5
$\begingroup$

I'd like to investigate into the ring opening, the migratory reverse reaction of the electrocyclic ring-closure of Dimethyldioxirane:

enter image description here

As a first question I'd like to ask if I have drawn the through bond interaction correctly?

enter image description here

Following scheme is wrong, because the p orbitals are perpendicular to the O-C-O as @ron clearly pointed out in his answer in Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?

enter image description here

Next I tried to draw the orbital correlation diagram. I guess for this correlation diagram I don't have to draw the through bond interaction, yes?

enter image description here

  • In the dioxo biradical product I've drawn a smaller energy gap between $\pi$ and $\pi^*$ orbitals as for $\sigma$ and $\sigma^*$. I guess this should be correct.
  • Because there is an additional through bond interaction for the product I've drawn a small exothermicity. I see no activation energy for the disrotatory ring opening. So the disrotatory ring opening must happens very fast (without a lot of thermal energy, may be even at room temperature).

Finally, I tried to draw the state correlation diagram:

enter image description here

  • For the photochemical reaction there is a transition from $S_0$ to $S_2$. Because of the avoided crossings the conrotatory correlation is steeper falling down, therefore I expect most of the electrons direcetly going in the pericyclic minimum and do there Interal Conversion to the $S_1$ surface and afterwards the excited product gives off a photon going back to the $S_0$ state. I expect products in excited states are not very long living and therefore undergoes emission of a photon as Internal Conversion under fluorescence here.
  • Another way out from the pericyclic minimum can be the directly falling to the transition state of the $S_0$ state and giving the product to a good quantum yield.
  • As third option some part of the $S_2$ states can go through the disrotatory pathway and directly emitting light from $S_2$ to $S_0$. As written in the first point I think this way is less probable because of the less steeper falling down the energy surface.

Is this all correct?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

I would like to help here, but I am not sure where to begin. First of all, I want to point out that the reaction you have shown is not an electrocyclic reaction, because it does not involve the central saturated carbon atom. It is rather a simple homolysis reaction. The terms "conrotatory" and "disrotatory" are misapplied, since the two oxygen atoms do not have functional groups on them, and the two modes of ring opening are indistinguishable.

Dioxirane does have an electrocyclic ring opening reaction available to it - namely cleavage of a carbon-oxygen bond to form "acetone-O-oxide". Perhaps you should consider that as an alternative pathway. Does that help?

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes. It's a homolysis reaction. I guess the first picture with the through bond interaction ist correct. For this old exam exercise only the homolysis reaction has to be considered. I know O-O bonds are weaker than C-O bonds. In the exam it was required to draw the correlation diagram. I therefore thought, everytime I draw an orbital correlation diagram, I can also draw a state correlation diagram. The question now is, to which symmetry the reaction coordinate belongs to? $\endgroup$
    – laminin
    Jun 18, 2015 at 18:38

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.