I was just doing some practice midterms and I came across this question. The answers were posted, but there are no explanations for anything. I wasn't able to find anything about this online, either.

Here are pictures of the question.

Here's the first part:

enter image description here

Here's the second part:

enter image description here

Part e) is what is throwing me off. I was flitting back and forth between O and N, and O, N, and C, but in the end got it wrong. It says to base the answer off of part d), but there are two filled orbitals! How do I know which pair of electrons occupies which orbital?

Does the sp3 lone pair of the nitrogen occupy the lowest occupied MO or the highest occupied MO? Because it appears from the correct answer that it occupies the higher one - but why is that? The other delocalized pair of electrons is the pi electrons in the C=O double bond, and I get that maybe they have different energies, but I can't think of any reason one pair should occupy one particular orbital over the other.

Furthermore, according to the movement of electron pairs in the resonance structures of f) (which were all correct according to the answer key), the third structure suggests the lone N pair is shared between the carbon and the nitrogen, not the oxygen and the nitrogen. What gives?

Also, as a side question - the reason why the solvent is polar aprotic, despite having the capacity to shed a hydrogen from the carbon to form an induction-stabilized carbanion, is because protic solvents have the labile hydrogen attached exclusively to an oxygen or a nitrogen, correct?

  • 2
    $\begingroup$ You can add images to your post with this button. $\endgroup$ Oct 31, 2016 at 21:02
  • $\begingroup$ I think what the (bad) test question is suggesting is that HOMO electrons are delocalized and the bulk of the density is on O and N because there is a node at C. I don't think they should refer to it as a "lone pair" for the purposes of this question... $\endgroup$
    – Zhe
    Oct 31, 2016 at 21:17
  • $\begingroup$ So it is inaccurate altogether to think that particular delocalized electrons present within the LOMO and HOMO are sourced from particular regions of the molecule? That any electron could occupy those orbitals, and that statistically speaking, it just would largely spatially correspond to the areas around O and N? $\endgroup$
    – leadgame
    Nov 1, 2016 at 3:13
  • $\begingroup$ Thank you for your help - I will use the image button next time. I would ask my professor, but he has some policy against answering questions... Nonetheless, maybe I should suck it up and risk drawing his ire anyway. I really want to know the answer to this. I will leave this question as unanswered for now though, if anyone can chime in with a more concrete answer. $\endgroup$
    – leadgame
    Nov 1, 2016 at 3:15

1 Answer 1


This is a common attempt to visualise what π systems actually mean for the overall Lewis structure. But let me note a couple of things first:

  • Obviously, there is no $\mathrm{sp^3}$ lone pair on DMF’s nitrogen. Instead, the atom is $\mathrm{sp^2}$ hybridised and the lone pair (if it is there) occupies a p-type orbital.

  • Remember that the π analysis only deals with the π orbitals (obviously); there is an underlying σ system which is implied but hardly mentioned. Thus, if ‘one bond’ is created, it is actually a double bond, while ‘no bond’ means ‘only the underlying single bond’.

If you want, in simplistic terms, you can add the π orbitals one after the other. The first is fully bonding and has contributions by all three atoms. Since it is fully bonding across two bonds but only contains two electrons, we can say that it contributes $\frac22$ bonding electrons to each bond or a bond order of ½.

The second orbital is not really bonding as it has a perpendicular nodal plane through the central atom (carbon). Thus, we cannot really analyse it in terms of the $\ce{C-O}$ or $\ce{C-N}$ bond order directly. However we can say that the electron density will be stronger at the ends, i.e. on nitrogen and oxygen. Each of these atoms practically experiences $\frac22$ lone pair electrons (do not participate in bonding) or ½ a lone pair.

Consider what would happen if you had filled the third orbital. It would have been antibonding with respect to both the $\ce{C-O}$ and the $\ce{C-N}$ bond. It would have contributed the same bond order of ½ to both — but this time in negative. That would contribute to an overall bond order of zero or the bonding electrons being on either side of the bond as lone pairs.

For your question, this means:

  • the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one

  • the nonbonding orbital contributes the lone pair. Since it only experiences contributions from oxygen and nitrogen, this is where the lone pair will be located.

  • it would be wrong to look at the Lewis structure, see a lone pair on nitrogen and immediately associate that with the ‘lone pair’ of the question. If anything, consider all relevant resonance forms and all their lone pairs. Also, as soon as it has resonanted to form a bond, it is no longer a lone pair.

  • finally, as per your side question, note that the $\ce{C-H}$ bond in DMF has a very, very low tendency to deprotonate. That proton is not even mentioned on both the Evans table and the Bordwell data. While it might lead to a carbene, the carbene is not exactly stable. A much more likely deprotonation would be that of a methyl group which would extend the π system. Note that the sequential treatment with butyllithium and DMF is a method to convert electron-rich aromatic rings (e.g. thiophene) into their corresponding carbaldehydes — in this reaction, the proton of DMF is less acidic than that of a thiophene or proton exchange would occur in place of nucleophilic attacks.

  • $\begingroup$ I understood the first part about the respective electron densities of each molecular orbital. But when you say that the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one - what does it mean to be compensated for? $\endgroup$
    – leadgame
    Nov 2, 2016 at 5:16
  • $\begingroup$ Is the idea that since its asking about a lone pair, you need to pick the MO that even allows for such a configuration? And that since the bonding MO involves electrons shared among all three atoms, it couldn't possibly house the electrons of the lone pair? $\endgroup$
    – leadgame
    Nov 2, 2016 at 5:19
  • $\begingroup$ @leadgame Well, think linear combinations. Consider a hydrogen molecule. $\sigma_1$ is the bonding MO and $\sigma_2$ is the antibonding MO. If you linearcombine $\sigma_1$ and $\sigma_2$ you get $\Psi_3 = \sigma_1 + \sigma_2$ and $\Psi_4 = \sigma_1 - \sigma_2$. $\Psi_3$ is localised only on one hydrogen, $\Psi_4$ on the other. If both $\sigma$ orbitals were filled, then we can consider it the same as a lone pair on both sides. That’s also the formal explanation for why number of lone pairs increases when going from $\ce{N2}$ over $\ce{O2}$ to $\ce{F2}$ and finally $\ce{[Ne2]}$. $\endgroup$
    – Jan
    Nov 2, 2016 at 15:35
  • $\begingroup$ (Note that $\ce{Ne2}$ doesn’t exist, hence the square brackets.) $\endgroup$
    – Jan
    Nov 2, 2016 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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