyl groups $(-CH_2CH_3)$,making it achiral. Thus,the product is optically inactive.

Catalytic hydrogenation of an optically active hydrocarbon $(X)$ yields an optically inactive compound $(Y), C_6H_{14}$. What is the hydrocarbon $(X)$?

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