ChemistryISC Class 12 · Aldehydes, Ketones and Carboxylic Acids

Carboxylic Acids: Preparation, Acidity and Reactions (ISC Class 12)

Updated 15 October 2026 · Fact-checked

Carboxylic acids are made by oxidising primary alcohols or aldehydes, hydrolysing nitriles, or carbonating Grignard reagents. They are stronger acids than phenols because the carboxylate ion is resonance-stabilised; electron-withdrawing groups increase acidity. They form salts, acid chlorides, anhydrides, esters and amides, lose CO₂ with soda lime, and undergo HVZ halogenation.

Understand Carboxylic Acids: Preparation, Acidity and Reactions

Structure. The carboxyl group, –COOH, combines C=O and O–H on one carbon. The carbonyl carbon is sp², and resonance with the –OH lone pair makes the C=O less reactive than in aldehydes and ketones (carboxylic acids do not give typical carbonyl tests). Monocarboxylic acids include methanoic (formic), ethanoic (acetic) and benzoic acid; dicarboxylic acids include ethanedioic (oxalic) acid.

Preparation (ISC list). (i) Oxidation of primary alcohols and aldehydes with acidified KMnO₄ or K₂Cr₂O₇. (ii) Hydrolysis of nitriles with dilute acid (or alkali, then acid): $C H_{3} C N + 2 H_{2} O + H C l \to C H_{3} C O O H + N H_{4} C l$ ; this adds one carbon to an alkyl halide (RX → RCN → RCOOH). (iii) Grignard reagent with CO₂ (dry ice), then acid: $C_{2} H_{5} M g B r + C O_{2} \to C_{2} H_{5} C O O M g B r$ ; $C_{2} H_{5} C O O M g B r + H_{2} O \to C_{2} H_{5} C O O H + M g (O H) B r$ . Aromatic acids are made similarly, or by oxidising alkylbenzenes (toluene → benzoic acid).

Physical properties. Lower acids are liquids with sharp smells and are miscible with water (hydrogen bonding); solubility falls as the chain grows. Boiling points are higher than those of alcohols of similar mass, because two molecules form a hydrogen-bonded dimer, which persists even in the vapour.

Acidity. $R C O O H + H_{2} O ⇌ R C O O^{-} + H_{3} O^{+}$ . The carboxylate ion is stabilised by two equivalent resonance structures (the negative charge is shared equally by both oxygens), so carboxylic acids (pKa about 4–5) are much stronger than phenols (pKa 10), whose phenoxide resonance places charge on less electronegative carbons. Effect of substituents (dealt with in detail by CISCE): electron-withdrawing groups (–NO₂, –CN, –F, –Cl, –Br, C₆H₅) increase acidity by stabilising the anion (−I effect); electron-releasing alkyl groups decrease it (+I effect). The effect grows with the electronegativity and number of substituents and falls with distance from –COOH. Examples: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH; Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH; HCOOH > CH₃COOH > CH₃CH₂COOH; benzoic acid (pKa 4.19) is stronger than ethanoic acid (4.76). On the benzene ring, electron-withdrawing groups at ortho/para increase acidity; electron-releasing groups (–CH₃, –OCH₃) decrease it.

Reactions showing acidity. With active metals, $2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}$ ; with alkalis, $C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$ ; with carbonates and hydrogencarbonates, CO₂ is evolved with effervescence: $C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$ . Phenols do not liberate CO₂ from NaHCO₃: this is the standard test that separates acids from phenols.

Formation of acid derivatives (replacing –OH). Acid chloride: $C H_{3} C O O H + S O C l_{2} \to C H_{3} C O C l + S O_{2} + H C l$ (also PCl₅ or PCl₃). Anhydride: heating with P₂O₅ (or a strong dehydrating agent), $2 C H_{3} C O O H \to (C H_{3} C O)_{2} O + H_{2} O$ . Ester: with an alcohol and conc. H₂SO₄ (esterification). Amide: with ammonia, the ammonium salt forms and loses water on heating, $C H_{3} C O O H + N H_{3} \to C H_{3} C O O N H_{4}$ ; $C H_{3} C O O N H_{4} \to C H_{3} C O N H_{2} + H_{2} O$ .

