Aldehydes, Ketones and Carboxylic Acids · Lesson 6 of 15
Reduction, Oxidation and Carbonyl Tests
“Use reaction selectivity and observations to distinguish aldehydes, ketones and methyl ketones.”
• Predict alcohols and hydrocarbons from carbonyl reduction. • Balance aldehyde oxidation tests in alkaline medium. • Identify methyl ketones using the haloform reaction. • Combine test results while recognising their limitations.
Reduction can retain oxygen or remove it
NaBH₄ or LiAlH₄ followed by appropriate work-up reduces aldehydes to primary alcohols and ketones to secondary alcohols. Catalytic hydrogenation with a suitable metal catalyst is another route. Hydride adds to the electrophilic carbon and the oxygen becomes an alkoxide before protonation. Thus CH₃CHO gives CH₃CH₂OH and cyclohexanone gives cyclohexanol. The oxygen is retained. NaBH₄ often permits selectivity for aldehyde or ketone over an ordinary ester; LiAlH₄ is more reactive and normally reduces esters too.
Clemmensen reduction uses zinc amalgam and strongly acidic HCl conditions to replace carbonyl oxygen overall by hydrogen. Wolff–Kishner reduction first forms a hydrazone with NH₂NH₂ and then uses strong base and heat in a high-boiling solvent such as ethylene glycol. An aldehyde RCHO gives RCH₃; a ketone RCOR′ gives RCH₂R′. Substrate sensitivity to strong acid or strong base helps choose the route. These are net deoxygenations; do not draw NaBH₄ as if it performed the same transformation.
Problem
Compare benzaldehyde with NaBH₄ and with Clemmensen conditions.
- 1.NaBH₄ followed by work-up adds hydrogen across C=O and produces C₆H₅CH₂OH, benzyl alcohol. The oxygen remains.
- 2.Zn(Hg)/HCl removes the carbonyl oxygen overall and gives C₆H₅CH₃, toluene. The original aldehyde H is retained and the carbon becomes CH₃.
Aldehydes oxidise readily; ordinary ketones resist mild tests
Oxidation of an aldehyde to an acid can occur without breaking its carbon skeleton. Ordinary ketone oxidation requires more demanding conditions and commonly cleaves C–C bonds adjacent to C=O, producing smaller acids or further oxidation products. This distinction underlies mild diagnostic tests. A negative mild test does not mean a ketone cannot ever be oxidised.
Tollens’ reagent contains the diamminesilver(I) complex in alkaline solution. On warming with a typical aldehyde, Ag(I) is reduced to metallic silver, observed as a silver mirror or dark silver deposit, while RCHO becomes RCOO⁻. Acid is obtained only after acidifying the carboxylate solution. Benzaldehyde responds, so aromatic character does not prevent this test.
Fehling’s reagent combines aqueous CuSO₄ with alkaline sodium potassium tartrate, which complexes copper. Many aliphatic aldehydes reduce Cu(II) on warming to a brick-red Cu₂O precipitate. Ordinary aromatic aldehydes such as benzaldehyde generally give a negative result under these conditions. The acid product again exists as carboxylate in the alkaline mixture. Therefore Tollens’ positive and Fehling’s negative is compatible with an aromatic aldehyde.
Haloform cleavage identifies a methyl carbonyl pattern
Methyl ketones have R–CO–CH₃. In I₂/NaOH the three α-hydrogens on the methyl are replaced by iodine, then hydroxide attacks the carbonyl and cleaves off the triiodomethyl group, which becomes CHI₃. A yellow precipitate of iodoform is the observation. The other organic fragment is RCOO⁻, with one carbon fewer than the starting ketone. This is not simple oxidation of the ketone to an acid with the same skeleton.
