# Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine?

I wish to know which effect out of steric inhibition of resonance (SIR) or steric inhibition of protonation (SIP) is dominant over the other when comparing basicities of o-toluidine and aniline:

Case 1, using only SIR effect
We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. Hence, we can say o-toluidine is more basic than aniline (as nitrogen's lone pairs are in resonance with benzene).

Case 2, using only SIP effect
If we consider that methyl's bulkiness shields the lone pair of nitrogen from protonation in o-toluidine and no such inhibition or shielding happens in aniline, we can conclude that aniline is more basic than o-toluidine.

The reason for the decreased basicity is essentially the ortho-effect as outlined here: Ortho-effect in substituted aromatic acids and bases.

We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. [...]

I don't see how resonance could be inhibited, as that essentially requires that the lone pair orbital of nitrogen would have to rotate significantly towards an in-plane arrangement. The methyl group is simply not big enough for that to happen, and if it were, the lone pair would more likely rotate toward the more shielded side, further reducing acidity.

The highest occupied molecular orbitals (HOMO) of the molecules show this quite nicely, as they are very, very similar.

On the DF-B97D3/def2-SVP level of theory the RMSD of the core structure (excluding the hydrogen and the methyl group) is negligible with only 0.03 Å.

From this I would conclude that there is no steric inhibition of resonance.

On the other hand, steric inhibition of protonation is given in 2-methyl aniline, where it is obviously missing in aniline.
Additionally there is a stabilising non-covalent interaction between the methyl and amine group, which can be analysed with NCIPLOT, and then displayed. The green blob is weakly attractive, red is strongly repulsive, blue would indicate strongly attractive interactions.

TL;DR: The dominant effect is steric inhibition of protonation, as steric inhibition of resonance is negligible (if possible at all).

Nitrogen’s lone pair can be in a number of different orbital types depending on the geometry of the protons around it. The most important are $\mathrm{sp^3}$ and pure $\mathrm p$. The interconversion between the two can lead to the substituents switching sides; therefore, this process is known as nitrogen inversion.

Nitrogen inversion means that the lone pair is accessible from both sides, so you need to block both sides sterically if you want to inhibit protonation.

To inhibit resonance sterically, introducing an inhibitor on one side is enough since the group must be perfectly in plane for resonance to be significant.

Thus, aniline is less basic than 2-methylaniline.

• But I have found out that pKb value of 2-methylaniline and aniline are 9.56 and 9.4 respectively, making aniline more basic. sources:en.wikipedia.org/wiki/Aniline and microkat.gr/msds/o-toluidine.html – Shrish Shankar Dec 16 '17 at 15:18
• @ShrishShankar Interesting. Although Wikipedia does not quote a $\mathrm pK_\mathrm a$ value for 2-methylaniline or ortho-toluidine, I have confirmed the prediction elsewhere. I am at a loss how to explain that, since also the inductive effect of the methyl group should make 2-methylaniline more basic. We may argue that the effect is extremely small, though; there may be a countereffect I can’t think of right now. – Jan Dec 16 '17 at 15:33
• There are two methyl groups needed to really block conjugation . – Mithoron Dec 16 '17 at 16:51
• @ShrishShankar Because resonance isn't really inhibited in 2-methylaniline, it's acid-base properties are connected with so-called ortho effect. – Mithoron Dec 29 '17 at 17:40
• To inhibit resonance sterically you have to include a huge™ moiety, because you have to essentially rotate the lone pair into the plane of the aromatic ring. "Inhibition of protonation" (as I read it: hindering protonation) is rather easy, just put a propeller right next to it, blocking one of the entry paths. – Martin - マーチン Feb 23 '18 at 9:21