OCR A Level Chemistry A: Carboxylic acids and esters (H432)

This study guide teaches section 6.1.3, Carboxylic acids and esters, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The section sits in Module 6, Organic chemistry and analysis, and its lettered learning outcomes, 6.1.3(a) to (f), are all covered below with worked examples. H432 is the full A level. The specification states that Module 6 content is assessed in Synthesis and analytical techniques (02) and Unified chemistry (03). Outcome 6.1.3(b) is linked to PAG7, qualitative analysis of organic functional groups.

Every H432 topic is linked from the OCR chemistry hub; tick outcomes off on the printable checklist. For a quick check of where you stand across the course, try the free diagnostics.

Outcomes in 6.1.3

Outcome	What you must be able to do	Additional guidance in the specification
6.1.3(a)	Explain why carboxylic acids dissolve in water, using hydrogen bonding	none
6.1.3(b)	Write equations for carboxylic acids reacting in aqueous conditions with metals, carbonates, metal oxides and alkalis	Comparing the acidity of different carboxylic acids is not required. PAG7 (see 6.3.1 c)
6.1.3(c)(i)	Make esters from carboxylic acids and alcohols with an acid catalyst such as concentrated H₂SO₄	none
6.1.3(c)(ii)	Make esters from acid anhydrides and alcohols	none
6.1.3(d)(i)	Hydrolyse esters in hot aqueous acid to carboxylic acids and alcohols	none
6.1.3(d)(ii)	Hydrolyse esters in hot aqueous alkali to carboxylate salts and alcohols	none
6.1.3(e)	Form acyl chlorides from carboxylic acids using SOCl₂	none
6.1.3(f)	Use acyl chlorides to make esters, carboxylic acids, and primary and secondary amides	Includes esterification of phenol, which is not readily esterified by carboxylic acids

The functional groups in this section

A carboxylic acid contains the carboxyl group, –COOH. Name it by counting all the carbons, including the carboxyl carbon, and ending in -oic acid. CH₃CH₂CH₂CH₂COOH is pentanoic acid; (CH₃)₂CHCH₂COOH is 3-methylbutanoic acid.

The derivatives you meet here replace the –OH of the carboxyl group:

• ester, –COOR’: named alkyl alkanoate. The alkyl part comes from the alcohol and is written first; the alkanoate part comes from the acid.
• acid anhydride, (RCO)₂O: two acyl groups joined through one oxygen, e.g. propanoic anhydride, (CH₃CH₂CO)₂O.
• acyl chloride, –COCl: named with -oyl chloride, e.g. pentanoyl chloride.
• amide: –CONH₂ is a primary amide; –CONHR’ is a secondary amide.

6.1.3(a): Why carboxylic acids dissolve in water

The carboxyl group can hydrogen bond to water in two ways:

• the δ+ hydrogen of the acid’s O–H bonds to a lone pair on a water oxygen;
• a δ+ hydrogen of a water molecule bonds to a lone pair on the acid’s C=O oxygen (and on the O–H oxygen).

So the short-chain acids, methanoic to butanoic acid, are miscible with water.

As the hydrocarbon chain gets longer, solubility falls. The non-polar chain cannot hydrogen bond to water and disrupts water’s hydrogen-bond network. Hexanoic acid is only slightly soluble.

In an answer, name the bond and say which atoms take part: δ+ H to a lone pair on O.

6.1.3(b): Reactions in aqueous conditions

Carboxylic acids are weak acids. In water they partly ionise:

RCOOH(aq) ⇌ RCOO⁻(aq) + H⁺(aq)

They still give the usual acid reactions, forming carboxylate salts. Name the salt with the metal first and the acid ending -oate: sodium pentanoate, magnesium propanoate.

Reacts with	General equation (monoprotic acid)	What you see
Reactive metal, e.g. Mg	2RCOOH + Mg → (RCOO)₂Mg + H₂	fizzing; metal dissolves
Carbonate, e.g. Na₂CO₃	2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂	fizzing; gas turns limewater milky
Metal oxide, e.g. CuO	2RCOOH + CuO → (RCOO)₂Cu + H₂O	black solid dissolves on warming; blue solution
Alkali, e.g. KOH	RCOOH + KOH → RCOOK + H₂O	no visible change

Ionic equations keep the weak acid as RCOOH, because it is mostly un-ionised:

2RCOOH + CO₃²⁻ → 2RCOO⁻ + H₂O + CO₂ RCOOH + OH⁻ → RCOO⁻ + H₂O

The carbonate test (PAG7)

Fizzing with a carbonate, giving CO₂, identifies a carboxylic acid on a test-tube scale (6.3.1(c)(vii)). Phenols are also weakly acidic, but they do not react with carbonates (6.3.1(c)(iii)). So a compound that dissolves in NaOH(aq) but gives no fizzing with Na₂CO₃(aq) is a phenol, not a carboxylic acid.

