5
$\begingroup$

In Pericyclic reactions, it is shown that the terminal bonds rotate (conrotation or disrotation) to allow for allignment of the FMO's, thereby forming the sigma bond

Pericyclic bond formation

However, a detail that most of my textbooks and references miss out on is how is the double bond rotating about it's axis? As far as I know, pi bonds are restricted in their rotation which is what gives rise to cis-trans isomerism. If this is the case, how are the bonds rotating about their axis to align the orbitals?

One possibility I can think of is that the double bond is broken before/during rotation, but I haven't seen this explicitly mentioned anywhere. Any insights are appreciated.

$\endgroup$
5
  • 2
    $\begingroup$ Well, it is a double bond at the start and a single bond at the end. So as the reaction proceeds, this restricted rotation becomes less and less restricted. $\endgroup$ – orthocresol Dec 18 '20 at 10:37
  • $\begingroup$ @orthocresol that sounds plausible... Could you provide a more formal answer, preferably with some references/diagrams? I did read this related answer of yours, but I couldn't comprehend it or find anything relevant there. $\endgroup$ – Aniruddha Deb Dec 18 '20 at 12:30
  • $\begingroup$ I don't have a formal answer, sorry, but it seems intuitive to me that as electrons flow out of the old π bonds (to form the new π and σ bonds) the restrictions on rotation decrease. I would guess that it is a continuous process in which the dihedral angle smoothly goes from 0° to 90° as the reaction proceeds. One way to verify this would be computation, but not something I can do now. $\endgroup$ – orthocresol Dec 18 '20 at 12:40
  • $\begingroup$ One way to maybe think about this: if the FMO's didn't rotate, the reaction would not proceed and the two double bonds would stay rigidly as it is in the π overlap. But because we are heating up the substrate, the molecule will start becoming excited and therefore, bond breakage will be initiated. So it's more like a case of availability of higher degrees of freedom for the molecular components with increased temperature, which allows such rotatory overlaps to occur. This bond rotation of course breaks the π overlap, but it's entropically favorable as heating is going on $\endgroup$ – Yusuf Hasan Dec 18 '20 at 16:21
  • $\begingroup$ Do remember that the pi orbitals on the central carbons will be forming a pi bond as the sp2 becomes more like sp3. Personally, how I was taught is that the drawing above is incorrect. There should be no line bonds drawn if you are representing the p-orbitals. $\endgroup$ – Beerhunter Dec 18 '20 at 21:57

Your Answer

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

Browse other questions tagged or ask your own question.