AQA A-Level Chemistry: Carboxylic acids and derivatives (7405)

This guide teaches section 3.3.9 Carboxylic acids and derivatives 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. It covers every point in 3.3.9.1 Carboxylic acids and esters and 3.3.9.2 Acylation, plus Required practical 10. The whole section is marked A-level only, so it is not on the AS papers. At A-level, the specification places organic chemistry (3.3) on Paper 2, and Paper 3 can test any content and any practical skill.

For quick recall, use the revision notes. To test yourself, use the 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

Specification section	What you must be able to do	Status
3.3.9.1 Carboxylic acids and esters	Draw and name carboxylic acids and esters; show they are weak acids that release CO₂ from carbonates; make esters from acids and alcohols with an acid catalyst; state uses of esters; describe fats and oils as esters of glycerol; hydrolyse esters in acid and alkali; explain soap and biodiesel	A-level only
3.3.9.2 Acylation	Draw acid anhydrides, acyl chlorides and amides; write equations for reactions of acyl chlorides and acid anhydrides with water, alcohols, ammonia and primary amines; outline the nucleophilic addition–elimination mechanism for acyl chlorides; give the industrial advantages of ethanoic anhydride for making aspirin	A-level only
Required practical 10	Prepare a pure organic solid and test its purity; prepare a pure organic liquid	A-level only

Related pages: making carboxylic acids by oxidation is in alcohols, and the carbonate test is in organic analysis. Weak-acid behaviour links to acids and bases.

3.3.9.1 Carboxylic acids and esters

Structures and names

A carboxylic acid contains the –COOH group: a C=O and an O–H on the same carbon. The carbon of –COOH counts as carbon 1 in the name, and the suffix is -oic acid.

• HCOOH: methanoic acid
• CH₃CH₂COOH: propanoic acid
• CH₃CH(CH₃)COOH: 2-methylpropanoic acid

An ester contains –COO– joining two carbon chains. It forms from an acid and an alcohol. The name has two words: the alcohol part first (as an alkyl group, such as methyl or ethyl) and the acid part second (ending -oate).

Worked example 1: naming esters

Name CH₃CH₂COOCH₂CH₃ and HCOOCH₃.

1. Find the C=O. The chain that contains the C=O carbon is the acid part.
2. In CH₃CH₂COOCH₂CH₃ the acid part is CH₃CH₂CO– (3 carbons, propanoate). The part on the other O is –CH₂CH₃ (ethyl).
3. Name: ethyl propanoate.
4. In HCOOCH₃ the acid part is HCO– (1 carbon, methanoate); the other part is methyl. Name: methyl methanoate.

Carboxylic acids are weak acids

A carboxylic acid only partially dissociates in water, so it is a weak acid:

CH₃CH₂COOH ⇌ CH₃CH₂COO⁻ + H⁺

Even so, it is strong enough to release carbon dioxide from carbonates. You see effervescence, and the gas turns limewater cloudy.

Worked example 2: an acid with a carbonate

Write an equation for ethanoic acid reacting with calcium carbonate.

1. Each CO₃²⁻ needs two H⁺, so you need two acid molecules.
2. The salt is calcium ethanoate, (CH₃COO)₂Ca.

2CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + CO₂ + H₂O

Making esters

A carboxylic acid and an alcohol react, in the presence of an acid catalyst, to give an ester and water. Concentrated sulfuric acid is the usual catalyst. The reaction is reversible, so the yield is never 100%.

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

Worked example 3: esterification yield

7.40 g of propanoic acid is warmed with excess ethanol and a few drops of concentrated sulfuric acid. 6.63 g of ethyl propanoate is collected. Use Ar values H = 1.0, C = 12.0, O = 16.0. Calculate the percentage yield.

1. Equation: CH₃CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂COOCH₂CH₃ + H₂O (1 : 1).
2. Mr of propanoic acid (C₃H₆O₂) = 74.0. Moles = 7.40 ÷ 74.0 = 0.100 mol.
3. Mr of ethyl propanoate (C₅H₁₀O₂) = 102.0. Theoretical mass = 0.100 × 102.0 = 10.2 g.
4. Percentage yield = 6.63 ÷ 10.2 × 100 = 65.0%.

