an organic compound "X" with a molecular formula C2H6O undergoes oxidation in the presence of alkaline KMNO4 and forms the compound 'Y',that has molecular formula C2H4O2.Identify 'X' AND 'Y'.
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Answer:
X:C2H5OH (ethanol or ethyl alcohol)
Y:CH3COOH (ehanoic acid or acetic acid)
Explanation:
Ethanol has a molecular formula of C2H6O. We know that when we do full oxidation of any alcohol with alkaline KMnO4 then it converts into carboxylic acid group. Here ethanol is belonging to alcohol group whereas ehanoic acid is belonging to carboxylic acid group. Therefore when we react ethanol with KMnO4(full oxidation process) then it changes into ethanoic acid which has a molecular formula of C2H4O2.
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