As Borazole ($\ce{B3N3H6}$) is aromatic, it can show electrophilic aromatic substitution. Now, my question is when borazole undergoes electrophilic aromatic substitution reaction, on which atom does the electrophile attack? Is it boron or nitrogen? Also, how to compare it's rate with the same type of reaction when carried on benzene? Should I consider stability of arenium ion (Wheland intermediate) formed when comparing the rates?

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    $\begingroup$ Borazole isn't really that aromatic if compared to benzene. I'd expect electrophiles to add to nitrogen. $\endgroup$ Aug 5, 2020 at 10:12

1 Answer 1


Chiavarino, et al. [1] report that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen. Nucleophiles such as methanol prefer boron.

We can explain that result in terms of both molecular orbitals and the Wheland intermediate. In the molecular orbital explanation, recall the familiar delocalized pi orbitals in the benzene ring. With the polar bonds of borazole the occupied bonding orbitals whose electrons will add to an electrophile are shifted towards the nitrogen atoms, so the electrophile will tend to bond there rather than on the boron (the latter has instead mostly the empty antibonding pi orbitals and is thus favored in nucleophilic substitution).

In terms of the intermediate, adding the electrophile to the ring leaves the remaining pi electrons in a deficient condition -- only four electrons cover the remaining five conjugated atoms, with most of the electron deficiency falling on atoms ortho and para to the attack site. Attacking on nitrogen puts the less electronegative boron atoms at these deficient sites leading to a more stable EAS intermediate.


  1. Barbara Chiavarino, Maria Elisa Crestoni, Simonetta Fornarini, "Electrophilic Substitution of Gaseous Borazine," J. Am. Chem. Soc. 1999, 121(11), 2619-2620 (https://doi.org/10.1021/ja983799b).

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