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Can rearrangement reactions in carbocations cause inversion?

If you have a carbocation rearrangement (i.e. a methyl shift or a hydride shift), is it possible for inversion of stereo chemistry to take place?

The carbocation has the free p-lobe, so is it possible for the hydride/methyl group to bond to either lobe, such that one can cause retention of stereochemistry while the other can cause inversion of stereochemistry? Does this logic even make sense?

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  • $\begingroup$ Firstly, I think you have to clarify some stuff. 1) Which carbon are you looking at the configuration of? The carbon undergoing migration, or something else? 2) By inversion, do you only mean a change in the Cahn-Ingold-Prelog descriptor from R to S (or vice versa), or do you mean an inversion in stereochemistry at the migrating centre (such as that seen in an SN2)? Concerted 1,2-alkyl shifts always proceed with retention of configuration at the migrating center. Whether this leads to a change in the CIP descriptor depends on the substituents present. $\endgroup$
    – orthocresol
    Nov 22, 2016 at 0:46
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    $\begingroup$ [...] The phrase "inversion" in organic chemistry typically has absolutely nothing to do with the CIP descriptor. In that diagram, any and all of the compounds can be either R or S, depending on the identity of the groups A, B, C, X, Y. For example, SN2 reactions always proceed with inversion, but the CIP descriptor may or may not change: see this question. $\endgroup$
    – orthocresol
    Nov 22, 2016 at 0:50
  • $\begingroup$ Hmm thanks for the link. Uhm, so R/S designations just depend on the group acting as the nucleophile and whether it will change the priorities of groups. Thanks. $\endgroup$
    – sdx
    Nov 22, 2016 at 1:01
  • $\begingroup$ I edited the question so it is a little bit clearer now that I've remedied some of the misunderstandings that I had... $\endgroup$
    – sdx
    Nov 22, 2016 at 1:11
  • $\begingroup$ I think that the question you're asking about inversion of stereochemistry upon migration is a good one, and for that general topic I would point you to this surprisingly good (and picturesque) wikipedia page about sigmatropic migrations. I don't think what you're specifically asking works as well though because if a migration occurs due to the formation of a carbocation, that carbocation has a pure p orbital on it, and thus anything that adds there will add 50/50 to either side. You'd need to migrate into a stereocenter. $\endgroup$
    – jheindel
    Nov 22, 2016 at 3:49

2 Answers 2

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Originally this was a comment, but I figure it's good enough for an answer.

As far as I can tell, you seem to be confusing the notion of suprafacial/antarafacial with retention/inversion.

You're asking whether the migrating group can loop around to the other side of the carbocation, right? That's known as an antarafacial process, not an inversion.

enter image description here

Retention, or inversion, refer to the stereochemistry at the migrating centre. If it is a hydride shift, then the migrating centre is a hydrogen, and there is no stereochemistry to speak of. The terms retention or inversion are therefore not applicable to hydride shifts.

enter image description here

In any case, the crux of the matter is that this migration is a concerted process (it occurs in one step), and an antarafacial migration cannot occur due to geometric reasons. The lobe of the p orbital on the other side is way too far away, and the $\ce{C-R}$ $\sigma$ orbital (which contains the electrons that you are trying to migrate) is not going to be able to reach it.

enter image description here

[1,2]-alkyl (or hydride) shifts are always going to be suprafacial!

(As a side note, the suprafacial [1,2] shift has to proceed with retention, as per Woodward-Hoffmann rules.)

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  • $\begingroup$ "Originally this was a comment, but I figure it's good enough for an answer." More than an year later, do you mind deleting your original comments on the question, if they are obsolete now? Thank you. $\endgroup$
    – Gaurang Tandon
    Mar 13, 2018 at 6:20
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Supplement to o-cresol's answer

The Woodward-Hoffman rules are central to this discussion, and succinctly put concern themselves with the conservation of orbital symmetry. Conservation of orbital symmetry requires the transformation of the molecular orbitals of reactants into those of products to proceed continuously by following a reaction path along which the symmetry of these orbitals remains unchanged.

First, a note about notations. In a sigmatropic rearrangement are concisely described by an order term [i,j], which is defined as the migration of a $\sigma$-bond adjacent to one or more $\pi$ systems to a new position (i−1) and (j−1) atoms removed from the original location of the $\sigma$-bond. When the sum of i and j is an even number, this is an indication of the involvement of a neutral, all C atom chain.

Now, given below is a visual summary of how a suprafarcial and an antarafarcial process takes place.

visual summary

Now, let's look at it the Wagner-Meerwein rearrangement i.e a 1,2-migration of an alkyl or aryl residue to a carbocation center. It represents the most simple and smallest system in which a sigmatropic rearrangement can take place.

reaction [1,2] alkyl rearrangment

Now if this reaction is to take place with stereochemical retention, the migrating group has to translate, (without rotation), into the new bonding position. The image given below illustrates this process.

enter image description here

Now turns out conservation of orbital symmetry in the transition state, demands that the process take place this way.

Imagine, if the migrating group were to rotate, we will have inversion (however, we can't form a new bond because the symmetry requirement is violated).

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