$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\text { (iii) } \mathrm{HBr}{(iv) \mathrm{Mg}, ether, then \mathrm{HCHO} / \mathrm{H}_3 \mathrm{O}^{+}}]{{(i)BH_3}{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2,{ }^{\text {(-) }} \mathrm{OH}}} \mathrm{A}$
$C{H_2} = C{H_2}\xrightarrow[{oxid}]{{Hypochloro}}$ $M\xrightarrow{R}\begin{array}{*{20}{c}}
{C{H_2} - OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2} - OH}
\end{array}$
$1.\,\,CH_3-C \equiv C -CH_3$
$2.\,\,CH_3 - CH_2 - CH_2 - CH_3$
$3. \,\,CH_3 - CH_2C \equiv CH$
$4.\,\,CH_3 - CH = CH_2$