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The carbonyl group (C=O) is the structural keystone of the second half of A-Level organic chemistry. In the previous lesson you met aldehydes (–CHO) and ketones (–CO–) and the nucleophilic addition reactions of HCN and NaBH₄. The next lesson develops the carboxylic acid derivatives — esters, amides, acyl chlorides, anhydrides. The present lesson sits between those two and focuses on the carboxylic acid family: compounds with the –COOH (carboxyl) functional group. We will recap the anatomy of –COOH, then build the four pillars of its A-Level chemistry — acidity (pKa typically 4-5, ~10¹¹ times stronger than alcohols), salt-forming reactions with bases and carbonates, esterification with alcohols (full nucleophilic acyl substitution mechanism plus the ¹⁸O-labelling evidence), and reduction with LiAlH₄ to primary alcohols. A short section on thermal decarboxylation closes the chemistry. By the end you should write balanced equations for every key reaction, explain why electron-withdrawing groups make carboxylic acids stronger acids using inductive and resonance arguments, and recognise the practical-skills protocol for preparing pure carboxylic acid by reflux and distillation.
Spec mapping (AQA 7405): This lesson maps to §3.3.9 (carboxylic acids and derivatives) of the AQA A-Level Chemistry specification, with cross-references to L0 of this course (§3.3.8 aldehydes and ketones — the oxidation source for carboxylic acids), L2 (§3.3.9 esters and amides — the carboxylic acid derivatives), the foundations course L7 (§3.3.5 alcohols — the oxidation route from primary alcohol → aldehyde → carboxylic acid), and §3.1.12 (acids and bases — pKa, Brønsted-Lowry equilibria, buffer reasoning). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall items include the definition of a carboxylic acid and the recall of its key reactions (with NaOH, Na₂CO₃ and NaHCO₃, with alcohols under acid catalysis, with LiAlH₄). AO2 features in every Paper 2 of the recent AQA series: writing balanced equations for neutralisation and esterification, drawing the curly-arrow mechanism for esterification via the tetrahedral intermediate, and computing the effect of substituent pKa differences on equilibrium position. AO3 evaluative questions ask candidates to rationalise the relative acidities of substituted carboxylic acids using inductive and resonance arguments, to interpret ¹⁸O isotope-labelling data to deduce the mechanism of esterification, and to predict reaction products in unfamiliar contexts.
Carboxylic acids are formed by oxidation of a primary alcohol (via the aldehyde intermediate) using acidified potassium dichromate(VI) under reflux, or directly by oxidation of an aldehyde with Tollens', Fehling's or KMnO₄. The general route is:
R–CH₂OH → R–CHO → R–COOH
The arrival point — the carboxyl group –COOH — is the focus of this lesson.
The carboxyl group –COOH is a compact functional group with rich chemistry, because it contains two oxygen atoms in different bonding environments: a carbonyl oxygen (C=O, double bond) and a hydroxyl oxygen (–O–H, single bond) on the same carbon. The carbon is sp² hybridised, giving a trigonal planar geometry with bond angles of approximately 120°. Crucially, the two C–O bonds are not equivalent in the free acid: the C=O bond is shorter (~123 pm) than the C–OH bond (~134 pm), reflecting the double-bond/single-bond distinction. After deprotonation, the two C–O bonds become identical (~127 pm, intermediate between single and double) — this is the experimental fingerprint of resonance, and we return to it under acidity.
Each –COOH group possesses both a strong hydrogen-bond donor (the O–H) and two strong hydrogen-bond acceptors (the two oxygens). In the pure liquid and even in non-polar solvents, carboxylic acids exist predominantly as cyclic hydrogen-bonded dimers, in which two carboxyl groups associate head-to-head with two simultaneous O–H···O=C hydrogen bonds. The thermodynamic consequence is that boiling points of carboxylic acids are anomalously high. Ethanoic acid (M = 60.05) boils at 118 °C — significantly higher than ethanol (78 °C, M = 46.07) or propanone (56 °C, M = 58.08). Short-chain carboxylic acids (C₁–C₄) are fully miscible with water because they can hydrogen-bond directly to the solvent; solubility falls sharply from C₅ upwards as the alkyl tail dominates.
Key Definition: A carboxylic acid is a compound containing the –COOH (carboxyl) functional group, in which a carbonyl group (C=O) and a hydroxyl group (–OH) are bonded to the same sp² carbon.
Carboxylic acids are weak acids: in aqueous solution they undergo only partial dissociation into a carboxylate anion and a proton.