The specification states that comparing the acidity of different carboxylic acids is not required.

Worked example 1: a metal and a carboxylic acid

0.851 g of magnesium ribbon reacts completely with excess aqueous propanoic acid. Use Ar(Mg) = 24.3, Ar(C) = 12.0, Ar(H) = 1.0, Ar(O) = 16.0 and a gas molar volume of 24.0 dm³ mol⁻¹.

(a) Write the equation. (b) Calculate the volume of hydrogen, in cm³. (c) Calculate the mass of salt formed.

(a) 2CH₃CH₂COOH + Mg → (CH₃CH₂COO)₂Mg + H₂

(b) n(Mg) = 0.851 ÷ 24.3 = 0.0350 mol. One Mg gives one H₂, so n(H₂) = 0.0350 mol. Volume = 0.0350 × 24.0 = 0.840 dm³ = 840 cm³.

(c) Magnesium propanoate is (C₃H₅O₂)₂Mg. Mr = 2(36.0 + 5.0 + 32.0) + 24.3 = 170.3. Mass = 0.0350 × 170.3 = 5.96 g.

6.1.3(c): Making esters

(i) Carboxylic acid + alcohol, acid catalyst

Heat the acid and alcohol with a few drops of concentrated sulfuric acid. The reaction is reversible, so the mixture reaches equilibrium and the yield is less than 100%.

CH₃CH₂CH₂CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂CH₂CH₂COOCH₂CH₃ + H₂O pentanoic acid + ethanol ⇌ ethyl pentanoate + water

To name the ester, take the alcohol’s chain as the alkyl group (ethanol → ethyl) and the acid’s name with -oic acid changed to -oate (pentanoic acid → pentanoate).

Worked example 2: yield of an ester

A student heats 5.10 g of pentanoic acid with excess ethanol and concentrated H₂SO₄. After purification, 4.55 g of ethyl pentanoate is collected. Calculate the percentage yield. Use Ar(C) = 12.0, Ar(H) = 1.0, Ar(O) = 16.0.

1. Mr(pentanoic acid, C₅H₁₀O₂) = 60.0 + 10.0 + 32.0 = 102.0
2. n(acid) = 5.10 ÷ 102.0 = 0.0500 mol
3. One acid molecule gives one ester molecule: at most 0.0500 mol of ester
4. Mr(ethyl pentanoate, C₇H₁₄O₂) = 84.0 + 14.0 + 32.0 = 130.0
5. Theoretical mass = 0.0500 × 130.0 = 6.50 g
6. Percentage yield = 4.55 ÷ 6.50 × 100 = 70.0%

(ii) Acid anhydride + alcohol

An acid anhydride reacts with an alcohol on warming to give an ester and a carboxylic acid (not water). No acid catalyst is needed and the reaction is not reversible, so the yield is better than with the acid.

(CH₃CH₂CO)₂O + CH₃CH₂CH₂CH₂OH → CH₃CH₂COOCH₂CH₂CH₂CH₃ + CH₃CH₂COOH propanoic anhydride + butan-1-ol → butyl propanoate + propanoic acid

6.1.3(d): Hydrolysis of esters

Both versions use heat, usually under reflux.

(i) Hot aqueous acid, e.g. dilute H₂SO₄ or HCl. This is the reverse of esterification and is reversible, so hydrolysis is incomplete:

CH₃CH₂COOCH₂CH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃CH₂OH

(ii) Hot aqueous alkali, e.g. NaOH(aq). This goes to completion. The products are the carboxylate salt and the alcohol:

CH₃CH₂COOCH₂CH₃ + NaOH → CH₃CH₂COONa + CH₃CH₂OH

The carboxylate ion does not react with the alcohol, so the reaction cannot reverse. Adding a strong acid afterwards gives the free carboxylic acid.

Worked example 3: identifying an ester from its hydrolysis products

An ester with formula C₇H₁₄O₂ is refluxed with excess KOH(aq). One product is potassium butanoate. The other is an alcohol that acidified dichromate oxidises to propanone. A 1.95 g sample of the ester is used.

(a) Identify the ester. (b) Find the mass of potassium butanoate made. Use Ar(K) = 39.1, Ar(C) = 12.0, Ar(H) = 1.0, Ar(O) = 16.0.

