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Question

(B) The molecular formula $C_5H_9Cl$ indicates a degree of unsaturation of $2$. $X$ is $3$-chloropent-$1$-ene $(CH_3CH_2CH(Cl)CH=CH_2)$,which is optic

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$3-$chloropent$-1-$ene and $4-$chloropent$-1-$ene

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(B) The molecular formula $C_5H_9Cl$ indicates a degree of unsaturation of $2$. $X$ is $3$-chloropent-$1$-ene $(CH_3CH_2CH(Cl)CH=CH_2)$,which is optically active due to the chiral center at $C_3$. Upon hydrogenation with $1 \ mole$ of $H_2$,$X$ forms $3$-chloropentane $(CH_3CH_2CH(Cl)CH_2CH_3)$,which is optically inactive $(Z)$ due to the presence of a plane of symmetry. $Y$ is $4$-chloropent-$1$-ene $(CH_2=CHCH_2CH(Cl)CH_3)$,which is also optically active due to the chiral center at $C_4$. Upon hydrogenation with $1 \ mole$ of $H_2$,$Y$ forms $2$-chloropentane $(CH_3CH_2CH_2CH(Cl)CH_3)$,which is still optically active $(P)$ because the chiral center at $C_2$ remains intact. Thus,$X$ is $3$-chloropent-$1$-ene and $Y$ is $4$-chloropent-$1$-ene.

$X$ and $Y$ are optically active isomers having the molecular formula $C_5H_9Cl$. When treated with one mole of $H_2$,$X$ gets converted to an optically inactive compound $Z$,but $Y$ gives an optically active compound $P$. The structures of $X$ and $Y$ respectively are:

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