Class 12 Carboxylic Acids and Their Derivatives — Reactions and Acidity
The carboxylic acid group is the most acidic functional group you meet in school organic chemistry, and the questions on it split neatly into two kinds. One kind asks you to compare — which of these acids is stronger, and why. The other asks you to convert — turn this acid into an ester, an amide, an acid chloride or an alkane. This article handles both, with the numbers actually worked out so you can see how big the effects are rather than only memorising an order.
The functional group and why it is acidic
- R — alkyl or aryl group; in methanoic acid it is simply H.
- Ka — the acid dissociation constant, in mol/L. Bigger Ka means a stronger acid.
- pKa — the negative logarithm. Smaller pKa means stronger acid. Students reverse this constantly, so write it on your formula sheet.
The acidity comes from what happens after the proton leaves. In the carboxylate ion the negative charge is shared equally over two oxygen atoms by resonance, and both C–O bonds become identical in length. A phenoxide spreads its charge over carbon atoms in a ring — less effective, because carbon holds negative charge poorly. An alkoxide cannot delocalise at all. That is the whole ladder: carboxylic acid > phenol > water > alcohol.
What substituents do to acidity
| Acid | Approximate pKa (25 °C) | Reason |
|---|---|---|
| Trichloroethanoic acid, CCl₃COOH | 0.65 | Three −I chlorine atoms pull charge away from the carboxylate |
| Dichloroethanoic acid, CHCl₂COOH | 1.29 | Two chlorines |
| Chloroethanoic acid, CH₂ClCOOH | 2.86 | One chlorine |
| Methanoic acid, HCOOH | 3.75 | No electron-donating alkyl group at all |
| Benzoic acid, C₆H₅COOH | 4.20 | sp² ring carbon is mildly electron-withdrawing |
| Ethanoic acid, CH₃COOH | 4.76 | The +I methyl group destabilises the anion |
Values are rounded and vary slightly between sources and temperatures — quote them as approximate. Two patterns to state in an answer: electron-withdrawing groups increase acidity, and their effect falls off sharply with distance, because the inductive effect is transmitted through σ bonds and weakens with every bond it passes through. So 2-chlorobutanoic acid is markedly stronger than 4-chlorobutanoic acid.
Worked example 1 — comparing two acids by calculation
Question. Calculate the pH of 0.10 M ethanoic acid (Ka = 1.8 × 10⁻⁵) and of 0.10 M chloroethanoic acid (pKa = 2.86). Comment.
Ethanoic acid. Using x² ÷ (0.10 − x) = Ka with x small:
x = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/L
pH = −log(1.34 × 10⁻³) = 3 − 0.13 = 2.87
Cross-check with the full quadratic x² + 1.8 × 10⁻⁵x − 1.8 × 10⁻⁶ = 0:
x = 1.333 × 10⁻³, pH = 2.88. The approximation was safe (α ≈ 1.3%).
Chloroethanoic acid. First convert:
Ka = 10−2.86 = 1.38 × 10⁻³.
The quick route gives x = √(1.38 × 10⁻³ × 0.10) = √(1.38 × 10⁻⁴) = 1.17 × 10⁻², i.e. pH 1.93 —
but α would then be 11.7%, which is not small, so the approximation is unsafe here.
Do it properly. x² + (1.38 × 10⁻³)x − 1.38 × 10⁻⁴ = 0
Discriminant = (1.38 × 10⁻³)² + 4(1.38 × 10⁻⁴) = 1.90 × 10⁻⁶ + 5.52 × 10⁻⁴ = 5.539 × 10⁻⁴
√(5.539 × 10⁻⁴) = 2.3535 × 10⁻²
x = (−1.38 × 10⁻³ + 2.3535 × 10⁻²) ÷ 2 = 1.108 × 10⁻² mol/L
pH = −log(1.108 × 10⁻²) = 1.96
Comment. One chlorine atom drops the pH from 2.87 to 1.96 — the hydrogen ion concentration rises by a factor of about 8. That is the inductive effect made numerical. The second lesson is procedural: check α before trusting the √(KaC) shortcut. As a working rule, if α exceeds about 5%, solve the quadratic.
