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I'm trying to draw the state correlation diagram for the reaction of cyclopent-2-enyl anion to pentadienyl anion but I don't know which symmetry element is preserved in this reaction. What could it be?

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Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion. Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about. So here's a picture of the reaction we're talking about.

enter image description here

And here is a picture of the 5 molecular orbitals for the pentadienyl system

enter image description here

Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons. That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO). The HOMO is so important because it is the orbital that will control our process. If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other. If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap. The type of rotation that leads to bonding overlap in this example is termed "disrotatory". It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen. This is the symmetry element that is preserved during the ring closure. The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.

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    $\begingroup$ I would suggest that you phrase your answer in terms of overall symmetry of the state wave function (capital \Psi) rather than the individual molecular orbitals (lowercase \psi). Also, upon change in geometry, the molecular orbitals themselves also change, and so the frozen orbital approximation may not be appropriate. $\endgroup$ – Eric Brown May 24 '14 at 23:39
  • $\begingroup$ Thank you for the suggestion, but orbital symmetry, Frontier Molecular Orbital theory, the Woodward-Hoffman rules - whatever name you choose - focuses upon the transformation of the HOMO during the process under consideration, so focusing only upon $\ce{\Psi_3}$ seems appropriate. You are correct that the orbitals transform during the process. In this case the 3 MOs of the allyl system and the 2 MOs of the newly formed sigma bond are created. For brevity I left out the MO's of the final product. Although it is nice to view them, they are not necessary to explain the transformation when $\endgroup$ – ron May 24 '14 at 23:53
  • $\begingroup$ you start from the pentadienyl side. Just google "Woodward-Hoffmann rules" for more detail. $\endgroup$ – ron May 24 '14 at 23:54
  • $\begingroup$ Maybe you should have done the same. A quick look through "Woodward-Hoffman Rules" on Wikipedia reveals: "However, as reactions do not take place between disjointed molecular orbitals, but electronic states, the final analysis involves state correlation diagrams." That is state (capital Psi) and not molecular orbitals (lowercase psi). Your answer does not correct symbology; and given that there are MO considerations and state symmetry considerations, this could be confusing. $\endgroup$ – Eric Brown May 25 '14 at 1:18
  • $\begingroup$ Please post an answer so that I can better understand your comment. As to correlation diagrams, in MO theory they show how each of the pentadienyl MOs transform (correlate) into their respective cyclopentenyl MOs. Again it would better help me understand your comments if you posted more detail. $\endgroup$ – ron May 25 '14 at 1:27
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The allyl anion is delocalized and planar. You have a mirror plane through its middle carbon normal to the plane of the ring, or a ${C_2}$ axis through that carbon depending on how the other two carbons are canted.

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