Aldehydes and Ketones
Quick Notes
- Aldehydes can be made by the oxidation of a primary alcohol with acidified K2Cr2O7 or KMnO4 (aldehyde collected by distillation).
- Ketones can be made by the oxidation of a secondary alcohol with acidified K2Cr2O7 or KMnO4
- Reactions of Aldehydes and Ketones:
- Reduction to Alcohols:
Use NaBH4 or LiAlH4.
Aldehydes form primary alcohols
Ketones form secondary alcohols - Reaction with HCN/KCN:
Forms hydroxynitriles (adds –OH and –CN across the C=O group).
Occurs via nucleophilic addition.
- Reduction to Alcohols:
- Tests for Carbonyl Compounds:
- 2,4-DNPH Test:
Reacts with both aldehydes and ketones → orange precipitate. - Fehling’s Solution / Tollens’ Reagent:
Distinguish between aldehydes and ketones.
Aldehydes are easily oxidised → give a positive result.
(Fehling’s: red ppt, Tollens’: silver mirror).
Ketones give no reaction. - Iodoform Test (I2/OH−):
Detects a CH3CO– group (typically methyl ketones or ethanal).
Gives a yellow precipitate of CHI3 (iodoform) with antiseptic smell.
- 2,4-DNPH Test:
Full Notes
Aldehydes and ketones and their reactions have been outlined in more detail here.
This page is just what you need to know for CIE A-level Chemistry :)
Aldehydes and ketones contain a carbonyl (C=O) functional group.
Making Aldehydes
Oxidation of Primary Alcohols
Aldehydes can be formed by the oxidation of Primary Alcohols

- Reagents: Acidified K2Cr2O7 or acidified KMnO4
(oxidising agents provide the oxygen required and are shown as [O] in equations). - Conditions: Heat and distillation
Note: Distillation prevents further oxidation to a carboxylic acid.
Example CH3CH2OH + [O] → CH3CHO + H2O
Oxidation of Secondary Alcohols
Ketones can be made by the oxidation of Secondary Alcohols

- Reagents: Acidified K2Cr2O7 or acidified KMnO4
(oxidising agents that provide oxygen required, shown as [O] in equations). - Conditions: Heat under reflux or distil
Note: Ketones do not oxidise further under normal conditions.
Example CH3CH(OH)CH3 → CH3COCH3 + H2O
Key Reactions
Aldehydes and ketones react with nucleophiles as the C=O bond is polar and nucleophiles are attracted to the Cδ+.
Nucleophiles are electron pair donors.
Reduction to Alcohols
Aldehydes and Ketones can be reduced to alcohols

- Reagents: NaBH4 (aqueous) or LiAlH4 (dry ether)
- Aldehydes form Primary alcohols
- Ketones form Secondary alcohols
Explanation: H− ion from the reducing agent adds to the carbonyl carbon, shown as [H] in equation.
Note - LiAlH4 is a more powerful reducing agent than NaBH4, because of this, if LiAlH4 is used no water can be present and the reaction must be carried out in dry ether.
Example CH3CHO + 2[H] → CH3CH2OH

Remember reduction in organic chemistry is the gaining of a carbon-hydrogen bond. To provide the hydrogen needed, we use reducing agents (such as NaBH4 and LiAlH4) and show hydrogen from a reducing agent in equations as [H].
Reaction with HCN/KCN (Forming Hydroxynitriles)
Carbonyls react with KCN in H2SO4 to form hydroxynitriles.

- Reagents: HCN, KCN as catalyst, heat
Explanation: Adds CN− and H+ across the C=O group to give an extra C–C bond.
Note: HCN is generated in situ by reacting KCN with H2SO4 because pure HCN is too toxic and volatile to handle safely. The KCN supplies CN− ions, while H2SO4 provides H+ ions. This combination produces the same product — a hydroxynitrile — without needing to use HCN directly.
Example CH3CHO + HCN → CH3CH(OH)CN
Nucleophilic Addition Mechanism (HCN)
Aldehydes and ketones react by nucleophilic addition mechanisms - a nucleophile starts the reaction by attacking the Cδ+ in the C=O bond.

Steps:
- CN− attacks the electrophilic carbon in the carbonyl group.
- Intermediate forms with a negative charge on oxygen.
- H+ (from HCN or acid) adds to O−, forming the hydroxynitrile.
Test with 2,4-DNPH
Carbonyl containing compounds (aldehydes and ketones) form an orange precipitate when added to a solution of 2,4-DNPH (Brady’s reagent).

- Reagent: 2,4-dinitrophenylhydrazine
- Observation: Bright orange or yellow precipitate
- Explanation: Forms a hydrazone derivative with C=O compounds (both aldehydes and ketones).

Distinguishing Aldehydes from Ketones
Because aldehydes can be further oxidised, they can be distinguished from ketones using oxidation reactions.
The two main tests used are Fehling's Solution and Tollens' Reagent.
Test | Aldehyde (CHO) | Ketone |
---|---|---|
Fehling’s | Red ppt of Cu2O | No change |
Tollens’ | Silver mirror forms (Ag0) | No silver formed |
Fehling’s Solution Test:

Aldehyde: Forms a brick-red precipitate (Cu2O).
Ketone: No change.
Tollens’ Silver Mirror Test:

Aldehyde: Forms a silver mirror (Ag).
Ketone: No change.
Iodoform Test (CH3CO– group)
The iodoform test detects compounds with the CH3CO– group (methyl ketones) or ethanal.

- Reagent: I2 in alkali (e.g. NaOH)
- Positive test: Yellow precipitate of CHI3 (iodoform)
Example CH3COCH3 + 3I2 + 4OH− → CHI3 + CH3COO− + 3I− + 3H2O
Summary
- Aldehydes can be made by oxidising primary alcohols and collected by distillation to prevent further oxidation.
- Ketones can be made by oxidising secondary alcohols; do not oxidise further under normal conditions.
- Both reduce to alcohols with NaBH4 or LiAlH4 (aldehydes form primary alcohols, ketones form secondary alcohols).
- HCN/KCN adds across C=O via nucleophilic addition to form hydroxynitriles (adds –CN and –OH).
- Tests:
- 2,4-DNPH: orange/yellow precipitate with carbonyls.
- Fehling’s/Tollens’: aldehydes positive (red ppt / silver mirror), ketones no reaction.
- Iodoform (I2/OH−): yellow CHI3 with CH3CO– (methyl ketones) or ethanal.