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Chemistry · Aldehydes, Ketones and Carboxylic Acids

Reactions of Carboxylic Acids

The carboxyl group is the busiest functional group in the chapter: it forms salts, builds an entire family of acyl derivatives, resists mild reduction, loses carbon dioxide, and even reshapes the benzene ring it sits on. NCERT §8.9 organises these reactions by which bond breaks — O–H, C–OH, or the whole −COOH group — and that classification is the cleanest way to learn them for NEET, where conversion and reaction-matching questions appear every year.

How the reactions are classified

Every reaction of a carboxylic acid breaks one of three things, and NCERT §8.9 groups them accordingly. Holding this map in mind keeps a long list of reagents organised.

Bond cleavedReaction classRepresentative products
O–HAcidity / salt formationCarboxylate salts + H₂ or CO₂
C–OHFormation of acyl derivativesEsters, acid chlorides, anhydrides, amides
Whole −COOHReduction, decarboxylation1° alcohols, hydrocarbons
α C–H / ring C–HSubstitution in the hydrocarbon partα-halo acids, meta-substituted aromatics
Figure 1

Reaction-derivative map — the carboxylic acid at the centre and the products that radiate from each reagent.

R–COOH carboxylic acid R–COONa (salt) NaOH / NaHCO₃ R–COOR′ (ester) R′OH / H⁺ R–COCl SOCl₂ / PCl₅ anhydride P₂O₅ / Δ R–CONH₂ (amide) NH₃ then Δ R–CH₂OH (1° alcohol) LiAlH₄ / B₂H₆ α-halo acid X₂ / red P (HVZ)

Salt formation and the effervescence test

Cleaving the O–H bond, carboxylic acids behave like the acids they are. With electropositive metals they evolve hydrogen, and with alkalis they form salts:

$$\ce{2CH3COOH + 2Na -> 2CH3COONa + H2 ^}$$

$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

Crucially, unlike phenols, carboxylic acids are acidic enough to react with the weak bases sodium carbonate and sodium bicarbonate, liberating carbon dioxide as visible effervescence:

$$\ce{R-COOH + NaHCO3 -> R-COONa + H2O + CO2 ^}$$

This brisk bubbling is the standard laboratory test for the carboxyl group. Phenols, being weaker acids, give no reaction with bicarbonate — so the test cleanly distinguishes a carboxylic acid from a phenol. (Soaps, incidentally, are simply the sodium salts of long-chain carboxylic acids such as stearic acid.)

NEET Trap

Bicarbonate test: acid yes, phenol no

Both phenol and a carboxylic acid will react with NaOH, so an NaOH test does not separate them. Only the carboxylic acid fizzes with NaHCO₃. Reach for sodium bicarbonate, not sodium hydroxide, when asked to distinguish the two.

Effervescence with NaHCO₃ = −COOH present. No effervescence = phenol or other weak acid.

Esterification and its mechanism

Cleaving the C–OH bond, carboxylic acids react with alcohols (or phenols) in the presence of a mineral-acid catalyst — concentrated H₂SO₄ or dry HCl gas — to give esters. This is the Fischer esterification:

$$\ce{CH3COOH + C2H5OH <=>[\text{conc. } H2SO4] CH3COOC2H5 + H2O}$$

The double arrow is essential: every step is an equilibrium, so the reaction is reversible. Water hydrolyses the ester back to the acid and alcohol, which is why the ester yield is improved by using excess alcohol or by removing the water as it forms.

Mechanistically, esterification is a nucleophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl carbon towards attack by the alcohol; a proton transfer in the tetrahedral intermediate converts −OH into a $\ce{-OH2+}$ group, a far better leaving group, which departs as neutral water; loss of a proton then unveils the ester.

Because the equilibrium constant is close to one, both the forward and reverse directions matter in practice. Le Chatelier's principle is the lever a chemist pulls to push it forward: adding a large excess of the cheaper reactant (usually the alcohol) or continuously distilling out the water as it forms shifts the position of equilibrium towards the ester. The same reasoning, run in reverse with excess water, is exactly how an ester is hydrolysed back to its parent acid.

