JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Carboxylic Acids: Acidity and Reactions

Carboxylic acids are made stronger by electron-withdrawing groups and weaker by donors; they are made by oxidation, by hydrolysis or from a Grignard reagent and CO₂, and they turn into esters, acid chlorides, amides, alcohols and alkanes.

All Aldehydes, Ketones and Carboxylic Acids notes JEE Mains Chemistry strategy

Why this matters

Twenty-four PYQs, none numerical, four from 2026. Ten rank acid strength or ask which compounds release CO₂ from sodium hydrogencarbonate; four choose the route that ends at a carboxylic acid; ten ask what an acid or one of its derivatives does with a reagent.

Concept 1 of 3: Ranking the strength of carboxylic acids

An acid is as strong as its anion is stable. The carboxylate ion already spreads its charge over two oxygens. A group that pulls electrons away spreads it further and strengthens the acid; a group that pushes electrons in weakens it.

Definition

−I groups strengthen the acid. The more electronegative the group, the stronger (F > Cl > Br > I); the more such groups, the stronger ( $\mathrm{Cl_3C > Cl_2CH > ClCH_2}$ ); the closer to COOH, the stronger, because the inductive effect fades along the chain.
Alkyl groups (+I) weaken it: methanoic acid (pKa 3.75) > ethanoic acid (4.76) > propanoic acid (4.88).
Benzoic acid (4.19) is stronger than ethanoic acid. A para $\mathrm{NO_2}$ strengthens it; a para $\mathrm{CH_3}$ or $\mathrm{OCH_3}$ weakens it. Almost any ortho group strengthens it (the ortho effect).
Carboxylic acids are stronger than phenols (pKa about 10). An acid stronger than carbonic acid (pKa about 6.4) releases $\mathrm{CO_2}$ from $\mathrm{NaHCO_3}$ ; phenols do not, except picric acid (2,4,6-trinitrophenol, pKa 0.38).

pKa of some acids (lower pKa, stronger acid)

\mathrm{CF_3COOH}\ (0.23) < \mathrm{CCl_3COOH}\ (0.65) < \mathrm{ClCH_2COOH}\ (2.86) < \mathrm{HCOOH}\ (3.75) < \mathrm{C_6H_5COOH}\ (4.19) < \mathrm{CH_3COOH}\ (4.76)

Worked example

Arrange in decreasing order of acid strength: butanoic acid, 2-chlorobutanoic acid, 3-chlorobutanoic acid, 4-chlorobutanoic acid.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q137Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Carboxylic Acids: Acidity and Reactions

The descending order of acidity for the following carboxylic acid is : (A)

CH_{3}COOH

(B)

F_{3}C - COOH

(C)

ClCH_{2}- COOH

(D)

FCH_{2}- COOH

(E)

BrCH_{2}- COOH

Choose the correct answer from the options given below:

Distance weakens the inductive effect

A halogen two or three carbons away from COOH has a small effect. Rank by position before counting halogens along a chain.

Picric acid behaves like a carboxylic acid with NaHCO₃

Three nitro groups make 2,4,6-trinitrophenol a strong acid, so it releases

\mathrm{CO_2}

from

\mathrm{NaHCO_3}

. Other phenols do not.

Drill 9 more on ranking the strength of carboxylic acids

Concept 2 of 3: Routes that end at a carboxylic acid

A carboxylic acid is made either by oxidising a carbon that already carries oxygen or a benzylic hydrogen, or by hydrolysing a group that is already at the acid oxidation level, or by adding

\mathrm{CO_2}

to a Grignard reagent. Note whether each route keeps, adds or removes a carbon.

Definition

Oxidation of a 1° alcohol or an aldehyde gives the acid with the same number of carbons. PCC is the exception: it stops at the aldehyde.
Hot alkaline $\mathrm{KMnO_4}$ oxidises any alkyl side chain with a benzylic hydrogen down to COOH on the ring.
Nitriles, amides, esters, acid chlorides and anhydrides all hydrolyse to the acid. A nitrile passes through the amide, and mild conditions stop there.
A Grignard reagent with $\mathrm{CO_2}$ gives an acid one carbon longer. The haloform reaction gives an acid one carbon shorter than the methyl ketone.

Starting compound	Reagents	Product
1° alcohol $\mathrm{RCH_2OH}$	Alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$ ; or Jones reagent	$\mathrm{RCOOH}$ , same carbons
Aldehyde $\mathrm{RCHO}$	Tollens' reagent, $\mathrm{K_2Cr_2O_7/H^+}$ or bromine water	$\mathrm{RCOOH}$ , same carbons
Alkylbenzene with a benzylic H	Hot alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$	Benzoic acid, whatever the chain length
Nitrile $\mathrm{RCN}$	$\mathrm{H_3O^+}$ and heat (or $\mathrm{OH^-}$ , then acid)	$\mathrm{RCOOH}$ , through the amide $\mathrm{RCONH_2}$
Grignard reagent $\mathrm{RMgX}$	Dry ice $\mathrm{CO_2}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon more
Methyl ketone $\mathrm{RCOCH_3}$	$\mathrm{I_2}$ and NaOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon fewer, and $\mathrm{CHI_3}$
1,1,1-Trihalide $\mathrm{RCCl_3}$	Aqueous KOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , same carbons
Ester, acid chloride or anhydride	Water with acid or alkali, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ (an ester also gives the alcohol)

Grignard plus CO₂ adds a carbon; the haloform reaction removes one; the rest keep the count.

