# Organic problem: acidity of benzene substituents

Ortho nitro benzoic acid is more acidic than para benzoic acid since intermolecular H bonding comes into play in the latter compound. But why is that among para nitro phenol and ortho nitro phenol, para compound is more acidic than that of ortho? There is intermolecular H bonding in para nitro phenol too, right?

• – user7951 Dec 20 '16 at 19:34

This isn't the answer you were looking for probably, but this is the answer you should take to heart as a chemist.

I want us to look at the difference in free energy changes between the $\mathrm{p}K_\mathrm{a}$'s in question:

\begin{array}{|c|c|} \hline \text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline \text{2-nitrophenol} & 7.23 \\ \text{4-nitrophenol} & 7.15 \\ \hline \text{2-nitrobenzoic acid} & 2.16 \\ \text{4-nitrobenzoic acid} & 3.44 \\ \hline \end{array}

$$K_\mathrm{eq} = \frac{K_\mathrm{a}}{\ce{[H2O]}}$$

$$\Delta \Delta G = -RT\ln K_{\mathrm a1} + RT \ln\ce{[H2O]} - (-RT\ln K_{\mathrm a2} + RT \ln\ce{[H2O]}) = -RT\Delta\ln K_{\mathrm a}$$ $$=-2.303 RT\Delta \log K_{\mathrm a} = 2.303 RT \Delta \mathrm{p}K_\mathrm{a}$$

For $298\,\mathrm{K}$, we can compute the following $\Delta \Delta G$ values: \begin{array}{|c|c|} \hline \text{Series} & |\Delta \Delta G| \\ \hline x\text{-nitrophenol} & 456\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline x\text{-nitrobenzoic acid} & 7303\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline \end{array}

For reference, the standard entropy of solvation for acetate is $-192\ \mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}$, which at $298\ \mathrm{K}$ is equivalent to $-57\,216\ \mathrm{J}\,\mathrm{mol}^{-1}$.

Now, personally, I don't have a great handle on how the solvation of the conjugate bases are going to be relative to the benzoic acids or phenols. But you can see that solvation effects can play a huge role (and these are aqueous dissociation constants).

My personal opinion is that it's not particularly useful to understand where such a small difference comes from, particularly for the phenol series. We're talking about one-tenth of an order of magnitude. Perhaps, as you say, the benzoic acids are different because of intermolecular hydrogen bonding. I'm a bit suspicious since I don't see this acting in a differential way between the two species.

Nevertheless, the main point still stands. Acid-base equilibria in aqueous medium is a non-trivial process to understand because water is a complicated solvent. You can try to rationalize this, but you may well be wrong without very cleverly designed experiments.

And the fact of the matter is that rationalizing the difference in acidity is not that useful. Maybe if you were using linear free energy relationships, but in that case, you know there is a difference, and that's more important.

Again, as I said at the beginning, this is probably not the answer you wanted, but I think you should be trying to understand the limits of what is useful/practical to know.

When I checked the pKas for o- and p-substituted nitrophenols, p-substituted nitrophenol is slightly more acidic (7.1) than o-nitrophenol (7.2).

In general, to judge the relative acidity of substituted phenols, you should first look out for how resonance can stabilize the conjugate base (e.g. the deprotenated molecule). Since the difference in acidity is so slight, you might expect that there shouldn't be big difference in the resonance stabilization, as that would usually result in a larger pKa difference. This website shows the resonance for o-,p-,and m-substituted nitrophenols. Indeed, In the o-nitrophenol, the electron lone-pair on the deprotenated phenol can only delocalize onto the nitrogen, resulting in N having a +1 charge and both O having -1 charges (as shown in the link above). This is the same for the p-nitrophenol.

However, the reason for the slight difference between the p- and o-substituted nitrophenols is due to the OH hydrogen being able to also hydrogen bond with the the ortho-nitro group. This additional hydrogen bonding interactions makes it harder for the molecule to completely 'let go' of the hydrogen. This shifts the acid-base equilibria towards the unprotenated form of the acid, lowering the acidity/raising the pKa (when compared to the p-substituted compound)

• So resonance of phenolate ion in the para form makes it more acidic where as in para nitro benzoic acid , delocalization of electron is not possible? – Harini Dec 20 '16 at 19:23
• @user129402 In the o-nitrophenol, the electron lone-pair on the deprotenated phenol can only delocalize onto the nitrogen, resulting in N having a +1 charge and both O having -1 charges (as shown in the link above). This is a higher formal charge than the delocalization shown for the p-nitrophenol. Indeed, when I checked the pKas for o- and p-substituted nitrophenols, p-substituted nitrophenol is more acidic. Why do you think it's the opposite? (I've edited my answer to reflect this--sorry, I couldn't comment previously) – NMJD Dec 20 '16 at 19:33
• @user129402 Oh sorry, I made a mistake about the relative formal charges: they are actually the same. The slight difference in the acidity does in fact come from hydrogen-bonding interactions in the o-substituted molecule. I've clarified this in my edited answer. – NMJD Dec 20 '16 at 19:39