$I. \,CH_3 - H$
$II.\, CH_3CH_2-H$
$III.\, CH_2 = CH - CH_2-H$
$IV.\, C_6H_5$
${\left( {C{H_3}} \right)_3}C - CH = C{H_2}\xrightarrow[{\left( {ii} \right)\,NaB{H_4} + NaOH}]{{\left( i \right)\,Hg{{\left( {C{H_3}COO} \right)}_2};\,THF}}\,?$