$Br - C{H_2} = CH = CH - C{H_3} + \begin{array}{*{20}{c}}
{C{H_2} = CH - CH - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,Br}
\end{array}$
${C_3}{H_7}OH\xrightarrow{{conc\,{H_2}S{O_4}}}X\xrightarrow{{B{r_2}}}Y\xrightarrow[{alkolic\,KOH}]{{high\,level}}Z$
$[Figure]$ $\xrightarrow{{HBr}}$