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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 $\ce{COO-}$.

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

The first two ionizations have a pKa of around 5, while the next ionization quite likely has a pKa 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.

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  • $\begingroup$ 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. $\endgroup$ – Nicolau Saker Neto Dec 5 '14 at 19:02
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    $\begingroup$ 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. $\endgroup$ – DavePhD Dec 10 '14 at 18:02
  • $\begingroup$ @DavePhD Once again you are quite correct. I shall leave a recent related answer of yours here for others to see. $\endgroup$ – Nicolau Saker Neto Dec 10 '14 at 21:48
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The third pka would refer to the deprotonation of the amino group of the 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 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 (BuLi or similar) might be able to deprotonate this compound.

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  • $\begingroup$ 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.) $\endgroup$ – SendersReagent Apr 7 '16 at 4:39

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