R–COOH(aq) ⇌ R–COO⁻(aq) + H⁺(aq)
For ethanoic acid (CH₃COOH), the acid dissociation constant Kₐ = 1.74 × 10⁻⁵ mol dm⁻³ at 25 °C, giving pKa = 4.76. For comparison: hydrochloric acid pKa ≈ −7 (strong, fully dissociated); ethanol pKa ≈ 16 (essentially neutral); water pKa = 14.0. So carboxylic acids sit firmly in the "weak acid" band — about 10¹¹ times stronger than the corresponding alcohol, but 10¹¹ times weaker than HCl.
This is the canonical A-Level reasoning question, and the answer has two interlocking parts.
Part 1: Resonance stabilisation of the carboxylate anion. When R–COOH loses a proton, the resulting carboxylate ion R–COO⁻ has the negative charge delocalised equally over both oxygens by resonance:
R–C(=O)–O⁻ ⇌ R–C(–O⁻)=O
The two canonical structures are identical in energy and contribute equally to the resonance hybrid. The hybrid has half a negative charge on each oxygen, and the two C–O bonds are of equal length (~127 pm, as the X-ray crystallography of sodium ethanoate confirms). Spreading the charge over two electronegative atoms is energetically favourable.
Part 2: Alkoxide ions have no equivalent stabilisation. When R–O–H (an alcohol) loses a proton, the negative charge sits localised on the single oxygen of the alkoxide R–O⁻. There is no resonance partner, no second oxygen, no charge delocalisation. The alkoxide is therefore much higher in energy than a carboxylate, and the deprotonation equilibrium for alcohols sits far to the left.
Conclusion: the equilibrium R–COOH ⇌ R–COO⁻ + H⁺ lies further to the right than R–OH ⇌ R–O⁻ + H⁺, because the carboxylate product is stabilised by resonance and the alkoxide is not.
Exam Tip: This question appears on almost every A2 organic paper in some form. The mark-scheme answer always demands the words "resonance" (or "delocalisation") and "negative charge spread over two oxygens". A diagram of the two resonance structures (or the hybrid with partial bonds and δ− charges) scores full marks.
Carboxylic acid strength is tunable across more than five orders of magnitude of Kₐ by changing the alkyl/substituent group R. The principle: any substituent that stabilises the carboxylate anion makes the parent acid stronger, and conversely any substituent that destabilises the anion makes the acid weaker.
| Acid | Formula | pKa | Kₐ / mol dm⁻³ | Relative strength |
|---|---|---|---|---|
| Trichloroethanoic acid | CCl₃COOH | 0.66 | 0.22 | ~13 000× CH₃COOH |
| Trifluoroethanoic acid | CF₃COOH | 0.23 | 0.59 | ~35 000× CH₃COOH |
| Dichloroethanoic acid | CHCl₂COOH | 1.29 | 0.051 | ~3000× CH₃COOH |
| Chloroethanoic acid | ClCH₂COOH | 2.87 | 1.35 × 10⁻³ | ~80× CH₃COOH |
| Methanoic acid | HCOOH | 3.75 | 1.78 × 10⁻⁴ | ~10× CH₃COOH |
| Ethanoic acid | CH₃COOH | 4.76 | 1.74 × 10⁻⁵ | 1 (reference) |
| Propanoic acid | CH₃CH₂COOH | 4.87 | 1.35 × 10⁻⁵ | 0.78× CH₃COOH |
| 2,2-dimethylpropanoic acid | (CH₃)₃CCOOH | 5.03 | 9.33 × 10⁻⁶ | 0.54× CH₃COOH |
Two trends emerge.
Electron-withdrawing substituents make the acid stronger. Halogens (Cl, F) and especially fluorinated alkyl groups (CF₃) withdraw electron density through the C–C σ bond by the inductive effect (−I). This pulls negative charge away from the COO⁻ oxygens of the conjugate base, spreading it over a larger volume and lowering its energy. The result: a more stable carboxylate, and therefore a stronger acid. Each chlorine in CCl₃COOH contributes additively, hence the dramatic 10⁴-fold strength enhancement from CH₃COOH to CCl₃COOH.
Worked example. Compare CH₃COOH (pKa 4.76) and ClCH₂COOH (pKa 2.87). The difference ΔpKa = 1.89 corresponds to a ratio of Kₐ values of 10¹·⁸⁹ ≈ 78. So chloroethanoic acid is about 80 times stronger than ethanoic acid. The single chlorine, two bonds away from the dissociating O–H, withdraws enough electron density to stabilise the conjugate base substantially. This is the inductive effect quantified.
