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My professor mentioned that borazine is aromatic, but not as much as, say, benzene. The reason he gave for this was that the B-N bonds in borazine, being polar, inhibit resonance. He also mentioned that the same inhibition of resonance occurs in h-BN. Why is this so? What difference does bond polarity make?

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Let us understand and try to answer your question with a few simple points.

First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons. According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for allowing the electrons of the side atom to delocalize into it needs to be satisfied. Along with this, the lone pair which needs to be delocalized should be in kept in a pure p-orbital. Boron and Nitrogen both facilitate these requirements and thus the lone pair delocalization takes place. Thus in a way, we could say that the aromaticity in the borazine moeity is due the pi-backdonation of unpaired electrons.

Borazine/Borazole Cannonical Resonance structures

Understanding that the main reason is sideways pi-backdonation for the cyclic delocalization of electrons, one also needs to keep in mind that Boron and Nitrogen have a significant difference in their electronegativities. The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene. You could similarly compare h-BN and grpahite; in the former case the delocalisation is less pronounced due to difference in electronegativity than in the later case where a very well pronounced delocalisation is observed.

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  • $\begingroup$ "The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene." I was asking for the explanation of this in the question. Could you help me out with that? $\endgroup$ May 28 at 7:17
  • $\begingroup$ The diffference in electronegativity seemed to me the prominent point in not letting the electrons fully delocalise $\endgroup$ May 28 at 7:19
  • $\begingroup$ The point which I made, @Draculin has mentioned the same in a brief and precise manner. So if you didn't get what I meant, refer his answer and the research paper as well. $\endgroup$ May 29 at 5:28
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You can think of it as: polarity causes slight concentration of pi electrons on nitrogen atoms in contrast to boron in the resonance hybrid due to which cyclic delocalization does not occur "completely"; hence the aromatic character is hindered

Further reference: https://www.researchgate.net/publication/225792945_Borazine_To_be_or_not_to_be_aromatic

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