Whiteboard Chemistry with Joe White

Carboxylic Acids & Derivatives

Carboxylic acids as weak acids, esters and esterification, the hydrolysis of esters, and the reactive acylating agents — acyl chlorides and acid anhydrides — through the nucleophilic addition–elimination mechanism.

AQA 7404/7405 Paper 2 A-level only
Hero motif · to come One acyl group, four derivatives
Where this sits

Oxidising an aldehyde (from 3.3.8) gives a carboxylic acid, and this page follows the family that grows from it. The unifying idea is the acyl group, R–C(=O)–: swap what is attached to it and you move between the acid, ester, acyl chloride, acid anhydride and amide. One mechanism — nucleophilic addition–elimination — links most of them.

Carboxylic acids

Carboxylic acids contain the –COOH group. They are weak acids — they only partially dissociate in water — but they are acidic enough to show the usual reactions:

  • with carbonates and hydrogencarbonates they release CO2 (effervescence) — the test that distinguishes a carboxylic acid from a phenol or alcohol;
  • with bases they form carboxylate salts + water.

Warmed with an alcohol and a concentrated sulfuric acid catalyst, a carboxylic acid forms an ester — the reaction is reversible (esterification), so it reaches an equilibrium:

Figure F1 · illustration to come Esterification & the acid test Acid + alcohol ⇌ ester; acid + carbonate → CO2.
Only a carboxylic acid (not a phenol or alcohol) fizzes with a carbonate.
🧪 Exam-style questions
Q1 [3 marks]

An alcohol is warmed with ethanoic acid and a few drops of concentrated sulfuric acid, then poured into sodium hydrogencarbonate solution. Suggest a simple way to detect if a reaction occurred, and give a reason for pouring the mixture into sodium hydrogencarbonate solution.

Show answer

A reaction has occurred if there is a sweet / fruity smell (an ester forms). 1 mark

The sodium hydrogencarbonate reacts with / removes (neutralises) the excess acid… 1 mark

…so that the ester smell can be detected clearly. 1 mark

Source: AQA A-Level Chemistry past papers.

Esters & hydrolysis

Esters have the group –COO– and are named alkyl alkanoate — the alkyl part comes from the alcohol, the alkanoate part from the acid. So ethanoic acid + ethanol gives ethyl ethanoate.

Naming, in two halves

Ethyl ethanoate, CH3COOC2H5: ethyl = the C2H5 from ethanol; ethanoate = the CH3COO from ethanoic acid. Write the alcohol-derived group first, the acid-derived group second.

Hydrolysis — the two routes

Hydrolysis splits the ester back apart. Which products you get depends on the conditions:

  • Acid hydrolysis (dilute acid, heat under reflux) is the reverse of esterification, so it is reversible and reaches an equilibrium — giving the carboxylic acid + alcohol.
  • Base hydrolysis (saponification) (aqueous NaOH, heat under reflux) goes to completion because the carboxylate salt formed does not react back — giving the carboxylate salt + alcohol.
Figure F2 · illustration to come Acid vs base hydrolysis · triglycerides One reversible, one one-way; fats are triesters of glycerol.
Base hydrolysis (saponification) goes to completion because the carboxylate salt cannot re-esterify.

Fats, oils & biodiesel

Fats and oils are triesters (triglycerides) of the triol glycerol (propane-1,2,3-triol) with three long-chain fatty acids. Biodiesel is a mixture of methyl esters of those fatty acids, made by reacting the triglyceride with methanol and a catalyst (transesterification), which releases glycerol as a by-product.

Uses of esters

Their pleasant smells and low reactivity make esters useful as solvents, plasticisers, and in perfumes and flavourings.

🧪 Exam-style questions
Q2 [1 mark]

Which compound is formed when phenyl benzenecarboxylate (C6H5COOC6H5) is hydrolysed under acidic conditions?

  1. C6H5CH2OH
  2. C6H5CHO
  3. C6H5COCH3
  4. C6H5COOH
Acid hydrolysis of an ester gives the carboxylic acid + alcohol. Here the acid part is benzenecarboxylic (benzoic) acid, C6H5COOH; the alcohol part is phenol, C6H5OH.
Q3 [3 marks]

Ethyl ethanoate is heated under reflux with aqueous sodium hydroxide. Name this type of reaction, write an equation, and state why it goes to completion rather than reaching an equilibrium.