Electron-donating substituents make the acid weaker. Alkyl groups donate electron density into the carboxyl carbon by the +I inductive effect (also termed hyperconjugation in some texts). This pushes electron density onto the COO⁻ oxygens of the conjugate base, raising its energy and destabilising the anion. The result: a weaker acid. Going from CH₃COOH (one α-methyl) → CH₃CH₂COOH (propanoic, one α-CH₂CH₃) → (CH₃)₃CCOOH (pivalic, three α-methyls) the pKa rises monotonically from 4.76 → 4.87 → 5.03.
Position matters. Inductive effects fall off rapidly with distance. 2-chlorobutanoic acid (Cl on Cα) has pKa ~2.85; 4-chlorobutanoic acid (Cl on Cγ, three carbons from COOH) has pKa ~4.5, almost back to that of butanoic acid itself. The −I effect attenuates with each intervening σ bond.
Exam Tip: When a question asks you to rank a set of carboxylic acids by acid strength, look at (i) the electronic nature of the substituent (−I withdraws, +I donates), (ii) the number of withdrawing substituents (more = stronger), and (iii) the distance from the COOH carbon (closer = stronger effect). Always relate strength back to the stability of the conjugate base, not the acid itself.
Carboxylic acids undergo straightforward neutralisation with strong bases such as NaOH to give a salt (sodium carboxylate) and water:
CH₃COOH(aq) + NaOH(aq) → CH₃COO⁻Na⁺(aq) + H₂O(l)
The ionic equation strips out the spectator sodium:
CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)
This is a Brønsted-Lowry acid-base reaction: the carboxylic acid donates a proton, the hydroxide accepts it. The driving force is the formation of water (pKₐ 14) from the much less stable hydroxide-plus-acid combination. The equilibrium lies overwhelmingly to the right (Keq ≈ 10⁹) and the reaction is treated as going to completion at A-Level.
The sodium carboxylate salt is ionic, fully water-soluble, and crystallises on evaporation of the solvent. Sodium ethanoate (CH₃COONa) is a familiar laboratory reagent and a component of acetate buffers (pKa 4.76 makes it ideal for buffering near pH 5).
Carboxylic acids are stronger acids than carbonic acid (pKa₁(H₂CO₃) ≈ 6.4) but weaker than HCl. This means they can displace CO₂ from carbonates and hydrogencarbonates:
2 CH₃COOH(aq) + Na₂CO₃(aq) → 2 CH₃COONa(aq) + H₂O(l) + CO₂(g)
CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O(l) + CO₂(g)
The observation is effervescence — visible fizzing as CO₂ gas evolves. The CO₂ can be tested by passing it through limewater (Ca(OH)₂ solution), which turns cloudy (milky) as insoluble CaCO₃ precipitates.
This reaction is the standard A-Level qualitative test for a carboxylic acid. Phenols (pKa ~10) and alcohols (pKa ~16) are too weak to displace CO₂ from carbonates — they give no fizzing. The test therefore distinguishes carboxylic acids from phenols, alcohols, and almost all other neutral organic compounds.
Practical-skills box — qualitative test protocol. Place ~1 cm³ of the unknown organic liquid in a clean dry test tube. Add a spatula tip of solid sodium carbonate (or 1 cm³ of saturated NaHCO₃ solution). Observe for effervescence. If gas evolves, bubble through limewater in a second tube. A positive result — fizzing and cloudy limewater — confirms the presence of a –COOH group. A negative result (no fizzing) rules out a carboxylic acid but does not distinguish phenol from alcohol.
Carboxylic acids react with ammonia in a simple acid-base step to give the ammonium carboxylate salt:
CH₃COOH + NH₃ → CH₃COO⁻NH₄⁺
On heating, ammonium carboxylates eliminate water to form the amide:
CH₃COO⁻NH₄⁺ —(heat)→ CH₃CONH₂ + H₂O
This is a slow, low-yield route to amides; the acyl chloride or anhydride route (covered in L2) is much faster.
The reaction of a carboxylic acid with an alcohol, catalysed by concentrated sulfuric acid, gives an ester and water. This is the Fischer esterification:
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
(conditions: conc. H₂SO₄ catalyst, gentle reflux, typically 60-80 °C)
The reaction is reversible and the equilibrium constant Kc is typically around 4 — meaning that without intervention only about 65% conversion is reached. To push the equilibrium towards the ester, chemists either use a large excess of one reactant (usually the alcohol, which is cheap), or remove water as it forms (using a Dean-Stark trap or molecular sieves), or distil off the ester (if it is volatile) as it forms.
This is one of the most important mechanisms in A-Level organic chemistry. There are six discrete steps. Drawing all the curly arrows correctly is worth four marks on a typical paper.