Uses of esters

The specification lists common uses: solvents, plasticisers, perfumes and food flavourings. Many small esters have sweet, fruity smells, which is why they appear in perfumes and flavourings.

Fats and oils

Vegetable oils and animal fats are esters of propane-1,2,3-triol (glycerol), HOCH₂CH(OH)CH₂OH. Each of the three –OH groups forms an ester link with a long-chain carboxylic acid, so one fat molecule contains three ester groups. These are often called triglycerides.

Hydrolysis of esters

Esters can be hydrolysed in two ways.

Condition	Products	Notes
Acid: heat under reflux with dilute acid	Carboxylic acid + alcohol	Reversible, so incomplete
Alkaline: heat under reflux with NaOH(aq)	Salt of the carboxylic acid + alcohol	Goes to completion

For ethyl propanoate:

CH₃CH₂COOCH₂CH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃CH₂OH (acid catalyst)

CH₃CH₂COOCH₂CH₃ + NaOH → CH₃CH₂COONa + CH₃CH₂OH

Alkaline hydrolysis goes to completion because the carboxylate ion formed does not react with the alcohol to re-form the ester. To get the free acid afterwards, add a strong acid.

Soap and biodiesel

Soap. Hydrolysing a vegetable oil or animal fat in alkaline conditions gives soap (salts of long-chain carboxylic acids) and glycerol.

Biodiesel is a mixture of methyl esters of long-chain carboxylic acids. It is made by reacting vegetable oils with methanol in the presence of a catalyst. Glycerol forms as well.

Worked example 4: soap and biodiesel from one fat

Glyceryl tridodecanoate, (C₁₁H₂₃COO)₃C₃H₅, has Mr 638.0. A 31.9 g sample is treated in two separate ways. Calculate the maximum mass of (a) sodium dodecanoate (Mr 222.0) from hydrolysis with NaOH, and (b) methyl dodecanoate (Mr 214.0) from reaction with methanol.

1. Moles of fat = 31.9 ÷ 638.0 = 0.0500 mol.
2. Soap equation: (C₁₁H₂₃COO)₃C₃H₅ + 3NaOH → 3C₁₁H₂₃COONa + C₃H₅(OH)₃.
3. Moles of soap = 3 × 0.0500 = 0.150 mol. Mass = 0.150 × 222.0 = 33.3 g.
4. Biodiesel equation: (C₁₁H₂₃COO)₃C₃H₅ + 3CH₃OH → 3C₁₁H₂₃COOCH₃ + C₃H₅(OH)₃.
5. Moles of methyl ester = 0.150 mol. Mass = 0.150 × 214.0 = 32.1 g.

The 1 : 3 ratio is the step to watch. One fat molecule holds three acid chains.

3.3.9.2 Acylation

Structures

Type	Group	Example
Acyl chloride	–COCl	CH₃COCl, ethanoyl chloride
Acid anhydride	–CO–O–CO–	(CH₃CO)₂O, ethanoic anhydride
Amide	–CONH₂ (or –CONHR)	CH₃CONH₂, ethanamide

An amide made from a primary amine has the alkyl group on nitrogen, for example CH₃CONHCH₃, N-methylethanamide.

The four reactions

Acyl chlorides and acid anhydrides both react with water, alcohols, ammonia and primary amines. In each case the acyl group (CH₃CO–) replaces an H on the nucleophile. The acyl chloride gives HCl as the by-product; the anhydride gives ethanoic acid.

Nucleophile	Product	Ethanoyl chloride	Ethanoic anhydride
Water	Carboxylic acid	CH₃COCl + H₂O → CH₃COOH + HCl	(CH₃CO)₂O + H₂O → 2CH₃COOH
Alcohol	Ester	CH₃COCl + CH₃OH → CH₃COOCH₃ + HCl	(CH₃CO)₂O + CH₃OH → CH₃COOCH₃ + CH₃COOH
Ammonia	Amide	CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl	(CH₃CO)₂O + 2NH₃ → CH₃CONH₂ + CH₃COONH₄
Primary amine	N-substituted amide	CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl	(CH₃CO)₂O + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃COONH₃CH₃

With ammonia and amines you need two molecules: one acts as the nucleophile, and the second reacts with the acid by-product to form a salt.