Ethanal is the important aldehyde exception because it contains CH₃CO– when written CH₃C(=O)H. Ethanol and secondary alcohols of the type CH₃CH(OH)R can also respond after oxidation to ethanal or a methyl ketone. Not every alcohol or ketone is positive. Pentan-2-one is positive; pentan-3-one is not. Benzophenone is negative; acetophenone is positive. An isolated positive test identifies a structural possibility, not a complete molecule.
| Test or reagent | Positive observation | What the result supports |
|---|---|---|
| 2,4-Dinitrophenylhydrazine | Coloured hydrazone solid | Ordinary aldehyde or ketone |
| Tollens’ reagent | Silver mirror/deposit | A reducible substrate, commonly aldehyde in this scope |
| Fehling’s reagent | Brick-red Cu₂O | Many aliphatic aldehydes; benzaldehyde usually negative |
| I₂/NaOH | Yellow CHI₃ | Methyl ketone, ethanal, or a suitable oxidisable alcohol |
| NaHCO₃ solution | CO₂ effervescence | An acid stronger than carbonic acid, often COOH |
These are comparisons within a defined set of possible compounds. Some other reducing molecules can respond to silver or copper tests, so do not turn “commonly aldehydes” into an unrestricted proof. A negative bromine-water or dilute permanganate result does not prove a molecule has no π bonds: benzene rings often persist under those mild conditions.
Problem
A compound C₈H₈O gives a 2,4-DNP derivative and iodoform, but not Tollens’ reagent. Strong oxidation gives C₇H₆O₂. Identify it.
- 1.The derivative supports C=O; the negative Tollens’ test within ordinary aldehydes/ketones favours a ketone. Iodoform then requires a methyl ketone RCOCH₃.
- 2.C₈H₈O can accommodate phenyl plus COCH₃: acetophenone, C₆H₅COCH₃. Its iodoform cleavage produces benzoate plus CHI₃. Acidification gives benzoic acid, C₇H₆O₂, consistent with the oxidation evidence.
- 3.The combined evidence supports acetophenone; no one observation on its own specifies it. Aromatic unsaturation also explains why it need not behave like an isolated alkene in mild decolourisation tests.
Quick Quiz
Quiz
What does cyclohexanone give with NaBH₄ then work-up?
Explanation: Hydride reduction retains oxygen as OH.
What is the aldehyde oxidation product in alkaline Tollens’ solution?
Explanation: The reaction medium keeps the product deprotonated until acidification.
Which compound gives an iodoform precipitate?
Explanation: Pentan-2-one contains CH₃CO–.
Which result is typical of benzaldehyde?
Explanation: The two oxidation tests have different substrate scope.
Which conditions deoxygenate a suitable ketone to a hydrocarbon?
Explanation: Wolff–Kishner reduction proceeds beyond hydrazone formation under basic heating.
Practice Problems
- Distinguish propanal and propanone, then ethanal and propanal.
- Distinguish acetophenone and benzophenone, and pentan-2-one and pentan-3-one.
- Distinguish phenol and benzoic acid, and benzoic acid and ethyl benzoate.
- Predict cyclohexanecarbaldehyde with Tollens’ reagent, excess ethanol/H⁺, and Zn(Hg)/HCl.
- Write the iodoform products of acetophenone with a balanced ionic equation.
- Reduce a molecule containing both ketone and ester selectively at the ketone.
Tollens’ reagent oxidises propanal with silver formation while ordinary propanone is negative. For ethanal versus propanal, use I₂/NaOH: ethanal gives yellow CHI₃ but propanal lacks the required CH₃CO pattern. A test must differ for the specific pair.
Key Takeaways
• NaBH₄ and LiAlH₄ reduce carbonyls to alcohols with different selectivity. • Clemmensen and Wolff–Kishner reactions remove carbonyl oxygen overall. • Alkaline oxidation produces carboxylate before acid work-up. • Tollens’ and Fehling’s tests have different scopes. • Iodoform formation requires a suitable CH₃CO pattern or its precursor. • Haloform cleavage reduces the acid-fragment carbon count by one. • Combine observations with formula and connectivity constraints.