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In the last exam we were asked to rationalize the mechanism of the cyclopropylmethylene carbene cleavage into acetylene and ethylene.

enter image description here

In the lecture we discussed pericyclic reactions (cycloadditions, electrocyclic reactions, sigmatropic reactions, ene reactions). I don't think it's a con-/disrotatory ring opening, because in my view only a $\sigma$-plane is conserved over the entire reaction coordinate. What type of ring opening is this; is it a concerted or radical based C-C bond cleavage? Because there is highly probable an elongation of the bond length (as drawn in the stretched cyclopropane moiety), I don't assume it to be concerted but rather radical based.

I tried to classify the reaction in terms of symmetry along its entire reaction coordinate. Because from the task, a singlet carbene, I know it's a $sp^2$ hybrid, so I strongly assume the empty p-orbital to be in conjugation with the Walsh orbitals, more precisely with the Möbius type orbitals.

enter image description here

The problem is very similar to cyclopropylcarbinyl cation:

enter image description here

In the exam I was asked to draw the orbital correlation diagram. For me it looks like two radicals forms the double bond of ethylene and the "biradical" like centers recombines with the "biradicaloid" (in $sp^2$ orbital). I call a carbene a diradicaloid or biradicaloid because the two radical electrons share the same orbital.

To correlate reactants with products I need to combine orbitals with the same symmetry. We only draw the orbitals which are involved either in bond braking or in bond making. How should I account for the biradicaloid singlet $sp^2$ orbital in the orbital correlation diagram? (It is forming the triple bond of acetylene, so there must be a change of hybridization.) For me it looks like to have no symmetry (neither A' nor A'' because it lies in the mirror plane, because it is eclipsed with the $\alpha$-carbon-H.

If it's a radical mechanism I assume the Möbius orbital of A'' symmetry (depicted in green) to correlate with the $\pi^*$ of ethylene.

I assume, the cleavage happens preferently with light, because in this way I can put a photon from A'' (green) into the orbital of A'' symmetry on the reactant site and therefore lowering the activation energy for the reaction.

I assume the orbital correlation diagram to look something like this:

enter image description here

There are some points unclear to me:

  • Has the empty p-orbital A'' symmetry ?

  • For me it looks somehow ugly to correlate orbitals which don't have symmetry labels (I think this procedure is not allowed and I need to build a proper set.) How can I build a linear combination in this particulate case, so that the orbital correlation from red to red makes sense?

  • There is a "$\pi^*$ orbital acetylene,2" unused (not correlated). Did I miss an orbital from the reactant site? (because there is one orbital too many on the product site)

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What is the type of the ring opening; is it a concerted or radical based C-C bond cleavage?

This is a chelotropic reaction, a subset of cycloaddition\fragmentation reactions. As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide. For example, if the reaction had been run with the cis-2,3-dimethylcyclopropane analogue and both cis- and trans-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process. When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we expect the reaction to be concerted.

I strongly assume the empty p-orbital to be in hyperconjugation with the walsh orbitals

There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the "bisected" conformation) should be favored. Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.

I was asked to draw the orbital correlation diagram.

I have difficulty reading your diagram, so I created a new orbital correlation diagram.

enter image description here

Let me say a few things about my figure in order to make it easier to understand.

  • The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.
  • The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.
  • The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.
  • There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.
  • This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.
  • I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.
  • $\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.
  • I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.
  • The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.
  • The "A" and "S" labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.

Now, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry. In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals. This analysis suggests that the thermal reaction can proceed in a concerted fashion.

Miscellaneous

  • Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur. The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene. Other concerted reactions may be preferred because they happen to be of lower energy.
  • Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction. But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not? Well, these are all equilibria and the forward reaction is extremely exothermic. We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds. Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic. The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria. It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed. If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.

Response to OP's comments:

How can we see in general from an orbital symmetry correlation diagram if a reaction is concerted or not?

If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion. If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.

In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively. All ground state reactant orbitals correlate with ground state product orbitals. Therefore the reaction is allowed to proceed in a concerted manner.

Is it okay to say biradicaloid instead of carbene?

No, not really. "Biradical" is generally used in cases where the electrons are on different atoms. Like when we break a bond in cyclopropane we have a 1,3-biradical.

In the orbital correlation diagram I would predict the acetylene a little bit lower because of more s character (sp) than ethylene (sp2). May be the energy difference is roughly as high as the difference between carbene p orbital A and carbene sp2 orbital.

That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals. My guess was that they would have roughly the same energy.

For the retro cheletropic reaction I assume a photolytical pathway

The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.

According to the paper, I've found here...

I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state. I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures. The photochemical route just allows them to generate the same ground state carbene at very low temperatures.

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  • $\begingroup$ That's a very nice answer! 1) How can we see in general from an orbital symmetry correlation diagram if a reaction is concerted or not? 2) Is it okay to say biradicaloid instead of carbene? 3) In the orbital correlation diagram I would predict the acetylene a little bit lower because of more s character ($\ce{sp}$) than ethylene ($\ce{sp^2}$). May be the energy difference is roughly as high as the difference between carbene p orbital A and carbene $\ce{sp^2}$ orbital. 4) For the retro cheletropic reaction I assume a photolytical pathway: $\endgroup$ – laminin Aug 14 '15 at 11:56
  • $\begingroup$ According to the paper, I've found here: pubs.acs.org/doi/pdf/10.1021/jo00297a057 "The photochemical decomposition of 2 gave only ethylene, acetylene, and cyclobutene as expected from Scheme I." This would make sense for me if I correlate the carbene p orbital A with acetylene $\ce{\pi_{x}}$ and $\ce{\sigma_3}$ A with an antibonding ethylene $\ce{\pi^{\star}}$ S because this Walsh orbital $\ce{\sigma_3}$ A looks similar as the antibonding $\ce{\pi^{\star}}$ orbital. I promote one electron from A symmetry into carbene p orbital A photolytically and therefore lower the activation energy. $\endgroup$ – laminin Aug 14 '15 at 11:58
  • $\begingroup$ Thank you for your "Response to OP's comments". Conc. 2) I see now that two interacting radicals which are called biradicaloids must be located at two different centers (goldbook.iupac.org/DT07358.html). Conc. 4) Because acetylene lies a little bit more down in energy, the two $\Pi$ orbitals of acetylene with A and S symmetry are correlating with $\sigma_2$ and $\sigma_3$. I'm not sure if we can correlate e.g. the sp2 acetylene part (S) from reactant with the ethylene (S) green line in my drawing here: imgur.com/W8eKLs2 , because they belong to different molecule parts? $\endgroup$ – laminin Aug 17 '15 at 15:07
  • $\begingroup$ 5) I tried to rationalize the mechanism with a frontier molecular orbital picture. Do you agree with the choosen HOMO/LUMO pairs? I've set acetylene's LUMO or HOMO with one of the two degenerate non-bonding orbitals: imgur.com/TdMfNFa $\endgroup$ – laminin Aug 17 '15 at 15:42
  • $\begingroup$ 4) I agree that we cannot correlate the sp2 orbital with ethylene because (like you said) they are in different parts of the molecule and do not directly interact. However, I disagree that the 2 acetylene pi orbitals correlate with $\sigma_2$ and $\sigma_3$, if you did this then you're forced to correlate the carbenic sp2 with the ethylene pi orbital. As I noted above I think that the carbenic sp2 correlates with the acetylenic pi_z. 5) Yes, I agree with your HOMO-LUMO pairs. $\endgroup$ – ron Aug 17 '15 at 18:25

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