\(Ph - CH _{2}- CH = CH - CH _{3}\) \(\xrightarrow{{(i)\,B{r_2}}}\)\(\begin{array}{*{20}{c}}
{\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}
\end{array}} \\
{Ph - C{H_2} - CH - CH - C{H_3}} \\
{|\,\,\,\,\,\,} \\
{Br\,\,\,}
\end{array}\) \(\xrightarrow[{ - 2HBr}]{{{\text{Alc}}{\text{. KOH}}}}\) \(Ph - CH _{2}- C \equiv C - CH _{3}\)
$C{H_2} = C{H_2}\,\xrightarrow{{{O_2}/Ag}}\,X\,\xrightarrow{{473\,K}}\,Y$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{C{H_3} - CH = CH - C{H_2} - CH - C{H_3}}
\end{array}$ $\longrightarrow $ $C{H_3} - CH = CH - C{H_2}C{O_2}H$