Chemistry Carboxylic Acids BUMHS 2023
PMDC Verified Question 39 of 94
Which of the following acids has the smallest dissociation constant?
A
\( \text{CH}_3\text{CHFCOOH} \)
B
\( \text{FCH}_2\text{CH}_2\text{COOH} \)
C
\( \text{BrCH}_2\text{CH}_2\text{COOH} \)
D
\( \text{CH}_3\text{CHBrCOOH} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( \text{BrCH}_2\text{CH}_2\text{COOH} \)
Concept:

The acid dissociation constant (\( \text{K}_a \)) is a measure of acid strength. The "smallest dissociation constant" implies the weakest acid. Acidity increases with highly electronegative substituents (Fluorine > Bromine) and their proximity to the carboxyl group (\( \alpha \)-position > \( \beta \)-position).

Solution:

  • To find the weakest acid, we want the weakest halogen placed furthest away from the \( -\text{COOH} \) group to minimize the electron-withdrawing inductive (\( -I \)) effect.


  • Option A: Fluorine on the \( \alpha \)-carbon (Very Strong).


  • Option D: Bromine on the \( \alpha \)-carbon (Strong).


  • Option B: Fluorine on the \( \beta \)-carbon (Moderate).


  • Option C: Bromine on the \( \beta \)-carbon. Bromine is less electronegative than Fluorine, and it is located far away on the \( \beta \)-carbon. This exerts the weakest \( -I \) effect, yielding the least stable anion and thus the smallest \( \text{K}_a \).


Why other options are incorrect:

All other options feature either a more electronegative halogen (F) or a closer proximity (\( \alpha \)-position), both of which result in a larger dissociation constant.

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