OxfordAQA A-Level Chemistry: Aldehydes and ketones (9620)

This study guide teaches section 3.3.8, Aldehydes and ketones, from the OxfordAQA International AS and A-level Chemistry (9620) specification (Version 5.3; International AS and A-level exams May/June 2020 onwards). Only one subsection, 3.3.8.1, makes up the section. The specification places the topic among the International A2 organic sections, which makes every point here International A-level only. The unit that examines it is Unit 4 (Organic 2 and Physical 2, covering 3.3.7–3.3.16), while Unit 5, the practical and synoptic unit, can assess anything in the course.

Use it with the revision notes and the practice questions. The course hub links every OxfordAQA Chemistry page, the printable checklist sets out every specification point, and the free 10-minute diagnostics show where to focus next.

What this section covers

Section	What the specification lists	Stage
3.3.8.1	Aldehydes are readily oxidised to carboxylic acids	International A-level only
3.3.8.1	Chemical tests to tell aldehydes from ketones, including Fehling’s solution and Tollens’ reagent	International A-level only
3.3.8.1	NaBH₄ in aqueous solution reduces aldehydes to primary alcohols and ketones to secondary alcohols (nucleophilic addition); write overall equations using [H]; outline the mechanism with the nucleophile shown as H⁻	International A-level only
3.3.8.1	KCN followed by dilute acid gives hydroxynitriles; write overall equations using HCN; outline the mechanism	International A-level only
3.3.8.1	Aldehydes and unsymmetrical ketones give mixtures of enantiomers with KCN then dilute acid; explain why	International A-level only
3.3.8.1	The hazards of using KCN	International A-level only

Section 3.3.8.1 lists no practical activity. The same tests appear in section 3.3.5.1 (see the alcohols guide) and in Required practical 5 (see organic analysis).

The carbonyl group and nucleophiles

Aldehydes, ketones, carboxylic acids and their derivatives all contain the carbonyl group, C=O, which nucleophiles attack.

• Aldehyde: the C=O carbon carries at least one hydrogen atom. The group is written –CHO and the name ends in -al.
• Ketone: the C=O carbon is bonded to two other carbon atoms. The name ends in -one.

Oxygen pulls the shared electrons towards itself, so the bond is polarised as Cδ+=Oδ−. A nucleophile (a lone-pair donor) bonds to the δ+ carbon. At the same time one of the carbon–oxygen bonds breaks and its electron pair moves on to the oxygen, which becomes O⁻. Adding H⁺ to that oxygen finishes the reaction. Two groups have joined across the old C=O, so the process is nucleophilic addition.

The carbonyl carbon has three regions of electron density around it, so it is trigonal planar (bond angles near 120°). Keep this in mind for the enantiomer section.

Oxidation of aldehydes

An aldehyde has a hydrogen on its carbonyl carbon. Oxidation replaces that C–H with C–OH, so aldehydes are readily oxidised to carboxylic acids:

RCHO + [O] → RCOOH
(CH₃)₂CHCH₂CH₂CHO + [O] → (CH₃)₂CHCH₂CH₂COOH     4-methylpentanal → 4-methylpentanoic acid

A ketone has no hydrogen on its carbonyl carbon, and oxidising it would mean breaking a C–C bond, so mild oxidising agents leave it unchanged. The two tests rely on this.

Tests: Fehling’s solution and Tollens’ reagent

Both are mild oxidising agents. Warm the sample with the reagent in a hot water bath rather than over a flame, since many carbonyl compounds are flammable.

Reagent	Active species	Aldehyde	Ketone
Fehling’s solution	Blue copper(II) complex in alkaline solution	Red precipitate of copper(I) oxide, Cu₂O	Remains a blue solution
Tollens’ reagent	Diammine silver(I) ion, [Ag(NH₃)₂]⁺	Silver mirror on the inside of the tube	Stays colourless; no silver forms

In a positive test the aldehyde is the reducing agent. The reagents are alkaline, so the acid formed is present as its carboxylate ion. Section 3.2.5.5 of the specification also names the reduction of [Ag(NH₃)₂]⁺ to silver as the way Tollens’ reagent separates aldehydes from ketones; the half-equation is [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃. The transition metals guide covers that redox side.

