An alkene "A" on reaction with `O_3` and Zn gives propanone and ethanol in equimolar Addition of HCl to alkene "A" gives "B" as the product. The structure of product "B" is:

Options

A. `H_(3)C-underset(Cl)underset(|)(CH)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(CH)`
B. `Cl-CH_(2)-CH_(2)-underset(CH_3)underset(|)overset(CH_(3))overset(|)(CH)`
C.
D. `H_(3)C-CH_(2)-underset(Cl)underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`

Answer

D

Explanation

`A=CH_(3)-underset(CH_(3))underset(|)(C)=CH-CH_(3)`
`B=CH_(3)-underset(CH_(3))underset(|)overset(Cl)overset(|)(C)-CH_(2)-CH_(3)`
Formation of B from A is markonikoff rule addition by E.A.R mechanism