In pyrrole, which carbon atoms have the greatest electron density?

In the resonance structures of pyrrole, all carbons can formally bear a negative charge, so it's not immediately obvious which one is most electron-rich:

Resonance structures of pyrrole

My book claims that because resonance forms 1c and 1d are greater contributors to the resonance hybrid than 1b and 1e, the α-carbons (which bear the negative formal charge in 1c and 1d) are more electron-rich. Is this really the case?


3 Answers 3


Interesting question, more subtle than I realized!

Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must also consider the inductive effect of the nitrogen atom, which draws some electron density away from C2 and C5 through the sigma bonds. This would imply that the two proximal carbons (closest to N) are more electron-deficient, and the distal (more distant) pair of carbons are more electron-rich.

Experimental considerations: Orthocresol has a good point that pyrroles are observed to form bonds with electrophiles at the proximal carbons (2/5). However, I think this trend may be due to the superior stability of the intermediate produced by that pathway (3 significant resonance forms) versus a distal (3/4) attack (2 significant forms). The cation from a proximal attack is better stabilized by resonance than the distal-attack cation. I expect that pyrrole forming C2 bonds with electrophiles has more to do with the stability of the intermediate of that pathway than necessarily greater electron density on C2. What I did not expect, but read on Wikipedia, is this:

Unlike furan and thiophene, [pyrrole] has a dipole in which the positive end lies on the side of the heteroatom, with a dipole moment of $1.58\ \mathrm{D}$.

This is experimental evidence demonstrating that the electron density is lower on the proximal side (C2,5) and greater on the distal side (C3,4). Besides molecular modeling of the electron density clouds, I think the molecular dipole moment is about the most conclusive evidence one can hope for.

It really comes down to the context: in an undergrad first-semester course, you might be expected to say the distal carbons have more density because the nitrogen withdraws from the proximal carbons. In more advanced courses, you might take orthocresol's perspective and base your answer in reactivity, concluding the proximal have more electron density. But I expect a doctorate researcher would probably conduct molecular modeling and compute the electron distributions using quantum mechanics, and perhaps conduct a few experiments to measure the dipole. And at that level, I think C3 and C4 have the honey.

  • $\begingroup$ But the key has the answer as C2 and C5 by following the inductive effect (may be 'cause I'm still in high school C3 and C4 and dipole moment were all neglected or remain unnoticed) so shall I take C2,5 or C3,4? And thank you for the detailed answer, it helped a lot :) $\endgroup$
    – Harini
    Dec 14, 2016 at 16:07
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    $\begingroup$ That is bizarre. Unless there are more than college-level concepts at play here that I'm not aware of, I would say with certainty that the inductive effect would /lessen/ the electron density at C2/5. The nitrogen withdraws electrons from the proximal carbons due to its higher electronegativity. I cannot see how any "inductive effect" here could make C2/5 more electron-rich. This is highschool, you say? Perhaps your teacher is mistaken. I would be interested to hear their logic. $\endgroup$ Dec 15, 2016 at 3:48
  • $\begingroup$ There is no 'most stable resonance structure'. These are all part of the same description as a singular structure is an incomplete description. $\endgroup$ Dec 20, 2016 at 20:02
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    $\begingroup$ Of course Martin, it is only a colloquial phrase for intuitive rationalizations. Would you prefer "the relative weights of the molecular orbital wavefunction vector components are likely nearly equal in magnitude"? $\endgroup$ Dec 20, 2016 at 20:16
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    $\begingroup$ @electronpusher That's a silly question: Yes, of course. A bit simpler would be all resonance forms have a relative similar contribution to the overall bonding situation. I find such colloquialism very bad, as for the untrained eye and mind they suggest something that isn't true at all. And especially this association continues to linger around. And that is very bad, since it is much harder to unlearn something than learning something new. $\endgroup$ Dec 21, 2016 at 15:31

That is really not a fair question and even with a computational approach the differences are more or less marginal. The approach by electron pusher is certainly the way to go, and here is the computations to back it up.
See if you can find the carbon with the highest electron density in the following plot showing the value of the electron density for the calculation on the DF-M06L/def-TZVPP level of theory.

electron density mapping

I have then performed an analysis of the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). The following shows the Laplacian of the electron density, with the zero-flux surfaces in blue that divide the molecule.

qtaim plot

When we integrate over the electron density in these basins, we know at which atom "are the most electrons", subtracting the nuclear charge, we get the overall charge. This should tell us, which carbons have the higher overall density.

  1 (C )    Charge:    0.015912     Volume:    83.414 Bohr^3
  2 (C )    Charge:    0.406209     Volume:    76.122 Bohr^3
  3 (N )    Charge:   -1.248690     Volume:    95.264 Bohr^3
  4 (C )    Charge:    0.406204     Volume:    76.122 Bohr^3
  5 (C )    Charge:    0.015911     Volume:    83.414 Bohr^3
  6 (H )    Charge:   -0.012007     Volume:    48.338 Bohr^3
  7 (H )    Charge:    0.004155     Volume:    47.506 Bohr^3
  8 (H )    Charge:    0.004162     Volume:    47.506 Bohr^3
  9 (H )    Charge:   -0.012006     Volume:    48.338 Bohr^3
 10 (H )    Charge:    0.420149     Volume:    28.032 Bohr^3

And for reference we need the numbering in the table:

numbering in pyrrole

TL;DR (1) The α-carbons to the nitrogen have a higher positive charge, hence the lower total electron density than the β-carbons.

Let's have another look at the π-orbitals to see where the reactivity towards electrophiles comes from.

mo14 mo17 mo18 (HOMO)
mo19 (LUMO) mo21

TL;DR (2) The highest electron density of the HOMO is at the α-carbons; electrophiles react there.

Please don't nag about the numbering, I recycled the calculation from another post, where I did not pay much attention to that. Sorry.

Ad. I agree with the analysis of the dipole moment since all $\ce{C/N-H}$ bonds are in plane. The calculation predicts 1.9 D (away from nitrogen, see here for the direction confusion). This is quite nicely explained in Ron's answer to Pyrrole and furan dipole moments.


The electron density on a particular atom not only depends on resonance but also on inductive-effect. And in case when resonance structures doesn't give a clear answer to the question we go for other effects like inductive effect, no bond resonance etc. So, here by just looking at resonating structures you will not get the answer — for that go for inductive effect and then you will see a clear difference between the electron-densities on different carbons of pyrrole.

  • $\begingroup$ Out of curiosity, how do you explain that when pyrrole donates electrons to electrophiles, the bond forms on C2/5? $\endgroup$ Dec 16, 2016 at 3:44
  • $\begingroup$ It has been already explained in any of the above comments...the intermediate carbocation is resonance stabilised to a greater extent than that in case of an attack at 3/4 or can say at beta position. For more you can go through Gilchrist's book on Heterocycles.Thank You. $\endgroup$
    – user38876
    Dec 18, 2016 at 20:29
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    $\begingroup$ So you say the distal carbon will be more electron densed as cited in electronpusher's answer? $\endgroup$
    – Harini
    Dec 20, 2016 at 19:09
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    $\begingroup$ Yeaah! Two factors are just confusing you...in case of electron density only resonance can't give you a clear picture hence you will go for inductive effect... But when you will start looking at the site of attack you will have to look at the stability of the resulting intermediate. $\endgroup$
    – user38876
    Dec 20, 2016 at 19:14
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    $\begingroup$ Wah!! Thanks I got it cleared :) but do give a look at Martin's answer for this question.he has explained it in a new perspective.. :) $\endgroup$
    – Harini
    Dec 21, 2016 at 18:21

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