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

The correct answer is (D) < (A) < (B) < (C).

To understand the increasing order of acidity of these carboxylic acids, we need to analyze the factors that stabilize the carboxylate conjugate base (R-COO^-) formed after deprotonation. The more stable the conjugate base, the stronger the acid, and the lower its pK_a value. Consequently, a stronger acid has a lower pK_a value, which means that the increasing order of acidity based on pK_a values (from lowest acidity / highest pK_a to highest acidity / lowest pK_a) corresponds to:
Weakest acid (highest pK_a) < ... < Strongest acid (lowest pK_a)

Let us analyze the inductive effect (-I effect) of the substituents attached to the carboxylic acid group (-COOH):

1. Halogen substituents: Halogens (F, Cl, Br) are highly electronegative and exert an electron-withdrawing inductive effect (-I effect). This electron withdrawal disperses the negative charge on the carboxylate ion (R-COO^-), stabilizing it and increasing the acidity of the parent acid.
2. Order of electronegativity: The electronegativity of the halogens decreases down the group:
F > Cl > Br
Therefore, the magnitude of the -I effect decreases in the order:
-F > -Cl > -Br
This means FCH₂COOH (C) is the most acidic because fluorine is the most electronegative and stabilizes the conjugate base the most. ClCH₂COOH (B) is less acidic than (C), and BrCH₂COOH (A) is less acidic than (B).

3. Formic acid (HCOOH): In formic acid (D), there is only a hydrogen atom attached to the -COOH group. Hydrogen has negligible inductive effect compared to the strongly electron-withdrawing halogens. Thus, there is no halogen-induced stabilization of the conjugate base, making HCOOH the weakest acid among the four.

Comparing the acidities, we have:
HCOOH (D) is the weakest acid (highest pK_a).
BrCH₂COOH (A) is stronger than formic acid due to the -I effect of bromine.
ClCH₂COOH (B) is stronger than (A) due to the higher electronegativity of chlorine compared to bromine.
FCH₂COOH (C) is the strongest acid (lowest pK_a) due to the highest electronegativity of fluorine.

Thus, the increasing order of acidity (and therefore the decreasing order of pK_a values) is:
(D) < (A) < (B) < (C)