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The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q109Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Carboxylic Acids: Acidity and Reactions

Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product ? (A)

R - C \equiv N

\underset{mild\ Condition}{\overset{{(i)H}^{+}/H_{2}O}{\rightarrow}}

(B)

R - \underset{(ii)H_{3}O^{+}}{MgX\overset{(i){CO}_{2}}{\rightarrow}}

(C)

R - C \equiv N\underset{\text{~(ii)~}H_{3}O^{+}\ }{\overset{\text{(i)~}{SnCl}_{2}/HCl\ }{\rightarrow}}

(D)

R - {CH}_{2} - OH\overset{PCC\ }{\rightarrow}

(E)

R - COCl\underset{\text{(ii)~}{Br}_{2}\text{~water}}{\overset{\text{(i)~}H_{2}/Pd - {BaSO}_{4}}{\rightarrow}}

Choose the correct answer from the options given below :

Mild hydrolysis of a nitrile stops at the amide

A nitrile needs vigorous hydrolysis to reach the acid. Under mild conditions the product is

\mathrm{RCONH_2}

, not RCOOH.

Count the carbons

Grignard carboxylation adds one carbon and the haloform reaction removes one. A route that gives the right functional group but the wrong chain length is the wrong answer.

Drill 3 more on routes that end at a carboxylic acid

Concept 3 of 3: Reactions of carboxylic acids and their derivatives

Most reactions of an acid replace its OH (nucleophilic acyl substitution): with an alcohol it becomes an ester, with

\mathrm{SOCl_2}

an acid chloride, with ammonia an amide. A few remove the COOH altogether, and one halogenates the carbon next to it.

Definition

Acid derivatives react with nucleophiles by addition, then loss of the leaving group. The better the leaving group, the faster: $\mathrm{Cl^- > RCOO^- > R'O^- > NH_2^-}$ .
So the rate of hydrolysis is acid chloride > anhydride > ester > amide.
Fischer esterification (acid, alcohol, concentrated $\mathrm{H_2SO_4}$ ) is reversible. The OH of the water comes from the acid, the OR of the ester from the alcohol.
The COOH group deactivates a benzene ring and directs to the meta position. Benzoic acid does not undergo Friedel–Crafts reactions.

Reagent	Product from RCOOH	Remember
$\mathrm{NaHCO_3}$ solution	$\mathrm{RCOONa + CO_2 + H_2O}$	Effervescence separates acids from phenols
$\mathrm{R'OH}$ , conc. $\mathrm{H_2SO_4}$ , heat	Ester $\mathrm{RCOOR'}$	Reversible; nucleophilic acyl substitution
$\mathrm{SOCl_2}$ (or $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ )	Acid chloride $\mathrm{RCOCl}$	With $\mathrm{SOCl_2}$ the by-products $\mathrm{SO_2}$ and HCl are gases
$\mathrm{P_2O_5}$ , heat; or heat alone for a suitable diacid	Anhydride $\mathrm{(RCO)_2O}$	cis-Butenedioic (maleic) acid gives a cyclic anhydride on heating; the trans acid cannot
$\mathrm{NH_3}$ , then heat	Amide $\mathrm{RCONH_2}$	Through the ammonium salt $\mathrm{RCOONH_4}$
$\mathrm{LiAlH_4}$ or $\mathrm{B_2H_6}$ , then $\mathrm{H_3O^+}$	1° alcohol $\mathrm{RCH_2OH}$	$\mathrm{NaBH_4}$ does not reduce COOH
Sodium salt with NaOH and CaO (soda lime), heat	Alkane $\mathrm{RH}$	Decarboxylation: one carbon fewer
Electrolysis of the sodium salt (Kolbe)	Alkane $\mathrm{R{-}R}$	Two R groups join
$\mathrm{X_2}$ and red phosphorus, then water (Hell–Volhard–Zelinsky)	α-Halo acid $\mathrm{RCH(X)COOH}$	Only the α-carbon is halogenated; it needs an α-hydrogen
Conc. $\mathrm{HNO_3}$ and conc. $\mathrm{H_2SO_4}$ (on benzoic acid)	3-Nitrobenzoic acid	COOH is meta-directing and deactivating

The first six change only the COOH group; soda lime and Kolbe remove it; HVZ acts at the α-carbon and nitration on the ring.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 12 · Q27Moderate

Example 3 · Aldehydes, Ketones and Carboxylic Acids · Carboxylic Acids: Acidity and Reactions

The correct order of the rate of hydrolysis of an ester, an acid chloride and an acid anhydride is:

HVZ halogenates only the α-carbon

The halogen goes to the carbon next to COOH, never further along the chain. An acid with no α-hydrogen, such as 2,2-dimethylpropanoic acid, does not react.

