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How do carboxylic acids and their derivatives react?

Carboxylic acids and their acidity, esters, acyl chlorides and acid anhydrides, and the reactions and interconversions of these derivatives.

A focused answer to WJEC A-Level Chemistry Unit 4, covering the acidity and reactions of carboxylic acids, esters, acyl chlorides and acid anhydrides, and how these derivatives interconvert and react with nucleophiles.

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  1. What this dot point is asking
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What this dot point is asking

WJEC wants you to describe the acidity and reactions of carboxylic acids, and the structures, reactions and interconversions of their derivatives: esters, acyl chlorides and acid anhydrides.

The answer

Carboxylic acids

The derivatives

Interconversion and acylation

Reactivity of the derivatives explained

The reactivity order acyl chloride > anhydride > ester > acid follows the leaving group released when a nucleophile attacks the carbonyl carbon. An acyl chloride loses chloride, a weak base and excellent leaving group, so it reacts fastest. An anhydride loses a carboxylate ion, a slightly poorer leaving group. An ester loses an alkoxide and an acid loses hydroxide, both poor leaving groups, so they react slowest and often only as reversible equilibria. The carbonyl carbon in every derivative is electron-deficient because of the electronegative oxygen, but the more electron-withdrawing the attached group, the more δ+\delta+ the carbon and the faster the attack.

Distinguishing and interconverting

Reactant Reagent Product Observation
Acyl chloride Water Carboxylic acid Steamy HCl fumes
Acyl chloride Alcohol Ester Steamy HCl fumes
Acyl chloride Ammonia Amide White solid, HCl
Carboxylic acid Carbonate Salt Effervescence
Ester Aqueous acid (reflux) Acid plus alcohol Reversible
Ester Aqueous base (reflux) Carboxylate plus alcohol Saponification

This map lets you move between any two derivatives in one or two steps, the skill assessed in synthesis questions.

Examples in context

Aspirin synthesis. Aspirin is made by acylating salicylic acid with ethanoic anhydride, the preferred industrial acylating agent because it is cheaper and less corrosive than the acyl chloride. Polyesters. Esterification of diacids with diols produces polyesters such as Terylene, used in fabrics and bottles, an industrial-scale application of the ester linkage.

Try this

Q1. State the observation when a carboxylic acid is added to sodium carbonate. [1 mark]

  • Cue. Effervescence (carbon dioxide gas is released).

Q2. Name the product of the reaction between ethanoyl chloride and water. [1 mark]

  • Cue. Ethanoic acid (plus hydrogen chloride).

Q3. State which is the more reactive acylating agent: an ester or an acyl chloride. [1 mark]

  • Cue. The acyl chloride.

Q4. Explain why an acyl chloride reacts faster with an alcohol than a carboxylic acid does. [2 marks]

  • Cue. The acyl chloride loses chloride, an excellent leaving group, and its carbonyl carbon is more δ+\delta+; the acid would lose hydroxide, a poor leaving group, so it reacts only slowly and reversibly.

Q5. Name the product of refluxing ethyl ethanoate with aqueous sodium hydroxide. [1 mark]

  • Cue. Sodium ethanoate and ethanol (saponification).

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20204 marksDescribe the reaction of ethanoyl chloride with ethanol, naming the organic product and the type of reaction, and give an equation.
Show worked answer →

Acyl chlorides react vigorously with alcohols in a nucleophilic substitution (addition-elimination): the alcohol oxygen attacks the δ+\delta+ carbonyl carbon and chloride is displaced.

CH3COCl+C2H5OH→CH3COOC2H5+HCl\text{CH}_3\text{COCl} + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{CH}_3\text{COOC}_2\text{H}_5 + \text{HCl}.

The organic product is the ester ethyl ethanoate, and steamy white fumes of hydrogen chloride are released.

Markers reward the ester product, the type of reaction, the equation, and the observation of HCl fumes.

WJEC 20184 marksA 0.100 mol dm−30.100\ \text{mol dm}^{-3} solution of a weak monoprotic carboxylic acid has a pH of 2.88. Calculate the acid dissociation constant KaK_a.
Show worked answer →

[H+]=10−pH=10−2.88=1.318×10−3[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.88} = 1.318 \times 10^{-3} mol dm-3.

For a weak monoprotic acid, [A−]=[H+][\text{A}^-] = [\text{H}^+] and [HA]≈0.100[\text{HA}] \approx 0.100 mol dm-3.

Ka=[H+][A−][HA]=(1.318×10−3)20.100=1.737×10−60.100=1.74×10−5K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \dfrac{(1.318 \times 10^{-3})^2}{0.100} = \dfrac{1.737 \times 10^{-6}}{0.100} = 1.74 \times 10^{-5} mol dm-3.

Markers reward converting pH to [H+][\text{H}^+], the weak-acid assumptions, and the value of KaK_a with units.

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