Which one of the following on treatment with $50 \%$ aqueous sodium hydroxide yields the corresponding…
- $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$
- $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}$

- $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CHO}$
Solution
$2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}+\mathrm{NaOH} \stackrel{50 \% \mathrm{NaOH}}{\rightleftharpoons} \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$
Related Theory
Aldehydes containing no $\alpha$-hydrogen atom on warming with $50 \% \mathrm{NaOH}$ or $\mathrm{KOH}$ undergo disproportionation i.e. self oxidation reduction known as Cannizzaro's reaction.
Caution
Aldehydes that contains alpha hydrogen atoms are not involved in Cannizzaro reaction.
Asked in: NEET 2007
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