An unusually small (doubly) aromatic molecule: C4.

When you talk π-aromaticity, benzene is the first molecule that springs to mind. But there are smaller molecules that can carry this property; cyclopropenylidene (five atoms) is the smallest in terms of atom count I could think of until now, apart that is from H3+ which is the smallest possible molecule that carries σ-aromaticity. So here I have found what I think is an even smaller aromatic molecule containing only four carbon atoms. And it is not only π-aromatic but σ-aromatic.

Let me go through the analysis (using a CCSD(T)/Def2-TZVPPD calculation, DOI: 10.14469/hpc/10226).

  1. Four carbons contain 16 valence electrons for bonding.
  2. Eight of these are conventional, forming four C-C single bonds around the 4-ring.
  3. Eight are left over, and these partition into a set of six and a set of two.
  4. The set of two are in p-π atomic orbitals and form a 4n+2 (n=0) aromatic system
  5. The set of six are in σ-sp AOs and form a 4n+2 (n=1) aromatic system.
  6. The three σ-MOs all contribute to the central C-C bond, particularly σ3 and σ2 in different ways.
  7. σ2 also reminds of [1.1.1]-propellane, where the two σ-electrons are in effect external to the central C-C bond, but spin coupled to form what might be called a σ exo-bond. There is also similarity to the exo bond in C2.
  8. The dissociation energy of the central bond can be estimated at 28 kcal/mol from the triplet state energy.
Bonding MOs for C4.
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π1
σ3 σ2
σ1

So this little molecule carries a lot of diversity in its chemical bonding; an ideal candidate perhaps for a tutorial in bonding theory of organic molecules?


The post has DOI: 10.14469/hpc/10252

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5 Responses to “
An unusually small (doubly) aromatic molecule: C4.

  1. Carl Feynman says:

    Would it be lower energy folded into a tetrahedron? Then there would be six sigma bonds. Would the angular strain energy be too large?

    • Henry Rzepa says:

      The form of C4 with Td symmetry has two electrons in a triply degenerate Td orbital and so must undergo Jahn-Teller distorsion to the lower C2v symmetry. This species is 106.0 kcal/mol higher in free energy than the bicyclic aromatic form and is a second order transition state. The first of these -ve force constants has vectors that distort back to the bicyclic form.

      The second has vectors that distort to a “methylenecyclopropene” like structure (but without the hydrogens). So the answer to your question is that the tetrahedral form is not stable.

  2. Mike Turner says:

    That’s a very neat electronic structure – does it make this the most stable isomer of C4?

  3. Henry Rzepa says:

    I show here the calculated IR spectrum.

    Who knows, it might exist in inter-stellar space!

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