Year 12

AQA A level Chemistry Carboxylic Acids and Esters Notes

Carboxylic Acids & Esters

— Chemistry · AS Level (7405) · Crash Course —
★ Exam Tip: This chapter is mostly revision of reactions you have already met (oxidation, nitrile hydrolysis, esterification) plus ester hydrolysis (acidic vs alkaline), LiAlH₄ reduction and hydroxy-acids forming cyclic esters. Sir: “organic — revise it all, learn it properly, and keep practising past papers.”

1. Carboxylic Acids

Definition: organic compounds containing the functional group –C(=O)–O–H (–COOH). Naming: the –COOH carbon is always C1; suffix -oic acid; an OH elsewhere is written as the prefix hydroxy- (e.g. 2-hydroxypropanoic acid = lactic acid).

OO–HRcarboxylic acid R–COOH

Physical properties

  • Higher m.p. and b.p. than alcohols/alkanes of similar size — hydrogen bonding (between –C=O and –O–H) plus permanent dipoles; acids often exist as hydrogen-bonded dimers.
  • Small carboxylic acids are soluble in water (hydrogen bonding with water); solubility falls as the chain gets longer. They are only weak acids — partially dissociated, unlike HCl or H₂SO₄.

Formation

Method Conditions Where
Oxidation of a primary alcohol or an aldehyde acidified K₂Cr₂O₇, reflux (alcohol) / [O] (aldehyde) done in Alcohols & Carbonyls
Hydrolysis of a nitrile (R–CN → R–COOH) dil. H₂SO₄ or HCl, heat under reflux (alkali then dil. HCl also works) done in Halogenoalkanes (board says “Alcohols”)
Hydrolysis of an ester dil. acid or alkali, then acid section 3 below

2. Reactions of Carboxylic Acids

① All the “acid-like” reactions — salt formation

Reacts with Gives Example Observation
With a metal (Na, Mg…) salt + H₂ CH₃COOH + Na → CH₃COONa + ½H₂ effervescence (H₂)
With a base / metal hydroxide salt + H₂O CH₃CH₂COOH + KOH → CH₃CH₂COOK + H₂O neutralisation (no visible change)
With a metal oxide salt + H₂O 2CH₃CH₂COOH + CaO → (CH₃CH₂COO)₂Ca + H₂O oxide dissolves
With a carbonate / hydrogencarbonate salt + H₂O + CO₂ 2CH₃COOH + CaCO₃ → (CH₃COO)₂Ca + H₂O + CO₂ effervescence (CO₂) — test for –COOH
Exam tip: Test for a carboxylic acid: add Na₂CO₃ or NaHCO₃ ⇒ effervescence of CO₂ (turns limewater milky). Alcohols and phenols don’t do this — that is why Na₂CO₃ distinguishes an acid from an alcohol (MCQ 2: MCPA has –COOH, 2,4-D has an alcohol ⇒ Na₂CO₃). Na reacts with both –COOH and –OH (MCQ 7: all three O–H in malic acid react with Na).

② Esterification

Already discussed in Alcohols: carboxylic acid + alcohol ⇌ ester + water, conc. H₂SO₄ + heat.

③ Reduction

CH₃CO₂H + 4[H] → CH₃CH₂OH + H₂O (LiAlH₄ in dry ether)
  • LiAlH₄ only (in dry ether). NaBH₄ cannot be used — it is not strong enough to reduce a carboxylic acid (nor an ester).
  • The reaction always gives the primary alcohol — it cannot be stopped at the aldehyde.
  • Sir: sometimes the question doesn’t ask for reagent/solvent — then just say “reduction”.

④ Chlorination — acyl chloride

CH₃COOH + PCl₅ → CH₃COCl + HCl + POCl₃

PCl₅ is also the test for the –OH group (steamy fumes of HCl). Product: an acyl chloride (R–COCl). (PCl₃ and SOCl₂ do the same — but PCl₅ is the one usually asked at AS.)

R–COOHsalt R–COO⁻M⁺ (Na, base, carbonate)primary alcohol R–CH₂OH (LiAlH₄)acyl chloride R–COCl (PCl₅)ester R–COOR′ (R′OH, conc. H₂SO₄)
carboxylic acid hub: four families of reaction

3. Esters

Definition: organic compounds containing the group –C(=O)–O– (R–COO–R′). Uses: perfumes, glues, artificial food flavourings (fruity smells — e.g. pineapple, banana).

Careful (sir): Naming and drawing: alkyl (from the alcohol) + alkanoate (from the acid); the acid part is written acyl-first, alcohol part is the alkyl. Formulae like CH₃CO₂CH₂CH₃ are fine. Never write an ester as “O–C–O” or “CO₂” attached wrongly — the O of the alcohol is bonded to the alkyl group and to the C=O carbon; check each atom’s bonds.

