$Ph - CH _{2}- CH = CH - CH _{3}$ $\xrightarrow{{(i)\,B{r_2}}}$$\begin{array}{*{20}{c}}
  {\,\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} 
\end{array}} \\ 
  {Ph - C{H_2} - CH - CH - C{H_3}} \\ 
  {|\,\,\,\,\,\,} \\ 
  {Br\,\,\,} 
\end{array}$ $\xrightarrow[{ - 2HBr}]{{{\text{Alc}}{\text{. KOH}}}}$ $Ph - CH _{2}- C \equiv C - CH _{3}$