b
(b) The formation of \(n-\)propyl bromide in presence of peroxide can be explained as follows.

Step\(-1:\) Peroxide undergo fission to give free radicals \(R - O - O - R \to 2 - R - \dot O\)

Step\(-2 :\) \(HBr\) combines with free radical to form bromine free radical \(R - \dot O + HBr \to R - OH + B\dot r\)

Step\(-3 :\) \(B\dot r\) attacks the double bond of the alkene to form a more stable free radical

Step\(-4 :\) More stable free radical attacks the \(HBr\)
\(C{H_3} - \dot CH - C{H_2} - Br + HBr \to \mathop {C{H_3}C{H_2}C{H_2}Br}\limits_{{\rm{n - propyl bromide}}} + B\dot r\)

Step\(-5 :\) \(B\dot r + B\dot r \to B{r_2}\)

\(C{H_3}CH = C{H_2} + Br \to \mathop {C{H_3} - CH - C{H_2}Br}\limits_{(more\,stable)} \)

\(C{H_3}CH = C{H_2} + Br \to \mathop {\begin{array}{*{20}{c}}
  {Br} \\ 
  | \\ 
  {C{H_3} - CH - C{H_2}} 
\end{array}}\limits_{(less\,stable)} \)