\(C_{6} H_{5} O H \frac{a q^{N a O H}}{C H C l_{3}} 2-O H-C_{6} H_{4}-C H O\)
$C{H_2} = C{H_2}\, \xrightarrow{{HOCl}} \,A\xrightarrow{{_R}}\begin{array}{*{20}{c}}
{C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_2}OH}
\end{array}$