AQA A-Level Chemistry: Aldehydes and ketones (7405)

This guide teaches section 3.3.8 Aldehydes and ketones of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The whole section is A-level only: it is not part of the AS content. The specification places organic chemistry (3.3) on A-level Paper 2, and Paper 3 can test any content and any practical skill. The section lists no required practical of its own.

For quick recall, use the aldehydes and ketones revision notes. To test yourself, use the aldehydes and ketones practice questions. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. To find your weak spots across the course, try one of the free 10-minute diagnostics.

What this unit covers

Spec point (3.3.8)	What you must be able to do	Status
Oxidation	State that aldehydes are readily oxidised to carboxylic acids	A-level only
Tests	Use Fehling’s solution and Tollens’ reagent to tell aldehydes from ketones	A-level only
Reduction with NaBH₄	Reduce aldehydes to primary alcohols and ketones to secondary alcohols; write equations with [H]; outline the nucleophilic addition mechanism with H⁻ as the nucleophile	A-level only
KCN then dilute acid	Form hydroxynitriles; write overall equations using HCN; outline the nucleophilic addition mechanism	A-level only
Enantiomers	Explain why aldehydes and unsymmetrical ketones give a mixture of enantiomers	A-level only
Hazards	State the hazards of using KCN	A-level only

You need the earlier sections first. Naming and functional groups come from introduction to organic chemistry. Making aldehydes and ketones from alcohols, and the distil-or-reflux choice, are in the alcohols study guide. Test-tube tests and IR are in organic analysis.

The carbonyl group

Aldehydes and ketones both contain the carbonyl group, C=O.

• In an aldehyde the C=O is at the end of the chain, written –CHO. Names end in -al: ethanal, CH₃CHO.
• In a ketone the C=O is joined to two carbon atoms. Names end in -one: propanone, CH₃COCH₃.

Oxygen is more electronegative than carbon, so the C=O bond is polar: Cδ+=Oδ−. The carbon is electron deficient. A nucleophile (an electron-pair donor) is attracted to it and forms a new bond. At the same time one pair of electrons in the C=O double bond moves onto the oxygen. Overall, the nucleophile adds across the C=O. This is nucleophilic addition, and every reaction in this section except oxidation works this way.

The carbon of the C=O has three bonding regions, so it is trigonal planar, with bond angles of about 120°. This flat shape matters later, when you explain why some products are racemic.

Oxidation of aldehydes

Aldehydes are readily oxidised to carboxylic acids. The carbon of the –CHO group carries a hydrogen atom, and oxidation turns C–H into C–OH:

RCHO + [O] → RCOOH
CH₃CH₂CHO + [O] → CH₃CH₂COOH        (propanal → propanoic acid)

Ketones are not easily oxidised. Their C=O carbon has no hydrogen atom, and oxidising them would mean breaking a C–C bond. Acidified potassium dichromate(VI) oxidises aldehydes too (orange to green), but it also oxidises primary and secondary alcohols, so it is not a specific test for aldehydes. Mild oxidising agents are used instead.

Tests: Tollens’ reagent and Fehling’s solution

Both reagents are mild oxidising agents. They oxidise an aldehyde but not a ketone. Warm the test tube in a hot water bath, not over a flame, because the organic compounds are flammable.

Test	What the reagent contains	Aldehyde	Ketone
Tollens’ reagent	[Ag(NH₃)₂]⁺, made from silver nitrate and aqueous ammonia	Silver mirror on the tube (silver(I) reduced to Ag)	No change
Fehling’s solution	A blue copper(II) complex in alkali	Blue solution gives a red precipitate of Cu₂O (copper(I) oxide)	Stays blue

In each positive test the aldehyde is the reducing agent. It is oxidised to a carboxylic acid. The reagents are alkaline, so the acid ends up as its carboxylate ion.

Each silver(I) ion gains one electron, Ag⁺ + e⁻ → Ag. Oxidising one aldehyde molecule to the acid releases two electrons, so one mole of aldehyde reduces two moles of silver(I) ions.

