I just had a question in a test asking to compare the relative acidity of picric acid (TNP) and carbonic acid, $\ce{H2CO3}$ (not phenol). It turns out that TNP is more acidic. I have no idea why? Is there some rule I'm missing or is this one of those exceptions that I just have to remember?

I have been told that alcohols are generally less acidic than carboxyl groups. But in this case, doesn't the hydroxyl group form a chelate ring already? If the stability of the reactant is much greater than the the stability of the product, why would the reaction even take place? Sure, there is cross conjugation in carbonic acid, but shouldn't it be extremely hard to break the chelation in picric acid?

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    $\begingroup$ strongly related chemistry.stackexchange.com/questions/9074/… $\endgroup$
    – Mithoron
    May 4, 2015 at 15:16
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    $\begingroup$ To me, the asker just wants to know how and why a hydroxyl group can be more acidic than a carboxyl group. $\endgroup$ May 5, 2015 at 3:27
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    $\begingroup$ Perhaps you misread what I wrote; carbonic acid ($\ce{H2CO3}$) contains a carboxyl! The acidic group in carbonic acid is the carboxyl group, R-COOH (for carbonic acid, R = OH). You've probably been told that hydroxyl groups (C-OH, present in alcohols and phenols) are weaker acids than carboxyl groups/carboxylic acids, yet picric acid is a substituted phenol and is quite a bit more acidic than carbonic acid and simple carboxylic acids. This is the source of your confusion? $\endgroup$ May 5, 2015 at 19:39
  • $\begingroup$ Short answer could be: picryl is stronger stabilizer of negative charge on oxygen then carboxyl. $\endgroup$
    – Mithoron
    May 6, 2015 at 0:36
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    $\begingroup$ Oh, you mean the intramolecular hydrogen bond between the hydroxyl and the nitro group? Indeed, that slightly lowers the energy of the reactant, which would make ionization a bit more difficult. However, the three nitro groups strongly lower the energy of the ionization product by an electron-withdrawing mesomeric effect, making deprotonation far more favourable. Have you studied this sort of stuff? Take a look at this section from the UC Davis ChemWiki, especially the part just above example 7.10. Does that help? $\endgroup$ May 6, 2015 at 19:22

1 Answer 1


So you want to compare the acidity of different Brønsted–Lowry acids. This means ultimately you want to compare the equilibrium constants for reactions of the type $\ce{HX <=> H+ + X-}$. To do so, you'll have to analyse what stabilisation/destabilisation effects are present in each acid $\ce{HX}$ and each conjugate base $\ce{X-}$. This is easier said than done, especially quantitatively. There are many sorts of possible effects, often acting in a compounded fashion. Fortunately, there are some basic principles to help guide us.

Your question specifically asks to contrast carbonic acid, $\ce{H2CO3}$ and picric acid, more descriptively called 2,4,6-trinitrophenol. This already starts off as being a bit tricky, because they are not very similar substances; of course, the more similar the two substances are when being compared, the easier it will be to pick out the differences in stabilisation/destabilisation effects and the safer it will be to extrapolate from qualitative reasoning. With this in mind, it is best if we build up to the answer by making comparisons between several molecules, making small changes at each step.

First off, let us start with alcohols. These are substances containing hydroxyl groups ($-\ce{OH}$) linked to carbon atoms which are saturated (carbon atoms with four single bonds, i.e. $\rm{sp^3}$-hybridised carbon) and whose three other bonds are only to carbon or hydrogen atoms (in other words, linked to no heteroatoms other than the hydroxyl oxygen). This class represents neither of your substances, but is simple and is related to phenols, so it will be a useful benchmark. A simple alcohol is methanol, $\ce{CH3OH}$, which can act as an acid by the equilibrium:

$$\ce{H3C-OH_{(aq)} <=> H+_{(aq)} + H3C-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=3\times10^{-16}$$

The small equilibrium constant establishes that methanol is a very weak acid in water, and exists almost completely as the unionised reactant. Values of $\mathrm pK_\mathrm a$ around 15 are typical for simple alcohols.

Now let us take a step towards the structures we want, by examining at a hydroxyl linked to an unsaturated carbon atom, specifically a $\rm{sp^2}$-hybridised carbon atom linked to no heteroatoms other than the hydroxyl oxygen. What we have here is an enol. Simple enols are not very stable molecules, but for simplicity we shall disregard that. For the simple molecule propen-2-ol, the equilibrium is:

$$\ce{H2C=C(OH)-CH3_{(aq)} <=> H+_{(aq)} + H2C=C(O^{-})-CH3_{(aq)}}\ \ \ \ \ K_\mathrm a=1.3\times 10^{-11}$$

Evidently, enols are quite a bit more acidic than alcohols. There are two new effects at play here. First, $\rm{sp^2}$-hybridised carbon atoms show an electron-withdrawing effect compared to $\rm{sp^3}$-hybridised carbon. This ability to "suck away" electron density helps stabilise molecules when close to electron-rich regions such as the oxygen atom. This effect stabilises the reactant, but it stabilises the enolate product more (as it has extra electron density to be pulled), which drives the equilibrium to the right. A second effect is resonance. Some of the electrons from the oxygen atom are said to conjugate with the carbon-carbon double bond, creating resonance or canonical structures, which contribute to the true description of bonding in the molecule. Once again, spreading the electrons out contributes to decrease the energy of a molecule. The resonance structures for the reactant and the product are similar, but one of the resonance structures in the reactant contains a formal positive charge on oxygen (a very electronegative atom) and a formal negative charge on carbon (a not very electronegative atom), so this form of conjugation is not quite as favourable. In the product, though there is still a formal negative charge on carbon in one of the canonical structures, there is no formal positive charge on oxygen, making conjugation more favourable. Thus the resonance argument is also a part of the explanation for the acidity.

