$(x)\begin{array}{*{20}{c}}
{O\,\,\,}\\
{||\,\,\,}\\
{C{H_3} - S - O - H}\\
{||\,\,\,\,}\\
{O\,\,\,\,}
\end{array}$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,O}\\
{\,\,\,\,\,\,||}\\
{(y)\,\,\,C{H_3} - C - O - H}
\end{array}$
$(z)\,\, CH_3 -OH$
\(\mathop {\begin{array}{*{20}{c}}
O \\
{||} \\
{Me - S - {O^\Theta }} \\
{||} \\
O
\end{array}}\limits_{3 - eq.\,R.S.} \) \(>\) \(\mathop {\begin{array}{*{20}{c}}
O \\
{||} \\
{Me - C - {O^\Theta }}
\end{array}}\limits_{2 - eq.\,R.S.} \) \(>\) \(Me - {O^\Theta }\)
So Acidic strength order is \(x>y>z\) so \(\mathrm{pK}_{\mathrm{a}}\) is \(z>y>x\)
$a$ અને $b$ ની બંધ લંબાઈની તુલના કરો
$\begin{matrix}
\overset{\Theta }{\mathop{\overset{\centerdot \,\centerdot }{\mathop{C}}\,}}\,{{H}_{2}}-C-C{{H}_{3}} \\
|| \\
O \\
\end{matrix}$ અને $\begin{matrix}
C{{H}_{2}}=C-C{{H}_{3}} \\
| \\
:\underset{\Theta }{\mathop{\underset{\centerdot \,\centerdot }{\mathop{O}}\,}}\,: \\
\end{matrix}$