`3`-Phenylpropene on reaction with `HBr` gives (as major product)

Options

A. `C_(6)H_(5)CH_(2)CH(Br)CH_(3)`
B. `C_(6)H_(5)CH(Br)CH_(2)CH_(3)`
C. `C_(6)H_(5)CH_(2)CH_(2)CH_(2)Br`
D. `C_(6)H_(5)CH(Br)CH=CH_(2).`

Answer

B

Explanation

According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.
`C_(6)H_(5)CH=CH-CH_(3)+Hbrrarroverset(Br)overset(|)(C_(6)H_(5)CHCH_(2)CH_(3)`