$C{H_3}C{H_2}OH\mathop {\xrightarrow{{{I_2}}}}\limits_{NaOH} C{H_3}CHO\mathop {\xrightarrow{{{I_2}}}}\limits_{NaOH} C{I_3}CHO$ $\xrightarrow[{NaOH}]{{{H_2}O}}CH{I_3} + HCOONa$
ટોલ્યુઇન $\xrightarrow{{KMn{O_4}}}A\xrightarrow{{SOC{l_2}}}$ $B\xrightarrow[{BaS{O_4}}]{{{H_2}/Pd}}C$
તો નીપજ $C$ શું હશે ?
$A\xrightarrow[{(ii)\,\,Conc.\,{H_2}S{O_4}/\Delta }]{{{\text{(i)}}\,{\text{C}}{{\text{H}}_3}MgBr/{H_2}O}}$
$B\xrightarrow[{(ii)\,Zn/{H_2}O}]{{(i)\,{O_3}}}C + D$
$D\xrightarrow[\Delta ]{{Ba\left( {OH} \right)}}\begin{array}{*{20}{c}} {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ {{H_3}C - C} \end{array}$$\begin{array}{*{20}{c}} {\,\,\,\,\,\,\,O} \\ {\,\,\,\,\,||} \\ { = CH - C - C{H_3}} \end{array}$
$\begin{array}{*{20}{c}}
O\\
{||}\\
{C{H_3} - C - C{H_3}}
\end{array}$ $ + \begin{array}{*{20}{c}}
{C{H_2}OH}\\
{\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_2}OH}
\end{array}$ $\overset{HCl}{\longleftrightarrow}$ ?