Q 3.97: Which of the following compounds is the strongest Bronsted acid? A CH
FCH
CO
H B CH
BrCH
CO
H
The Correct Answer and Explanation is:
The question compares two compounds:
A. CH₂FCH₂COOH
B. CH₂BrCH₂COOH
These are both carboxylic acids (–COOH group) with a halogen atom (fluorine or bromine) attached two carbon atoms away from the carboxylic acid group. We are to determine which is the stronger Brønsted acid.
Correct Answer: A. CH₂FCH₂COOH
Explanation:
A Brønsted acid is a substance that donates a proton (H⁺). The strength of a Brønsted acid depends on how easily it can donate this proton. This is largely determined by the stability of its conjugate base—the more stable the conjugate base, the stronger the acid.
The conjugate base of a carboxylic acid is a carboxylate ion (RCOO⁻). When a proton is removed from the –COOH group, the remaining negative charge can be stabilized by electron-withdrawing groups through a process called inductive effect.
Inductive Effect:
- Electron-withdrawing groups, like halogens, pull electron density through the sigma bonds, stabilizing the negative charge on the conjugate base.
- The stronger the electron-withdrawing group, the greater the acid strength.
- Fluorine is more electronegative than bromine, meaning it pulls electron density more effectively.
- Even though both halogens are two carbons away, fluorine’s effect is stronger than bromine’s.
Therefore, CH₂FCH₂COOH has a more stabilized conjugate base due to the stronger inductive effect of fluorine, making it a stronger Brønsted acid.
Summary:
- Fluorine is more electronegative than bromine.
- Fluorine stabilizes the conjugate base better.
- CH₂FCH₂COOH donates H⁺ more easily.
Thus, CH₂FCH₂COOH is the stronger Brønsted acid
