- ✓Acetone
- BAcetaldehyde
- CPropionaldehyde
- DFormaldehyde
$\begin{array}{*{20}{c}}
{C{H_3}} \\
{C{H_3}}
\end{array} > C = O\, + \,$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{H_2} - CH - C - C{H_3}}
\end{array}$ $\xrightarrow[{ - \,{H_2}O}]{{HCl}}$ $\mathop {{{(C{H_3})}_2}C = CH.COC{H_3}}\limits_{Mesityl\,\,oxide} $
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$C{{H}_{3}}COOEt+{{H}_{2}}O\xrightarrow{{{H}^{+}}}C{{H}_{3}}COOH+EtOH$

$\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
The standard reduction potentials are given as below $\left(\mathrm{E}_{\mathrm{red}}^{\circ}\right)$
$\mathrm{E}_{\mathrm{MmO}_4^{-} / \mathrm{Mm}^{2+}}^{\circ}=+1.51 \mathrm{~V}$
$\mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}$
If the equilibrium constant of the above reaction is given as $K_{\text {eq }}=10^x$, then the value of $x=$____ (nearest integer)