Show answer

Base hydrolysis / saponification. 1 mark

CH3COOC2H5 + NaOH → CH3COONa + C2H5OH 1 mark

The ethanoate ion (carboxylate salt) formed does not react back with the alcohol, so the reaction is not reversible / goes to completion. 1 mark

Source: AQA A-Level Chemistry past papers.

Acyl chlorides & acid anhydrides

Replace the –OH of a carboxylic acid with –Cl and you get an acyl chloride (e.g. ethanoyl chloride, CH3COCl); join two acid molecules with loss of water and you get an acid anhydride (e.g. ethanoic anhydride, (CH3CO)2O). Both are acylating agents: they hand the acyl group CH3CO– to a nucleophile. Acyl chlorides are the more reactive of the two.

The four nucleophiles — one pattern

Each of the four common nucleophiles attacks the acyl group and displaces the leaving group. The product is set by the nucleophile:

Reactions of ethanoyl chloride, CH3COCl, with the four nucleophiles
NucleophileOrganic productEquation
Watercarboxylic acidCH3COCl + H2O → CH3COOH + HCl
Alcohol (e.g. ethanol)esterCH3COCl + C2H5OH → CH3COOC2H5 + HCl
AmmoniaamideCH3COCl + 2NH3 → CH3CONH2 + NH4Cl
Primary amine (e.g. CH3NH2)N-substituted amideCH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3Cl
With ammonia and amines, two equivalents are needed

One molecule of the nucleophile forms the product; a second mops up the HCl released (as NH4+Cl or the alkylammonium salt). Balance the equation with 2NH3 / 2 amine, not one.

The mechanism: nucleophilic addition–elimination

Every one of those reactions goes by the same two-stage mechanism. The carbonyl carbon is Cδ+ (electron-poor), so:

  1. Addition: the nucleophile’s lone pair attacks Cδ+; the C=O π bond breaks onto the oxygen, giving a tetrahedral intermediate.
  2. Elimination: the C=O reforms and pushes out the leaving group (Cl from an acyl chloride, or the carboxylate from an anhydride). A proton is then lost from the nucleophile to give the neutral product.
Figure F3 · mechanism · illustration to come Nucleophilic addition–elimination + the four products Nu adds to Cδ+, C=O reforms and ejects the leaving group, then −H+.
The name to write is nucleophilic addition–elimination — not substitution.

Interactive nucleophile switcher loads here. The worked pathway below (water → carboxylic acid) is the static fallback.

Acyl chlorides vs anhydrides — and aspirin

Acyl chlorides react violently with water, giving off misty fumes of HCl; the Cl released also gives a white precipitate with acidified silver nitrate. Acid anhydrides are gentler and easier to control. That is why the industrial synthesis of aspirin uses ethanoic anhydride rather than ethanoyl chloride:

Why the anhydride is preferred for aspirin

Ethanoic anhydride is cheaper, less corrosive and less vulnerable to hydrolysis than the acyl chloride, and its reactions are easier to control. Crucially it does not release corrosive HCl gas — the only by-product is ethanoic acid.

🧪 Exam-style questions
Q4 [1 mark]

Which reaction involves addition–elimination?

  1. (CH3)2CHBr + KOH → CH3CH=CH2 + KBr + H2O
  2. CH3COCl + C6H5OH → CH3COOC6H5 + HCl
  3. CH3CH=CH2 + Cl2 → CH3CHClCH2Cl
  4. CH3CH2CH2Br + NaOH → CH3CH2CH2OH + NaBr
Only the acyl chloride reacts by nucleophilic addition–elimination. The others are elimination, electrophilic addition and nucleophilic substitution respectively.
Q5 [1 mark]

Which compound reacts with warm dilute aqueous sodium hydroxide?

