Question:
Find the strongest carboxylic acid from the following:
Find the strongest carboxylic acid from the following:
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Stronger –I effect → stronger acid due to stabilization of conjugate base.
Updated On: Jun 19, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding acidity trend.
Carboxylic acid strength increases with electron-withdrawing groups because they stabilize the carboxylate ion through the inductive effect.
Step 2: Effect of halogens.
Among halogens, fluorine is the most electronegative, so it shows the strongest \(-I\) effect. Chlorine and bromine show weaker electron-withdrawing effects comparatively.
Step 3: Comparing given acids.
- Benzoic acid: no strong \(-I\) group → weakest - Cl\(_2\)CCOOH: moderate \(-I\) effect - Br\(_3\)CCOOH: weaker than chlorine - F\(_3\)CCOOH: strongest \(-I\) effect
Step 4: Stability of conjugate base.
CF\(_3\) group strongly stabilizes the carboxylate ion, making trifluoroacetic acid the strongest acid among the options.
Step 5: Final conclusion.
Thus, F\(_3\)CCOOH has the highest acidity.
Final Answer: \[ \boxed{F_3CCOOH} \]
Carboxylic acid strength increases with electron-withdrawing groups because they stabilize the carboxylate ion through the inductive effect.
Step 2: Effect of halogens.
Among halogens, fluorine is the most electronegative, so it shows the strongest \(-I\) effect. Chlorine and bromine show weaker electron-withdrawing effects comparatively.
Step 3: Comparing given acids.
- Benzoic acid: no strong \(-I\) group → weakest - Cl\(_2\)CCOOH: moderate \(-I\) effect - Br\(_3\)CCOOH: weaker than chlorine - F\(_3\)CCOOH: strongest \(-I\) effect
Step 4: Stability of conjugate base.
CF\(_3\) group strongly stabilizes the carboxylate ion, making trifluoroacetic acid the strongest acid among the options.
Step 5: Final conclusion.
Thus, F\(_3\)CCOOH has the highest acidity.
Final Answer: \[ \boxed{F_3CCOOH} \]
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