Propene on reaction with `Br_(2) "in" C Cl_(4)` yields

Options

A. `3`-bromopropane
B. cis-`1,2`-dibromopropane
C. trans-`1,2`-dibromopropane
D. `(+-)1,2`-dibromopropane

Answer

D

Explanation

`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.