Redox Reactions of Aldehydes and Ketones Revision Notes | A Level AQA Chemistry

1. The Carbonyl Group: Why Aldehydes and Ketones React Differently

Picture the carbonyl group, C=O, at the heart of both aldehydes and ketones. Oxygen is far more electronegative than carbon, so the pi electrons are pulled towards oxygen, leaving the carbonyl carbon with a significant partial positive charge ($\delta+$). That electron-poor carbon is an open invitation to any electron-rich species — a nucleophile.

Here is where aldehydes and ketones part ways. In an aldehyde, the carbonyl carbon carries one hydrogen and one carbon group (or just a hydrogen in methanal). In a ketone, it carries two carbon groups. That structural difference has two major consequences: it controls how easily the compound is oxidised, and it determines which type of alcohol forms on reduction.

With the carbonyl group established, the next question is: what happens when a reducing agent attacks that $\delta+$ carbon? The answer lies in the mechanism of nucleophilic addition — and NaBH4 is the reagent AQA focuses on.

2. Reduction Using NaBH4 — Overall Equations

NaBH4 in aqueous solution acts as a source of hydride ions, H$^-$. In overall equation terms, AQA uses [H] to represent the reductant — this is a shorthand for the hydrogen delivered by NaBH4, not a separate reagent you add to a flask. Every mole of carbonyl compound gains two [H] atoms across the C=O bond.

The product type follows directly from the structure. An aldehyde has one carbon substituent on the carbonyl carbon, so after reduction the carbon bearing the OH group is attached to one carbon chain and one hydrogen — that is a primary alcohol. A ketone has two carbon substituents, so after reduction the carbon bearing the OH group is attached to two carbon chains — that is a secondary alcohol.

Worked example: Writing Reduction Equations Using [H]

Write the overall equation for the reduction of propanal (CH3CH2CHO) and propanone (CH3COCH3) using NaBH4. Identify the product type in each case.

1. Propanal: CH3CH2CHO — aldehyde (CHO at chain end) Propanone: CH3COCH3 — ketone (C=O flanked by two CH3 groups) — Recognising the functional group tells you the product type before you write a single equation.
2. Aldehyde + 2[H] → primary alcohol Ketone + 2[H] → secondary alcohol — [H] represents the hydrogen delivered by NaBH4 in aqueous solution. Always use this notation in overall equations at A Level.
3. $$\text{CH}_3\text{CH}_2\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$$ — The CHO group becomes CH2OH. The product is propan-1-ol — a primary alcohol because the OH-bearing carbon is attached to only one other carbon group.
4. $$\text{CH}_3\text{COCH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_3$$ — The C=O becomes CHOH. The product is propan-2-ol — a secondary alcohol because the OH-bearing carbon is attached to two carbon groups (both CH3).
5. Reagent: NaBH4 in aqueous solution — AQA requires you to state NaBH4 in aqueous solution as the conditions. Writing just 'NaBH4' without specifying aqueous solution may lose a mark.

Answer: Propanal → propan-1-ol (primary alcohol): CH3CH2CHO + 2[H] → CH3CH2CH2OH. Propanone → propan-2-ol (secondary alcohol): CH3COCH3 + 2[H] → CH3CH(OH)CH3. Conditions: NaBH4 in aqueous solution.

3. The Nucleophilic Addition Mechanism for NaBH4 Reduction

The overall equation tells you what forms, but the mechanism explains how. NaBH4 dissociates in aqueous solution to release BH4$^-$ ions. Each BH4$^-$ ion delivers a hydride ion, H$^-$, which acts as the nucleophile. AQA specifies that you show the nucleophile as H$^-$ in the mechanism — not as a hydrogen atom or as part of the BH4 structure.