Hydrolysis of esters

Hydrolysis = splitting by water — the reverse of esterification. Two ways:

Condition Reaction Notes
① Acidic conditions ester + H₂O ⇌ alcohol + carboxylic acid; catalyst H⁺ (dil. H₂SO₄), heat under reflux reversible — the equilibrium mixture contains all four, hard to separate
② Alkaline conditions ester + NaOH → alcohol + sodium carboxylate; heat under reflux goes to completion (irreversible); alcohol is easily distilled off; then add dil. HCl to convert the salt to the carboxylic acid, and distil that off
CH₃CH₂COOCH₂CH₃ + H₂O ⇌ CH₃CH₂OH + CH₃CH₂COOH (H⁺)CH₃CH₂COOCH₂CH₃ + NaOH → CH₃CH₂OH + CH₃CH₂COONa (alkali)CH₃CH₂COONa + HCl → CH₃CH₂COOH + NaCl (to get the acid)
esterH⁺ / H₂O: reversibleNaOH: completealcohol + acidalcohol + saltthen dil. HCl ⇒ acid
acid hydrolysis gives an equilibrium mixture; alkaline hydrolysis goes to completion
Why two methods: Sir: examiners usually ask for the acidic one in the paper (“dilute H₂SO₄, heat”), but alkaline hydrolysis is better in practice because it isn’t reversible. For separating: the alcohol and acid in an equilibrium mixture can be distilled apart if their boiling points are very different (e.g. octanoic acid vs ethanol). Ester + water: the ester doesn’t dissolve (only temporary dipoles) while the alcohol and acid do (hydrogen bonding).

4. Hydroxy-carboxylic Acids

Definition: organic compounds containing both –OH and –COOH groups. Heated with conc. H₂SO₄ they form esters with themselves — cyclic esters — in two ways:

Type What happens Board examples
① Self-esterification the –OH and the –COOH of the same molecule react (H₂O lost; ring closes) HOCH₂CH₂COOH → cyclic ester (ring: –CH₂–CH₂–O–C(=O)–); CH₃CH(OH)COOH → cyclic ester (board draws a 3-membered ring)
② With another unit two molecules each use their OH on the other’s COOH ⇒ a cyclic diester (2 ester links) HOCH₂CH₂COOH (×2) → 8-membered ring; CH₃CH(OH)COOH (×2) → lactide (6-membered ring)
OOCCH–CH₃OOCCH₃–CH
lactide: 6-membered ring made from two molecules of lactic acid (two ester links)
2 hydroxy-acidscyclic diester+ 2H₂O releasedconc. H₂SO₄ + heat(esterification / condensation)
Note: The three- and four-membered rings the board draws for self-esterification are very strained — in practice the two-molecule product (lactide, MCQ 82) is the one that forms. Whatever the board shows, the exam idea: each –OH + –COOH pair loses H₂O to form an ester link, and the product is cyclic. Polyesters (e.g. “soluble stitches”: lactic + glycolic acid) are the same reaction continued.

5. Past-Paper MCQs (board)

Q Question Working / answer
2 reagent to distinguish MCPA (–COOH) from 2,4-D (–CH₂CH₂OH) Na₂CO₃(aq): effervescence only with the acid ⇒ D ✓ (Na reacts with both; AgNO₃ — both have aryl Cl; Fehling’s — no aldehyde)
3 lardolure RCH(CH₃)O–C(=O)–H hydrolysed with acid a methanoate ester: HCOO–R′ ⇒ methanoic acid + alcohol RCH(CH₃)OH ⇒ D ✓
4 banana ester CH₃CO₂CH₂CH(CH₃)CH₂CH₃ — which pair produces it? acyl part = ethanoic acid, alkyl part = 2-methylbutan-1-ol ⇒ C ✓ (A, B, D have the wrong acid/alcohol)
7 malic acid: substance reacting with all three –OH groups Na reacts with both –COOH and the alcoholic OH ⇒ C ✓ (KOH and Na₂CO₃ react only with COOH; ethanol/H₂SO₄ esterifies only COOH)
25 strobilurin ester warmed with dil. H₂SO₄, then H₂/Pd the ester is hydrolysed by aqueous acid ⇒ –COOH (+ methanol); H₂/Pd then saturates the C=C bonds ⇒ saturated acid ⇒ A ✓ (the acid is not reduced to an alcohol — no LiAlH₄)
Q Question Working / answer
39 santonin: warm dil. H₂SO₄ then cold acidified KMnO₄ ⇒ X; H atoms displaced by Na? hydrolysis of the lactone ⇒ COOH + OH; cold KMnO₄ turns each of the 2 C=C into a diol (4 OH) ⇒ COOH (1) + OH (1) + 4 OH = 6 ⇒ D ✓
46 ethanedioic acid + ethanol (conc. H₂SO₄) ⇒ diester C₆H₁₀O₄ both –COOH esterify ⇒ CH₃CH₂O–C(=O)–C(=O)–OCH₂CH₃ ⇒ B ✓
70 caprolactone (cyclic ester) from which compound by one reaction ring closes from a hydroxy-acid: HO(CH₂)₅COOH (6-hydroxyhexanoic acid) ⇒ C ✓
82 lactide formed on heating with acid catalyst from which compound? 2-hydroxypropanoic acid (lactic acid) — two molecules form the cyclic diester ⇒ C ✓
86 chlorogenic acid + excess Na₂CO₃(aq) at room temperature only the –COOH reacts (Na₂CO₃ is too weak for alcohols; room temperature doesn’t hydrolyse the ester) ⇒ A ✓ (the sodium carboxylate)

