MCQ
The reaction, which forms nitric oxide, is
- A$C$ and ${N_2}O$
- B$Cu$ and ${N_2}O$
- C$Na$ and $N{H_3}$
- ✓$Cu$ and $HN{O_3}$
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$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_2} - C{H_2} - OH}
\end{array}\,$ ${\xrightarrow[{Pyridine{\kern 1pt} cold}]{{Cr{O_3}}}}$ Product


$(R = 8.314\,J\,{\rm{degre}}{{\rm{e}}^{{\rm{ - 1}}}}mo{l^{ - 1}})$
Compound $(A)$ exist in geometrical isomers and $(B)$ gives Cannizaro reaction. $(A)$ will be