Step 1 — Protonation of the carbonyl oxygen. H₂SO₄ donates a proton to the C=O oxygen of CH₃COOH. The carbonyl oxygen becomes positively charged, which activates the carbonyl carbon by making it even more strongly δ+ than in the neutral acid. (Curly arrow: from a lone pair on the C=O oxygen to the incoming H⁺.)
Step 2 — Nucleophilic attack by the alcohol. Ethanol's oxygen lone pair attacks the activated carbonyl carbon. The C=O π bond breaks and the electrons move onto oxygen, generating a tetrahedral intermediate: a carbon bonded to two –OH groups, one –OC₂H₅ group (still bearing a positively-charged H from ethanol), and the methyl R group. (Curly arrows: alcohol O lone pair → C; π bond → O.)
Step 3 — Proton transfer. The positively-charged –OEt⁺ loses its proton (to a solvent water molecule or to HSO₄⁻), becoming a neutral –OEt group. (Curly arrow: O–H bond breaks, electrons go to O.)
Step 4 — Protonation of one of the original –OH groups. A second protonation step now activates one of the two original hydroxyls towards leaving as water. (Curly arrow: lone pair on –OH → H⁺.)
Step 5 — Loss of water. The protonated –OH⁺ departs as a neutral water molecule. The departing electrons go back into the C–O bond and reform the C=O π bond, regenerating a carbonyl group — but now the substituent on the carbonyl carbon is the ester –OC₂H₅ rather than the original –OH. (Curly arrows: C–O bond → O of departing water; lone pair on the other O → forms the new C=O π bond.)
Step 6 — Deprotonation. The newly-formed protonated ester loses its proton (to solvent or HSO₄⁻), regenerating the catalyst H⁺ and giving the neutral ester CH₃COOC₂H₅.
The overall mechanism is acid-catalysed nucleophilic acyl substitution: a nucleophile (alcohol O) attacks the acyl carbon, a tetrahedral intermediate forms, and a leaving group (water) departs.
A famous mechanistic experiment by Roberts and Urey (1938) used isotopically labelled ¹⁸O in the alcohol — i.e. ethanol where the hydroxyl oxygen was ¹⁸O instead of the normal ¹⁶O. Two possible outcomes were imaginable in advance:
Experimentally, the ester was found to contain the ¹⁸O label and the water by-product contained only ¹⁶O. Outcome A is correct. This pins down which oxygen ends up where: the ester's bridging oxygen comes from the alcohol, and the water by-product's oxygen comes from the carboxylic acid's –OH. This is consistent only with the tetrahedral-intermediate mechanism drawn above.
Exam Tip: Isotope-labelling questions appear regularly on A-Level papers as AO3 evaluative items. The expected reasoning is: identify which bond breaks (and which is retained), then map that conclusion onto the mechanism. A diagram with the ¹⁸O traced from the alcohol all the way into the ester always scores full marks.
Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride (LiAlH₄), commonly written as LiAlH₄ in dry ether (typically diethyl ether or tetrahydrofuran), at room temperature, followed by an aqueous acidic workup.
CH₃COOH + 4[H] → CH₃CH₂OH + H₂O
The "4[H]" notation hides the fact that two equivalents of hydride are required (one to deprotonate the –COOH, one to reduce the C=O), but A-Level mark schemes accept the simplified [H] notation.
The mechanism involves hydride (H⁻) transfer from the AlH₄⁻ ion to the carbonyl carbon in two stages, going through an aldehyde intermediate that is itself further reduced under the reaction conditions; LiAlH₄ is so reactive that the aldehyde never accumulates and the product isolated is the primary alcohol. The hydride attacks the δ+ carbonyl carbon (nucleophilic addition), the C=O π bond breaks, and the resulting alkoxide is protonated on workup.
Crucially, NaBH₄ does not reduce carboxylic acids. NaBH₄ is a milder hydride source and is only kinetically competent to reduce aldehydes and ketones (whose carbonyl carbon is more electrophilic, because there is no adjacent –OH to delocalise charge). Carboxylic acids and esters require the more powerful LiAlH₄.
| Carbonyl compound | NaBH₄ | LiAlH₄ |
|---|---|---|
| Aldehyde | Reduces to 1° alcohol | Reduces to 1° alcohol |
| Ketone | Reduces to 2° alcohol | Reduces to 2° alcohol |
| Carboxylic acid | No reaction | Reduces to 1° alcohol |
| Ester | No reaction | Reduces to 1° alcohol + 1° alcohol |
| Amide | No reaction | Reduces to amine |
Common Misconception: Students sometimes write that NaBH₄ reduces –COOH. It does not. The marking system at A-Level will treat "NaBH₄ reduces carboxylic acid" as a factual error and lose the mark.
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