The mechanism has two key stages. First, H$^-$ attacks the $\delta+$ carbonyl carbon using a curly arrow drawn from the H$^-$ to the carbon. Simultaneously, the pi bond electrons shift onto oxygen, forming an alkoxide ion intermediate (RO$^-$). Second, the alkoxide ion is protonated by water in the aqueous solution: a curly arrow is drawn from the lone pair on the negatively charged oxygen of the alkoxide to the hydrogen atom of a water molecule, breaking the O–H bond of water heterolytically and transferring H$^+$ to the alkoxide oxygen. This gives the alcohol product and releases OH$^-$.

Notice that this is classified as nucleophilic addition because a nucleophile (H$^-$) adds across the C=O double bond — nothing is eliminated. The pi bond breaks: H forms a new bond to the carbonyl carbon, and the pi bond electrons move onto oxygen, giving the alkoxide intermediate. The oxygen was already sigma-bonded to the carbonyl carbon via the C=O single bond component; it does not form a new bond to carbon at this stage. This distinguishes nucleophilic addition from substitution, where one group replaces another.

4. Oxidation of Aldehydes to Carboxylic Acids

Reduction is only half the story. Aldehydes are also readily oxidised — and this is a property that ketones do not share under mild conditions. The C-H bond on the carbonyl carbon of an aldehyde can be broken oxidatively, converting the CHO group into a COOH group. The product is a carboxylic acid.

Ketones cannot be oxidised in the same way because they have no C-H bond on the carbonyl carbon — both substituents are carbon groups. To oxidise a ketone, you would need to break a C-C bond, which requires far more forcing conditions than are used in standard A Level reactions. This difference in oxidisability is the basis of the chemical tests used to distinguish aldehydes from ketones.

The practical consequence of this difference is that reagents such as Tollens' reagent (ammoniacal silver nitrate) and Fehling's solution oxidise aldehydes but leave ketones unchanged. These are the chemical tests AQA uses to distinguish between the two functional groups. Tollens' reagent produces a silver mirror with an aldehyde; Fehling's solution produces a brick-red precipitate. Ketones give no reaction with either reagent.

5. Connecting Reduction and Oxidation: The Full Redox Picture

Now that both reactions are established, it is worth seeing them as part of a connected oxidation state picture. In general terms, the carbonyl carbon in a carboxylic acid is more oxidised than in an aldehyde, which is in turn more oxidised than in a primary alcohol. However, it is important to note that NaBH4 only reduces aldehydes and ketones — it does not reduce carboxylic acids. Reducing a carboxylic acid to an aldehyde or alcohol requires a much stronger reductant such as lithium aluminium hydride (LiAlH4). The oxidation state ladder (carboxylic acid → aldehyde → primary alcohol) is a useful conceptual tool, but each step requires different reagents.

For ketones, the picture is simpler: ketone → secondary alcohol on reduction, and the reverse (secondary alcohol → ketone) on oxidation. There is no carboxylic acid equivalent because the ketone cannot be oxidised further under mild conditions. This asymmetry between the aldehyde and ketone pathways is a recurring theme in AQA questions that ask you to compare or evaluate the chemistry of carbonyl compounds.

The chart above uses relative values to show the trend — the exact oxidation state numbers are less important than the direction. Moving from carboxylic acid to aldehyde to primary alcohol, the carbonyl carbon becomes progressively more reduced (gains hydrogen, loses oxygen). The ketone sits between aldehyde and carboxylic acid in terms of oxidation state, but its reduction leads to a secondary alcohol rather than a primary one.

6. Exam Technique: Writing Mechanisms and Equations for Full Marks

AQA questions on this topic fall into three main types: writing overall equations using [H], outlining the nucleophilic addition mechanism, and explaining or comparing the oxidation behaviour of aldehydes and ketones. Each type has specific mark-winning requirements that are worth knowing before you sit the paper.

