I came across a problem which stated that borole isn't aromatic. I thought that borole is similar to pyrrole so it should be aromatic, but that isn't the case. Can I know why isn't it aromatic?
I was also wondering if we replace the nitrogen in pyridine with boron then will the compound remain aromatic?

  • 1
    $\begingroup$ Draw the pi orbital system, then fill in the available electrons. You will hopefully then be able to answer your questions. $\endgroup$ – Martin - マーチン Aug 25 '20 at 9:48
  • $\begingroup$ I see that boron has 2 electron available. So won't they go into resonance? $\endgroup$ – Priyank Aug 25 '20 at 10:32
  • $\begingroup$ @Safdar Boron has empty p orbitals, but can resonance of borole not involve delocalisation of electrons in the vacant p orbital from the neighbouring pi bonds. Won't this cause charge seperation and negative charge formation on boron. This idea is different from pyrrole where the ring has negative charge but can it not occur? $\endgroup$ – Aditya Roychowdhury Aug 25 '20 at 12:21
  • $\begingroup$ Three words: because it isn't. $\endgroup$ – Oscar Lanzi Aug 27 '20 at 1:02

A very basic way to do this would be to draw a diagram that shows the p-orbitals going above and below the plane.

For Borole, the bonding orbitals would look like as shown below (taken from the wikipedia page on Borole):

Borole pi-orbital system Natural Bonding Orbitals of Borole. Structure optimised using ORCA BP86-D3BJ and def2- TZVPP basis set. The calculated occupencies of the obitals going from left to right are $0.13$, $1.9$ and $1.9$ respectively.

[emphasis mine]

The point to be taken here is the sentence in bold, the occupany in the empty p-orbital of boron [left most diagram and first value given] is small compared to the other two π-bond orbitals. This means that there is no complete delocalisation in borole. Only the double bonds are in conjugation.

A reason for this is that the orbital of Boron is an empty p-orbital and so the total number of electrons that can potentially delocalise is $4$ and so not of the form $4n+2$. If anything, the compound would become "anti-aromatic" if conjugation were to happen and so that delocalisation doesn't take place.

Now, on your second question, the lone pair of nitrogen in pyridine doesn't contribute to the aromaticity since the lone pair is out of plane of the ring. So adding placing boron shouldn't affect the aromaticity.


Simply for the beginners in organic chemistry, this explanation of aromaticity is good enough:

An aromatic (or aryl) compound contains a set of covalently bound atoms with specific characteristics:

  1. A delocalized conjugated $\pi$ system, most commonly an arrangement of alternating single and double bonds.
  2. Coplanar structure, with all the contributing atoms in the same plane.
  3. Contributing atoms arranged in one or more rings.
  4. A number of $\pi$ delocalized electrons that is even, but not a multiple of $4$. That is, $4n + 2$ $\pi$-electrons, where $n = 0, 1, 2, 3,$ and so on. This is known as Hückel's rule.

Hence, borole ($\ce{C4H5B}$), which has mimicked pyrrole ring follows first three rules but the forth. It is planner with contributing atoms arranged in a single ring in same plane. It has a conjugated $\pi$-system with consecutive $\mathrm{p}$-orbitals along the ring, but one of them is empty (still valid to count for aromaticity; e.g., tropylium ion, $\ce{C7H7+}$). However, borole has missed the most important rule: Hückel's rule. It does not contain $4n + 2$ $\pi$-electrons, therefore not aromatic.


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