4
$\begingroup$

My teacher had once told me that graphite was just an insanely huge polymer of benzene (Those weren't his exact words but the gist). Today While we were studying aromaticity in organic compounds we steadily advanced from benzene to napthalene the annulenes and so on. What really stuck me was how as the number of benzene rings increased the stuff tended towards graphite (atleast thats what the structure seemed to indicate). So the main question is why isn't graphite aromatic if its just a big collection of benzene molecules

$\endgroup$
0
9
$\begingroup$

Graphite/graphene is aromatic. The first image below is a common crystallographic representation of graphite and does not account for covalent bonding. The second one shows the molecular structure of graphite as a kekule representation. Graphite is made of planar sheets of carbon atoms in 6-atoms rings with conjugated double bonds. This also satisfies the 4n+2 π electrons criteria needed within the aromatic ring for graphite to be aromatic. This aromaticity is why graphite is not only stable to high temperatures but also a more thermodynamically favorable allotrope than diamond under typical conditions.

Crystallogrphic Representation Crystallogrphic Representation of Graphite.

Kekule REpresentation of Graphite
Kekule Representation of Graphite.

$\endgroup$
2
  • $\begingroup$ +1 (possibly helpful ion.chem.usu.edu/~boldyrev/graphene.php) and so I've just asked How is the aromaticity in graphene different from the aromaticity in benzene? $\endgroup$ – uhoh May 13 '19 at 0:57
  • $\begingroup$ +1. To OP: I copy a comment from the thread mentioned by uhoh. Indeed related arguments. Benzene is one cycle, no other cores have the same aromaticity. "I might be too naive but at the end it means that while in benzene there is a sextet delocalised in a ring, in graphene the delocalisation over the neighbours results in just two pi electrons delocalised but pinned onto a single ring. The other are described by extended orbitals and are those responsible for the in plane electrical conductivity. I am aware that worded this way might even sound trivial." Replace graphene with graphite. $\endgroup$ – Alchimista Mar 14 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.