The increasing order of acidity of the following compounds based on \(pK_a\) values is

(A) \(BrCH_2COOH\)

(B) \(ClCH_2COOH\)

(C) \(FCH_2COOH\)

(D) \(HCOOH\)

Choose the correct answer from the options given below:

D < A < B < C

The correct answer is option 1. D < A < B < C.

Let us delve deeper into the concept of acidity and how substituents affect the acidity of carboxylic acids.

Carboxylic acids (\(R-COOH\)) are acidic compounds due to the presence of the carboxyl group (\(-COOH\)). In aqueous solution, carboxylic acids partially dissociate to yield hydrogen ions (\(H^+\)) and carboxylate ions (\(RCOO^-\)). The equilibrium constant for this dissociation is represented by the acid dissociation constant (\(K_a\)).

\(R-COOH \rightleftharpoons RCOO^- + H^+\)

The \(K_a\) value represents the tendency of the acid to donate a proton (H⁺) in solution. The lower the \(K_a\) value, the weaker the acid. The \(pK_a\) value is the negative logarithm of the \(K_a\) value, often used to compare the relative acidity of different acids.

Effect of Substituents on Acidity:
The acidity of carboxylic acids can be influenced by substituents attached to the carbon atom adjacent to the carboxyl group (\(R\)). Substituents can have either electron-donating or electron-withdrawing effects, which affect the stability of the conjugate base formed upon dissociation.

Electron-Withdrawing Substituents: Substituents such as halogens (\(F\), \(Cl\), \(Br\)), nitro (\(NO_2\)), and cyano (\(CN\)) groups are electron-withdrawing. They stabilize the carboxylate ion (\(RCOO^-\)) through inductive and resonance effects, making the corresponding carboxylic acid more acidic.

Electron-Donating Substituents: Substituents such as alkyl (\(R'\)) groups are electron-donating. They destabilize the carboxylate ion, reducing the acidity of the carboxylic acid.

Analysis of Given Compounds:

Let us consider the given compounds:

A. \(BrCH_2COOH\) (Bromine-substituted acetic acid): Bromine (\(Br\)) is electron-withdrawing.

B. \(ClCH_2COOH\) (Chlorine-substituted acetic acid): Chlorine (\(Cl\)) is also electron-withdrawing.

C. \(FCH_2COOH\) (Fluorine-substituted acetic acid): Fluorine (\(F\)) is the most electron-withdrawing among the halogens.

D. \(HCOOH\) (Formic acid): No substituent. Serves as the reference.

Considering the electron-withdrawing nature of halogens, with fluorine being the most electron-withdrawing, the acidity order based on \(pK_a\) values should follow the trend: \(D < A < B < C\)

Therefore, the correct explanation is: option 1. D < A < B < C