Question

Ethylene glycol gives oxalic acid on oxidation with

• A. acidified K\(_2\)Cr\(_2\)O\(_7\)
• B. acidified KMnO\(_4\)
• C. alkaline KMnO\(_4\)
• D. periodic acid

Hint:

Alkaline KMnO\(_4\) causes vigorous oxidation, converting diols into acids like oxalic acid.

Answer 1

The Correct Option is C

Step 1: Identify ethylene glycol.
Ethylene glycol is a dihydric alcohol:
\[ HO-CH_2-CH_2-OH \]
Step 2: Strong oxidation converts glycol to dicarboxylic acid.
Under strong oxidizing conditions, both terminal \(-CH_2OH\) groups are oxidized to \(-COOH\).
Step 3: Oxidant required for complete oxidation.
Alkaline KMnO\(_4\) is a strong oxidizing agent that oxidizes ethylene glycol to oxalic acid:
\[ HOCH_2CH_2OH \xrightarrow{alk. KMnO_4} HOOC-COOH \]
Step 4: Conclusion.
Thus alkaline KMnO\(_4\) gives oxalic acid.
Final Answer:
\[ \boxed{\text{(C) alkaline KMnO\(_4\)}} \]