The increasing order of acidity of the following compounds based on pKa values is
(A) BrCH2COOH
(B) ClCH2COOH
(C) FCH2COOH
(D) HCOOH
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (D) < (A) < (B) < (C)
A carboxylic acid is stronger when its carboxylate ion is more stable, and a stronger acid has a lower pKa. A halogen on the α-carbon is an electron withdrawing group: by its −I effect it pulls electron density away from the –\( \ce{COO^{-}} \) group, spreads the negative charge and stabilises the anion.
The −I effect of a halogen grows with its electronegativity:
F > Cl > Br
Formic acid carries only a hydrogen atom next to the –COOH group, so its anion gets none of this stabilisation; it is the weakest of the four.
With the given labels:
(D) \( \ce{HCOOH} \): no halogen: first (weakest acid)
(A) \( \ce{BrCH2COOH} \): the least electronegative of the three halogens: second
(B) \( \ce{ClCH2COOH} \): third
(C) \( \ce{FCH2COOH} \): the most electronegative halogen: last (strongest acid)
Shortcut: HCOOH (D) is the weakest and \( \ce{FCH2COOH} \) (C) is the strongest. Only Option (1) starts with (D) and ends with (C).
NCERT (2026-27 edition):
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. Conversely, electron donating groups decrease the acidity by destabilising the conjugate base. … The effect of the following groups in increasing acidity order is Ph < I < Br < Cl < F < CN < NO2 < CF3”