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(D) The oxidation of acetylene $(HC \equiv CH)$ with alkaline $KMnO_4$ proceeds through the formation of glyoxal $(CHO-CHO)$ as an intermediate. Furth

Vedclass · Accepted Answer

oxalic acid

Vedclass · Answer

(D) The oxidation of acetylene $(HC \equiv CH)$ with alkaline $KMnO_4$ proceeds through the formation of glyoxal $(CHO-CHO)$ as an intermediate. Further oxidation of glyoxal leads to the formation of oxalic acid $(HOOC-COOH)$ as the major product. The reaction is: $HC \equiv CH + 2[O]$ $\xrightarrow{Alk. KMnO_4} CHO-CHO$ $\xrightarrow{2[O]} HOOC-COOH$ Therefore,the correct option is $(D)$.

The major product formed in the oxidation of acetylene by alkaline $KMnO_4$ is:

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