Got a few issues with this:

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Here is the Lewis Structure:

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As we can see, we have five lone pairs on which a hydrogen proton can land and form the solvent's conjugate acid. Let's consider them left to right.

1) Oxygen lone pair. Adding a hydrogen proton here would make the oxygen have a formal charge of +1. Not exactly ideal for an atom which is so electronegative (even more EN than nitrogen is).

2) sp3 nitrogen. This seems like a good site; it parallels that the ammonia molecule. However, we would be creating a +1 formal charge.

3) sp2 nitrogen. We again would be creating a +1 formal charge.

4) sp nitrogen. And finally, we would be creating a +1 formal charge.

I know that the more s-character, the more able is the hybridized atom able to stabilize negative charge. It isn't called electronegativity for nothing.

That being said, is electronegativity inversely related to electropositivity?

So should the sp nitrogen be least able to stabilize a positive formal charge?

From a Lewis acid/base consideration, we would say that each of these lone pair sites are pretty nucleophilic. So which is the most nucleophilic. We need the strongest electron donating tendency. And that appears to be the sp3 nitrogen because in an sp3 nitrogen, the electrons are further from the nucleus on a time-average basis than the electrons in an sp2 or sp nitrogen.


1 Answer 1


Your reasoning is not entirely correct, as you consider that the charge is only located on the atom which bears the proton upon protonation. But nearby atoms, bonds and molecules will also participate in the stabilization of the positive charge, through several factors.

In the gas phase (so considering no solution phase effects), you should at least consider that, in addition to the instrinsic capability to bear a proton of the atom (which goes as the inverse of electronegativity, as you state in your discussion), one has also to consider the following factors:

  1. Resonance effects (delocalization of the charge on other atoms, that may be less electronegative than the atom directly bearing the proton). In the molecule given above, this would lead to a stabilization of the $sp$ and $sp^{2}$ hybridized nitrogens, as there would be a partial positive charge on the neighbouring carbon atom, as: $$\ce{RCH=N+H-R' <-> RC+H-NH-R'} $$ or $$\ce{RC#N+H <-> RC^{+}=NH}$$

  2. Inductive effects (or if you prefer availability of electrons in nearby atoms that can be displaced towards the charged site). In your constrained textbook example, this would be hard to indicate: there are a lot of heteroatoms (nitrogens and oxygens) and electron withdrawing groups (CN) which are likely to remove electrons from the carbon atoms, and in the central part of the molecule, these are likely to contribute to destabilize the molecule.

  3. Intramolecular hydrogen bonds. Although sometimes neglected, these can contribute quite a lot to the stabilization of the charge when mutiple protonation sites involving heteroatoms are present. Just as a reminder, in the gas phase, a hydrogen bond in a protonated molecule can reach up to 120 kJ/mol. In your example they would be hard to predict without modelling the complete structure, but one could consider a possible H-bond between the protonated imine ($sp^{2}$ nitrogen) and the alcohol oxygen.

Then if we were to consider solution phase, the interactions between neighbouring molecules and the propensity to form hydrogen bonds with the protonated molecule will also play a role on the possible protonation sites.

Based on the above considerations, one should consider that the most likely stable forms would be protonation on either the nitrile group ($sp$ nitrogen) or the imine group ($sp^{2}$ nitrogen), as both allow to put a partial positive charge on the neighbouting carbon atom. The protonated nitrile group is a protonation on an $sp$ nitrogen, with charge delocalization on an $sp$ carbon which means that the charge stabilization will be less efficient than for an imonnium ion.

If one considers purely gas phase consideration, one can use the proton affinity scale to rank these various contributions.

If we look for proton affinities in the NIST database we have:

  • for a secondary amine $\ce{H3C-NH-CH3}$ 929 kJ/mol

  • for a nitrile $\ce{H3C-C#N}$ 779 kJ/mol

Unfortunately little experimental data is available for imines. But some theoretical data is available in this article by Hammerum et al.. If one considers the methylimine $\ce{Z-CH3-CH=N-CH3}$ proton affinity is calculated at 943 kJ/mol, higher than dimethylamine. But on the other hand if one considers $\ce{H2C=N-CH3}$, its proton affinity is of 880 (exp) or 902 (theo), lower than dimethylamine.

So inductive effects clearly play a very important role in the order between amine protonation and imine protonation in the gas phase. In a really constrained molecule as above, they are likely to play a major role, as is the possibility to form hydrogen bonds in the solution.

If one now puts in the possible formation of intermolecular hydrogen bonds, protonated secondary amines can make two hydrogen bonds with two hydrogen bond acceptors, whereas protonated secondary imines can only make one hydrogen bond. This will likely tip the balance towards amines and would make answer (4) most likely in solution, whereas I would be quite at loss to give an answer in the gas phase.


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