Topic 22 of 34
First detect a carbonyl, then distinguish its class and methyl-carbonyl unit
No single test provides the complete structure.
A-Level 9476 (2026-2027)
First detect a carbonyl, then distinguish its class and methyl-carbonyl unit
No single test provides the complete structure.
| Test and conditions | Positive observation | What it supports |
|---|---|---|
| 2,4-DNPH reagent | Yellow/orange precipitate of a hydrazone derivative. | Aldehyde or ketone carbonyl; ordinary carboxylic acids, esters and amides do not give the same test. |
| Warm with Tollens reagent | Silver mirror or silver deposit. | An aldehyde among the ordinary aldehyde/ketone comparison set. |
| Warm with Fehling reagent | Blue solution gives brick-red Cu2O precipitate. | An ordinary aliphatic aldehyde; aromatic aldehydes such as benzaldehyde usually do not give this test. |
| Acidified oxidant with suitable warming | Dichromate orange to green or acidified manganate(VII) decolourises. | Aldehydes oxidise readily to acids; ordinary ketones resist mild oxidation. |
| Warm alkaline aqueous iodine | Yellow CHI3 precipitate. | A CH3CO- unit in the carbonyl compound, including ethanal where the other substituent is H. |
Aldehyde oxidation is RCHO + [O] → RCOOH in acidic conditions; in alkaline test reagents, the carboxylate is the organic oxidation product. Ketones do not have the aldehydic H and resist these mild oxidations without carbon-carbon cleavage. Tollens and Fehling observations are therefore complementary evidence, not universal tests for every possible reducing organic compound.
Worked example
Distinguish the three named carbonyl examples
Compare ethanal, propanone and phenylethanone using 2,4-DNPH, Tollens and alkaline iodine.
- All three contain aldehyde/ketone carbonyls, so all give a 2,4-DNPH precipitate.
- Ethanal is the aldehyde and gives the Tollens silver result; the two ketones do not under the ordinary test conditions.
- All three have a CH3CO unit: CH3CHO, CH3COCH3 and C6H5COCH3.
All three are iodoform-positive. A positive iodoform test therefore cannot by itself distinguish an aldehyde from a ketone.
The carbonyl iodoform transformation can be represented as RCOCH3 + 3I2 + 4OH- → RCOO- + CHI3 + 3I- + 3H2O. The methyl carbon becomes CHI3; the other carbonyl fragment becomes a carboxylate. Use that carbon accounting when deducing an unknown.