$\mathop {{H_3}C\mathop C\limits^ \ominus {H_2}}\limits_{(i)} \,,$$\mathop {{H_2}C = \mathop C\limits^ \ominus H}\limits_{(ii)} $ અને $\mathop {H - C \equiv \mathop C\limits^ \ominus }\limits_{(iii)} $
Among alkane, alkene and alkyne are most acidic and alkanes are least acidic, so the order of base strengths
alkane \(>\) alkene \(>\) alkyne
\(C H_{3} \stackrel{\ominus}{C} H_{2}>H_{2} C=\stackrel{\ominus}{C} H>H-C \equiv {C}^{\ominus}\)

$(x)\begin{array}{*{20}{c}}
{O\,\,\,}\\
{||\,\,\,}\\
{C{H_3} - S - O - H}\\
{||\,\,\,\,}\\
{O\,\,\,\,}
\end{array}$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,O}\\
{\,\,\,\,\,\,||}\\
{(y)\,\,\,C{H_3} - C - O - H}
\end{array}$
$(z)\,\, CH_3 -OH$

$\mathop {C{H_3} - \mathop C\limits^ \oplus H - C{H_3}}\limits_I $
$\mathop {C{H_3} - \mathop C\limits^ \oplus H - OC{H_3}}\limits_{II} $
$\mathop {C{H_3} - \mathop C\limits^ \oplus H - C{H_2} - OC{H_3}}\limits_{III} $
