Steric inhibition of protonation in o-substituted anilines

[Vishesh Jain]

Homework Statement

According to my textbook (Organic Chemistry by Solomon & Fryhle) "When an o-substituted aniline is protonated (at the N), increase of H-N-H bond angle leads to steric hindrance with the group at the o-position. This makes this protonated conjugate acid unstable and makes the amine less basic." But why dosen't it apply when the o-substituent is -OH or -OCH3 ?

The attempt at a solution

Because this effect operates when a methyl group is at ortho (as shown in textbook photo) but it dosen't for a methoxy (much bulkier) group, i presumed it's due to the O in -OCH3 which forms a H-bond with the protonated amine (thereby stabilizing it). But this steric inhibition effect operates in o-nitro aniline, (according to this image) where also there should be H-bonding, and o-nitroaniline is the weakest base. Why steric inhibition effect applies in case of o-nitroaniline but not o-methoxy aniline ..?

Explanations involving computational chemistry & molecular orbitals are most welcome.

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[TeethWhitener]

This is a question where access to either a molecular modeling kit or modeling software helps out a lot. Maybe take a look at Avogadro:
https://avogadro.cc
or similar software.

To address your question more directly (and which becomes immediately obvious with a model), the lowest energy geometry of nitrobenzene is planar, but constraining the nitro group to be coplanar with the aryl ring in o-nitroaniline puts one of the oxygens extremely close to the protons in the anilinium group. This is where the steric hindrance comes into play. In the hydroxyl and methoxy analogs, the lone pair on the oxygen instead can point toward the anilinium.

 

[Vishesh Jain]

Thank you for your reply sir, and also for your software tip ...