Acidity order of carboxylic acids — effect of electron-withdrawing groups

Question

Arrange the following in decreasing order of acidity: $\text{CH}_3\text{COOH}$ , $\text{ClCH}_2\text{COOH}$ , $\text{Cl}_2\text{CHCOOH}$ , $\text{Cl}_3\text{CCOOH}$ , $\text{HCOOH}$ . Explain the effect of electron-withdrawing groups on acid strength.

(JEE Main 2023, similar pattern)

Solution — Step by Step

A stronger acid donates its proton more easily, meaning its conjugate base (carboxylate ion) is more stable. Anything that stabilises the carboxylate ion (by dispersing the negative charge) increases acidity.

Electron-withdrawing groups (EWGs) pull electron density away from the $\text{COO}^-$ group, spreading the negative charge over a larger region and stabilising the anion.

Chlorine is an electron-withdrawing group that exerts a strong -I (negative inductive) effect. More Cl atoms = stronger electron withdrawal = more stable carboxylate = stronger acid.

\text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH} > \text{HCOOH} > \text{CH}_3\text{COOH}

The pKa values confirm this: $\text{Cl}_3\text{CCOOH}$ (0.65) < $\text{Cl}_2\text{CHCOOH}$ (1.29) < $\text{ClCH}_2\text{COOH}$ (2.86) < $\text{HCOOH}$ (3.75) < $\text{CH}_3\text{COOH}$ (4.76).

In $\text{CH}_3\text{COOH}$ , the methyl group has a +I effect (electron-donating), which pushes electron density toward the $\text{COO}^-$ group and destabilises the anion. In $\text{HCOOH}$ , there is no such effect. So formic acid is stronger than acetic acid.

\boxed{\text{Cl}_3\text{CCOOH} > \text{Cl}_2\text{CHCOOH} > \text{ClCH}_2\text{COOH} > \text{HCOOH} > \text{CH}_3\text{COOH}}

Why This Works

Acid strength is fundamentally about the stability of the conjugate base. EWGs stabilise the carboxylate anion by two mechanisms:

Inductive effect (-I): Through-bond electron withdrawal. Falls off rapidly with distance — substituents on the alpha carbon have the strongest effect.
Field effect: Through-space electrostatic interaction. This is why the position of the substituent matters — closer to the $\text{COOH}$ means stronger effect.

Both effects work in the same direction for halogens, making haloacids significantly stronger than the parent acid.

Alternative Method

You can also compare pKa values directly. Lower pKa = stronger acid. For quick ranking in exams, use these rules:

More EWGs = stronger acid
Closer EWG to $\text{COOH}$ = stronger acid
Stronger EWG (F > Cl > Br > I by -I effect) = stronger acid

For JEE, the most tested comparison is between positional isomers: $\text{F-CH}_2\text{COOH}$ vs $\text{CH}_3\text{CHF-COOH}$ . The one with F closer to $\text{COOH}$ is more acidic. Also remember that among halogens, F has the strongest -I effect but Cl has a stronger effect through resonance in aromatic systems (where -I and +R compete).

Common Mistake

Students sometimes rank $\text{HCOOH}$ as weaker than $\text{CH}_3\text{COOH}$ , thinking “more carbon = more acid.” The opposite is true — $\text{CH}_3$ is electron-donating (+I), which destabilises the conjugate base and weakens the acid. Formic acid (pKa 3.75) is about 10 times stronger than acetic acid (pKa 4.76). Never assume bigger molecule = stronger acid without checking the electronic effects.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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