I was given the first structure, and then drew the other 5 resonance structures:

resonance structures of 1-(cyclopenta-2,4-dien-1-ylidene)-N,N-dimethylmethanamine

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity.

My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There is charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule.

Hope I'm correct so far; here's my question:

In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised?


2 Answers 2


First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.

Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).

In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$-hybridised?

If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$-hybridized so that the the lone pair is in a $\mathrm{p}$-like orbital to ensure appropriate symmetry for $\pi$-overlap. Put another way, if an atom is $\mathrm{sp^2}$-hybridized in one resonance structure, then it is $\mathrm{sp^2}$-hybridized in all of them. Atoms that are $\mathrm{sp^2}$-hybridized and $\mathrm{sp^3}$-hybridized have differing geometries, which is not permitted in the resonance phenomenon.

In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$-hybridized (not the other way around).

In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems, which is not very good but will give you the idea.

My reasoning is that the first structure contributes more.

The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list.

  • $\begingroup$ Thanks for the comprehensive and lucid answer. Is M.O. more descriptive? Why use resonance and hybridisation at all? $\endgroup$
    – Brian
    Mar 16, 2013 at 13:23
  • 8
    $\begingroup$ Hybridization and resonance represent a "quick and dirty" or "back of the envelope" method that you could do anywhere. Generating good MO diagrams and representations for molecules even as complex as yours requires calculus, linear algebra, and a computer with appropriate software. The answer generated generated by hybridization/resonance is almost always the same as the MO theory answer, so choose the quick and easy option first. Use the more difficult theory only if the first approximation is not good enough. $\endgroup$
    – Ben Norris
    Mar 16, 2013 at 13:38
  • $\begingroup$ @BenNorris If an atom participates in resonance why its hybridization must be $\mathrm{sp^2}$? I mean shouldn't be the weighted sum of the hybridizations in each resonance structure? $\endgroup$
    – ado sar
    Nov 11, 2020 at 13:49
  • 1
    $\begingroup$ @HarryHolmes - Orbitals are mathematics constructs defining regions of space with high electron probability density. Geometry is an empirically measured property. $\endgroup$
    – Ben Norris
    May 6, 2021 at 13:00
  • 1
    $\begingroup$ @HarryHolmes - That's a "yes." $\endgroup$
    – Ben Norris
    May 6, 2021 at 13:10

I wanted to add a perspective to this since this can be a contentious issue, and I would still say knowing whether a nitrogen is $sp^2$ or $sp^3$ hybridized is not straightforward. I was originally hoping to add this under this question specifically about aniline, but that question is now closed and marked as duplicate, with a link to this question.

In a simple molecule like aniline, one might think that the hydrogens of the amine would be planar with the ring. However, we know from crystal structures of aniline that it is indeed pyramidal. If you have access to the Cambridge Structural Database (CSD), you can see this in structure BAZGOY. The improper dihedral angle of the hydrogens is ~26 degrees in that structure.

I was also curious to see what QM had to say about this. I performed some optimizations of analine with the improper dihedral angle of the hydrogens constrained at various angles. Here is a drawing of the improper dihedral I constrained.


I used the MP2 method and cc-pVTZ(-F) basis set, in implicit solvent (PBF) using the Jaguar software from Schrodinger. You can see in this plot and gif that the lowest energy angle was at ~35 degrees. That's pretty much completely pyramidal (in ethane, an equivalent angle is 33.4).



However, that being said, I also wouldn't say this is dogmatic. This is an implicit solvent calculation, and in real water with real hydrogens and oxygens to hydrogen bond to, things may be still different. NMR might be able to help figure this out, or it could further confuse if the timescale of flipping smears the signal to give a flat angle when it is not.

I should also point out that there are other crystal structures that contradict this. If you look at the crystal structure VOMFOS, which is of an adenosine analog, the amine hydrogens are planar with the ring. Is that just an artifact of the crystallization packing network, or a true representation of the hybridization of that nitrogen? Hard to say.

I think my main point here is to be careful what you assume. If you asked me on a test in undergrad whether the amine on aniline is $sp^2$ or $sp^3$, I would have said $sp^2$ with confidence. Experimental and computational evidence suggests that might not be the case.


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.