$(I) \;CH_2 = CH - OH $
$(II)\; CH_2 = CH - CH_2OH$
$(III)\;CH_3 - CH_2 - OH$
$(IV)\; \begin{matrix} C{{H}_{3}}-CH-C{{H}_{3}} \\ |\,\,\,\, \\ OH \\ \end{matrix}$
\(CH _{2}=C H-\stackrel{\oplus}C H_{2}>C H_{3}-\stackrel{\oplus}C H-C H_{3}>\) \(C H_{3}-\stackrel{\oplus}C H_{2}>C H_{2}=\stackrel{\oplus}C H\)




$1.$ $\begin{array}{*{20}{c}}
{C{H_3}C{H_2}CHC{H_3}{\mkern 1mu} } \\
{{\mkern 1mu} \,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\,^ + }N{{(C{H_3})}_3}^ - OH}
\end{array}$ $\xrightarrow{{heat}}$ $\mathop {C{H_3}CH = CHC{H_3}}\limits_X $ $+$ $\mathop {C{H_3}C{H_2}CH = C{H_2}}\limits_Y $
$2.$ $\begin{array}{*{20}{c}}
{C{H_3}C{H_2}CHC{H_3}} \\
{\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,Br}
\end{array}$ $\xrightarrow{{heat}}$ $\mathop {C{H_3}CH = CHC{H_3}}\limits_X $ $+$ $\mathop {C{H_3}C{H_2}CH = C{H_2}}\limits_Y $
$\begin{align}
(A)\,Nu+R-L\to \overset{\oplus }{\mathop{Nu}}\,-R+{{L}^{-}} \\
(B)\,R-{{L}^{\oplus }}\to {{R}^{\oplus }}+:L \\
\end{align}$