Next up is phenol, the parent compound of 2,4,6-trinitrophenol. The effects present here are much the same as in the enol above, except that conjugation happens in a larger extent in both the reactant and the product. Experimentally:

$$\ce{C6H5-OH_{(aq)} <=> H+_{(aq)} + C6H5-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=1.1\times10^{-10}$$

This represents only a small increase in acidity compared to propen-2-ol, which suggests conjugation is not the main factor in phenol acidity, but rather the inductive effect of the $\rm{sp^2}$ carbon.

Finally, we've built up to our first structure, 2,4,6-trinitrophenol. The new bit here is the addition of nitro groups ($\ce{-NO2}$), which are strongly electron-withdrawing both by induction and mesomerism (conjugation). Again, these effects decrease the energy of both the reactant and the product, but stabilize the product more because there is extra density to be spread around. As a quick interesting aside, the acidities of 2-nitrophenol and 4-nitrophenol are similar, and both are somewhat larger than 3-nitrophenol, which suggests most of the increase in acidity comes from electron-withdrawing by mesomerism rather than induction, represented by canonical structures similar to these. You also make a good point in mentioning the intramolecular hydrogen bond present only in 2,4,6-trinitrophenol and not its ionised form. Indeed, this helps stabilize the reactant a slight amount, and decreases its acidity. Let's see how all these effects add up:

$$\ce{(O2N)3C6H2-OH_{(aq)} <=> H+_{(aq)} + (O2N)3C6H2-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=0.4$$

The acidity increase from the electron-withdrawing effects of the three nitro groups far overwhelms the acidity decrease due to the formation of the intramolecular hydrogen bond. How could you have known this beforehand? Strictly speaking, you couldn't have deduced this from thin air (not without a computer, at any rate!), but generally you are expected to know the situation in similar cases and work out the answer based on previous knowledge. Intramolecular hydrogen bonding usually doesn't play a huge role in increasing or decreasing acidity, at least in simple molecules, as they are relatively weak interactions and so don't contribute massively to increasing/decreasing the energy of a structure. Furthermore, nitro groups are among the most electron-withdrawing there are, so if pulling electrons is going to have any role, it'll show some of its strongest effects with nitro groups.

To prove this answer comes to an end, we reach our final structure, carbonic acid. First, we notice it is related to carboxylic acids, and its acidic group is the $\ce{-COOH}$ in the structure. If you think about it, the carboxyl group looks like an enol where a $\ce{=CH2}$ group is replaced by an oxygen atom. Thus the analysis is similar to the enol. However, there is an increase electron-withdrawing effect due to induction (oxygen being more electronegative than carbon) and mesomerism (in the canonical structures, the negative formal charges are now on an oxygen atom, rather than a carbon atom as was the case for the enol). The instability of carbonic acid interferes, but leaving that aside, its dissociation would be something like:

$$\ce{H2CO3_{(aq)} <=> H+_{(aq)} + HCO3^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=2.5\times 10^{-4}$$

Comparing everything, the acidity increases in the order:

Simple alcohols < Simple enols and phenols < Carbonic acid and simple carboxylic acids < Picric acid

How would you know carbonic acid is a weaker acid than picric acid? Once again, it's not something you can deduce by pure logic; it requires some benchmarking and comparison with previous data and knowledge. The analysis above would have been helpful, but has its limits. In Chemistry, it is often far easier to explain than predict. Whether you should have been expected to know the answer to this question, I'm not sure. It would depend on how your chemistry course is conducted.

  • $\begingroup$ So, in general can I make an assumption that a strong withdrawing group, like SO3H or NO2, present at ortho or para positions of phenol is more acidic than carbonic acid? I ask this because I'm studying for a competitive exam and questions like this are sometimes asked. $\endgroup$ May 9, 2015 at 7:10
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    $\begingroup$ If there is a $\ce{-SO3H}$ group, that itself will make the molecule a strong acid. A single $\ce{-NO2}$ group is not enough to make phenol more acidic than carbonic acid, and 2,4-dinitrophenol is also slightly weaker than carbonic acid. It takes three nitro groups to beat carbonic acid. Unfortunately when comparing sufficiently different molecules, there's no easy way to find the answer. Perhaps taking some time to acquaint yourself with the pKa table I linked here will help. $\endgroup$ May 9, 2015 at 11:06

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