The specification suggests recording observations of these reactions. Ethanoyl chloride reacts vigorously with water, and the hydrogen chloride produced appears as misty fumes.

Mechanism: nucleophilic addition–elimination

You must be able to outline this mechanism for acyl chlorides with water, alcohols, ammonia and primary amines. The carbonyl carbon is δ+ because both O and Cl pull electron density away from it.

Worked example 5: ethanoyl chloride and water

Step 1 (addition)
  H2O:  →  C(δ+) of CH3–C(=O)–Cl     lone pair on O attacks the carbonyl C
  C=O  →  O                          the π electrons move onto O

  Intermediate:  CH3–C(O⁻)(Cl)–O⁺H2

Step 2 (elimination)
  O⁻ lone pair → C–O                  C=O re-forms
  C–Cl → Cl                           Cl⁻ leaves
  Cl⁻ removes H⁺ from O⁺H2            HCl forms

  Product: CH3COOH + HCl

In the intermediate, show the negative charge on O and the positive charge on the oxygen that came from water. Every arrow starts at a lone pair or a bond.

With an alcohol the attacking atom is the alcohol O, and the product is an ester. With ammonia or a primary amine the attacking atom is N, and the intermediate carries N⁺. For ammonia and amines, the H⁺ is removed by a second NH₃ or amine molecule.

Aspirin: ethanoic anhydride versus ethanoyl chloride

Aspirin is made by acylating 2-hydroxybenzoic acid. Industry uses ethanoic anhydride rather than ethanoyl chloride:

HOC₆H₄COOH + (CH₃CO)₂O → CH₃COOC₆H₄COOH + CH₃COOH

Advantages you can explain from the chemistry:

• The by-product is ethanoic acid, not hydrogen chloride. Hydrogen chloride is a corrosive, toxic gas.
• Ethanoic anhydride reacts less vigorously than ethanoyl chloride, so the reaction is easier to control.
• Ethanoic anhydride is hydrolysed less readily by water, including moisture in the air, so it is easier to store and handle.

Required practical 10

Preparation of a pure organic solid and test of its purity. A typical example is aspirin.

1. React, then collect the crude solid by filtration under reduced pressure.
2. Recrystallise: dissolve the solid in the minimum volume of hot solvent, so the product crystallises on cooling.
3. Filter the hot solution if insoluble impurities are present.
4. Cool so crystals form. Soluble impurities stay in solution.
5. Filter under reduced pressure, wash with a little cold solvent, and dry.
6. Measure the melting point. A pure solid melts sharply at the value given in a data source. An impure solid melts lower and over a wider range.

Preparation of a pure organic liquid. A typical example is an ester.

1. Heat the acid, alcohol and acid catalyst under reflux.
2. Distil off the crude ester.
3. Shake with sodium hydrogencarbonate solution in a separating funnel to remove acid. Release the pressure often, because CO₂ forms. Keep the organic layer.
4. Dry with an anhydrous drying agent, such as anhydrous sodium sulfate, then decant or filter.
5. Redistil, collecting the fraction that boils at the ester’s boiling point.

Common errors

• Naming esters backwards. CH₃COOCH₂CH₃ is ethyl ethanoate; CH₃CH₂COOCH₃ is methyl propanoate.
• Calling a carboxylic acid “strong” because it fizzes with carbonates. It is weak: partial dissociation.
• Forgetting that acid hydrolysis is reversible and alkaline hydrolysis gives a salt, not the acid.
• Using a 1 : 1 ratio for fats. A fat gives three soap or biodiesel molecules.
• Writing CH₃COCl + NH₃ → CH₃CONH₂ + HCl. In excess ammonia the HCl becomes NH₄Cl, so use 2NH₃.
• Calling the mechanism “nucleophilic substitution”. The AQA name is nucleophilic addition–elimination.
• Missing the charges on the intermediate, or showing Cl⁻ leaving before the C=O re-forms.
• Using a large volume of hot solvent in recrystallisation, which loses product.

Next steps

For revision planning across the course, see the AQA A-Level Chemistry exam preparation guide.

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. Published by AQA. This guide covers section 3.3.9 and Required practical 10.