17.1 Aldehydes and ketones
- Syllabus
- 9701–2028–2029
- Topic
- 17.1
- Level
- AS
| Alcohol starting material | Oxidant and apparatus | Carbonyl product |
|---|---|---|
| primary, RCH₂OH | acidified K₂Cr₂O₇ or acidified KMnO₄; heat and distil | aldehyde, RCHO |
| secondary, R₂CHOH | acidified K₂Cr₂O₇ or acidified KMnO₄; heat and distil | ketone, R₂CO |
RCHX2OH+[O]RCHO+HX2O
RX2CHOH+[O]RX2CO+HX2O
Distillation separates the carbonyl product as it forms. This is essential for an aldehyde because leaving it in hot oxidising mixture allows further oxidation to a carboxylic acid; it also matches the stated preparation condition for ketones.
A tertiary alcohol does not produce an aldehyde or ketone by this oxidation pattern. Do not reflux a primary alcohol when the target is the aldehyde.
| Carbonyl starting material | Reagent and conditions | Product class | Structural change |
|---|---|---|---|
| aldehyde | NaBH₄ or LiAlH₄ | primary alcohol | C=O → CH–OH |
| ketone | NaBH₄ or LiAlH₄ | secondary alcohol | C=O → CH–OH |
| aldehyde or ketone | HCN, KCN catalyst, heat | hydroxynitrile | CN and OH add to the former carbonyl carbon |
CHX3CHO+2[H]CHX3CHX2OH
CHX3CHO+HCNCHX3CH(OH)CN
CHX3COCHX3+HCN(CHX3)X2C(OH)CN
The CN carbon becomes part of the product skeleton, so HCN addition increases the carbon count by one. Reduction adds no carbon: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol, not a tertiary alcohol.
The C=O bond is polar: carbon is δ+ and oxygen is δ−. CN⁻ is the nucleophile because the carbon lone pair can attack the electron-deficient carbonyl carbon.
Step 1: draw a full curly arrow from the lone pair on the carbon end of :C≡N⁻ to the carbonyl carbon, and simultaneously draw another from the C=O π bond to oxygen. This forms a tetrahedral alkoxide intermediate bearing O⁻ and CN.
Step 2: the O⁻ lone pair takes H from HCN; draw O⁻ → H and H–CN → CN. The hydroxynitrile forms and CN⁻ is regenerated, which explains KCN's catalytic role.
CHX3CHO+HCNKCN, heatCHX3CH(OH)CN
Attack occurs through carbon, giving a nitrile –C≡N, not through nitrogen. Curly arrows begin at electron pairs—a lone pair or π bond—not at the δ+ carbon or proton.
Add 2,4-dinitrophenylhydrazine reagent (2,4-DNPH) to the sample. Formation of a yellow or orange precipitate of a 2,4-dinitrophenylhydrazone is a positive result for an aldehyde or ketone C=O group.
The test establishes that one of these carbonyl classes is present, but both aldehydes and ketones respond. A second oxidation-based test is required to decide which class the unknown belongs to.
A positive 2,4-DNPH result does not identify the carbon skeleton and is not a general test for alcohols or carboxylic acids. Record the precipitate, not merely a vague colour change in solution.
| Warmed reagent | Aldehyde observation | Ketone observation | Meaning |
|---|---|---|---|
| Fehling’s solution | blue solution gives brick-red Cu₂O precipitate | remains blue; no precipitate | aldehyde is oxidised while Cu²⁺ is reduced |
| Tollens’ reagent | silver mirror / grey Ag deposit | no silver deposit | aldehyde is oxidised while Ag⁺ is reduced |
First establish a carbonyl with 2,4-DNPH, then use either mild oxidation test. A positive Fehling’s or Tollens’ result identifies an aldehyde; a valid negative result for a DNPH-positive compound supports a ketone.
Aldehydes are readily oxidised to carboxylates in these alkaline test reagents and to carboxylic acids on acidification. Ketones lack the aldehydic H and are not readily oxidised under the same mild conditions.
A negative result is useful only when fresh reagent, warming and a DNPH-confirmed carbonyl are established. It is not proof of a ketone for an arbitrary unknown.
Warm the aldehyde or ketone with alkaline aqueous iodine. A yellow precipitate of triiodomethane, CHI₃, is a positive result for the methyl-carbonyl group CH₃CO–R.
CHX3COR+3IX2+4OHX−CHIX3(s)+RCOX2X−+3IX−+3HX2O
The CH₃ side becomes CHI₃ and the remaining acyl fragment becomes RCO₂⁻. Propanone has R = CH₃, so it gives CHI₃ and ethanoate, CH₃CO₂⁻; ethanal has R = H and gives methanoate, HCO₂⁻.
Ethanal is the only aldehyde with the required CH₃CO– arrangement; methyl ketones also qualify. Propanal, CH₃CH₂CHO, lacks this connectivity and gives no yellow precipitate.