Explaining stereochemistry in a sigmatropic 1,3-alkyl migration

I have quite a bit of difficulty in explaining sigmatropic rearrangement reactions. Consider the following reaction:

How do I explain the formation of the shown product? When simply moving the electron pair in my head, I come to the other enantiomer, where the $\ce{-CH3}$ group shows in our direction.

I've repeated all my notes on sigmatropic rearrangements but unfortunately, I don't get it. I suppose that I should treat this like an alkyl migration. The latter should happen over a suprafacial intermediate state as we have an uneven number of electron pairs (1) and we have thermal conditions. This results in a suprafacial reaction. Knowing this, I went on and added the HOMO and LUMO orbitals and postulated intermediate steps. Note: Yes, I am aware, that pericyclic reactions happen in a concerted fashion, but somehow I need to see, why the groups on the bridging carbon have this conformation.

So, let's go on to my question...

Is my proposed explanation okay? If not, please provide some information. I really have trouble with this kind of reactions.

• Both enantiomers are produced in equal amounts. If you break the bond on the "other" side of the methylated carbon you produce the enantiomer. – ron Jan 29 '18 at 20:05

Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).

The analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of "like" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.

You don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.

Problem #2 is that you tried to connect the large unshaded lobe of the σ* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.

Problem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the reaction being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.

[Note that this has a different meaning from the suprafacial or antarafacial components in a Woodward–Hoffmann analysis.]

So what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ* LUMO.

Now one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ* can overlap with the shaded lobe of the π* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.

If you now "join" the shaded lobes then you can see how the correct stereochemistry arises.

In general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial components, and the number of $4r$ antarafacial components. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.

I am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew:

• Thank you for this very helpful answer! Your sketches did really help to understand this. One little question: How do you count the supra- and antarafacial components? I don't get it. My textbook says, that a [1,j] supra-supra sigmatropic rearrangement is thermally allowed if j + 1 = 4n + 2. In this case j should be 3, so we have 3 + 1 = 4 = 4n + 2. So n would be 1/2... – Sam Jan 29 '18 at 20:50
• You mean for the WH analysis at the bottom? – orthocresol Jan 29 '18 at 20:53
• Yes, exactly. I can't combine this with the definition I've read. – Sam Jan 29 '18 at 21:09
• I'll say that the way you and I have been taught pericyclics seems to be different. Anyway, I'll try to make some sense of it. The first WH analysis has two suprafacial components, which is probably what your book means by suprafacial-suprafacial. In this case, suprafacial component means that the dotted lines I draw go to the same "side". For the π bond the meaning of this is immediately obvious, for the σ less so: here since both lines to the σ bond go to the larger "inside" lobe, the σ bond is a suprafacial component. Now you see that for $j = 3$ this suprafacial+suprafacial – orthocresol Jan 29 '18 at 21:13
• component is thermally forbidden by the WH rules, just as your textbook predicted: since 4 cannot be expressed by $4n+2$ with $n$ an integer, it is not thermally allowed. Instead the thermally allowed version has two components, one suprafacial and one antarafacial. – orthocresol Jan 29 '18 at 21:14