Optically active isomer (A) of `(C_5H_9 Cl) `on treatment with one mole of H2 gives an optically inactive compound (B) compound (A) will be :

Options

A. `CH_3 - underset(CH_2 Cl) underset(|) (CH) - CH =CH_2`
B. `Cl- underset(CH_3)underset(|)(CH)= CH -CH_3`
C. `CH_3 - underset(Cl) underset(|)(CH) -CH_2 -CH=CH_2`
D. `CH_3 -CH_2- underset(Cl) underset(|)(CH) - CH =CH_2`

Answer

D

Explanation

`CH_3 - CH_2 - underset(Cl) underset(|)CH - CH =CH_2 underset("pt")overset(H_2)to Et - underset(Cl) underset(|)(CH)-Et `optically inactive.