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$\mathrm{CH}_{3}-\mathrm{CN} \stackrel{\mathrm{Na}_{2} \mathrm{C}_{5} \mathrm{OH}}{\longrightarrow} A \stackrel{\mathrm{HNO}_{2}}{\longrightarrow} B \underset{573 \mathrm{~K}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} C$ here $C$ is
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The correct answer is:
$\mathrm{CH}_{3}-\mathrm{CHO}$
The complete reaction takes place as follows :
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