Worked example 1: choosing and describing a test

Two colourless liquids are known to be 4-methylpentanal and 2-methylpentan-3-one. Both have the formula C₆H₁₂O. Describe how Tollens’ reagent tells them apart.

1. Warm a little of each liquid with fresh Tollens’ reagent in a hot water bath.
2. 4-methylpentanal: a silver mirror forms, because the aldehyde is oxidised to 4-methylpentanoate and Ag⁺ is reduced to Ag.
3. 2-methylpentan-3-one: no silver forms; the mixture stays colourless, because the ketone is not oxidised.

Give an observation for both liquids: “no silver mirror”, not “no reaction”.

Worked example 2: identifying an aldehyde from the silver formed

A saturated, non-cyclic aldehyde has the general formula CₙH₂ₙO. Warming 0.750 g of one such aldehyde with excess Tollens’ reagent gives 1.62 g of silver. Find its molecular formula. (Ar: Ag = 107.9; C = 12.0, H = 1.0, O = 16.0)

Each Ag⁺ gains one electron, and turning one –CHO group into –COOH releases two electrons. So 1 mol of aldehyde gives 2 mol of Ag.

n(Ag)       = 1.62 ÷ 107.9 = 0.01501 mol
n(aldehyde) = 0.01501 ÷ 2   = 0.007507 mol
Mr          = 0.750 ÷ 0.007507 = 99.9, so 100 to 3 s.f.
14n + 16 = 100  →  n = 6

The aldehyde is C₆H₁₂O. Any hexanal isomer fits, for example 2-ethylbutanal, (CH₃CH₂)₂CHCHO.

Reduction with NaBH₄

Aqueous NaBH₄, sodium tetrahydridoborate(III), reduces both kinds of carbonyl compound:

• an aldehyde gives a primary alcohol
• a ketone gives a secondary alcohol

Write overall equations with [H] standing for the reductant. Two [H] are needed, one for the carbon and one for the oxygen:

(CH₃CH₂)₂CHCHO + 2[H] → (CH₃CH₂)₂CHCH₂OH             2-ethylbutanal → 2-ethylbutan-1-ol
(CH₃)₂CHCOCH(CH₃)₂ + 2[H] → (CH₃)₂CHCH(OH)CH(CH₃)₂    2,4-dimethylpentan-3-one → 2,4-dimethylpentan-3-ol

The mechanism

Show the nucleophile as the hydride ion, drawn with its lone pair (H:⁻), as the specification asks.

Worked example 3: reduction of 2-methylpentan-3-one

Outline the mechanism when aqueous NaBH₄ reduces 2-methylpentan-3-one, CH₃CH₂COCH(CH₃)₂.

1. Draw the ketone with the C=O bond labelled Cδ+ and Oδ−.
2. First arrow: starts at the H⁻ lone pair and ends on the carbonyl carbon.
3. Second arrow: from the double bond on to the oxygen atom. This gives an alkoxide ion, with the negative charge on O.
4. Third arrow: an O⁻ lone pair bonds to a hydrogen of a water molecule (add an arrow from that O–H bond on to the water’s oxygen), or you can show H⁺ instead of water.

          Oδ−                        O⁻                         OH
          ‖                          |                          |
C₂H₅ – Cδ+ – CH(CH₃)₂   →   C₂H₅ – C – CH(CH₃)₂   →   C₂H₅ – C – CH(CH₃)₂
          ↑                          |            + H₂O         |
         :H⁻                         H                          H

The product is 2-methylpentan-3-ol, a secondary alcohol, as expected from a ketone.

Hydroxynitriles from KCN and dilute acid

Treat an aldehyde or ketone with potassium cyanide and then dilute acid and you get a hydroxynitrile. The new carbon carries both –OH and –C≡N, and the chain is one carbon longer than before. The specification asks for the overall equation written with HCN:

(CH₃CH₂)₂CHCHO + HCN → (CH₃CH₂)₂CHCH(OH)CN          3-ethyl-2-hydroxypentanenitrile
(CH₃)₃CCOCH₃ + HCN → (CH₃)₃CC(OH)(CH₃)CN            2-hydroxy-2,3,3-trimethylbutanenitrile

Naming tip. The –C≡N carbon is C1. In the first product the chain is C≡N, CH(OH), CH, then one ethyl group (C4–C5); the other ethyl is a substituent on C3.

