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In pyrrole, the lone pair of electrons belonging to the nitrogen is part of the aromatic ring. However, in pyridine it is part of an sp2-hybridized orbital. Why can't it be in the p-orbital and take part in the aromatic ring?

I mean, why can't the lone pair in pyrrole be in the p-orbital, with the sp2-hybridized orbital (which is not bonded to anything) sporting a single electron?

I'm not talking about the nitrogen atom having two p-orbitals participating in the aromatic ring. That would be strange. I wonder why the existing p-orbital of N cannot carry two electrons, the way it does in pyrrole.

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P.S. A related question, with a very interesting answer: sp2-hybpidization in pyridine and pyrrole

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    $\begingroup$ Because there is already a $p$ orbital of nitrogen which takes part in the aromatic ring. You can't have two $p$ orbitals on one atom pointing in the same direction. And you can't have orbital take part in the ring if it is oriented differently. $\endgroup$
    – Ivan Neretin
    Apr 27, 2016 at 15:59
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    $\begingroup$ Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7 \neq 4n +2$ for $n \in \mathbb{Z}$ $\endgroup$
    – orthocresol
    Apr 27, 2016 at 16:06
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    $\begingroup$ Yes, the fact that you put one electron in the p orbital and two in the sp2 orbital is solely to satisfy Huckel's rule. For pyrrole, there are only four other carbons which contribute one electron each, and the nitrogen therefore has to give two of its own in order for it to reach $4n+2$. If the lone pair sticks out, the aromatic ring would not be "disrupted". The ring would just not be aromatic. Also, the sp2 orbital in pyrrole is bonded to a hydrogen 1s. $\endgroup$
    – orthocresol
    Apr 27, 2016 at 16:13
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    $\begingroup$ Well, then it's Huckel's rule alright. Also, if you'd have a pair on $p$, you'd have an unpaired electron on that $sp^2$, which is not nice. $\endgroup$
    – Ivan Neretin
    Apr 27, 2016 at 16:15
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    $\begingroup$ Might find useful? chem.ucla.edu/~harding/tutorials/lone_pair.pdf $\endgroup$
    – ipcamit
    Apr 27, 2016 at 16:22

2 Answers 2

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As follows from our discussion in the comments to the question, pyridine uses only a single electron in the p-orbital in order to comply with Huckel's rule. Kudos to Orthocresol:

Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7≠4n+27≠4n+2$ for $n ∈Z$

Of course, electrons know nothing of Huckel and his rule, but apparently they position themselves into the most energetically suitable configuration. I am not yet qualified to explain why exactly it is the most suitable one.

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    $\begingroup$ The key point is mentioned here! Electrons position themselves in the most suitable configuration, and if you look at the MO Diagram of Pyridine it becomes clear why the line pair will sit in Nitrogen rather than in the p Orbital $\endgroup$
    – IT Tsoi
    Apr 28, 2016 at 10:17
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    $\begingroup$ @ITTsoi - that may well be so, and I've been shown an MO diagram in the chat, but I haven't yet mastered the MO theory. If I do that, I'll expand my answer. Or maybe someone else will post an additional answer with an MO-based explanation. $\endgroup$
    – CowperKettle
    Apr 28, 2016 at 14:46
  • $\begingroup$ Huckels rule is irrelevant and the orbitals are not "single" electrons. $\endgroup$
    – matt_black
    Dec 7, 2018 at 10:02
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It has the wrong geometry

The other P orbital can't participate in the aromatic ring because it has the wrong geometry: it points in a direction orthogonal to the orbitals involved in the aromatic part of the ring. The picture in the question shows this clearly.

There is already a P-orbital on nitrogen involved in the aromatic system of the ring (the one that points "up"). The orbital coloured yellow-ish in the picture doesn't overlap with the aromatic system at all as it points "out" from the ring and (if you did calculations) would have no overlap with the other orbitals in the the aromatic system. No overlap means it can't be involved in aromaticity. It is, ultimately, simple geometry and no other explanations are required.

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