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While studying the reasons behind the stereospecificity of [3,3] Sigmatropic reactions, I ran into an image portraying both possible transition states for the reaction: The Chair TS and Boat TS. The complete FMO orbitals were only drawn for the Chair TS and the final geometries of the products was shown for both transition states:

Incomplete FMO Analysis

I find it a little bit hard sometimes to visualize the three dimensional transitions that result in a specific configuration, so I tried to complete the FMO analysis with the orbitals for the Boat TS and to draw the sigma-bond rotations necessary to create constructive overlap between the lobes. I came up with the following:

Complete FMO Analysis

On the Chair TS we see the σ* shaded lobe is perpendicular to the shaded lobe of the π HOMO from the double bond next to it (having a 3D model would make it easier to see because in the drawing they seem aligned, but they are actually perpendicular). The sigma-bond would then have to rotate 90º inward for effective constructive interaction, "throwing" the R-Group outwards and resulting in the E-configuration seem in the product. The other σ* lobe also rotates 90º in the same direction (I forgot to draw the arrow there, but it would be the same as the one beside the shaded lobe), but for some reason this rotation result in the regular "cyclic-like" geometry for that double bond (Z-configuration?).

If it wasn't hard enough to visualize it on the Chair TS, it gets worse on the Boat TS. I came up with the orbital configuration as seem in the image. First, I got a little confused about the σ* shading, because the whole premise for the shading had to do with constructive overlap where the double bonds would occur. But there I had to draw opposite phases on the lower part. That, I believe, is because initially they were aligned with the same phases (as in the Chair TS), but the conformation twisting threw the constructive phases in opposite directions. Anyways, to reach constructive overlap the sigma-bond next to the substituted carbon would have to rotate close to 90º counter-clockwise, while the other sigma-bond would have to rotate almost 180º. That explains why this transition state is so unfavorable, but I'm still struggling a bit with the geometry. The 180º rotation sounds like it would result in the unsubstituted double bond pointing out of the "cyclic-like" shape as seem in the product. But I can't clearly see how the 90º rotation throws the R-Group inward in a Z-configuration geometry. Instead, I came up with another orbital configuration possible that might be the correct one:

BoatTS

In that orbital configuration the sigma-bond of the substituted carbon wouldn't have to rotate, and we would have the R-Group pointing inward like in the product.

Which orbital configuration is correct, if any? Is my reasoning about the sigma-bond rotations and the resulting products geometry correct? Do you have any tips when it comes to visualizing geometrical issues on more complicated molecules?

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    $\begingroup$ Remember that the p-orbitals of the double bonds are orthogonal to the plane of the carbon framework. The Cope rearrangement you show will be an equilibrium mixture of the E-isomer (major) and the starting material (minor) because the activation energy for the chair transition state is lower than for the boat. At more elevated temperature some "leakage" to the Z-isomer may be seen because the three dienes are in equilibrium with each other. Your best bet is to make a molecular model and traslate what you see to paper. $\endgroup$ – user55119 Jun 6 at 22:18
  • $\begingroup$ I see, it's just a little harder to visualize where the plane of the carbon framework is on those conformations, because the molecule as a whole doesn't have a single plane I guess. That makes me think the second Boat TS orbitals drawing is a little closer to reality than the first. Also makes me think the HOMO orbitals (blue) point outside of the "paper" plane, like the lobes where coming towards us. $\endgroup$ – IanC Jun 7 at 11:48
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For chair-like transition states the angle at which you draw your chair can make a significant difference to how easy it is to visualise stereoelectronic interactions. Sadly, there is no real way to know which is the best way to draw it, except by either trial and error, or prior experience (or leveraging on other people's experience, i.e. copying it from somebody else!).

If you rotate the chair a bit so that the forming bond is in front (as opposed to at the side), then you can have something like this:

Chair TS for Cope rearrangement leading to (E)-alkene

You don't need to overthink how the sigma bond rotates; this is a sigmatropic rearrangement, not a electrocyclic reaction where there are conrotatory/disrotatory modes. The geometry at the substituted carbon definitely changes (from tetrahedral to trigonal planar), so there is some kind of "flattening" as the reaction proceeds, but all you really need to pay attention to is the relative disposition of the R substituent and the rest of the carbon chain: here they are clearly trans to one another, which is an (E)-alkene. This is stereospecific, so if you start with the conformation shown in the left, you will end up with a trans alkene. There isn't a different "rotation mode" that leads to the cis alkene.

The main way of getting a cis, or (Z)-alkene, is not actually via a boat TS, but rather via a chair TS with the R substituent axial instead of equatorial. Note that this is a different conformation from the previous image, so again it's not about which direction the sigma bond rotates in, but rather the original location of the substituent (axial/equatorial). The orbital diagram is exactly the same, so there's no need to redraw it, but it should be easy enough to see that this leads to the reverse stereochemistry.

The boat TS is trickier to draw. This is my preferred representation:

Boat TSs

but I am not entirely sure which TS is higher in energy (i.e. whether the substituent prefers to point up or point down), even after consulting several references. I don't think it's a matter of great significance; after all, the boat TS usually only plays a very minor role, as far as I know.

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  • $\begingroup$ Thanks, things are a bit clearer with your answer. I was wondering, considering the stereospecificity of the reaction, can I conclude that only the equatorial form of the Chair TS exists for that specific reactant (with a dashed bond to the R substituent)? Analogously, the Boat TS for that specific reactant would always be axial? $\endgroup$ – IanC Jul 10 at 22:51
  • $\begingroup$ No, you can always redraw it such that it is axial in a chair TS. You need to flip the entire structure to do that. $\endgroup$ – orthocresol Jul 11 at 4:19
  • $\begingroup$ Sorry, I think I'm still a bit confused, because the course material I have states the Chair TS leads to a trans product (stereospecificity). While the Boat TS would lead to a cis product, though this TS isn't favored due to the lack of constructive orbital overlap. If the substituent could be either in the axial or equatorial position wouldn't the final product be a mix of cis/trans conformations? Or maybe the equatorial TS is more stable and the course material is just not showing the axial TS because it's more unstable and thus unlikely to happen? $\endgroup$ – IanC Jul 11 at 23:48
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    $\begingroup$ “The chair TS leads to the trans product” Not exactly; the chair TS with the substituent equatorial leads to the trans product. “Wouldn’t the final product be a mix of cis/trans”? Yes. “Not showing... because it’s more unstable” It is a minor product, but it will be formed in nonzero quantities. I also don’t entirely agree with the use of the word “stereospecific”, but I’ll leave it for now. $\endgroup$ – orthocresol Jul 12 at 4:10
  • $\begingroup$ Thanks, I took a while to answer so I could research more. On the video explanation of the analysis there’s less ambiguity than the text I guess. It states clearly that the chair conformation can have the substituent in either position, but the equatorial is more stable one for the substituent resulting in the preferred isomer. I’ll look further on the subject but you made things much clearer! $\endgroup$ – IanC Jul 14 at 14:32

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