Hunsdiecker reaction involve free radical in intermediate steps.

Options

A. `CH_(3) - CH = CH_(2) + HBr to CH_(3)- underset(Br) underset(|) (C ) H- CH_(3)`
B.
C. `CH_(3) - CH = CH_(2) + HBr to CH_(3) - CH_(2) - CH_(2) - Br`
D. `CH_(3) CHO + NH_(2) OH overset(H^(+))(to) CH_(3) - CH = N - OH`

Answer

C

Explanation

Without intermediate reaction take place as under `CH_(3) - CH = CH_(2) + HBr to CH_(3) - overset(Br) overset(|) ( C) H - CH_(3)` (According to markownikoff.s rule)
But the halogen bonded with terminal carbon so it takes place in presence of peroxide by free radical mechanism
`R - underset("peroxide") (O - O) R to 2 R overset(.)( O) , HBr + overset(.) ( R) O to ROH + Br^(.)`
`CH_(3) - CH = CH_(2) + Br^(.) to CH_(3) - C overset(.) (H) - CH_(2) - Br`
`CH_(3) - overset(.) (C ) H - CH_(2) Br + HBr to CH_(3) - CH_(2) - CH_(2) Br + Br^(.)`