# Which lone pairs/bonding pairs go into which pi molecular orbitals?

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:

Here's the second part:

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?

• You can add images to your post with this button. – orthocresol Oct 31 '16 at 21:02
• 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... – Zhe Oct 31 '16 at 21:17
• 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? – leadgame Nov 1 '16 at 3:13
• 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. – leadgame Nov 1 '16 at 3:15

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.

• 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.
• @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 $\ce{[Ne2]}$. – Jan Nov 2 '16 at 15:35
• (Note that $\ce{Ne2}$ doesn’t exist, hence the square brackets.) – Jan Nov 2 '16 at 15:36