Decarboxylation. (i) Chemical: the sodium salt heated with soda lime (NaOH + CaO) loses CO₂ as Na₂CO₃ and gives an alkane with one carbon fewer: $C H_{3} C H_{2} C O O N a + N a O H \to C_{2} H_{6} + N a_{2} C O_{3}$ . (ii) Kolbe's electrolysis: electrolysis of a concentrated aqueous solution of the sodium salt gives an alkane with twice the alkyl group at the anode, with CO₂, while H₂ and NaOH form at the cathode: $2 C H_{3} C O O N a + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 N a O H + H_{2}$ .

Hell–Volhard–Zelinsky (HVZ) reaction. Acids with an α-hydrogen react with Cl₂ or Br₂ in the presence of a little red phosphorus to give α-halo acids: $C H_{3} C H_{2} C O O H + C l_{2} \to C H_{3} C H C l C O O H + H C l$ . Repeated treatment replaces further α-hydrogens.

Ring substitution in benzoic acid. –COOH is electron-withdrawing, deactivating and meta directing: nitration gives 3-nitrobenzoic acid, $C_{6} H_{5} C O O H + H N O_{3} \to C_{6} H_{4} (N O_{2}) C O O H + H_{2} O$ , and sulphonation gives 3-sulphobenzoic acid. Benzoic acid does not undergo Friedel–Crafts reactions.

Tests. All three common acids give effervescence with NaHCO₃. Formic acid also reduces Tollens' reagent and Fehling's solution, because it contains an aldehyde-like H (H–COOH). Acetic acid with neutral FeCl₃ gives a red colour (a reddish-brown precipitate on boiling). Benzoic acid with neutral FeCl₃ gives a buff (pale brown) precipitate. Uses: formic acid in leather tanning and coagulating rubber latex; acetic acid as vinegar and as a solvent and raw material for plastics and drugs; benzoic acid and sodium benzoate as food preservatives and in medicines.

Carboxylic Acids: Preparation, Acidity and Reactions: formulas and key facts

Acid dissociation: $R C O O H + H_{2} O ⇌ R C O O^{-} + H_{3} O^{+}, p K_{a} = - lo g K_{a}$; Smaller pKa means stronger acid.
Grignard route: $R M g X + C O_{2} \to R C O O M g X H_{2} O / H^{+} R C O O H$; Adds one carbon to the alkyl group of RX.
Soda lime decarboxylation: $R C O O N a + N a O H CaO, heat R H + N a_{2} C O_{3}$; Alkane with one carbon fewer.
Kolbe's electrolysis: $2 R C O O N a + 2 H_{2} O \to R - R + 2 C O_{2} + 2 N a O H + H_{2}$; Alkane with twice the alkyl group, at the anode.
HVZ reaction: $R C H_{2} C O O H X_{2}, red P R C H X C O O H + H X$; Needs an α-hydrogen; X = Cl or Br.
Substituent effect on acidity: $- I (EWG) ↑ acidity; + I (alkyl) ↓ acidity$; Effect decreases with distance from –COOH.

How to solve Carboxylic Acids: Preparation, Acidity and Reactions questions

ISC sets carboxylic acid questions as acidity orders, conversion chains with A, B, C, D, reactions to name, and distinguishing tests. Keep a mental map of what each reagent does to –COOH.

1For acidity orders, judge the anion: more electron-withdrawing groups, closer to –COOH, more electronegative → more stable anion → stronger acid; alkyl groups do the opposite.
2Write the order in the direction asked (increasing or decreasing) with clear < or > signs.
3For conversions: SOCl₂/PCl₅ → acid chloride; P₂O₅ → anhydride; alcohol/H⁺ → ester; NH₃ then heat → amide; soda lime → alkane (one C fewer); Kolbe electrolysis → alkane (double); Cl₂/red P → α-chloro acid.
4To lengthen a chain by one carbon to an acid: RX → RCN (KCN) → RCOOH (hydrolysis), or RX → RMgX → RCOOH (CO₂).
5For tests, give NaHCO₃ effervescence for acids versus phenols, Tollens' for formic acid, and neutral FeCl₃ colours for acetic and benzoic acids.
6Balance every equation and write conditions over the arrow.

Carboxylic Acids: Preparation, Acidity and Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Neutralising an acid with sodium hydrogencarbonate

Suppose a hiker is stung by an ant, and imagine the sting injects about 0.023 g of methanoic (formic) acid (M = 46). A first-aid guide suggests dabbing the area with a paste of baking soda, NaHCO₃ (M = 84).

Write the equation and calculate the mass of NaHCO₃ needed to neutralise the acid exactly. Why is NaHCO₃ better than NaOH for this purpose?

