$(I)\,\,CH_2 =CH-CH_2Cl$ $(II)\,\,CH_2=CH-CH(CH_3) Cl$
$(III)\,\,CH_2 =C(CH_3)CH_2Cl$ $(IV)\,\, CH_3CH = CH-CH_2Cl$
$\begin{array}{*{20}{c}}
{C{H_3}} \\
|\,\,\,\,\,\, \\
{C{H_{_3}} - C - C{H_2}Br} \\
|\,\,\,\,\,\, \\
H\,\,\,\,
\end{array}\,\xrightarrow[{C{H_3}OH}]{{C{H_3}{O^ - }}}$