Which of the following has the highest pKa value?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Iodoacetic acid
\( \displaystyle pK_a=-\log K_a \), so the weakest acid has the highest pKa.
A halogen on the α-carbon is an electron withdrawing group. By its −I effect it pulls electron density from the carboxylate ion, spreads its negative charge and so stabilises it, which makes the acid stronger. The pull falls as the electronegativity of the halogen falls, F > Cl > Br > I, so the acid strength falls in the same order:
\( \ce{FCH2COOH > ClCH2COOH > BrCH2COOH > ICH2COOH} \)
Iodine withdraws electrons least, so iodoacetic acid is the weakest of the four and has the highest pKa.
Why the other options are incorrect
Option (1) Bromoacetic acid — Incorrect. Bromine is more electronegative than iodine, so bromoacetic acid is the stronger acid and has a lower pKa.
Option (2) Chloroacetic acid — Incorrect. Chlorine withdraws electrons more strongly than bromine and iodine, so its pKa is lower still.
Option (3) Fluoroacetic acid — Incorrect. Fluorine, the most electronegative halogen, makes this the strongest acid of the four, with the lowest pKa.
NCERT (2026-27 edition):
8.9.1 Reactions Involving Cleavage of O–H Bond – “… Smaller the pKa, the stronger the acid ( the better it is as a proton donor).”
“Effect of substituents on the acidity of carboxylic acids: Substituents may affect the stability of the conjugate base and thus, also affect the acidity of the carboxylic acids. Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects. …”
“The effect of the following groups in increasing acidity order is Ph < I < Br < Cl < F < CN < NO2 < CF3”