Question

Is the ortho isomer necessarily more acidic than the para isomer?

Answer 1

You mean for disubstituted benzene compounds? In general it is true, but sometimes there are exceptions, so be careful. o-nitrophenol has about the same pKa than p-nitrophenol (http://www.chem.ucla.edu/~cantrill/30A_S05/Acids_Bases_2_Ans.pdf).

Generally it would mean the resonance structure of the conjugate base stabilizes the para isomer more (= more weakly acidic or more strongly basic), so the original para isomer is less acidic.

Let's examine the conjugate base of o-nitroaniline and compare it to that of p-nitroaniline. `"NH"_2` is a strongly activating group, while `"NO"_2` is a strongly-deactivating group.

You should notice that the `pi` electrons are more delocalized throughout the para isomer than in the ortho isomer. The `pi` electrons were able to redistribute across 6 bonds in the para isomer, but only 4 bonds in the ortho isomer. Two of those bonds were within the benzene ring, which is important.

The greater delocalization of `pi` electrons in general stabilizes the structure more because the energy is more spread out between all the molecular orbitals, and thus is not piled up on a few orbitals. As a result, each orbital can be lower in energy overall.

Additionally, since both substituents are acting in the same resultant direction (`"NH"_2` activates and `"NO"_2` deactivates the ring), the electron donation/withdrawal effect is minimized for the para isomer.

Since this means the para isomer conjugate base is more stable (a stronger base) than the ortho isomer conjugate base, the original para isomer is a weaker acid than the original ortho isomer.

Therefore, the original ortho isomer is a bit more acidic. We can see that in the pKas:

`"pKa"_"ortho" ~~ -0.28`
(https://pubchem.ncbi.nlm.nih.gov/compound/2-nitroaniline)

`"pKa"_"para" ~~ 1.01`
(https://pubchem.ncbi.nlm.nih.gov/compound/7475)

(A lower `"pKa"` is more acidic.)