${\left( {C{H_3}} \right)_3}C - CH = C{H_2}\xrightarrow[{\left( {ii} \right)\,NaB{H_4} + NaOH}]{{\left( i \right)\,Hg{{\left( {C{H_3}COO} \right)}_2};\,THF}}\,?$
$1.\,\,CH_3-C \equiv C -CH_3$
$2.\,\,CH_3 - CH_2 - CH_2 - CH_3$
$3. \,\,CH_3 - CH_2C \equiv CH$
$4.\,\,CH_3 - CH = CH_2$
$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\text { (iii) } \mathrm{HBr}{(iv) \mathrm{Mg}, ether, then \mathrm{HCHO} / \mathrm{H}_3 \mathrm{O}^{+}}]{{(i)BH_3}{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2,{ }^{\text {(-) }} \mathrm{OH}}} \mathrm{A}$