Aldehydes, Ketones and Carboxylic Acids: Application Problems (5)

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Question

Arrange the following in increasing order of acidity and justify: (i) acetic acid, (ii) chloroacetic acid, (iii) trichloroacetic acid, (iv) formic acid, (v) benzoic acid.

Solution — Step by Step

Acidity depends on stability of the carboxylate anion (the conjugate base). Electron-withdrawing groups (EWG) stabilize negative charge, increasing acidity. Electron-donating groups (EDG) destabilize it, decreasing acidity.

(i) CH3_3COOH: methyl is a mild EDG, slightly destabilizes carboxylate. Ka1.8×105K_a \approx 1.8 \times 10^{-5}, pKa=4.76\text{p}K_a = 4.76.

(ii) ClCH2_2COOH: one Cl pulls electrons (-I effect), stabilizes carboxylate. pKa=2.86\text{p}K_a = 2.86.

(iii) Cl3_3CCOOH: three Cl’s pull strongly. pKa=0.65\text{p}K_a = 0.65 — very strong acid.

(iv) HCOOH: just H, no group to donate or withdraw. pKa=3.75\text{p}K_a = 3.75.

(v) C6_6H5_5COOH: phenyl is mildly EWG via inductive but EDG via resonance — overall slightly EWG. pKa=4.20\text{p}K_a = 4.20.

Acidity (lowest pKa\text{p}K_a first): (iii) Cl3_3CCOOH (0.65) > (ii) ClCH2_2COOH (2.86) > (iv) HCOOH (3.75) > (v) C6_6H5_5COOH (4.20) > (i) CH3_3COOH (4.76).

(i) < (v) < (iv) < (ii) < (iii).

Final answer: CH3_3COOH < C6_6H5_5COOH < HCOOH < ClCH2_2COOH < Cl3_3CCOOH.

Why This Works

The I-I effect of halogens delocalizes negative charge from the carboxylate oxygen to the rest of the molecule, stabilizing the anion. Each additional Cl adds more I-I, so trichloroacetate is most stabilized → trichloroacetic acid is most acidic.

Methyl, in contrast, donates electrons (+I), making CH3_3COO^- less stable than HCOO^- (no +I), so formic acid is more acidic than acetic acid.

Benzoic acid: phenyl’s net effect is slightly electron-withdrawing in the carboxylic-acid context, making it slightly more acidic than acetic acid.

Alternative Method

Compare KaK_a values directly if memorized. The pattern Cl3C>ClCH2>H>C6H5>CH3\text{Cl}_3 \text{C} > \text{ClCH}_2 > \text{H} > \text{C}_6\text{H}_5 > \text{CH}_3 for the alpha substituent is a standard JEE-NEET ranking.

Common Mistake

Students often place benzoic acid above formic acid in acidity, reasoning that phenyl is “aromatic and electron-withdrawing”. But phenyl’s resonance donation slightly counteracts its inductive withdrawal, leaving it weaker overall than HCOOH. Memorize: HCOOH > C6H5COOH > CH3COOH\text{HCOOH > C}_6\text{H}_5\text{COOH > CH}_3\text{COOH}.

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