Preparation — five routes
| Starting material | Reagent | Note |
|---|---|---|
| Primary alcohol or aldehyde | KMnO₄ / K₂Cr₂O₇, acidic or alkaline | Both stop at the acid |
| Alkylbenzene | Hot alkaline KMnO₄, then H₃O⁺ | The whole side chain becomes –COOH, whatever its length |
| Nitrile | Acid or alkaline hydrolysis | Via the amide; one more carbon than the parent halide |
| Amide | Hydrolysis | Same carbon count |
| Grignard reagent | Dry CO₂ (or solid CO₂), then H₃O⁺ | Adds exactly one carbon — a favourite conversion step |
Reactions you must be able to write
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄)
CH₃COOH + SOCl₂ → CH₃COCl + SO₂↑ + HCl↑
CH₃COONa + NaOH —(CaO, heat)→ CH₄ + Na₂CO₃
CH₃COOH + Cl₂ —(red P)→ ClCH₂COOH + HCl
Check the third one atom by atom. On the left, CH₃COOH is C₂H₄O₂ and SOCl₂ is S₁O₁Cl₂, giving 2 C, 4 H, 3 O, 1 S, 2 Cl. On the right, CH₃COCl is C₂H₃OCl, plus SO₂ and HCl, giving 2 C, 3 + 1 = 4 H, 1 + 2 = 3 O, 1 S, 1 + 1 = 2 Cl. Balanced. Notice that the oxygen from SOCl₂ counts too — forgetting a reagent's own atoms is the commonest balancing slip in organic equations.
Three points about these reactions:
- SOCl₂ is preferred over PCl₅ or PCl₃ for making acid chlorides because both by-products are gases and escape, leaving a pure product.
- Decarboxylation with soda lime (NaOH + CaO) removes the –COOH entirely and gives an alkane with one carbon less.
- The Hell–Volhard–Zelinsky reaction needs an α-hydrogen. Benzoic acid and methanoic acid have none, so it fails for both.
The derivatives and their reactivity order
The order follows two ideas that point the same way. First, the better the leaving group, the faster the substitution — Cl⁻ leaves easily, ⁻NH₂ does not. Second, the more the attached atom donates electron density back into the carbonyl by resonance, the less electrophilic the carbonyl carbon becomes; nitrogen donates strongly, chlorine barely at all. So an acid chloride reacts with water violently at room temperature, while an amide needs prolonged heating with acid or alkali.
Ester hydrolysis is worth a line of its own. In acid it is reversible — the reverse reaction is simply esterification. In alkali it is irreversible, because the carboxylate salt formed is deprotonated and no longer attackable. That irreversible version is saponification, the soap-making reaction:
Worked example 2 — esterification yield
Question. 6.00 g of ethanoic acid is refluxed with excess ethanol and a little concentrated H₂SO₄. 6.16 g of ethyl ethanoate is isolated. Find the percentage yield.
Step 1 — molar masses.
M(CH₃COOH) = 2(12.011) + 4(1.008) + 2(15.999) = 24.022 + 4.032 + 31.998 = 60.05 g/mol
M(CH₃COOC₂H₅) = 4(12.011) + 8(1.008) + 2(15.999) = 48.044 + 8.064 + 31.998 = 88.11 g/mol
Step 2 — moles of the limiting reactant.
n = 6.00 ÷ 60.05 = 0.0999 mol (ethanol is in excess)
Step 3 — theoretical mass.
0.0999 × 88.11 = 8.80 g
Step 4 — percentage yield.
(6.16 ÷ 8.80) × 100 = 70.0%
Why not 100%? Esterification is an equilibrium. Excess alcohol and removal of water push it forward, but it never runs to completion by itself. A yield well below 100% is expected here — that is chemistry, not carelessness.
Safety: concentrated sulphuric acid is added slowly with cooling, and the mixture is refluxed, never heated in an open tube.
