$Br - C{H_2} = CH = CH - C{H_3} + \begin{array}{*{20}{c}}
{C{H_2} = CH - CH - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,Br}
\end{array}$

$C{H_2} = C{H_2}\, \xrightarrow{{HOCl}} \,A\xrightarrow{{_R}}\begin{array}{*{20}{c}}
{C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2}OH}
\end{array}$