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

Updated On: Jun 1, 2025
  • (D) $<$ (A) $<$ (B) $<$ (C)
  • (A) $<$ (D) $<$ (C) $<$ (B)
  • (B) $<$ (A) $<$ (D) $<$ (C)
  • (C) $<$ (B) $<$ (D) $<$ (A)
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The Correct Option is A

Approach Solution - 1

To determine the increasing order of acidity of the compounds based on their pKa values, we need to understand how different substituents affect the acidity of carboxylic acids. Acidity is influenced by the electron-withdrawing or electron-donating nature of substituents:

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

The presence of electron-withdrawing groups, such as halogens, will stabilize the carboxylate anion formed upon deprotonation, thus increasing acidity. The more electronegative the substituent, the greater the acidity.

Let's analyze each compound's acidity based on the electronegativity of the substituent:

  • Compound (C) FCH2COOH has fluorine, the most electronegative element, increasing the acidity significantly.
  • Compound (B) ClCH2COOH has chlorine, which is less electronegative than fluorine, so its acidity is lower than compound (C).
  • Compound (A) BrCH2COOH has bromine, which is less electronegative than chlorine, resulting in even lower acidity than compound (B).
  • Compound (D) HCOOH (formic acid) has no halogen substituents, making it the least acidic of the group.

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

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Approach Solution -2

To determine the increasing order of acidity for the given compounds based on their pKa values, we analyze their structures and the effect of substituents on acidity.

The acidity of carboxylic acids is influenced by the electron-withdrawing or electron-donating nature of substituents attached to the carboxyl group. The more electron-withdrawing the substituent, the more it stabilizes the conjugate base (carboxylate ion), thereby increasing the acidity of the compound.

Let's examine the compounds:

  • (A) BrCH2COOH: Bromoacetic acid
  • (B) ClCH2COOH: Chloroacetic acid
  • (C) FCH2COOH: Fluoroacetic acid
  • (D) HCOOH: Formic acid

In terms of electron-withdrawing ability, the substituents can be ranked as follows: F > Cl > Br. Fluorine, being the most electronegative, is the strongest electron-withdrawing group among them, followed by chlorine and bromine.

Comparing the acidity:

  • FCH2COOH (fluoroacetic acid) is the most acidic due to the strong electron-withdrawing effect of fluorine.
  • ClCH2COOH (chloroacetic acid) is less acidic than fluoroacetic acid but more acidic than bromoacetic acid.
  • BrCH2COOH (bromoacetic acid) is less acidic than chloroacetic acid due to the weaker electron-withdrawing effect of bromine.
  • HCOOH (formic acid) is the least acidic as it lacks any additional electron-withdrawing substituent compared to the others.

Therefore, the increasing order of acidity, based on pKa values, is: (D) < (A) < (B) < (C)

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