Optically active isomer (A) of (C_5H_9Cl) on | vedclass

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(D) Compound $(A)$ is $CH_3-CH_2-CH(Cl)-CH=CH_2$ ($3$-chloro$-1-$pentene).It is optically active because the third carbon atom is a chiral center (att

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$CH_3-CH_2-CH(Cl)-CH=CH_2$

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(D) Compound $(A)$ is $CH_3-CH_2-CH(Cl)-CH=CH_2$ ($3$-chloro$-1-$pentene).It is optically active because the third carbon atom is a chiral center (attached to $H$,$Cl$,$-CH_2CH_3$,and $-CH=CH_2$).Upon hydrogenation with one mole of $H_2$,it forms $CH_3-CH_2-CH(Cl)-CH_2-CH_3$ ($3$-chloropentane).In $3-$chloropentane,the third carbon is attached to two identical ethyl groups $(-CH_2CH_3)$,making the molecule achiral and thus optically inactive.

Optically active isomer $(A)$ of $(C_5H_9Cl)$ on treatment with one mole of $H_2$ gives an optically inactive compound $(B)$. Compound $(A)$ will be:

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