A secondary alkanol can be oxidized to give an?

1. Analyze the question You're asked what product forms when a secondary alkanol (alcohol) is oxidized.

2. Explain the concept The product of oxidizing an alcohol depends on its class:

| Alcohol type | Structure | Oxidation product | |---|---|---| | Primary | R-CH₂OH | Alkanal (aldehyde), then alkanoic acid with further oxidation | | Secondary | R₂CH-OH | Alkanone (ketone) | | Tertiary | R₃C-OH | Not easily oxidized (no H on the carbon bearing the OH) |

Oxidation works by removing the hydrogen from the -OH group and the hydrogen on the carbon bearing the -OH. A secondary alcohol has exactly one hydrogen on that carbon, so once it is removed a C=O forms, and nothing more can be removed without breaking C-C bonds.

3. Step-by-step solution Take propan-2-ol, a typical secondary alkanol:

CH₃-CH(OH)-CH₃ --[K₂Cr₂O₇/H₂SO₄ or KMnO₄/H⁺]--> CH₃-CO-CH₃ + H₂O

• The product, propanone (acetone), is an alkanone.
• Alkynes (B) come from elimination or dehydrohalogenation, not oxidation.
• Alkanoic acids (C) come from oxidizing primary alcohols or aldehydes.
• Alkanals (D) come from the mild oxidation of primary alcohols.

4. Answer The correct option is A. alkanone.

5. Exam tip Remember: "1° → aldehyde → acid; 2° → ketone; 3° → no reaction." The number of hydrogens on the carbon carrying the -OH decides how far the oxidation can go.