Correct Answer - D
`CH_(3)CH=CH_(2)underset(C Cl_(4))overset(Br_(2))rarroverset(3)(C )H_(3)underset(Br)underset(|)overset(2)(C )Hoverset(1)CH_(2)Br`
Note that `C-2` is asymmetrical. Thus, the product is optically active. Since the bromonium ion is symmetrical, each `C` atom is attacked equally well and , consequently, results in the formation of enantiomers in equal amounts.