(a) Potassium butanoate tells you the acyl part is CH₃CH₂CH₂COO–. Propanone comes from oxidising a secondary alcohol, propan-2-ol, (CH₃)₂CHOH. So the ester is CH₃CH₂CH₂COOCH(CH₃)₂, propan-2-yl butanoate. Check: C₄H₇O₂ + C₃H₇ = C₇H₁₄O₂.

(b) Mr(C₇H₁₄O₂) = 84.0 + 14.0 + 32.0 = 130.0; n = 1.95 ÷ 130.0 = 0.0150 mol. Mr(CH₃CH₂CH₂COOK) = 48.0 + 7.0 + 32.0 + 39.1 = 126.1. Mass = 0.0150 × 126.1 = 1.89 g.

6.1.3(e): Making acyl chlorides with SOCl₂

Thionyl chloride, SOCl₂, converts a carboxylic acid into an acyl chloride. Both by-products are gases, which leave the mixture and make the acyl chloride easier to isolate:

RCOOH + SOCl₂ → RCOCl + SO₂ + HCl

For 3-methylbutanoic acid:

(CH₃)₂CHCH₂COOH + SOCl₂ → (CH₃)₂CHCH₂COCl + SO₂ + HCl

The product is 3-methylbutanoyl chloride. Keep everything dry: acyl chlorides react with water.

6.1.3(f): Acyl chlorides in synthesis

Acyl chlorides are far more reactive than carboxylic acids. The carbonyl carbon, bonded to both O and Cl, carries a large δ+ charge and is attacked readily by nucleophiles. Reactions happen at room temperature, are not reversible and release HCl.

Worked example 4: one acyl chloride, five products

Give the organic product when 3-methylbutanoyl chloride, written as RCOCl with R = (CH₃)₂CHCH₂–, reacts with each reagent.

Reagent	Equation	Organic product
Water	RCOCl + H₂O → RCOOH + HCl	3-methylbutanoic acid (carboxylic acid)
Propan-1-ol	RCOCl + CH₃CH₂CH₂OH → RCOOCH₂CH₂CH₃ + HCl	propyl 3-methylbutanoate (ester)
Phenol	RCOCl + C₆H₅OH → RCOOC₆H₅ + HCl	phenyl 3-methylbutanoate (ester)
Ammonia	RCOCl + 2NH₃ → RCONH₂ + NH₄Cl	3-methylbutanamide (primary amide)
Ethylamine	RCOCl + 2CH₃CH₂NH₂ → RCONHCH₂CH₃ + CH₃CH₂NH₃⁺Cl⁻	N-ethyl-3-methylbutanamide (secondary amide)

Notes on the table:

• With water the reaction is vigorous and misty fumes of HCl are given off.
• With ammonia or an amine, two molecules are needed. The first forms the amide; the second, being a base, removes the HCl as an ammonium salt.
• An amide made from ammonia is primary (–CONH₂). One made from a primary amine has one alkyl group on the nitrogen and is secondary (–CONHR’).

Why phenol needs an acyl chloride

The specification notes that phenol is not readily esterified by carboxylic acids. Heating phenol with a carboxylic acid and an acid catalyst gives very little ester. An acyl chloride reacts with phenol at room temperature, so this is the standard route to a phenyl ester.

Comparing routes to an ester

Route	Conditions	Reversible?	By-product
Carboxylic acid + alcohol	heat, conc. H₂SO₄ catalyst	yes	water
Acid anhydride + alcohol	warm	no	carboxylic acid
Acyl chloride + alcohol or phenol	room temperature	no	HCl

Common errors

• Naming the ester the wrong way round: the alcohol gives the first word. Propyl ethanoate comes from propan-1-ol and ethanoic acid, not ethanol and propanoic acid.
• Writing water as a product of an anhydride esterification. The second product is a carboxylic acid.
• Giving a carboxylic acid, rather than its salt, as the product of alkaline hydrolysis.
• Leaving out HCl when an acyl chloride reacts, or using only one NH₃ with an acyl chloride.
• Writing H⁺ for a carboxylic acid in an ionic equation as if it were fully ionised.
• Saying a phenol fizzes with sodium carbonate. It does not; that is the test that separates it from a carboxylic acid.

Where to go next

For fast recall before an exam, use the Carboxylic acids and esters revision notes, then test yourself with the Carboxylic acids and esters practice questions. The mole methods used above are in Amount of substance, and acid-base reactions of salts are in Acids.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.1.3 Carboxylic acids and esters, learning outcomes (a) to (f). Published by OCR.