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Are graphite and hexagonal boron nitride aromatic?

Graphite has a planar network of 6-membered rings with each carbon connected to three other carbons. Since the valency of carbon is not satisfied, there will be some $\pi$-bonds. Can graphite act like an aromatic system? Can I explain this using Hückel's rules?

Hexagonal boron nitride has a similar structure. So if graphite exhibits aromatic character, then so will boron nitride (like inorganic benzene and benzene).

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Graphite is definitely aromatic and boron nitride is at least partially aromatic. For instance, in this paper, the authors calculate the percent resonance energy (%RE) of graphite as a comparison to other known aromatic systems. The %RE is a measure of the total resonance stabilization per carbon atom relative to some reference system.

They find the %RE of graphite to be 3.061 and that of benzene to be 3.528. This value for graphite is larger than that calculated for naphthalene (2.924), thus for all practical purposes, graphite is essentially as aromatic as benzene.

In the case of hexagonal boron nitride(h-BN), I haven't been able to find any reference which says explicitly that the system is aromatic, but there certainly is some delocalization of $\pi$ electrons as I've seen a few papers which note that aromatic systems adsorb to BN nanotubes (which ought to be electronically similar to h-BN) quite strongly and a large portion of this binding interaction comes from dispersion (i.e. $\pi$ interactions). See here for instance.

You mention "inorganic benzene" which is the molecule borazine. Borazine actually only exhibits about half the total resonance stabilization per atom as benzene. This is attributed to the polarity of $\ce{B-N}$ bonds which inhibits delocalization of electrons. See this paper on the aromaticity of borazine for instance.

However, the first reference I gave found that graphite is nearly as aromatic as benzene, so I would be willing to speculate that h-BN is about as aromatic as borazine.

Can I explain this using Huckel's rules?

I would be hesitant to use Huckel's rules on a $\ce{B-N}$ system as Huckel's rules come from Huckel theory which is really only meant to describe fully conjugated hydrocarbons. Nonetheless, people will point out when other systems follow these rules just in case it turns out to be significant.

In contrast to what I just said, however, there are boron clusters which have been found to follow Huckel's rule, but this doesn't mean you can extend Huckel's rules to heteroatomic systems such as h-BN at will.

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Yes, both graphite and borazine are aromatic in nature as each ring of a plane have six π-electrons (similar to benzene). This aromaticity explains why graphite and borazine is unusually thermodynamically more stable than diamond.

That's why graphite have more meting point than diamond and on pyrolysis of diamond (unstable), it turns to graphite (more stable). Graphite is only a few $\pu{eV}$ more stable than diamond. Diamond does not contain any delocalised electrons. Graphite contains one delocalised electron per carbon. These cause greater attraction between carbon atoms hence giving stronger bonds, more stability to the structure. Graphite has greater van der Waals forces. Because of the oscillating delocalised electrons which induce temporary dipoles in graphite increasing attraction and stability. Diamond all delocalised electrons are used in covalent bonding.

The standard enthalpy of formation of diamond ($H^\circ_\mathrm{f} = \pu{2.425 kJ/mol}$) is slightly larger than the enthalpy of formation of graphite, which is the most stable form of carbon at $\pu{25 °C}$ and $\pu{1 atm}$ pressure. At very high temperatures and pressures, diamond becomes more stable than graphite (Source).

The decay of diamond is thermodynamically favorable $(ΔG = \pu{−2.99 kJ/mol})$ under normal conditions, it would take an extremely long time (possibly more than the age of the Universe) for diamond to decay into graphite.

Borazine, with a standard enthalpy change of formation $ΔH^\circ_\mathrm{f}$ of $\pu{−531 kJ/mol},$ is thermally very stable.

resonance of borazine

In the above pictures, each curved arrow represents the flow of two electrons. The structure on the far right has the electrons in the "double bonds"; all localized in the nitrogen p orbitals.

Borazine is made up of boron (electropositive: Lewis acid) and nitrogen (electronegative: Lewis base). Therefore, its pi electron cloud is "lumpy," with the electrons spending more time near the nitrogens than near the borons. Note that although the electrons in the cloud spend more time near the nitrogens, the nitrogens have a positive formal charge!

This should reinforce your understanding of formal charge as a book-keeping system only. It does give us valuable insight; nitrogen does get less electron density than it would like for having to share. Still, though, its natural electronegativity assures that it will get the lions share of the π-cloud.

We can see this from the resonance possibilities: one of the resonance structures has the electrons fully localized on nitrogen. Since benzene’s electron cloud is more fully delocalized, and since electron delocalization is what gives aromatic molecules their special stability, benzene would be more stable than borazine.

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  • $\begingroup$ what about boron nitride.Is it too aromatic? $\endgroup$ – Pink Jan 26 '17 at 4:15
  • $\begingroup$ Maybe you can answer this yourself. What is the structure of boron nitride? $\endgroup$ – Zhe Jan 28 '17 at 22:57
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Both compounds are aromatic.

To determine aromaticity, count all of the electrons that can contribute to a resonance structure. This includes electrons within a Pi-bond, and lone-pair electrons.

If they follow Huckel's rule of 4n+2, and if it is a planar molecules, then it is aromatic.

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  • $\begingroup$ Counting electrons does not determine everything. 10-annulene follows the electron count rule (10) but can't assume a planar structure unless modified; disulfur dinitride with six conjugated electrons is messed up by interactions between formally nonadjacent atoms; I could say more but am running low on comment space. $\endgroup$ – Oscar Lanzi Jul 1 at 0:40

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