Consider the two compounds $\ce{FCH2COOH}$ and $\ce{ClCH2COOH}$. Since fluorine is more electronegative than $\ce{Cl}$ it will show a greater –I effect. It will pull the the electrons closer towards it and give $\ce{CH2}$ a small positive charge and so on to $\ce{COO-}$. The $\ce{COO-}$ in compound 1 will be more stable but have a greater positive less negative charge than compound 2.

Since $\ce{COO-}$ in compound 2 has lesser charge it should attract $\ce{H+}$ better than compound 1

In short, why is the strength of an acid in an organic compound directly proportional to the stability of the acid's conjugate base?

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    $\begingroup$ I think this question is conflating a few concepts. Stability is a relative concept, so in order to compare two completely different compounds, you need to establish a common baseline first. Then it's depending on exactly that reference what behaves how. So the question you need answered is which state predominates in an acid base reaction to establish some kind of acid strength measure. For most acids and bases, water seems an excellent choice as reference. $\endgroup$ Mar 8, 2022 at 14:38
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    $\begingroup$ I fixed your mixup with the charges. Still unclear? $\endgroup$
    – Karl
    Mar 8, 2022 at 18:27

1 Answer 1


Less (absolute, i.e. less negative) local charge on the oxygen atoms of the conjugate base means the energetic penalty for removing the $\ce{H+}$ is lessened.

Or in other words: The -I effect of the halogen atoms weakens the COO-H bond.


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