Worked example 1. 1.16 g of propanal is warmed with excess Tollens’ reagent. Calculate the maximum mass of silver formed. (Ar: C = 12.0, H = 1.0, O = 16.0, Ag = 107.9)

Mr(propanal, C₃H₆O) = 3(12.0) + 6(1.0) + 16.0 = 58.0
n(propanal)         = 1.16 ÷ 58.0 = 0.0200 mol
n(Ag)               = 2 × 0.0200 = 0.0400 mol
mass of Ag          = 0.0400 × 107.9 = 4.316 g

Answer: 4.32 g of silver (3 significant figures).

Reduction with NaBH₄

Sodium tetrahydridoborate(III), NaBH₄, in aqueous solution reduces carbonyl compounds:

• aldehyde → primary alcohol
• ketone → secondary alcohol

Write overall equations with [H] for the reductant. Two [H] add across the C=O:

CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH         (butanal → butan-1-ol)
CH₃COCH₂CH₃  + 2[H] → CH₃CH(OH)CH₂CH₃        (butanone → butan-2-ol)

Mechanism: nucleophilic addition with H⁻

NaBH₄ acts as a source of the hydride ion, H⁻, which is the nucleophile. Show it as H⁻ with a lone pair.

Worked example 2. Outline the mechanism for the reduction of propanone by NaBH₄ in aqueous solution.

• Step 1: draw propanone with the C=O polarised, Cδ+ and Oδ−.
• Step 2: draw a curly arrow from the lone pair on H⁻ to the carbonyl carbon.
• Step 3: in the same step, draw a curly arrow from the C=O double bond to the oxygen. The oxygen now carries a negative charge and three lone pairs.
• Step 4: the intermediate is (CH₃)₂CH–O⁻. Draw a curly arrow from a lone pair on O⁻ to an H atom of a water molecule (or to H⁺), and an arrow from that O–H bond in water onto its oxygen.
• Answer: the product is propan-2-ol, (CH₃)₂CHOH.

        Oδ−                    O⁻                     OH
        ‖                      |                      |
CH₃ – Cδ+ – CH₃  + :H⁻  →  CH₃–C–CH₃   then +H⁺ →  CH₃–C–CH₃
                               |                      |
                               H                      H

Worked example 3. 3.60 g of butanone is reduced with NaBH₄ and 2.96 g of butan-2-ol is collected. Calculate the percentage yield. (Mr: butanone 72.0, butan-2-ol 74.0)

n(butanone)       = 3.60 ÷ 72.0 = 0.0500 mol
theoretical n(alcohol) = 0.0500 mol  (1 : 1)
actual n(alcohol) = 2.96 ÷ 74.0 = 0.0400 mol
percentage yield  = 0.0400 ÷ 0.0500 × 100 = 80.0 %

Answer: 80.0 %. Compare moles, not masses.

Forming hydroxynitriles: KCN followed by dilute acid

Carbonyl compounds react with potassium cyanide, KCN, followed by dilute acid. The product is a hydroxynitrile: an OH group and a C≡N group on the same carbon. The carbon chain grows by one carbon.

Write the overall equation using HCN:

CH₃CHO + HCN → CH₃CH(OH)CN              (ethanal → 2-hydroxypropanenitrile)
CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN          (propanone → 2-hydroxy-2-methylpropanenitrile)

Naming. The nitrile carbon is carbon 1 and is included in the chain length. The OH is then on carbon 2. From ethanal (2 carbons) the product has 3 carbons, so it is a propanenitrile.

Mechanism: nucleophilic addition with CN⁻

KCN provides cyanide ions, CN⁻. The nucleophile is the lone pair on the carbon of CN⁻, which is why the new bond is C–C and the product is a nitrile.

Worked example 4. Outline the mechanism for the reaction of ethanal with KCN followed by dilute acid.

