# In aminobenzoic acid, do we have three different pKa's?

In para-aminobenzoic acid, how can I know whether the amino group loses a proton or not when $\mathrm{pH}$ exceeds $8.5$? I have only found two values for $\mathrm{p}K_\mathrm{a}$ but I never found info about the pKa (in this case it would be the third $\mathrm{p}K_\mathrm{a}$ at $\mathrm{pH}>8.5$) of the amino group.

In any case I guess at basic $\mathrm{pH}$, negative charge over $\ce{NH2}$ would be stabilized by aromatic ring and $\ce{COO-}$.

The two acidity constants in 4-aminobenzoic acid are due to the loss of $$\ce{H^+}$$ from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of $$\ce{H^+}$$ turning it into the 4-aminobenzoate anion.

The first two ionizations have a $$\mathrm{p}K_\mathrm{a}$$ of around 5, while the next ionization quite likely has a $$\mathrm{p}K_\mathrm{a}$$ well below ~25 (compare with other substituted anilines), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.

• After a year and a half, I've realized there's a small but important mistake in my answer. The third ionization should have a $pKa$ above 25, not below it. Larger values of $pKa$ indicate a lower tendency for deprotonation. – Nicolau Saker Neto Dec 5 '14 at 19:02
• The two marcoscopic pKa for 4-aminobenzoic acid are 2.4 and 4.9. When the microscopic pKas are close to each other, it is not a particular group that deprotonates. The neutral form of 4-aminobenzoic acid will included the zwitterion and non-zwitterion forms. – DavePhD Dec 10 '14 at 18:02
• @DavePhD Once again you are quite correct. I shall leave a recent related answer of yours here for others to see. – Nicolau Saker Neto Dec 10 '14 at 21:48

The third $$\mathrm{p}K_\mathrm{a}$$ would refer to the deprotonation of the amino group of the $$\ce{NH2-Ph-COO-}$$ (singly deprotonated) species.

This would be very difficult to deprotonate since the molecule is already negatively charged. Deprotonation would make a doubly charged species which is normally unfavorable.

Contrary to your assertion, the $$\ce{COO-}$$ provides no stabilization for the negative charge on amine, and is likely to modestly inhibit delocalization into the benzene ring as well.

Very strong bases $$(\ce{BuLi}$$ or similar) might be able to deprotonate this compound.

• The carboxylate would still stabilize a negative charge on the nitrogen relative to, for instance, a hydrogen or methyl group. Otherwise, carboxylic acid dianion chemistry wouldn't work. (Three years later, I know, but if I found it while searching other may, too.) – SendersReagent Apr 7 '16 at 4:39