Figure 2

Fischer esterification mechanism — protonation, nucleophilic addition, and loss of water as the better leaving group.

R–COOH + H⁺ protonated carbonyl (activated) tetrahedral intermediate R–COOR′ ester R′OH –H₂O, –H⁺ Acid-catalysed, reversible nucleophilic acyl substitution

Acid chlorides, anhydrides and amides

The C–OH bond is also the gateway to the rest of the acyl derivatives. Each derivative replaces the −OH with a different leaving group, and all of them are interconvertible — an acid chloride, for instance, reacts readily with alcohols to give esters or with ammonia to give amides. The acid itself sits at the centre of this network because its −OH is the most stubborn leaving group; converting it into a better one (Cl, an acyloxy group, or an activated water) is what makes the downstream chemistry flow.

Acid chlorides

The hydroxyl group is replaced by chlorine on treatment with PCl₅, PCl₃ or SOCl₂:

$$\ce{CH3COOH + SOCl2 -> CH3COCl + SO2 ^ + HCl ^}$$

Thionyl chloride is the preferred reagent because both by-products, $\ce{SO2}$ and $\ce{HCl}$, are gases that escape the mixture, leaving a clean acid chloride that needs no further purification.

Anhydrides

Heating an acid with a dehydrating agent such as P₂O₅ (or concentrated H₂SO₄) removes a molecule of water between two acid molecules to give an anhydride:

$$\ce{2CH3COOH ->[P2O5][\Delta] (CH3CO)2O + H2O}$$

Dicarboxylic acids can lose water intramolecularly to give cyclic anhydrides such as succinic anhydride.

Amides

Carboxylic acids react with ammonia to give an ammonium salt, which on strong heating loses water to form the amide:

$$\ce{R-COOH + NH3 -> R-COONH4 ->[\Delta] R-CONH2 + H2O}$$

DerivativeReagentProductNote
EsterR′OH, conc. H₂SO₄R-COOR'Reversible (Fischer)
Acid chlorideSOCl₂ (or PCl₃/PCl₅)R-COClSOCl₂ preferred — gaseous by-products
AnhydrideP₂O₅, Δ(RCO)2ODehydration of two acids
AmideNH₃ then ΔR-CONH2Via ammonium salt
Why -COOH gives up its proton so readily

The salt-forming and substituent chemistry here is rooted in the carboxylate's stability — see Acidity of Carboxylic Acids.

Reduction to primary alcohols

Reactions involving the whole −COOH group begin with reduction. Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride, or better still by diborane:

$$\ce{R-COOH ->[(i) LiAlH4][(ii) H3O+] R-CH2OH}$$

Diborane ($\ce{B2H6}$) is valuable because it reduces the carboxyl group while leaving ester, nitro and halo groups untouched — useful selectivity in multi-functional molecules. The carboxyl group is otherwise stubborn: sodium borohydride does not reduce it at all.

NEET Trap

NaBH₄ cannot touch −COOH

NaBH₄ reduces aldehydes and ketones happily, which tempts students to apply it to acids too. It does not work. Only the stronger hydride sources LiAlH₄ or diborane reduce a carboxylic acid, and the product is always a primary alcohol.

Acid → 1° alcohol needs LiAlH₄ or B₂H₆; NaBH₄ leaves the acid unchanged.

Decarboxylation

Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime (NaOH and CaO in a 3:1 ratio). The −COONa group is replaced by −H, so the product has one fewer carbon than the salt:

$$\ce{CH3COONa + NaOH ->[CaO][\Delta] CH4 + Na2CO3}$$

Alkali-metal salts of carboxylic acids also decarboxylate on electrolysis of their aqueous solutions, giving hydrocarbons with twice the number of carbons in the alkyl chain — the Kolbe electrolysis. The contrast is worth fixing in memory: soda-lime decarboxylation shortens the chain by one carbon (R−COONa → R−H), whereas Kolbe electrolysis doubles the alkyl skeleton (two R• radicals couple to give R−R).