Try it first, then show the solutionSolution

$H C O O H + N a H C O_{3} \to H C O O N a + H_{2} O + C O_{2}$ .
Moles of HCOOH $= \frac{0.023}{46} = 5.0 \times 1 0^{- 4}$ mol; the ratio is 1 : 1.
Mass of NaHCO₃ $= 5.0 \times 1 0^{- 4} \times 84 = 0.042$ g.
NaHCO₃ is a weak base: it neutralises the acid (giving CO₂) without itself burning the skin, as strongly alkaline NaOH would.

Answer: About 0.042 g of NaHCO₃; it neutralises the acid without being caustic.

Memory hook: Acids fizz with bicarbonate: the bubbles are CO₂, and phenols cannot do it.

Tricks to remember Carboxylic Acids: Preparation, Acidity and Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Pull makes it stronger
Groups that pull electrons (F, Cl, NO₂) stabilise RCOO⁻ and make the acid stronger; groups that push (CH₃, C₂H₅) make it weaker. Closer pull, stronger acid.
Use it when: Acidity order questions.
Soda lime subtracts, Kolbe doubles
Soda lime: CH₃CH₂COONa → C₂H₆ (one carbon less). Kolbe electrolysis: CH₃CH₂COONa → C₄H₁₀ (alkyl group doubled).
Use it when: Decarboxylation products.
HVZ hits alpha
Hell–Volhard–Zelinsky: Halogen goes to the α-carbon (next to COOH), with red P as helper.
Use it when: Predicting the HVZ product.

Carboxylic Acids: Preparation, Acidity and Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: The acid that reduces Tollens' reagent · 1 mark · MCQBoard level

Which of the following carboxylic acids reduces Tollens' reagent?
(a) ethanoic acid
(b) benzoic acid
(c) methanoic acid
(d) propanoic acid

Try it first, then show the solutionSolution

Methanoic acid, H–COOH, has an H on the carbonyl carbon, like an aldehyde.
It is oxidised (to CO₂ and water) and reduces Ag⁺ to a silver mirror.

Answer: (c) methanoic (formic) acid.

Example 2: Propanoic acid from a Grignard reagent · 2 marks · Short answerBoard level

How is propanoic acid prepared from ethylmagnesium bromide? Write the equations.

Try it first, then show the solutionSolution

Pass dry CO₂ into the Grignard reagent (or pour it on dry ice): $C_{2} H_{5} M g B r + C O_{2} \to C_{2} H_{5} C O O M g B r$ (1 mark).
Hydrolyse with dilute acid: $C_{2} H_{5} C O O M g B r + H_{2} O \to C_{2} H_{5} C O O H + M g (O H) B r$ (1 mark).

Answer: Carbonation with CO₂, then acidic hydrolysis.

Example 3: Halogen and acidity · 3 marks · Short answerHarder board level

Arrange FCH₂COOH, ClCH₂COOH, BrCH₂COOH and CH₃COOH in decreasing order of acidic strength, and explain the order.

Try it first, then show the solutionSolution

FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH (1 mark).
A halogen withdraws electron density through the σ bonds (−I effect), dispersing the negative charge of the carboxylate ion and stabilising it, so the acid ionises more (1 mark).
The −I effect follows electronegativity, F > Cl > Br; the CH₃ group in ethanoic acid releases electrons (+I) and destabilises the anion (1 mark).

Answer: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH, following the −I effect of the halogen.

Example 4: Four derivatives from one acid · 3 marks · Short answerHarder board level

Write equations for converting ethanoic acid into (a) ethanoyl chloride, (b) ethanoic anhydride and (c) ethanamide, with reagents and conditions.

Try it first, then show the solutionSolution

(a) $C H_{3} C O O H + S O C l_{2} \to C H_{3} C O C l + S O_{2} + H C l$ (or PCl₅) (1 mark).
(b) Heat with P₂O₅: $2 C H_{3} C O O H \to (C H_{3} C O)_{2} O + H_{2} O$ (1 mark).
(c) With NH₃, then heat: $C H_{3} C O O H + N H_{3} \to C H_{3} C O O N H_{4}$ ; $C H_{3} C O O N H_{4} \to C H_{3} C O N H_{2} + H_{2} O$ (1 mark).

Answer: SOCl₂ (or PCl₅); P₂O₅ with heat; NH₃ then heat.