Worked example 3 — the acetic acid / acetate buffer
Because a carboxylic acid is weak, mixing it with its own sodium salt gives a buffer. The Henderson–Hasselbalch equation is:
pH = pKa + log([salt] ÷ [acid])
(a) 0.10 M ethanoic acid + 0.10 M sodium ethanoate:
pH = 4.76 + log(0.10 ÷ 0.10) = 4.76 + log 1 = 4.76 + 0 = 4.76
When salt and acid are equal, pH equals pKa exactly. That is worth remembering as a
check on any buffer answer.
(b) 0.20 M sodium ethanoate + 0.10 M ethanoic acid:
pH = 4.76 + log(0.20 ÷ 0.10) = 4.76 + log 2 = 4.76 + 0.301 = 5.06
Doubling the salt raises the pH by only 0.30 — buffers resist change, which is the point.
(c) What ratio gives pH 5.00?
5.00 = 4.76 + log(ratio) → log(ratio) = 0.24 → ratio = 100.24 = 1.74
So you need about 1.74 mol of ethanoate for every 1 mol of ethanoic acid. Check by substituting
back: 4.76 + log 1.74 = 4.76 + 0.2405 = 5.00. ✓
Telling an acid from a phenol from an alcohol
| Reagent | Carboxylic acid | Phenol | Alcohol |
|---|---|---|---|
| Blue litmus | Turns red | Turns red (faintly) | No change |
| NaHCO₃ solution | Brisk CO₂ effervescence | No effervescence | No reaction |
| NaOH solution | Dissolves | Dissolves | Does not react |
| Neutral FeCl₃ | Red colour; brown-red precipitate on boiling | Violet colour | No change |
| Ester test (alcohol + conc. H₂SO₄) | Fruity smell of the ester | Poor — needs an anhydride or acyl chloride | Fruity smell with an acid |
The sodium bicarbonate row is the single most useful test in the whole organic syllabus: only the carboxylic acid is strong enough to displace carbonic acid and release CO₂.
Common mistakes that cost marks
- Reading pKa the wrong way round. Lower pKa means stronger acid. Ethanoic acid (4.76) is weaker than benzoic acid (4.20).
- Calling a carboxylic acid strong. It is a weak acid — 0.10 M ethanoic acid is only about 1.3% ionised.
- Using NaBH₄ to reduce –COOH. It does not work. LiAlH₄ (or diborane) is needed, and both must be kept away from water.
- Forgetting that HCOOH gives a positive Tollens' test. It carries an H on the carbonyl carbon, so it reduces Tollens' reagent like an aldehyde. No other common carboxylic acid does.
- Writing acid hydrolysis of an ester as irreversible. Acid hydrolysis is reversible; alkaline hydrolysis (saponification) is not.
- Reversing the derivative reactivity order. Amides are the least reactive, not the most — nitrogen's lone pair stabilises the carbonyl.
- Applying the √(KaC) shortcut to a fairly strong weak acid. Check α first, as worked example 1 shows.
- Saying –COOH is ortho/para directing on a ring. It is deactivating and meta directing.
Why this chapter matters
| Where it appears | What is actually tested |
|---|---|
| CBSE / ICSE Class 12 organic unit | Acidity comparison with reasons, conversions, distinguishing tests |
| Physical chemistry crossover | Buffers, Ka and pH calculations, titration curves |
| Practical and viva | NaHCO₃ test, ester (fruity smell) test, functional group identification |
| NEET / JEE foundation | Ranking acid strengths, predicting derivative reactivity |
| IIT-JAM / CUET-PG later on | Nucleophilic acyl substitution mechanisms, enolate chemistry |
Syllabus documents change, so build your revision plan from the one your board has published for your own session rather than from a remembered pattern.
Buffer problems become quick once you can see the ratio move. The Henderson–Hasselbalch calculator takes pKa and the salt and acid concentrations and returns the pH — or works backwards from a target pH to the ratio you need, exactly as in part (c) above.
Open the Henderson–Hasselbalch Buffer Calculator →If acidity comparisons and conversions are where you lose marks, guided practice fixes it faster than re-reading. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — details at abcchemistry.in.