MCQ
Oxalic acid on being heated upto ${90\,^o}C$ with conc. ${H_2}S{O_4}$ forms
- A$HCOOH + C{O_2}$
- B$C{O_2} + {H_2}O$
- ✓$C{O_2} + CO + {H_2}O$
- D$HCOOH + CO$
✓
Answer
Correct option: C.
$C{O_2} + CO + {H_2}O$
c
$\begin{array}{*{20}{c}}
{COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{COOH}
\end{array}$ $ + {H_2}S{O_4}$ $\xrightarrow{{95\,{\,^o}C}}$ $CO + C{O_2} + {H_2}O$
$\begin{array}{*{20}{c}}
{COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{COOH}
\end{array}$ $ + {H_2}S{O_4}$ $\xrightarrow{{95\,{\,^o}C}}$ $CO + C{O_2} + {H_2}O$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Copper $(II)$ ions gives reddish brown precipitate with potassium ferrocyanide. The formula of the precipitate is
→In acidic medium, $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}$, shows oxidising action as represented in the half reaction
→$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{-} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O}$
$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and $\mathrm{A}$ are respectively are:
$N_2$ and $O_2$ are converted in to $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct ?
→Spectrum of $L{i^{2 + }}$ is similar to that of
→The compound ${C_4}{H_{10}}O$ can show
→The vapour pressure of two pure liquids $(A)$ and $(B)$ are $100$ and $80\,torr$ respectively. The total pressure of solution obtained by mixing $2\,mole$ of $(A)$ and $3\,mole$ of $(B)$ would be ........ $torr$.
→For the following graphs,
Choose from the options given below, the correct one regarding order of reaction is:
$(A)\xrightarrow[{\begin{array}{*{20}{c}}
{(2){\kern 1pt} A{g_2}O/{H_2}O} \\
{(3){\kern 1pt} \Delta }
\end{array}}]{{(1){\kern 1pt} \,excess\,{\kern 1pt} C{H_3}I/{K_2}C{O_3}}}(B)$ $\xrightarrow[{(iii){\kern 1pt} \Delta }]{{\begin{array}{*{20}{c}}
{(i){\kern 1pt} excess{\kern 1pt} C{H_3}I/{K_2}C{O_3}} \\
{(ii){\kern 1pt} A{g_2}O}
\end{array}}}$ $\begin{array}{*{20}{c}}
{{H_2}C = CH - CH - CH = C{H_2}} \\
{|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } \,\,\,\,\\
{{\mkern 1mu} {\mkern 1mu} C{H_3}}
\end{array}$
→{(2){\kern 1pt} A{g_2}O/{H_2}O} \\
{(3){\kern 1pt} \Delta }
\end{array}}]{{(1){\kern 1pt} \,excess\,{\kern 1pt} C{H_3}I/{K_2}C{O_3}}}(B)$ $\xrightarrow[{(iii){\kern 1pt} \Delta }]{{\begin{array}{*{20}{c}}
{(i){\kern 1pt} excess{\kern 1pt} C{H_3}I/{K_2}C{O_3}} \\
{(ii){\kern 1pt} A{g_2}O}
\end{array}}}$ $\begin{array}{*{20}{c}}
{{H_2}C = CH - CH - CH = C{H_2}} \\
{|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } \,\,\,\,\\
{{\mkern 1mu} {\mkern 1mu} C{H_3}}
\end{array}$
Identify $A$
If we use pyrene $(CCl_4)$ in the Riemer-Tiemann reaction in place of chloroform, the product formed is
→Which of the following compounds has a zero dipole moment ?