Question

The increasing order of acidity of the following compounds based on pKa values:
(A) BrCH$_2$COOH
(B) ClCH$_2$COOH
(C) FCH$_2$COOH
(D) HCOOH

• (D) $<$ (A) $<$ (B) $<$ (C)
• (A) $<$ (D) $<$ (C) $<$ (B)
• (B) $<$ (A) $<$ (D) $<$ (C)
• (C) $<$ (B) $<$ (D) $<$ (A)

Answer 1

The Correct Option is A

To rank the acidity of the given compounds in increasing order based on their pKa values, we examine the impact of substituents on carboxylic acid acidity. Acidity is modulated by electron-withdrawing or electron-donating substituent effects:

• (A) BrCH₂COOH
• (B) ClCH₂COOH
• (C) FCH₂COOH
• (D) HCOOH

Electron-withdrawing groups, such as halogens, enhance the stability of the carboxylate anion after deprotonation, thereby increasing acidity. Higher substituent electronegativity correlates with greater acidity.

Analyzing each compound's acidity by substituent electronegativity:

• Compound (C) FCH₂COOH, featuring fluorine (the most electronegative element), exhibits the highest acidity.
• Compound (B) ClCH₂COOH, with chlorine (less electronegative than fluorine), is less acidic than compound (C).
• Compound (A) BrCH₂COOH, containing bromine (less electronegative than chlorine), is less acidic than compound (B).
• Compound (D) HCOOH (formic acid), lacking halogen substituents, is the least acidic among the group.

Consequently, the order of increasing acidity is: (D) $<$ (A) $<$ (B) $<$ (C)