6. Long CIE-style Questions (board)

CIE Q9 · propanoic acid schemeCH₃CH₂CO₂H ⇄ CH₃CH₂CH₂OH → ester
(a) reaction 1: CH₃CH₂CO₂H + 4[H] → CH₃CH₂CH₂OH + H₂O(b) reaction 2 = oxidation: acidified K₂Cr₂O₇, heat under reflux(c) 2CH₃CH₂CO₂H + CaCO₃ → (CH₃CH₂CO₂)₂Ca + H₂O + CO₂ (three products)(d) reaction 3 (⇒ CH₃CO₂CH₂CH₂CH₃): ethanoic acid + conc. H₂SO₄, heat; other product: H₂O
CIE Q4 · V, W, X (all empirical formula CH₂O)C atoms: V = 1, W = 2, X = 3

V gives a brick-red precipitate with Fehling’s ⇒ aldehyde ⇒ HCHO (methanal, 1 C). W: fruity-smelling liquid ⇒ ester ⇒ HCOOCH₃ (2 C). X (3 C, C atoms bonded together, effervescence with Na₂CO₃ ⇒ –COOH; oxidation with reflux gives Y which does not react with 2,4-DNPH ⇒ not a carbonyl ⇒ primary alcohol + carboxylic acid): HOCH₂CH₂COOH.

(c)(ii) Y = HOOC–CH₂–COOH (propanedioic acid). (d) X + conc. H₂SO₄, warm ⇒ cyclic diester Z, C₆H₈O₄ (two X molecules; –H₂O × 2) — draw the 8-membered ring with two ester links. (ii) esterification.

CIE Q6 · ester Q with empirical formula C₂H₄OMr = 87.5
empirical mass C₂H₄O = 44n = 87.5 / 44 ≈ 2molecular formula = C₄H₈O₂

(b) four ester isomers: W HCOOCH₂CH₂CH₃ · X CH₃COOCH₂CH₃ · Y CH₃CH₂COOCH₃ · Z HCOOCH(CH₃)₂. Q is hydrolysed with aq. H₂SO₄, and the mixture is then refluxed with acidified dichromate to give a single product R. Tests on R: no reaction with Tollens’ ⇒ not an aldehyde; none with 2,4-DNPH ⇒ not a carbonyl; effervescence with Na₂CO₃ ⇒ –COOH. (d) R = CH₃COOH (so Q gives the same acid from both parts: ethanol → ethanoic acid, and ethanoic acid) ⇒ Q = X, ethyl ethanoate. (e) none of the esters is chiral — no carbon with 4 different groups.

✓ Quick Recap — Carboxylic Acids & Esters
  • R–COOH: –COOH carbon is C1, suffix -oic acid. High m.p./b.p. (H-bonding); small ones soluble; weak acids.
  • Acid reactions: + metal → salt + H₂; + base/oxide → salt + H₂O; + carbonate → salt + H₂O + CO₂ (effervescence = test for –COOH).
  • LiAlH₄/dry ether reduces acid → 1° alcohol (not NaBH₄; can’t stop at aldehyde). PCl₅ → acyl chloride (+ HCl + POCl₃); steamy fumes = –OH test.
  • Ester R–COO–R′: alkyl (alcohol) + alkanoate (acid); made by esterification (conc. H₂SO₄, heat, reversible). Uses: perfumes, glues, flavourings.
  • Hydrolysis: acid (H⁺/H₂O, reflux) ⇌ alcohol + acid; alkali (NaOH, reflux) → alcohol + carboxylate (irreversible), then dil. HCl → acid.
  • Hydroxy-acids (–OH + –COOH) with conc. H₂SO₄: cyclic ester (self) or cyclic diester with another molecule (lactide from lactic acid).