Soda lime removes a carbon

Decarboxylation of

\mathrm{RCOONa}

gives

\mathrm{RH}

, with one carbon fewer than the acid. Sodium propanoate gives ethane, not propane.

Drill 9 more on reactions of carboxylic acids and their derivatives

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Ranking the strength of carboxylic acids
pKa of some acids (lower pKa, stronger acid)
$\mathrm{CF_3COOH}\ (0.23) < \mathrm{CCl_3COOH}\ (0.65) < \mathrm{ClCH_2COOH}\ (2.86) < \mathrm{HCOOH}\ (3.75) < \mathrm{C_6H_5COOH}\ (4.19) < \mathrm{CH_3COOH}\ (4.76)$

Reference tables (2)

Routes that end at a carboxylic acid8 rows

Starting compound	Reagents	Product
1° alcohol $\mathrm{RCH_2OH}$	Alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$ ; or Jones reagent	$\mathrm{RCOOH}$ , same carbons
Aldehyde $\mathrm{RCHO}$	Tollens' reagent, $\mathrm{K_2Cr_2O_7/H^+}$ or bromine water	$\mathrm{RCOOH}$ , same carbons
Alkylbenzene with a benzylic H	Hot alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$	Benzoic acid, whatever the chain length
Nitrile $\mathrm{RCN}$	$\mathrm{H_3O^+}$ and heat (or $\mathrm{OH^-}$ , then acid)	$\mathrm{RCOOH}$ , through the amide $\mathrm{RCONH_2}$
Grignard reagent $\mathrm{RMgX}$	Dry ice $\mathrm{CO_2}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon more
Methyl ketone $\mathrm{RCOCH_3}$	$\mathrm{I_2}$ and NaOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon fewer, and $\mathrm{CHI_3}$
1,1,1-Trihalide $\mathrm{RCCl_3}$	Aqueous KOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , same carbons
Ester, acid chloride or anhydride	Water with acid or alkali, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ (an ester also gives the alcohol)

Grignard plus CO₂ adds a carbon; the haloform reaction removes one; the rest keep the count.

Reactions of carboxylic acids and their derivatives10 rows

Reagent	Product from RCOOH	Remember
$\mathrm{NaHCO_3}$ solution	$\mathrm{RCOONa + CO_2 + H_2O}$	Effervescence separates acids from phenols
$\mathrm{R'OH}$ , conc. $\mathrm{H_2SO_4}$ , heat	Ester $\mathrm{RCOOR'}$	Reversible; nucleophilic acyl substitution
$\mathrm{SOCl_2}$ (or $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ )	Acid chloride $\mathrm{RCOCl}$	With $\mathrm{SOCl_2}$ the by-products $\mathrm{SO_2}$ and HCl are gases
$\mathrm{P_2O_5}$ , heat; or heat alone for a suitable diacid	Anhydride $\mathrm{(RCO)_2O}$	cis-Butenedioic (maleic) acid gives a cyclic anhydride on heating; the trans acid cannot
$\mathrm{NH_3}$ , then heat	Amide $\mathrm{RCONH_2}$	Through the ammonium salt $\mathrm{RCOONH_4}$
$\mathrm{LiAlH_4}$ or $\mathrm{B_2H_6}$ , then $\mathrm{H_3O^+}$	1° alcohol $\mathrm{RCH_2OH}$	$\mathrm{NaBH_4}$ does not reduce COOH
Sodium salt with NaOH and CaO (soda lime), heat	Alkane $\mathrm{RH}$	Decarboxylation: one carbon fewer
Electrolysis of the sodium salt (Kolbe)	Alkane $\mathrm{R{-}R}$	Two R groups join
$\mathrm{X_2}$ and red phosphorus, then water (Hell–Volhard–Zelinsky)	α-Halo acid $\mathrm{RCH(X)COOH}$	Only the α-carbon is halogenated; it needs an α-hydrogen
Conc. $\mathrm{HNO_3}$ and conc. $\mathrm{H_2SO_4}$ (on benzoic acid)	3-Nitrobenzoic acid	COOH is meta-directing and deactivating

The first six change only the COOH group; soda lime and Kolbe remove it; HVZ acts at the α-carbon and nitration on the ring.

Watch out for (6)

Distance weakens the inductive effect
→ Ranking the strength of carboxylic acids
Picric acid behaves like a carboxylic acid with NaHCO₃
→ Ranking the strength of carboxylic acids
Mild hydrolysis of a nitrile stops at the amide
→ Routes that end at a carboxylic acid
Count the carbons
→ Routes that end at a carboxylic acid
HVZ halogenates only the α-carbon
→ Reactions of carboxylic acids and their derivatives
Soda lime removes a carbon
→ Reactions of carboxylic acids and their derivatives

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