Example 5: One acid, three products · 5 marks · Long answerHOTS / competency

An acid X (C₃H₆O₂) gives effervescence with NaHCO₃. Its sodium salt heated with soda lime gives gas A; electrolysis of a concentrated aqueous solution of the sodium salt gives hydrocarbon B at the anode. X with chlorine in the presence of red phosphorus gives C.
(a) Identify X, A, B and C. [2]
(b) Write the equations for the formation of A and B, and name the reaction that gives B. [2]
(c) Is C a stronger or weaker acid than X? Explain. [1]

Try it first, then show the solutionSolution

(a) X propanoic acid, CH₃CH₂COOH; A ethane; B butane; C 2-chloropropanoic acid, CH₃CHClCOOH (2 marks).
(b) $C H_{3} C H_{2} C O O N a + N a O H CaO, heat C_{2} H_{6} + N a_{2} C O_{3}$ (1 mark).
$2 C H_{3} C H_{2} C O O N a + 2 H_{2} O \to C_{4} H_{10} + 2 C O_{2} + 2 N a O H + H_{2}$ ; Kolbe's electrolysis (1 mark).
(c) Stronger: the α-chlorine withdraws electrons (−I) and stabilises the carboxylate ion (1 mark).

Answer: X propanoic acid, A ethane, B butane (Kolbe's electrolysis), C 2-chloropropanoic acid, a stronger acid.

Where students lose marks in Carboxylic Acids: Preparation, Acidity and Reactions

The slips examiners see most often, and what to write instead.

Mistake: Ignoring the molecular formula in identification problems
ISC examiners reported acetic acid written for an acid C₇H₆O₂, so every later compound was wrong (2025).
Do this: Check the formula first: C₇H₆O₂ with acidic properties is benzoic acid.
Usually costs: All the identification marks.
Mistake: Writing the salt instead of the ketone or amide
Examiners reported calcium acetate written for the product of heating it (acetone), and the ammonium salt written instead of the amide (2024, 2025).
Wrong working
$C_{2} H_{5} C O O H N H_{3} C_{2} H_{5} C O O N H_{4}$ (final answer)
Right working
$C_{2} H_{5} C O O N H_{4} heat C_{2} H_{5} C O N H_{2} + H_{2} O$
Do this: Follow each step to its final product: heated ammonium salts give amides; heated calcium salts give ketones.
Usually costs: 1 mark.
Mistake: Missing soda lime or heat in decarboxylation
Examiners noted that candidates did not mention soda lime for decarboxylation of sodium benzoate (2024, 2025).
Do this: Write 'NaOH + CaO (soda lime), heat' over the arrow.
Usually costs: 1 mark.
Mistake: Writing the order in the wrong direction
Examiners reported decreasing orders given when increasing was asked (2025).
Do this: Read 'increasing' or 'decreasing' twice and use '<' or '>' explicitly.
Usually costs: 1 mark.

Exam tips

Learn the CISCE preparation list (alcohols, aldehydes, nitriles, Grignard) and reaction list (metals, alkalis, carbonates, derivatives, decarboxylation, HVZ, ring substitution).
For acidity questions, name the effect (−I, +I, resonance) and its direction for each substituent.
Use NaHCO₃ effervescence to tell acids from phenols; mention that phenols give no effervescence.
Show the number of carbon atoms in decarboxylation answers: soda lime removes one; Kolbe electrolysis doubles the alkyl part.
In conversion chains, write structural formulae and IUPAC names for every intermediate.

