$C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$


$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C = CH}
\end{array}\xrightarrow[{{H_2}O}]{{HgS{O_4},{H_2}S{O_4}}}X$ $\xrightarrow[{(ii)\,conc.{H_2}S{O_4}/\Delta }]{{(i)\,{C_2}{H_5}MgBr,{H_2}O}}Y$