Reduction, oxidation and tests to distinguish aldehydes from ketones
Aldehydes and ketones are reduced to alcohols or hydrocarbons. Aldehydes are easily oxidised, so Tollens' and Fehling's tests tell them apart. The iodoform test picks out CH₃CO- compounds.
Reduction
| Reagent and condition | What happens | Example |
|---|---|---|
| NaBH₄ or LiAlH₄, or H₂ with Ni, Pt or Pd | Aldehyde → primary alcohol; ketone → secondary alcohol | CH₃COCH₃ → (NaBH₄) CH₃CH(OH)CH₃ |
| Clemmensen: Zn-Hg + conc. HCl | >C=O → >CH₂ (a hydrocarbon) | CH₃COCH₃ → CH₃CH₂CH₃ |
| Wolff-Kishner: NH₂NH₂, then KOH in ethylene glycol, heat | >C=O → >C=NNH₂ → >CH₂ + N₂ | C₆H₅COCH₃ → C₆H₅CH₂CH₃ |
Clemmensen vs Wolff-Kishner
Both give a hydrocarbon. NaBH₄ and LiAlH₄ give only an alcohol.
Oxidation
- 1★Aldehydes: RCHO → [O] RCOOHEasily oxidised by HNO₃, KMnO₄ and K₂Cr₂O₇, and even by mild oxidants such as Tollens' and Fehling's reagents. The H on the C=O carbon is replaced without breaking any C-C bond
- 2Ketones: R-CH₂-CO-CH₂-R' → (strong oxidant, heat) RCOOH + R'CH₂COOH, or RCH₂COOH + R'COOHOnly vigorous conditions oxidise ketones. A C-C bond next to the C=O breaks, giving a mixture of acids with fewer carbons
★ key step
Tests that tell aldehydes from ketones
- 1★Tollens' test: RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 2H₂O + 4NH₃Warm the aldehyde with freshly prepared ammoniacal silver nitrate: a bright silver mirror forms. Aliphatic and aromatic aldehydes respond; ketones do not
- 2★Fehling's test: RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O + 3H₂OMix Fehling's A and B in equal amounts, add the aldehyde and heat: a red-brown precipitate of Cu₂O. Aromatic aldehydes such as benzaldehyde do not respond
- 3Haloform reaction: R-CO-CH₃ + 3NaOX → R-COONa + CHX₃ + 2NaOHMethyl ketones (and ethanal) are oxidised by sodium hypohalite to the salt of an acid with one carbon fewer. A C=C double bond in the molecule is not attacked
- 4★Iodoform test: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂OA yellow precipitate of iodoform with a typical smell shows a CH₃CO- group, or a CH₃CH(OH)- group that is first oxidised to CH₃CO-
★ key step
Who gives the iodoform test?
| Compound | CH₃CO- or CH₃CH(OH)- present? | Yellow CHI₃? |
|---|---|---|
| Ethanal, CH₃CHO | Yes (CH₃CO-H) | Yes |
| Propanone, butan-2-one, pentan-2-one | Yes | Yes |
| Acetophenone, C₆H₅COCH₃ | Yes | Yes |
| Ethanol, propan-2-ol | Yes (CH₃CH(OH)-) | Yes |
| Methanal, propanal, benzaldehyde | No | No |
| Pentan-3-one, benzophenone | No | No |
| Methanol, propan-1-ol | No | No |
Worked example: identify the compound
Compound A (C₅H₁₀O) forms a 2,4-DNP derivative and does not reduce Tollens' reagent. It gives a yellow precipitate when warmed with iodine and NaOH. Clemmensen reduction of A gives n-pentane. Identify A.
- A forms a 2,4-DNP derivative, so it has a C=O group.
- A does not reduce Tollens' reagent, so it is a ketone, not an aldehyde.
- A gives iodoform, so it has a CH₃CO- group: a methyl ketone.
- Clemmensen reduction turns >C=O into >CH₂ without changing the carbon chain. n-Pentane is a straight chain, so A has no branch.
- A straight-chain C₅ methyl ketone is CH₃COCH₂CH₂CH₃. (3-Methylbutan-2-one would give 2-methylbutane instead.)
A is pentan-2-one, CH₃COCH₂CH₂CH₃.
Doctors once checked urine sugar with Fehling's (or Benedict's) solution: the aldehyde group of glucose reduces Cu²⁺ to red-brown Cu₂O. The same chemistry as Tollens' test was used to silver glass mirrors.
2 marks: “Write the Clemmensen and Wolff-Kishner reductions of propanone.” Clemmensen: CH₃COCH₃ → (Zn-Hg, conc. HCl) CH₃CH₂CH₃. Wolff-Kishner: CH₃COCH₃ → (NH₂NH₂) (CH₃)₂C=NNH₂ → (KOH, ethylene glycol, heat) CH₃CH₂CH₃ + N₂.
3 marks: “Give one chemical test to distinguish each pair: (i) butanal and butanone (ii) acetophenone and propiophenone, C₆H₅COCH₂CH₃ (iii) ethanal and benzaldehyde.” (i) Tollens' (silver mirror) or Fehling's (red-brown Cu₂O): butanal only. (ii) Iodoform (I₂/NaOH): yellow CHI₃ with acetophenone only; propiophenone has no CH₃CO- group. (iii) Fehling's: red-brown Cu₂O with ethanal only, as aromatic aldehydes do not respond (the iodoform test also works: ethanal only).
3 marks: “What is the haloform reaction? Write the iodoform reaction of propanone. Which of these give it: ethanol, methanol, pentan-2-one, pentan-3-one?” A methyl ketone is oxidised by NaOX to RCOONa + CHX₃. CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O. Ethanol and pentan-2-one give it; methanol and pentan-3-one do not.
Tollens' = silver mirror, Fehling's = red-brown Cu₂O, iodoform = yellow CHI₃. Ketones fail the first two; only CH₃CO- passes the third.