# Why is para-aminobenzoic acid more acidic than ortho-aminobenzoic acid?

Compare the acidic strength of o-, m-, p-aminobenzoic acids.

I got that meta will be the most acidic as it won't be able to show R+ effect of $$\ce{NH2}.$$

But among the other two, i.e. ortho and para, shouldn't ortho be more acidic as it will cause ortho effect? But according to my book and the data I found online:

$$\begin{array}{lc} \hline \text{Compound} & \mathrm{p}K_\mathrm{a}\text{(amino)} \\ \hline \textit{o}\text{-Aminobenzoic acid} & 4.89 \\ \textit{m}\text{-Aminobenzoic acid} & 4.79 \\ \textit{p}\text{-Aminobenzoic acid} & 4.77 \\ \hline \end{array}$$

• Do you think there may be hydrogen bonding between the amino group and the carboxylic acid? Feb 23 at 6:57
• Feb 24 at 0:03
• yes, I did thought about that but our chemistry teacher told us that ortho effect would dominate all other, so I got confused. Mar 8 at 3:10

I would like to add that since the hydrogen of the $$\ce{-COOH}$$ group is more acidic and $$\ce{N}$$ has a lone pair. This is why H-bond occurs with $$\ce{-NH2}$$ group as electron donor and not the other way round.
Also, I presume that by ortho effect you intended to tell that due to repulsion of $$\ce{-NH2}$$ group $$\ce{-COOH}$$ group would go out of the plane. However, as we have seen, it is difficult as it is already H-bonded with the nitrogen atom and moving out of the plane would put strain into the H-bond. Moreover, the repulsion by $$\ce{-NH2}$$ group is not too intense so it would (in most part) avoid the strain.