# Is the conjugate base of pyrrole resonance stabilised?

This is something I am conceptually struggling with, if I deprotonate pyrrole on the nitrogen in the ring, is it true that the conjugate base is not resonance stabilised, because the orbital is orthogonal to the ring? However, with a pKa of 23, I feel like it must have some resonance stabilisation.

• Yes - it isn't resonance stabilised, and 23 is sort of normal, what are you comparing it against? The pyrrole anion has the lone pair in an sp2 orbital which makes it slightly more acidic than a typical sp3-hybridised amine. For comparison you can look at the "amides and carbamates" section in the Evans pKa table. Also it would be helpful if you could use the full tag organic-chemistry instead of just organic, thanks. Jun 5 '17 at 17:53
• sorry I will do that in future! My organic tutor said that it was resonance stabilised and that concerns me greatly... Jun 5 '17 at 18:29
• No problem. That's odd - well, technically the conjugate base is resonance stabilised, in the same way as how pyrrole itself is resonance stabilised (aromatic), because it does have that lone pair in the pi system. However, the additional lone pair created on N by deprotonation is orthogonal. Maybe it was a wording issue? Jun 5 '17 at 18:46
• definitely a full conjugation was done. Same with imidazole-my tutor demonstrated full resonance, and argued that imidazole is more acidic because you can distribute the charge to the nitrogens in the ring (more electronegative than carbon so favourable) Jun 5 '17 at 19:02
• I see. Well, in the conjugate base of pyrrole (and imidazole), the nitrogen has two lone pairs. One is delocalised into the pi system, the other isn't. The intent was probably to illustrate the delocalisation of the one that is delocalised (the one in the p orbital), which is perfectly legal. Jun 5 '17 at 19:11

There is like you assumed no reasonable amount of resonance stabilisation. However, the aromatic π-system contributes to the stability of σ-skeleton.

Below you can find the MO of pyrrole and its N-deprotonated anion (click it to enlarge). You can see that in MO 17 the electron density is shifted further into the centre of the ring, accounting for the larger negative charge on the nitrogen. It is a bit more difficult to see, but that is also true for MO 18 and 14 (pyrrole), which becomes 15 (anion). The most significant change is that the upon deprotonation resulting lone pair becomes much higher in energy (12 to 16), and it is as you note perpendicular to the π-system in a σ-type orbital. Note that the image only shows the trend and has no information of the actual orbital energies, apart from that they increase from bottom to top.

• How do you get the MO diagram? Jul 13 '17 at 1:01
• @Mockingbird I have calculated the MO on the DF-BP86/def2-SVP level of theory. Jul 13 '17 at 4:02
• Is there any software? Jul 13 '17 at 4:06
• @Mockingbird Yes, I used the quantum chemistry software package Gaussian 09, but there are other, openly accessible software packages, too. The graphics I have produced with ChemCraft. Jul 13 '17 at 4:17
• Can you mention some of those softwaress? Jul 13 '17 at 5:21

As you pointed out, the lone pair is indeed orthogonal to the ring. Not that it would matter, since the aromatic ring already has 6 electrons, so you're not going to conjugate 2 more.

The lower pKa value is likely due to the face that $sp^{2}$ hybridized carbons are more electron withdrawing due to their higher $s$ character. That should be enough to help stabilize the conjugate base.