1
$\begingroup$

No nodes should be introduced in the rotational interconversion between the s-trans and s-cis forms of 1,3-butadiene.

What is the explanation for that, if one of the $\pi$-systems should rotate $180^\circ$? It seems to be that two positive and negative lobes turn upside down.

$\endgroup$
1
$\begingroup$

During the interconversion of s-cis and s-trans 1,3-butadiene, the molecule passes through a geometry where the two ethylenic arms are rotated 90 degrees one from the other. In this conformation the two ethylenic arms are orthogonal to each other and there is no overlap between them. In an MO sense they can be treated as two separate ethylenes. Rotation beyond this conformation allows overlap to reoccur and the phase of the HOMO will be correctly reestablished with only one node occurring between C2 and C3.

$\endgroup$
  • $\begingroup$ Thanks Ron for the explanation. Since s-trans is lower in energy and s-cis higher, I thought of an excited state to LUMO with one node between C2 and C3 in the transition state (90°), and then upon full rotation to s-cis, the node automatically disappears and hence fully symmetrical. How do you think is that realistic? $\endgroup$ – user40014 Feb 8 '14 at 21:27
  • $\begingroup$ In the 90 degree transition state, the two ethylenic branches are orthogonal to each other, there is no overlap between them at that point. Further rotation restores overlap and the 4 orbital HOMO is restored. The butadiene LUMO is not involved in the process. $\endgroup$ – ron Feb 8 '14 at 22:14

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.