The following reaction is an example of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Cannizzaro reaction
Aldehydes that have no \( \alpha \)-hydrogen atom undergo self oxidation and reduction (disproportionation) when heated with concentrated alkali: one molecule of aldehyde is reduced to the alcohol and another is oxidised to the salt of the carboxylic acid. This is the Cannizzaro reaction.
Benzaldehyde has no \( \alpha \)-hydrogen (the CHO group is attached directly to the ring), so with concentrated NaOH it gives benzyl alcohol and sodium benzoate:
\( \displaystyle \ce{2C6H5CHO + NaOH} \xrightarrow[\Delta]{\text{conc.}} \ce{C6H5CH2OH + C6H5COONa} \)
The reaction shown is therefore a Cannizzaro reaction.
Why the other options are incorrect
Option (2) Hell-Volhard Zelinsky reaction — Incorrect. In this reaction a carboxylic acid having an \( \alpha \)-hydrogen is halogenated at the \( \alpha \)-carbon by \( \ce{Cl2} \) or \( \ce{Br2} \) with red phosphorus: \( \ce{RCH2COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}/\text{red P}} \ce{RCH(X)COOH} \).
Option (3) Friedel-Craft reaction — Incorrect. This is the alkylation or acylation of an aromatic ring with an alkyl or acyl halide and anhydrous \( \ce{AlCl3} \), e.g. \( \ce{C6H6 + CH3COCl -> C6H5COCH3 + HCl} \).
Option (4) Gatterman-Koch reaction — Incorrect. This makes benzaldehyde from benzene, carbon monoxide and HCl with anhydrous \( \ce{AlCl3} \)/CuCl: \( \ce{C6H6 + CO} \xrightarrow{\ce{HCl},\ \text{anhyd. }\ce{AlCl3}/\ce{CuCl}} \ce{C6H5CHO} \). It forms an aldehyde rather than using one up.
NCERT (2026-27 edition):
Cannizzaro Reaction
