Methyl alcohol on oxidation with acidified K2Cr2O7 gives__________?
Correct answer: C. HCOOH
- A. CH3COOH
- B. CH3CHO
- C. HCOOH
- D. CH3COCH3
Explanation
Methyl alcohol, CH3OH, is a primary alcohol and strong oxidation with acidified K2Cr2O7 carries it through methanal to methanoic acid, HCOOH. The oxidation sequence is CH3OH -> HCHO -> HCOOH, so the one-carbon chain remains intact. Students may choose CH3CHO by applying the aldehyde product rule but overlook that further oxidation continues under these conditions.
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About Alcohols
Alcohols contain a hydroxyl group attached to a saturated carbon atom and are classified as primary, secondary or tertiary. Their nomenclature, hydrogen bonding, acidity, oxidation, dehydration, reaction with sodium and conversion to haloalkanes are covered, along with the distinction between alcohols and phenols, where the hydroxyl group is attached directly to an aromatic ring.
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