Oxidation of acetaldehyde with \(SeO_2\) forms:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Glyoxal
Selenium dioxide (\( \ce{SeO2} \)) is a selective oxidising agent. It does not attack the aldehyde group itself; it oxidises the CH3 or CH2 group next to a carbonyl group (the α-carbon) to a new C=O group, giving a 1,2-dicarbonyl compound. \( \ce{SeO2} \) is reduced to selenium.
In acetaldehyde the α-carbon is the \( \ce{CH3} \) group, which becomes a second CHO group:
\( \ce{CH3CHO + SeO2 -> OHC-CHO + Se + H2O} \)
The product, ethanedial (OHC–CHO), is called glyoxal.
Why the other options are incorrect
Option (1) Ethanoic acid — Incorrect. Formed when the CHO group is oxidised, for example by \( \ce{K2Cr2O7} \)/\( \ce{H^{+}} \), \( \ce{KMnO4} \) or Tollens’ reagent; \( \ce{SeO2} \) attacks the α-\( \ce{CH3} \) group instead.
Option (2) Methanoic acid — Incorrect. It has only one carbon; \( \ce{SeO2} \) oxidation does not break the C–C bond.
Option (4) Oxalic acid — Incorrect. \( \ce{(COOH)2} \) would need both carbons oxidised to the acid level; \( \ce{SeO2} \) stops at the dialdehyde.
Note: The question is from beyond the NCERT textbook.