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:

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


The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compare to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the distance between the two groups.

H-Bonding in ortho aminobenzoic acid.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.