When ethylene is passed through alkaline KMnO4 solution 1, 2-Ethanediol is formed. The Purple color of KMnO4 decolorizes
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Ethyne on reacting with cold KMnO4 adds 4OH- to the triple bond. Forming a very unstable compound (OH)2CH--CH(OH)2. 2 OH- on a single carbon are unstable and the molecule would lose 2H2O and form Ethan-1,2-dial. This compund further is unstable and undergoes further oxidation to form oxalic acid.