For overall equations, the mark scheme expects the [H] notation, the correct product (primary or secondary alcohol), and a balanced equation. State symbols are not always required for organic equations at this level, but check the question — if it says 'include state symbols', you must include them. For the mechanism, the mark scheme typically awards marks for: the curly arrow from H- to the carbonyl carbon, the curly arrow from the C=O pi bond to oxygen, the correct alkoxide intermediate with the negative charge on oxygen, and the protonation step showing water as the proton source.

Worked example: Exam-Style: Outlining the NaBH4 Reduction Mechanism for Propanone

Outline the mechanism for the reduction of propanone (CH3COCH3) by NaBH4 in aqueous solution. Show the nucleophile as H-. [4 marks]

1. Nucleophile: H- (hydride ion from NaBH4) Electrophilic site: carbonyl carbon of CH3COCH3 (delta+ due to C=O polarity) — [B1] Identifying H- as the nucleophile is the first mark. State this explicitly if the question asks you to 'identify the nucleophile'.
2. Arrow 1: from H- (lone pair / negative charge) → carbonyl carbon of C=O This represents the nucleophile donating a pair of electrons to form a new C-H bond. — [M1] The curly arrow must start from H- and point to the carbonyl carbon. Starting from anywhere else loses this mark.
3. Arrow 2: from C=O pi bond → oxygen atom This breaks the pi bond and places the negative charge on oxygen, forming the alkoxide intermediate: (CH3)2CHO- — [M1] Both curly arrows in Step 2 and Step 3 are needed together to show the concerted electron movement. The intermediate is (CH3)2CHO- — an alkoxide ion.
4. Arrow 3: from O-H bond of H2O → O- of (CH3)2CHO- Product: (CH3)2CHOH (propan-2-ol) + OH- — [M1] The aqueous solution provides water for protonation. Show the curly arrow from the O-H bond of water to the negatively charged oxygen of the alkoxide. The final product is propan-2-ol, a secondary alcohol.
5. Propanone has two CH3 groups on the carbonyl carbon. After reduction, the OH-bearing carbon is attached to two CH3 groups → secondary alcohol confirmed. — Always verify the product type matches the starting material. A ketone always gives a secondary alcohol on reduction with NaBH4.

Answer: Mechanism: H- attacks the delta+ carbonyl carbon of propanone (curly arrow from H- to C); simultaneously the C=O pi bond breaks onto oxygen (curly arrow from pi bond to O), forming the alkoxide intermediate (CH3)2CHO-. Water then protonates the alkoxide (curly arrow from O-H of H2O to O-), giving propan-2-ol (a secondary alcohol) and OH-.

Every AQA specification point for this topic maps onto a specific skill. Overall equations use [H] as the reductant and must show the correct alcohol product type. The mechanism requires H- as the nucleophile, two curly arrows in the addition step, the alkoxide intermediate, and protonation by water. Oxidation questions require you to explain the structural reason why aldehydes oxidise and ketones do not. Comparison questions need you to link structure to product type for both reduction and oxidation.

7. Summary and Spec Checklist

• Aldehydes + 2[H] → primary alcohols (NaBH4, aqueous solution)
• Ketones + 2[H] → secondary alcohols (NaBH4, aqueous solution)
• Both reactions proceed via nucleophilic addition; nucleophile is H-
• Mechanism: H- attacks delta+ carbonyl carbon → alkoxide intermediate → protonation by water → alcohol
• Aldehydes are readily oxidised to carboxylic acids (RCHO + [O] → RCOOH)
• Ketones resist oxidation under mild conditions — no C-H bond on carbonyl carbon
• Tollens' reagent and Fehling's solution distinguish aldehydes (positive result) from ketones (no reaction)

The molecule viewer above shows ethanal (CH3CHO). Compare the 2D skeletal formula — where the CHO group at the chain end is clearly visible — with the 3D view, which shows the planar geometry around the carbonyl carbon. That planar arrangement is a consequence of the sp2 hybridisation of the carbonyl carbon, and it is what makes the carbonyl carbon accessible to nucleophilic attack from above or below the plane of the molecule.