The product formed has two chirality centres, therefore, four stereoisomers \(\left(2^{2}\right)\) will exist.
\(\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C{H_2} - C = CH - C{H_3}}
\end{array}\) \(\xrightarrow[{Peroxide}]{{HBr}}\) \(\mathop {\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
|
\end{array}} \\
{C{H_3} - C{H_2} - {}_*CH - \mathop {CH}\limits^{*\,} - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br}
\end{array}}\limits_{Major\,\,product} \)
$\mathop {C{H_3} - }\limits_\delta \mathop {C{H_2} - }\limits_\gamma \mathop {CH = }\limits_\beta \mathop {C{H_2}}\limits_\alpha $
$(E)$

$(1) \,Cl_2 \to 2Cl^\bullet $
$(2)\, Cl^\bullet + CH_4 \to CH_3Cl + H^\bullet $
$(3)\, Cl^\bullet +CH_4 \to CH^\bullet _3 + HCl$
$(4)\, H^\bullet +Cl_2 \to HCl + Cl^\bullet $
$(5)\, CH^\bullet _3 + Cl_2 \to CH_3Cl + Cl^\bullet $
$n - Bu -\equiv\frac{(i) n-BuLi,n - C _{5} H _{11} Cl}{(ii) Lindlar\,\, cat, H _{2}}$