The correct order of increasing pKa values of the compounds :
(A) CH3COOH
(B) CH3OCH2COOH
(C) Cl3COOH
(D) (CH3)3CCOOH
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) < (B) < (A) < (D)
A stronger acid has a lower pKa. Electron withdrawing groups on the carbon next to –COOH stabilise the carboxylate ion and make the acid stronger; electron donating groups destabilise it and make the acid weaker.
(C) Trichloroacetic acid, \( \ce{Cl3CCOOH} \): three strongly electron withdrawing chlorine atoms; the strongest acid of the four: first.
(B) \( \ce{CH3OCH2COOH} \): the electronegative oxygen of the methoxy group withdraws electrons by its −I effect: second.
(A) \( \ce{CH3COOH} \): acetic acid, with a methyl group: third.
(D) \( \ce{(CH3)3CCOOH} \): the tert-butyl group pushes electrons (+I) more strongly than methyl and destabilises the anion; the weakest acid: last.
Shortcut: The trichloro acid has the lowest pKa and the tert-butyl acid the highest; only Option (4) starts with (C) and ends with (D).
Note: It seems that there is a mistake in the original NTA paper, which prints Cl3COOH when it should read Cl3CCOOH (trichloroacetic acid).
NCERT (2026-27 edition):
Effect of Substituents on the Acidity of Carboxylic Acids – …