Worked Example

What hydrocarbon results when sodium propanoate ($\ce{CH3CH2COONa}$) is heated with soda-lime, and how does it differ from the Kolbe product of the same salt?

Soda-lime. The −COONa is replaced by −H, removing one carbon: $\ce{CH3CH2COONa + NaOH ->[CaO][\Delta] C2H6 + Na2CO3}$, giving ethane.

Kolbe electrolysis. Two ethyl radicals couple, doubling the chain: the product is butane ($\ce{C4H10}$). Same starting salt, completely different carbon count — a classic NEET discriminator.

Hell-Volhard-Zelinsky (HVZ) reaction

Turning to the hydrocarbon part, a carboxylic acid bearing an α-hydrogen is halogenated at the α-position when treated with chlorine or bromine in the presence of a small amount of red phosphorus. This is the Hell-Volhard-Zelinsky reaction:

$$\ce{CH3CH2CH2COOH ->[(i) Br2,\ red\ P][(ii) H2O] CH3CH2CHBrCOOH}$$

The α-halo acids produced are versatile intermediates: the new C–X bond can be displaced by nucleophiles to make α-hydroxy acids, α-amino acids and substituted acids. An acid with no α-hydrogen, such as benzoic acid, does not undergo HVZ.

Ring substitution in benzoic acid

On an aromatic acid, the carboxyl group also controls the chemistry of the ring. In electrophilic aromatic substitution the −COOH group is deactivating and meta-directing, so reactions such as nitration give predominantly the meta product:

$$\ce{C6H5COOH ->[conc. HNO3][conc. H2SO4] 3\text{-}O2N\text{-}C6H4COOH}$$

Benzoic acid does not undergo Friedel-Crafts reactions: the ring is deactivated, and the Lewis-acid catalyst aluminium chloride bonds to the carboxyl group, shutting the reaction down.

Quick Recap

Reactions in one screen

  • O–H cleavage: salts with Na/NaOH; effervescence with NaHCO₃ distinguishes acid from phenol.
  • C–OH cleavage (derivatives): ester (R′OH/H⁺, reversible), acid chloride (SOCl₂ preferred), anhydride (P₂O₅/Δ), amide (NH₃ then Δ).
  • Whole −COOH: reduction to 1° alcohol by LiAlH₄ or B₂H₆ (never NaBH₄); decarboxylation with soda-lime gives a hydrocarbon with one fewer carbon.
  • Hydrocarbon part: HVZ α-halogenation with X₂/red P; on benzoic acid −COOH is deactivating, meta-directing, and blocks Friedel-Crafts.

NEET PYQ Snapshot — Reactions of Carboxylic Acids

Decarboxylation arithmetic and reaction-matching are the formats NEET keeps reusing for −COOH chemistry.

NEET 2023 · Q.69

Weight (g) of two moles of the organic compound obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is:

  1. 18
  2. 16
  3. 32
  4. 30
Answer: (3)

This is soda-lime decarboxylation: $\ce{CH3COONa + NaOH ->[CaO][\Delta] CH4 + Na2CO3}$. The organic product is methane (M = 16 g/mol), so two moles weigh $2 \times 16 = 32$ g.

NEET 2021 · Q.100

Match the reactions in List-I with their names in List-II (relevant rows): (a) $\ce{R-COCH3 + NaOX}$ → haloform; (c) $\ce{R-CH2OH + R'COOH}$, conc. H₂SO₄ → ?; (d) $\ce{R-CH2COOH}$ with (i) X₂/red P (ii) H₂O → ?

  1. (c) Esterification, (d) Hell-Volhard-Zelinsky reaction
  2. (c) Hell-Volhard-Zelinsky reaction, (d) Esterification
  3. (c) Haloform, (d) Gattermann-Koch
  4. (c) Gattermann-Koch, (d) Haloform
Answer: (1)

An alcohol plus a carboxylic acid with conc. H₂SO₄ is esterification (row c). A carboxylic acid treated with X₂/red P then water is α-halogenation, the Hell-Volhard-Zelinsky reaction (row d). The official key matches (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i).