Worked example 4: the mechanism with 4-methylpentanal

4-methylpentanal, (CH₃)₂CHCH₂CH₂CHO, is treated with KCN and then dilute acid. Outline the mechanism.

1. Show Cδ+=Oδ− on the aldehyde group.
2. KCN supplies cyanide ions. The attacking lone pair is the one on the carbon atom of :C≡N⁻; draw the arrow from it to the δ+ carbon.
3. Second arrow: double bond on to oxygen. The intermediate is (CH₃)₂CHCH₂CH₂CH(CN)O⁻.
4. Third arrow: the negatively charged oxygen uses a lone pair to pick up H⁺, which the dilute acid supplies.

     Oδ−                   O⁻                    OH
     ‖                     |                     |
R – Cδ+ – H   →   R – C – H   →(+H⁺)→   R – C – H        R = (CH₃)₂CHCH₂CH₂–
     ↑                     |                     |
    :CN⁻                   CN                    CN

The product is 2-hydroxy-5-methylhexanenitrile.

Worked example 5: yield of a hydroxynitrile

3.00 g of 3,3-dimethylbutan-2-one (C₆H₁₂O) gives 2.79 g of 2-hydroxy-2,3,3-trimethylbutanenitrile (C₇H₁₃NO). Find the percentage yield. (Ar: C 12.0, O 16.0, H 1.0, N 14.0)

Mr(ketone)  = 72.0 + 12.0 + 16.0 = 100.0;  n = 3.00 ÷ 100.0 = 0.0300 mol
Mr(product) = 84.0 + 13.0 + 14.0 + 16.0 = 127.0
maximum mass = 0.0300 × 127.0 = 3.81 g
yield = 2.79 ÷ 3.81 × 100 = 73.2 %

Why the product can be a pair of enantiomers

From optical isomerism (section 3.3.7.1): an asymmetric carbon (one bonded to four different groups) is a chiral centre, its two non-superimposable mirror-image forms are enantiomers, and a mixture of equal amounts of them is a racemate that has no overall effect on plane-polarised light.

The explanation, point by point:

1. The carbonyl carbon and the three atoms joined to it lie in one plane.
2. The cyanide ion can reach that carbon from above the plane or from below it.
3. Neither face is favoured, so both approaches happen equally often.
4. When the new C(OH)(CN) carbon ends up with four different groups, attack from the two faces gives the two enantiomers.
5. The two enantiomers are made in a 50 : 50 ratio. That racemate has no net effect on plane-polarised light.

This is why the specification names aldehydes and unsymmetrical ketones. In each, the two groups already on the C=O carbon differ, and adding OH and CN makes four different groups. (Methanal is the exception among aldehydes: its two groups are both H.)

Worked example 6: racemate or not?

Starting compound	Groups on the C=O carbon	Groups on the product carbon	Mixture of enantiomers?
2-ethylbutanal	H and CH(C₂H₅)₂	H, OH, CN, CH(C₂H₅)₂	Yes
3,3-dimethylbutan-2-one	CH₃ and C(CH₃)₃	CH₃, OH, CN, C(CH₃)₃	Yes
2,4-dimethylpentan-3-one	CH(CH₃)₂ and CH(CH₃)₂	two identical CH(CH₃)₂ groups	No: symmetrical ketone, no chiral centre

The hazards of using KCN

• Potassium cyanide is highly toxic by ingestion, skin contact or inhaled dust.
• Acid converts cyanide ions into hydrogen cyanide, a highly toxic gas that is very volatile.

So use a fume cupboard, gloves and eye protection, and add the dilute acid slowly. KCN is used rather than HCN because, as an ionic solid, it supplies plenty of CN⁻; HCN is a weak acid that ionises only slightly, and a dangerous gas.

Common errors

• Starting the cyanide arrow at N. Carbon’s lone pair does the attacking, which is why a new C–C bond forms.
• Leaving out the arrow from C=O to oxygen, which leaves carbon with five bonds.
• Writing 1[H] in reduction equations; the C=O needs 2[H].
• Using a 1 : 1 ratio of aldehyde to silver in calculations; it is 1 : 2.
• Saying a symmetrical ketone gives enantiomers.
• Writing “KCN is toxic” without the HCN released when acid is added.

Where to go next

For a shorter version, use the revision notes, then try the practice questions and the diagnostics.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.8 Aldehydes and ketones (3.3.8.1), International A2 content.