Why is benzoic acid a stronger acid than 3-aminobenzoic acid? Obviously at meta position, only negative inductive effect will operate. But, how does that make it a weaker acid than benzoic acid?


The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.2$. The $\mathrm{p}K_\mathrm{a}$ of aniline (aniline's conjugate acid really) is $4.8$.

The two macroscopic $\mathrm{p}K_\mathrm{a}$s of 3-aminobenzoic acid (aniline-3-carboxylic acid) are $3.1$ and $4.8$.

When the microscopic $\mathrm{p}K_\mathrm{a}$s of two groups in a molecule are similar, the macroscopic $\mathrm{p}K_\mathrm{a}$s are not reflective of one particular group protonating/deprotonating. The neutral form will include both a zwitterionic and an non-zwitterionic form in the case of 3-aminobenzoic acid.

When the groups are close in space within the molecule, there will be electrostatic interaction. If the amino group is protonated, it facilitates deprotonation of the carboxylic acid group. If the carboxylic acid group is deprotonated, it facilitates protonation of the amino group.

Addition: There is a very relavent article: Carbon-13 NMR determination of acid-base tautomerization equilibriums J. Org. Chem., 1984, 49 (4), pp 691–696. It reports that the neutral species is 59% zwitterion and 41% molecular (non-zwitterion).

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