# Why is 4-hydroxypyridine more acidic than benzoic acid?

4-hydroxypyridine (or 4-pyridone) has a $$\mathrm{p}K_\mathrm{a}$$ in the ballpark of $$3$$ to $$3.33\:^1$$ (in the pdf linked below, the $$\mathrm{p}K_\mathrm{a}$$ of 1-methyl-4-pyridone has been given to be $$3.33$$, however I suppose removing a methyl would only cause a slight decrease in the $$\mathrm{p}K_\mathrm{a}$$ value). This seems an unexpectedly low value to me. 4-pyridone seems more acidic than I expected it to be, since the $$\mathrm{p}K_\mathrm{a}$$ of other common acids are:$$\:^2$$

$$\begin{array}{lr} \hline \mathrm{Acid} & \mathrm{p}K_\mathrm{a} \\ \hline \text{Benzoic acid} & 4.20 \\ \text{Formic acid} & 3.75 \\ \text{Acetic acid} & 4.75\\ \text{Phenol} & 10.0 \\ \text{Carbonic acid} & 6.37\rlap{\:^1} \\ \hline \end{array}$$

Now I had expected 4-hydroxypyridine to be more acidic than phenol, because of the negative charge in the former's conjugate base being delocalised onto the electronegative nitrogen, yielding a more stable resonating structure than the latter's corresponding resonating structure. I thought that 4-pyridone's $$\mathrm{p}K_\mathrm{a}$$ might be somewhere around $$\ce{H2CO3}$$'s $$\mathrm{p}K_\mathrm{a}$$, but I was wrong. Also, I wasn't quite sure that it would be acidic enough to have a $$\mathrm{p}K_\mathrm{a}$$ value any close to other carboxylic acids. But it seems to me that it is more acidic than benzoic acid, formic acid and acetic acid.

Why is this so? I can't see any excetional characteristic in its structure which might impart it such a low $$\mathrm{p}K_\mathrm{a}$$. This does seem more of an exception to me, as of now, but it would be very helpful if someone clarified and explained such a trend.

### References:

$$[1]:$$ http://gissmo.nmrfam.wisc.edu/static/pKa_compilation_R_Williams.pdf

$$[2]:$$ https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/acidity2.htm

• chemicalbook.com/ProductMSDSDetailCB6303828_EN.htm 4-hydroxypyridine has a pKa of 2.8 according to this – Safdar Jul 22 at 13:13
• @Safdar However, this page from the same website shows a $\rm{pK_a}$ value of $3.2$, so I wan't really sure about the authenticity of the $\rm{pK_a}$ value mentioned there. However, my question still holds, in both the cases, the $\rm{pK_a}$ value is definitely lower than expected. – user91457 Jul 22 at 13:24
• Why is it lower than expected? The conjugate base is quite stabilized. – Zhe Jul 22 at 14:11
• @Zhe It is stabilised, but I don't see why it is so stabilised that 4-hydroxypyridine is so much more acidic than phenol. – user91457 Jul 22 at 14:51

The following are two resonance forms of 4-pyridone,

So, when 4-pyridone gets de-protonated, the negative charge on nitrogen gets delocalised on more electronegative atom i.e., oxygen, resulting in more stable resonance structure!

Note that, this explanation is identical to that one which is given when we compare $$\mathrm pK_{\mathrm a}$$ of benzoic acid and phenol.

• I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion. – user91457 Jul 22 at 16:15
• Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone? – user91457 Jul 22 at 16:15
• @FakeMod, extended conjugation and aromaticity in 4-pyridone is what making it more stable than formate anion – Rahul Verma Jul 22 at 16:37
• But aromaticity isn't new to the 4-pyridone anion. It was there even before deprotonation, so I don't think it can be a driving factor here, however, extended conjugation does seem reasonable. Thanks! – user91457 Jul 22 at 17:02
• Well, you too have wrong tautomer in there... – Mithoron Jul 22 at 21:50

First of all let us examine the structural formula of 4-pyridone:

Now, as you can see, when it acts as an acid a negative charge forms on the nitrogen atom. This negative charge is well stabilized as it undergoes extensive resonance which includes oxygen as well. The fact that the negative charge is formed on nitrogen also contributes to the acidity.

• "when the hydrogen is abstracted and negative charge comes , it acquires aromaticity": I don't think so. Wasn't it aromatic all along, even before deprotonation? Reference: chemistry.stackexchange.com/a/126146/91457 – user91457 Jul 22 at 14:58
• Wrong tautomer you have there... – Mithoron Jul 22 at 15:08
• @Safdar: I changed it back to tautomers. Also, followed andselisk♦'s instruction to make similar fond in both text and image. – Mathew Mahindaratne Jul 22 at 15:43
• I thought about the answer, and I am still wondering what makes the two equivalent resonating structures of the formate ions less stable then the conjugate base of 4-pyridone. In fact, oxygen is more electronegative than nitrogen, so having a negative charge delocalised on two oxygen seems like a much stable configuration than delocalisation on one oxygen and one nitrogen. Moreover, the resonating structures of formate are equivalent resonating structures, imparting an "extra" stability to the anion. – user91457 Jul 22 at 16:16
• Can the delocalisation over mere 3 extra carbons justify the increased acidic strength of 4-pyridone? – user91457 Jul 22 at 16:17