• Step 1: curly arrow from the lone pair on the carbon of :CN⁻ to the δ+ carbonyl carbon.
• Step 2: curly arrow from the C=O double bond to the oxygen, giving the intermediate CH₃CH(CN)–O⁻.
• Step 3: curly arrow from a lone pair on O⁻ to H⁺ from the dilute acid.
• Answer: 2-hydroxypropanenitrile, CH₃CH(OH)CN.

       Oδ−                       O⁻                    OH
       ‖                         |                     |
CH₃ – Cδ+ – H  +  :CN⁻  →   CH₃–C–H   then +H⁺ →  CH₃–C–H
                                 |                     |
                                 CN                    CN

Worked example 5. Calculate the maximum mass of hydroxynitrile from 2.90 g of propanone. (Ar: C = 12.0, H = 1.0, N = 14.0, O = 16.0)

n(propanone) = 2.90 ÷ 58.0 = 0.0500 mol
product (CH₃)₂C(OH)CN = C₄H₇NO, Mr = 48.0 + 7.0 + 14.0 + 16.0 = 85.0
maximum mass = 0.0500 × 85.0 = 4.25 g

Answer: 4.25 g.

The hazards of using KCN

• KCN is highly toxic if swallowed, inhaled as dust or absorbed.
• With acid, KCN releases hydrogen cyanide, HCN, which is highly toxic and very volatile, so it escapes easily as a gas.

So you wear gloves and eye protection, work in a fume cupboard, and add the acid in a controlled way. KCN is used rather than HCN itself because HCN is toxic and volatile, so it is hard to handle safely, and because HCN is a weak acid, so it supplies few CN⁻ ions. KCN is ionic and supplies plenty of CN⁻ nucleophile.

Why the products can be racemic

You met chirality in optical isomerism (3.3.7). A carbon with four different groups is a chiral centre, and its two non-superimposable mirror images are enantiomers. A 50 : 50 mixture is a racemic mixture (racemate) and is optically inactive.

The explanation for KCN followed by dilute acid, in steps:

1. The C=O carbon and the three atoms bonded to it lie in one plane (trigonal planar).
2. CN⁻ can attack the carbon from either side of this plane.
3. Attack from each side is equally likely.
4. If the new carbon has four different groups, the two directions of attack give the two enantiomers.
5. They form in equal amounts, so the product is a racemic mixture. The rotations of plane-polarised light cancel, so the mixture is optically inactive.

When does a chiral centre form? Look at the two groups already on the C=O carbon. The product carbon carries those two groups, plus OH and CN.

Carbonyl compound	Groups on C=O carbon	Product chiral?
Ethanal (aldehyde)	CH₃ and H	Yes: CH₃, H, OH, CN are all different, so racemic
Butanone (unsymmetrical ketone)	CH₃ and C₂H₅	Yes: racemic
Propanone (symmetrical ketone)	CH₃ and CH₃	No: two identical groups
Methanal	H and H	No

This is why the specification says aldehydes and unsymmetrical ketones form mixtures of enantiomers. Methanal is the one aldehyde that does not, because its carbonyl carbon carries two H atoms.

Common errors

• Drawing the CN⁻ arrow from the nitrogen. It starts at the lone pair on the carbon.
• Starting the H⁻ arrow at a B–H bond of BH₄⁻. The specification asks for H⁻ as the nucleophile.
• Forgetting the second arrow, from C=O to O. Without it the carbon would have five bonds.
• Writing reduction equations with one [H] instead of two.
• Naming CH₃CH(OH)CN as “2-hydroxyethanenitrile”. Count the nitrile carbon: it is a propanenitrile.
• Saying a racemic product forms “because the product is chiral”. The reason is equal attack from both sides of a planar C=O group.
• Saying propanone gives a racemic mixture. Its product has two identical CH₃ groups, so no chiral centre.
• Giving “toxic” for KCN without saying that acid releases HCN gas.

Where this links

Exam technique: exam preparation. Electron transfer in the Tollens’ and Fehling’s tests: redox equations.

Next: revision notes and practice questions.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.8 Aldehydes and ketones (A-level only).