12 min readPathway sheet1-mark MCQ on reagents; 2 or 3-mark equations for HVZ, decarboxylation, Kolbe electrolysis, esterification or SOCl₂; often inside a conversion; also a NEET favourite

Reactions and uses of carboxylic acids

Carboxylic acids lose -OH to give acid derivatives, lose CO₂ when heated with soda lime, and are α-halogenated in the Hell-Volhard-Zelinsky reaction.

HVZ reactionRCH₂COOH → (i) X₂, red phosphorus (ii) H₂O → RCH(X)COOH, where X = Cl or Br

DecarboxylationRCOONa + NaOH → (CaO, heat) RH + Na₂CO₃; soda lime is NaOH and CaO in a 3 : 1 ratio

Kolbe electrolysis2CH₃COONa + 2H₂O → (electrolysis) CH₃CH₃ + 2CO₂ + H₂ + 2NaOH

ReductionLiAlH₄ or diborane (B₂H₆) reduces -COOH to -CH₂OH; NaBH₄ does not

Reactions that break the C-OH bond

1
Anhydride: 2CH₃COOH → (H⁺ or P₂O₅, heat) (CH₃CO)₂O + H₂O
Ethanoic anhydride
2★
Esterification: RCOOH + R'OH ⇌ (conc. H₂SO₄ or dry HCl gas) RCOOR' + H₂O
Reversible. The acid protonates the C=O oxygen so the alcohol can attack. Removing water or the ester as it forms pushes the reaction forward (Le Chatelier's principle)
3★
Acyl chlorides: RCOOH + PCl₅ → RCOCl + POCl₃ + HCl; 3RCOOH + PCl₃ → 3RCOCl + H₃PO₃; RCOOH + SOCl₂ → RCOCl + SO₂ + HCl
Thionyl chloride is preferred: SO₂ and HCl are gases and escape, so the acyl chloride is easy to purify
4
Amides: CH₃COOH + NH₃ ⇌ CH₃COO⁻NH₄⁺ → (strong heat) CH₃CONH₂ + H₂O
The ammonium salt loses water. Benzoic acid gives benzamide, C₆H₅CONH₂. Phthalic acid gives phthalamide, which on strong heating loses NH₃ to give phthalimide

★ key step

Reactions of the -COOH group as a whole

1★
Reduction: RCOOH → (i) LiAlH₄ in ether, or B₂H₆ (ii) H₃O⁺ → RCH₂OH
Diborane does not easily reduce ester, nitro or halo groups. NaBH₄ does not reduce -COOH
2★
Decarboxylation: RCOONa + NaOH → (CaO, heat) R-H + Na₂CO₃
Heating the sodium salt with soda lime removes CO₂: the alkane has one carbon fewer. CH₃COONa gives CH₄
3
Kolbe electrolysis: 2CH₃COONa + 2H₂O → (electrolysis) CH₃CH₃ + 2CO₂ + H₂ + 2NaOH
Electrolysing a sodium or potassium salt in water gives an alkane with twice the carbons of the alkyl group. Ethane forms at the anode and H₂ at the cathode

★ key step

Substitution in the hydrocarbon part

1★
Hell-Volhard-Zelinsky (HVZ): R-CH₂-COOH → (i) X₂, red phosphorus (ii) H₂O → R-CH(X)-COOH
Acids with an α-hydrogen are halogenated at the α-carbon (X = Cl or Br), using a small amount of red phosphorus
2
Example: CH₃COOH + Cl₂ → (red P) ClCH₂COOH + HCl
Chloroethanoic acid
3
Ring substitution: C₆H₅COOH + HNO₃ → (conc. H₂SO₄) m-O₂NC₆H₄COOH + H₂O
-COOH is deactivating and meta-directing: 3-nitrobenzoic acid. Br₂ with FeBr₃ gives 3-bromobenzoic acid
4
Benzoic acid + CH₃Cl or CH₃COCl, with anhydrous AlCl₃ → no Friedel-Crafts product
-COOH deactivates the ring, and the catalyst AlCl₃ (a Lewis acid) gets bonded to the carboxyl group

