# Orbital and state correlation diagram for ring-closure of pentadienyl to a cyclic allyl

This is an exam question. I am to construct the orbital and state correlation diagram for one of the reaction below (disrotation or conratation). Also, I should explain whether the reaction is thermally or photochemically allowed. I know the procedure to these kinds of problems:

1. Draw (Hückel) π orbitals of the reactant
2. Draw relevant orbitals of the product. Here: π and newly formed σ orbitals
3. Identify the symmetry element that is preserved through the reaction (common for reactant and product)
4. Assign symmetry labels to orbitals
5. Correlate orbitals of same symmetry

I know what the molecular orbitals of the reactant looks like, but I am unsure about those of the product. Here's what I have Using the vertical reflection plane perpendicular the screen, I get the MO's of both reactant and product (in the order from bottom to top) to have the following symmetries: S, A, S, A, S. However, the orbital correlation diagram will be very boring if this is correct, and the reaction will be allowed both thermally and photochemically.

So are the product orbitals correct? I just assumed that an in-phase and out-of-phase combination of the π orbitals on C1 and C5 lead to the desired σ orbitals. Or perhaps my ordering is incorrect?

Final orbital correlation diagram: Concerning the state correlation diagram, I look at the electron configurations of the states of interest, which I take to be $S_0$, $S_1$, and $S_2$. I find the symmetry of each state by multiplying the symmetry of each electron in every MO. However, should I just draw straight lines between each state? The corresponding states are of the same symmetry, so no avoided crossings occur. I don't see whether the reaction is thermally or photochemically allowed from the state correlation diagram. Or did I do something wrong?

Electron configurations: Non-finished state correlation diagram: • Your assigned symmetries are correct, however the second cyclopentyl MO from the bottom should be the highest in energy since it is an antibonding sigma orbital. You've drawn it like a pi orbital, but is a sigma orbital. – ron Mar 8 '16 at 19:39
• Obviously, that was a stupid mistake! Thank you for pointing it out. I am confused about the disroratory and conrotatory pathways. As for which of the pathways that are photochemically allowed, it is fairly easy to see that only the disrotatory pathway is possible in the ground state; that's the only way to gain a bonding $\sigma$ orbital. Thus, the conrotatory ring-closure is photochemically allowed, which we also see from the LUMO orbitals. Is that an acceptable assessment, would you think? – Yoda Mar 8 '16 at 19:51
• I have been thinking about this, and the question specifies clearly the pentadiene anion. Hence, there are five electrons occupying the Hückel orbitals, so my figure is incorrect. Fixing this, I find that the $S_0$ state of the reactants becomes the $S_1$ state of the product, and vice versa, indicating that the photoinduced reaction is allowed, while the thermal reaction is not. – Yoda Mar 10 '16 at 18:57
• Also, what is the $S_1$ occupations? Is the HOMO excited to the LUMO, or does the HOMO become fully occupied, and the HOMO-1 becomes singly occupied? Both possibilities are a singlet state. And then, what about the $S_2$ state - which transition leads from $S_1$ to $S_2$? – Yoda Mar 10 '16 at 18:59