The $\ce{pK_a}$ values of benzoic acid and phenylacetic acid are around 4.2 and 4.31 respectively.

In benzoic acid, you have the resonance being the dominating effect, destabilizing a conjugate base anion. In phenylacetic acid, resonance doesn't work anymore, and only the -I inductive effect remains, which should stabilise the anion.

Why, then, is benzoic acid the stronger acid?

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    $\begingroup$ You considered the resonance, but neglected that phenyl rings are electron-withdrawing via the inductive effect. (An often underappreciated fact!) $\endgroup$ Commented May 2, 2021 at 1:04
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    $\begingroup$ Usually resonance effects are more important, but the keyword is usually. It's not always true: think for example about aromatic substitution of halobenzenes, there are competing effects which give them a "special" place as o/p-directing but deactivating substituents. I think with benzoic/formic acid it gets very messy because there are significant effects arising from solvation and it might not be easiest to compare these solely based on electronic effects. $\endgroup$ Commented May 2, 2021 at 1:31
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    $\begingroup$ If you want a fully theoretical, fully rigorous answer, then you need to solve the Schrodinger equation for these molecules and reactions. That’s extremely complicated, and doesn’t give you any real insight into the problem anyway. If you want to have simplified and understandable models, then you need to sacrifice something: and what you lose is the applicability or generality to every situation. The difficulty lies in striking a balance between these two. $\endgroup$ Commented May 2, 2021 at 2:45
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    $\begingroup$ In some sense, resonance and inductive effects are huge successes in this regard. They are easy to understand, and can be applied to almost every organic molecule. But you cannot make overgeneralised comments like resonance is always more important than inductive effects (how do you even quantify it? There isn’t a SI unit for resonance effects.), or if you do, you must bear in mind that there will be exceptions that don’t fit. Part of the problem is that these acidities are very close to one another, so the explanation needs a bit more subtlety than just some broad, sweeping statement. $\endgroup$ Commented May 2, 2021 at 2:49
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    $\begingroup$ TL;DR: You’re looking for an exact, accurate answer to a quantitative phenomenon (pKa’s), but we only have inexact, qualitative tools to use in the explanation. On top of that, we have multiple inexact tools which point in opposite directions, which makes it even less clear which one is more important. The way I see it, we have two choices: either use more exact tools (QM and computational chemistry) at the cost of insight, or fudge it with the inexact tools and accept that this is a bit of a limitation. $\endgroup$ Commented May 2, 2021 at 2:55

1 Answer 1


The acidity of benzoic acid $(\ce{C6H5-COOH})$ and its substituted derivatives, $\ce{R-C6H4-COOH}$ $(e.g., \ \ce{4-NO2-C6H4-COOH})$ have been studied extensively using the Hammett Equation. Similarly, acetic acid $(\ce{CH3-COOH})$ and its substituted derivatives, $\ce{R-CH2-COOH}$ $(e.g., \ \ce{Cl-CH2-COOH})$ have been studied extensively as well using similar Hammett Equation. You may read some brief discussion of Hammett Equation here.

Actually, you are comparing benzoic acid and a $\ce{R-CH2-COOH}$ where $\ce{R = Ph}$. Their is no relationship between these two acids except for both having $\ce{Ph}$ functionality. Benzoic acid has $\ce{Ph}$ directly connected to $\ce{COOH}$ group while acetic acid derivative has it connected through $\ce{-CH2\!-}$ group. Because of that, acidity of $\ce{Ph-CH2-COOH}$ is depend solely on inductive effect. It isworth noting that I cannot find any comparative studies of these two type of carboxylic aid in one single research.

The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20$ $(K_\mathrm{a} = 6.3 \times 10^{-5})$ while the $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.75$ $(K_\mathrm{a} = 1.77 \times 10^{-5})$. Thus, benzoic acid is about 3.6 times acidic than acetic acid. That is because when you consider $\ce{R-COOH}$ acid, it is more acidic if $\ce{R}$ group is electron withdrawing group (EWG) such as $\ce{Ph}$ group, compared to $\ce{R}$ group being electron donating group such as $\ce{CH3}$ group (EDG) like in the case of acetic acid.

Now, let's increase the electron withdrawing ability of $\ce{R-CH2-COOH}$ acid compared to acetic acid (where $\ce{R = H}$): The $\mathrm{p}K_\mathrm{a}$ value of 2-chloroacetic acid (where $\ce{R = Cl}$) is $2.87$ $(K_\mathrm{a} = 1.36 \times 10^{-3})$, which is about 76.8 times acidic than acetic acid. Thus, we can conclude that $\ce{ClCH2}$ group has larger inductive effect than that of $\ce{Ph}$ group. Thus, if we change $\ce{Cl}$ in $\ce{ClCH2}$ group to $\ce{Ph}$ group, then resulting carboxylic acid should be less acidic than that of 2-chloroacetic acid.

As given in OP's question, this carboxylic acid, 2-phenylacetic acid $(\ce{Ph-CH2-COOH})$ has a $\mathrm{p}K_\mathrm{a}$ value of $4.31$. It is more acidic than acetic acid, but not quite acidic as benzoic acid because of the extra $\ce{CH2}$ group. Remember, inductive effect is weaken by additional $\ce{C-C}$ bonds (c.f., ortho-, meta- and para-inductive effects).

Source of $\mathrm{p}K_\mathrm{a}$ values


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