  1. C6H6
  2. CH3CH=CH2
  3. CH3CH2CH2NH2
  4. (CH3CO)2O
The acid anhydride is hydrolysed / reacts with the alkali. Benzene, propene and the amine do not react with warm dilute NaOH.
Q6 [1 mark]

Which compound forms a white precipitate when added to aqueous silver nitrate?

  1. bromoethane
  2. ethanal
  3. ethanoic anhydride
  4. ethanoyl chloride
Ethanoyl chloride reacts vigorously with the water in the reagent, releasing Cl which gives an immediate white precipitate of AgCl. Bromoethane would need warming and would give a cream precipitate.
Q7 [3 marks]

Aspirin is produced by reacting salicylic acid with ethanoic anhydride. Give one reason why ethanoic anhydride is used rather than ethanoyl chloride, and outline how you would test the by-product to show that no acyl chloride was used.

Show answer

Ethanoic anhydride is cheaper / less corrosive / less readily hydrolysed / does not release corrosive HCl gas (any one). 1 mark

Add (acidified) silver nitrate solution to the reaction mixture. 1 mark

An acyl chloride would give a white precipitate (of AgCl); with the anhydride there is no white precipitate, confirming no chloride is present. 1 mark

Source: AQA A-Level Chemistry past papers.

The derivatives map

Pull the whole topic together around the acyl group. Reactivity runs acyl chloride > acid anhydride > ester > carboxylic acid/amide — the better the leaving group, the more reactive the acylating agent. Reading the map is the exam skill: name the reagent that makes each product, and vice versa.

Which reagent makes which — reactions of ethanoyl chloride / ethanoic anhydride
Add this nucleophileGet this productProduct type
WaterCH3COOHcarboxylic acid
AlcoholCH3COORester
AmmoniaCH3CONH2amide
Primary amineCH3CONHRN-substituted amide
The marks that get dropped
  • Naming the mechanism substitution — it is nucleophilic addition–elimination (the tetrahedral intermediate forms first).
  • Forgetting the second equivalent of NH3 / amine that removes the HCl — the equation must balance.
  • Confusing the hydrolysis routes: acid gives the acid + alcohol (reversible); base gives the carboxylate salt + alcohol (one-way).
  • For the acyl chloride, missing the observations: violent reaction with water, misty HCl fumes, white precipitate with AgNO3.

Examiner reports repeatedly note that the addition–elimination mechanism is drawn carelessly — the curly arrow must begin at the nucleophile’s lone pair and the elimination arrow must run from the C–Cl bond as the C=O reforms. Precise arrows earn the marks; vague ones do not.

🎯 Build the mark scheme — acylation of an amine

Ethanoyl chloride reacts with methylamine. Name the mechanism, give the balanced equation and the organic product, and give one observation if water were used instead. Award yourself a mark for each:

  • Mechanism: nucleophilic addition–elimination.
  • Equation: CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3Cl.
  • Organic product: N-methylethanamide.
  • With water instead: violent reaction / misty fumes of HCl / white precipitate with AgNO3.
3.3.9 Carboxylic acids & derivatives — Quick-reference summary
  • Carboxylic acids (–COOH) are weak acids: they fizz with carbonates (CO2 — the distinguishing test) and form salts with bases.
  • Esterification: carboxylic acid + alcohol ⇌ ester + water (conc H2SO4 catalyst, reversible).
  • Ester hydrolysis: acid hydrolysis is reversible (→ acid + alcohol); base hydrolysis (saponification) is one-way (→ carboxylate salt + alcohol). Fats/oils are triesters of glycerol; biodiesel is their methyl esters.
  • Acyl chlorides (RCOCl) and acid anhydrides ((RCO)2O) are reactive acylating agents. With a nucleophile they give: + H2O → acid; + alcohol → ester; + NH3 → amide; + amine → N-substituted amide.
  • Mechanism: nucleophilic addition–elimination. Aspirin is made from salicylic acid + ethanoic anhydride (cheaper and less corrosive than the acyl chloride).

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This topic rewards a clear grasp of the derivatives map and the addition–elimination mechanism: which reagent makes which product, and the arrows for the nucleophile adding then the leaving group departing. Sessions drill acylation and the aspirin synthesis on real AQA past questions.

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