12
$\begingroup$

Why does benzyne prefer to undergo dimerisation to stabilise itself, even though the product formed (biphenylene) is anti-aromatic?

dimerisation of benzyne

Why does it not undergo trimerisation through the following pathway, resulting in an aromatic and presumably more stable compound (triphenylene)?

enter image description here

Improve this question
$\endgroup$
7
  • 6
    $\begingroup$ Central rings of biphenylene and triphenylene are approximately non-aromatic, I can see why you could think otherwise though $\endgroup$ – Mithoron Sep 2 '17 at 15:00
  • 9
    $\begingroup$ Actually, trimerisation does happen. See e.g. Chem. Rev. 1962, 62, 81. Biphenylene as a whole is considered to be aromatic. See e.g. Z. Naturforsch. 1973, 28B, 478 or J. Phys. Org. Chem. 2010, 24, 263 or J. Am. Chem. Soc. 1996, 118, 2903. It is a pretty stable compound in all respects. $\endgroup$ – orthocresol♦ Sep 2 '17 at 15:04
  • 9
    $\begingroup$ Also, 3-body collisions are unlikely to occur, even more so when a reactive intermediate such as o-benzyne is involved. $\endgroup$ – ron Sep 2 '17 at 16:24
  • 4
    $\begingroup$ So, is bipheylene the kinetic product and triphenylene the thermodynamic one? $\endgroup$ – Ayushmaan Sep 3 '17 at 3:16
  • 5
    $\begingroup$ When reactive intermediates such as o-benzyne are involved, there is no equilibrium, hence thermodynamics are not involved. All products produced are kinetically derived $\endgroup$ – ron Sep 4 '17 at 1:58
8
$\begingroup$

As per Wikipedia,

Biphenylene is quite stable both chemically and thermally, and behaves in many ways like a traditional polycyclic aromatic hydrocarbon. However, both the spectral and chemical properties show the influence of the central [4n] ring, leading to considerable interest in the system in terms of its degree of lessened aromaticity.

This implies that biphenylene is rather stable, despite its anti-aromaticity.

As for triphenylene, I'm not sure, and it doesn't say so anywhere, but perhaps the free energy of formation of triphenylene is much more, as it requires more molecules to be put together, leading to the reaction being non-spontaneous. However, benzyne does trimerize as well, forming triphenylene.

So benzyne can actually undergo both dimerization and trimerization, forming two different, but stable products.

As requested, a more primary source is available here. This site cites other sources, whose links I have provided in the quotation.

These lines say more or less the same thing:

Several recent publications (2 3 4 5 6) concluded that biphenylene possesses significant strain energy of the order of 50–60 kcal/mol. These numbers are at least confusing since molecules that possess such tremendous strain can hardly exist. In the meantime, biphenylene is a stable molecule that has been synthesized a long time ago while its chemical properties show typical aromatic behaviour.

Improve this answer
$\endgroup$
1
  • 6
    $\begingroup$ Whether or not what's written in Wikipedia is accurate, please provide references to more reliable primary literature if possible. $\endgroup$ – Linear Christmas Sep 3 '17 at 17:01
2
$\begingroup$

Just to add to the above answer with a possible theory of why this might be happening-:

The central ring in biphenylene is majorly non aromatic as mithoron pointed out in his comment. My teacher says that the central ring has two bond lengths as a result of Jahn-Teller effect. Because of the longer bonds in between the rings, conjugation is broken between the p orbitals of the atoms in the central ring making it non-aromatic. Hence it does not share its properties with other anti-aromatic compounds.

And biphenylene being the major product during benzyne polymerisation reaction is due to the less probability of 3-body collisions as pointed by ron. Benzyne being a di-radical is highly reactive hence kinetics plays a major role as compared to thermodynamics.

Improve this answer
$\endgroup$

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.