★ key step

Reagent and product at a glance

Reagent	Product from RCOOH (or its salt)
NaHCO₃	RCOONa + H₂O + CO₂ (effervescence)
P₂O₅ or H⁺, heat	Anhydride, (RCO)₂O
R'OH + conc. H₂SO₄	Ester, RCOOR'
PCl₅, PCl₃ or SOCl₂	Acyl chloride, RCOCl
NH₃, then strong heat	Amide, RCONH₂
LiAlH₄ or B₂H₆, then H₃O⁺	Primary alcohol, RCH₂OH
Sodium salt + soda lime, heat	Alkane RH, one carbon fewer
Electrolysis of the sodium salt	Alkane R-R (Kolbe)
X₂ + red phosphorus	α-Halo acid, RCH(X)COOH

Worked example: write the products

Problem

What happens when: (i) ethanoic acid is treated with Cl₂ and red phosphorus (ii) sodium propanoate is heated with soda lime (iii) benzoic acid is treated with SOCl₂ (iv) 3-nitrobenzoic acid is reduced with diborane?

Working

(i) HVZ reaction: CH₃COOH + Cl₂ → (red P) ClCH₂COOH + HCl. Product: chloroethanoic acid.
(ii) Decarboxylation: CH₃CH₂COONa + NaOH → (CaO, heat) CH₃CH₃ + Na₂CO₃. Product: ethane, one carbon fewer.
(iii) C₆H₅COOH + SOCl₂ → C₆H₅COCl + SO₂ + HCl. Product: benzoyl chloride.
(iv) B₂H₆ reduces -COOH to -CH₂OH but leaves -NO₂ alone. Product: 3-nitrobenzyl alcohol, m-O₂NC₆H₄CH₂OH.

Answer

Chloroethanoic acid; ethane; benzoyl chloride; 3-nitrobenzyl alcohol

Uses of carboxylic acids

Methanoic acid: in the rubber, textile, dyeing, leather and electroplating industries.
Ethanoic acid: as a solvent, and as vinegar in the food industry.
Hexanedioic acid (adipic acid): to make nylon-6,6.
Esters of benzoic acid: in perfumery.
Sodium benzoate: a food preservative.
Higher fatty acids: to make soaps and detergents.

In real life

The vinegar in pickles and the sodium benzoate in bottled sauces both stop microbes from spoiling food. Hexanedioic acid is a raw material for nylon, used in ropes and fishing nets.

Exam angle

2 marks: “Write the Hell-Volhard-Zelinsky reaction of ethanoic acid. What structural feature must an acid have for this reaction?” CH₃COOH → (i) Cl₂, red P (ii) H₂O → ClCH₂COOH. The acid must have an α-hydrogen.

2 marks: “What is Kolbe electrolysis? Write the equation for sodium ethanoate.” Electrolysis of an aqueous solution of a sodium or potassium carboxylate gives an alkane with twice the alkyl carbons. 2CH₃COONa + 2H₂O → CH₃CH₃ (anode) + 2CO₂ + H₂ (cathode) + 2NaOH.

3 marks: “Give reasons: (i) SOCl₂ is preferred for making acyl chlorides (ii) water is removed during esterification (iii) benzoic acid does not undergo Friedel-Crafts reaction.” (i) Its by-products, SO₂ and HCl, are gases and escape, so purification is easy. (ii) Esterification is reversible; removing water shifts the equilibrium forward. (iii) -COOH deactivates the ring and AlCl₃ bonds to the carboxyl group.

Remember it

Acid tool kit: SOCl₂ → acyl chloride; R'OH + H⁺ → ester; soda lime → one carbon fewer; Kolbe → double up; X₂ + red P → α-halo acid; B₂H₆ → alcohol.