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Hi so right now I'm learning about the Cope and Claisen rearrangements, and often something is shown where the antibonding pi orbitals of the two alkenes have to overlap in a certain way, like in the image below:

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I guess I was wondering if a different molecule of 1,5-hexadiene could exist where only one of the the antibonding pi orbitals was flipped, for example if the signs of the lobes were changed in only the antibonding orbital corresponding to only one of the alkenes. If we flattened out the molecule in the image above into a planar hexagon, it would look like the right option in the image below, if we only considered the frontmost lobes of the two antibonding pi orbitals (couldn't find a better image). However, if we changed the - parts of the lobes to + and vise versa for only one of the antibonding orbitals, the molecule would look like the left option in the image below. Text

Since the pi bonds are rigid, there doesn't seem to be a way to flip only one of them upside down and get back the same octagon shape, and I'd assume it would take a lot of energy to "twist" one of the alkenes and briefly break its bonds such that its pi antibonding orbital was flipped. Therefore, would that mean that if I had a lot of 1,5-hexadiene, would I technically have two versions of it, one where the edge most lobes of the two pi antibonding orbitals aligned and another where they didn't? Or would one of these versions be more predominant than the other due to some sort of interaction between the two alkenes? Unlike a conjugated system there are two sp3 carbons in between the alkenes in 1,5-hexadiene, so I assumed the two double bonds' orbitals would be independent from each other.

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    $\begingroup$ There is no "if". Ask yourself a simpler question: what happens in ethylene? If the two $p$ orbitals are aligned the right way, they form the bonding $\pi$ orbital, and otherwise the antibonding one. Does that mean we have two kinds of ethylene? Absolutely not. How so? $\endgroup$ Apr 20 at 8:01

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