NEET 2022 · Q.53

$\ce{RMgX + CO2 ->[ether] Y ->[H3O+] R-COOH}$. What is Y in the above reaction?

  1. $\ce{R3CO^-Mg^+X}$
  2. $\ce{RCOO^-X+}$
  3. $\ce{(RCOO)2Mg}$
  4. $\ce{R-COO^-Mg^+X}$
Answer: (4)

A Grignard reagent adds to CO₂ to give the magnesium carboxylate salt $\ce{R-COO^-Mg^+X}$. Acidic work-up (H₃O⁺) then protonates this carboxylate salt to the free carboxylic acid — the reverse of the salt-formation chemistry of −COOH.

FAQs — Reactions of Carboxylic Acids

The reagent and selectivity points NEET tests most often.

How does the carbonate test distinguish a carboxylic acid from a phenol?
Carboxylic acids are strong enough acids to react with weak bases such as sodium bicarbonate, liberating carbon dioxide as visible effervescence: R-COOH + NaHCO3 gives R-COONa + H2O + CO2. Phenols are weaker acids and do not react with bicarbonate, so they give no effervescence. The brisk bubbling with NaHCO3 is therefore a reliable test for the carboxyl group.
Why is Fischer esterification carried out with a mineral-acid catalyst and why is it reversible?
Concentrated H2SO4 or dry HCl gas protonates the carbonyl oxygen, activating the carbon towards nucleophilic attack by the alcohol. Because every step in the mechanism is an equilibrium, the overall reaction is reversible: water can hydrolyse the ester back to the acid and alcohol. The ester is favoured by using excess alcohol or by removing water as it forms.
Which reagents convert a carboxylic acid into an acid chloride and why is SOCl2 preferred?
PCl3, PCl5 and SOCl2 all replace the -OH of the acid with -Cl. SOCl2 (thionyl chloride) is preferred because its by-products, SO2 and HCl, are both gases that escape the mixture, leaving a pure acid chloride and making purification easy.
Why does NaBH4 fail to reduce a carboxylic acid while LiAlH4 succeeds?
The carboxyl group is relatively resistant to reduction. Sodium borohydride (NaBH4) is too mild and does not reduce it. The stronger hydride donors lithium aluminium hydride (LiAlH4) or diborane (B2H6) reduce the acid to a primary alcohol; diborane is useful because it leaves ester, nitro and halo groups untouched.
What is decarboxylation and what product does soda-lime give?
Decarboxylation is the loss of carbon dioxide. When the sodium salt of a carboxylic acid is heated with soda-lime (NaOH and CaO in a 3:1 ratio), the -COONa group is replaced by -H, giving a hydrocarbon with one fewer carbon. Sodium ethanoate gives methane: CH3COONa + NaOH gives CH4 + Na2CO3.
What is the Hell-Volhard-Zelinsky (HVZ) reaction?
A carboxylic acid bearing an alpha-hydrogen is halogenated at the alpha-carbon when treated with Cl2 or Br2 in the presence of a small amount of red phosphorus, giving an alpha-halocarboxylic acid. This is the HVZ reaction and it provides useful synthetic intermediates.
How does the -COOH group direct electrophilic substitution on a benzene ring?
In benzoic acid the carboxyl group is deactivating and meta-directing, so electrophilic substitution gives mainly the meta product. Benzoic acid does not undergo Friedel-Crafts reactions because the ring is deactivated and the AlCl3 catalyst bonds to the carboxyl group.
Related Topics
Aldehydes, Ketones and Carboxylic Acids Back to the full chapter hub. Acidity of Carboxylic Acids Why −COOH gives up its proton, and substituent effects. Carboxylic Acid Nomenclature & Preparation Naming acids and the routes that make them. Oxidation & Reduction of Carbonyls The reduction chemistry that frames LiAlH₄ vs NaBH₄.