Practice questions on Carboxylic Acids: Preparation, Acidity and Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following liberates CO₂ from aqueous NaHCO₃?
(a) phenol
(b) ethanol
(c) ethanoic acid
(d) ethanal
Hint
It must be stronger than carbonic acid.
Answer
(c) ethanoic acid.
Q2 · 2 marks · Short answerBoard level
Give one use each of formic acid, acetic acid and benzoic acid.
Hint
Leather and rubber, vinegar, food preservation.
Answer
Formic acid: tanning of leather or coagulating rubber latex. Acetic acid: as vinegar (food), or as a solvent and raw material for plastics and drugs. Benzoic acid: as a food preservative (as sodium benzoate) or in medicines.
Q3 · 2 marks · Short answerHarder board level
Nitration of benzoic acid gives mainly 3-nitrobenzoic acid. Write the equation and explain why the meta product forms.
Hint
–COOH withdraws electrons by resonance.
Answer
$C_{6} H_{5} C O O H + H N O_{3} conc. H_{2} S O_{4} C_{6} H_{4} (N O_{2}) C O O H + H_{2} O$ . The –COOH group withdraws electrons from the ring by resonance, reducing electron density most at the ortho and para positions; the meta position is the least deactivated, so the electrophile attacks there.
Q4 · 3 marks · Short answerHarder board level
Write balanced equations for the reaction of ethanoic acid with (a) sodium, (b) sodium hydroxide, (c) sodium carbonate and (d) sodium hydrogencarbonate. Which of these reactions is also given by phenol?
Hint
Phenol is weaker than carbonic acid.
Answer
(a) $2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}$ . (b) $C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$ . (c) $2 C H_{3} C O O H + N a_{2} C O_{3} \to 2 C H_{3} C O O N a + H_{2} O + C O_{2}$ . (d) $C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$ . Phenol gives (a) and (b) but not (c) or (d) (it does not release CO₂).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Ethanoic acid has a higher boiling point than propan-1-ol, although their molar masses are equal.
Reason (R): Carboxylic acid molecules form strongly hydrogen-bonded dimers.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Both have M = 60.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Arrange HCOOH, CH₃COOH, CH₃CH₂COOH and (CH₃)₂CHCOOH in decreasing order of acidic strength, giving a reason.
Hint
Count and size of electron-releasing alkyl groups.
Answer
HCOOH > CH₃COOH > CH₃CH₂COOH > (CH₃)₂CHCOOH. Alkyl groups release electrons (+I), destabilising the carboxylate ion; the effect increases from no alkyl group (HCOOH) to methyl, ethyl and isopropyl.
Q7 · 3 marks · Short answerHOTS / competency
0.60 g of a monobasic carboxylic acid is exactly neutralised by 20.0 mL of 0.50 M NaOH. Find its molar mass and identify the acid if it gives a red colour with neutral FeCl₃.
Hint
Monobasic: 1 mol acid needs 1 mol NaOH.
Answer
Moles of NaOH $= 0.0200 \times 0.50 = 0.010$ mol = moles of acid. Molar mass $= \frac{0.60}{0.010} = 60$ g mol⁻¹. A monocarboxylic acid of M = 60 is ethanoic acid, CH₃COOH, consistent with the red colour with neutral FeCl₃.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student electrolyses a concentrated aqueous solution of sodium ethanoate between platinum electrodes and collects the gases at each electrode.
(i) Name the products at the anode and at the cathode. [1]
(ii) Write the overall equation. [1]
(iii) What would be obtained instead if dry sodium ethanoate were heated with soda lime? [1]
Hint
Kolbe's electrolysis doubles the alkyl group.
Answer
(i) Anode: ethane and carbon dioxide; cathode: hydrogen (and NaOH in solution). (ii) $2 C H_{3} C O O N a + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 N a O H + H_{2}$ . (iii) Methane: $C H_{3} C O O N a + N a O H CaO C H_{4} + N a_{2} C O_{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Why is a carboxylic acid more acidic than phenol?Show answer
The carboxylate ion has two equivalent resonance structures with charge on two oxygens; phenoxide puts charge on less electronegative ring carbons.
2.Product of propanoic acid with Cl₂/red P?Show answer
2-Chloropropanoic acid (HVZ reaction).
3.Product of sodium propanoate with soda lime?Show answer
Ethane.
4.Product of Kolbe electrolysis of sodium ethanoate at the anode?Show answer
Ethane and CO₂.
5.Directing effect of –COOH in benzoic acid?Show answer
Deactivating and meta directing.
6.Which acid gives a silver mirror?Show answer
Formic (methanoic) acid.

Carboxylic Acids: Preparation, Acidity and Reactions: frequently asked questions

How do electron-withdrawing groups affect the acidity of carboxylic acids?

They pull electron density away from the carboxylate ion (−I effect), stabilising it, so the acid becomes stronger; the effect is larger for more electronegative groups and closer positions.

What is the HVZ reaction?

The halogenation of a carboxylic acid with an α-hydrogen by chlorine or bromine in the presence of red phosphorus, giving an α-halo acid.

What is Kolbe's electrolysis?

Electrolysis of a concentrated aqueous solution of the sodium salt of a carboxylic acid, giving an alkane with twice the alkyl group at the anode, with CO₂.

How do you distinguish formic acid from acetic acid?

Formic acid reduces Tollens' reagent (silver mirror) and Fehling's solution; acetic acid does not.

Why do carboxylic acids have high boiling points?

They form hydrogen-bonded dimers, so more energy is needed to separate the molecules.