Comparing acidities of substituted and aromatic carboxylic acids

Question

When comparing the acidities of carboxylic acids, we primarily see the electropositivity of the carboxylic acid carbons, i.e. we see how effectively the negative charge on the carboxylate ion is dispersed upon ionization. Electron withdrawing groups increase the acidity by stabilising the anion via electron withdrawal.

Now consider the following two comparisons:-

1. Formic acid ($\ce{HCOOH}$) and benzoic acid($\ce{C6H5COOH}$)
2. Benzoic acid and 3-chloropropanoic acid ($\ce{ClCH2CH2COOH}$)

$$\begin{array}{c|c} \hline \text{Species} & \mathrm{p}K_\mathrm{a} \\ \hline \ce{HCOOH} & 3.75\mathrm{-}3.77 \\ \ce{PhCOOH} & 4.19 \\ \ce{ClCH2CH2COOH} & 3.98 \\ \hline \end{array}$$

(source)

My questions:

1. Why is 3-chloropropanoic acid less acidic than formic acid. The negative inductive effect of chlorine helps in dispersing the negative charge on the carboxylate ion. It is not conjugated and too far to exert positive mesomeric effect. Therefore, the negative inductive effect should increase the acidity by improving the stability of the conjugate base.

2. Why is benzoic acid weaker than formic acid? Unless I am mistaken, in several other places, the phenylic moiety is treated as slightly electron withdrawing, as in case of alcohol acidity, acidity of hydrogens in base catalysed reactions etc. I am unable to understand the resonance here or any other factor, that might cause this anomaly.

Answer 1

In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation:

1. Due to steric hindrance, it may be the case that the negative charge is less accessible to solvent molecules, and thus more weakly stabilized. I suspect the difference in enthalpy here is actually negligible, but I'm not able to find a reference at this time.
2. Perhaps more importantly, there is an entropic effect, in that larger molecules tend to require a larger number of solvent molecules to solvate them. These solvent molecules form highly ordered, cage-like structures, which is entropically unfavorable. I believe similar underlying mechanisms are at work in, e.g., the hydrophobic effect, micellization, protein folding, etc.

As Klaus Warzecha points out, if the chlorine were not as far-removed from the carboxylate moiety (since induction rapidly diminishes with growing distance), the strength of the inductive effect would trump the solvent effects.

In the case of benzoic acid, there are probably similar solvent effects occurring. Additionally, the benzene ring is actually electron-donating within the $\pi$-electron system of the molecule, which can only be destabilizing to the carboxylate conjugate base. Unlike in, e.g., phenol or aniline, the excess negative charge cannot actually be delocalized into the ring. Experimentally, the electron-donating effect of the ring (equivalently, the electron-withdrawing effect of the carboxylate) is confirmed by the fact that benzoic acid is severely deactivated in electrophilic aromatic substitution reactions by comparison to ordinary benzene. The benzene ring's electron-withdrawing effect within the $\sigma$-framework (rationalized, according to valence bond theory, by the greater s-character of $\mathrm{sp^2}$ carbons) of the molecule is probably quite weak. There may also be Coulombic destabilization between the strong $\delta^+$ of the carboxylate carbon atom and the $\mathrm{sp^2}$ carbons of the aromatic ring, again due to greater s-character, although that's speculative.

Answer 2

Actually, the chlorine atom in 3-chloropropanonic acid does have a significant -I effect; the $\mathrm{p}K_\mathrm{a}$ of propanoic acid is $4.86$. Your expectations were just a bit too high. If you want a stronger effect, you need to get a bit closer: the $\mathrm{p}K_\mathrm{a}$ of 2-chloropropanoic acid is $2.83$.