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Elimination Reactions

By James Ashenhurst

Elimination (E1) Reactions With Rearrangements

Last updated: August 27th, 2019 |

Elimination Reactions (E1) That Occur With Hydride Or Alkyl Shifts

One last (weird) reaction to show you with respect to elimination reactions. Can you see what’s weird about it?

How did that double bond get over there? Normally when elimination occurs, we remove a hydrogen from the carbon adjacent to the leaving group. But here, something extra has taken place.

Elimination (E1) With Hydride Shift

Let’s look at all the bonds that form and the bonds that break so we can track down exactly what happens:

Notice how it differs from a typical elimination reaction? Sure, we’re forming C-C (π), and breaking C-H and C-OH, but we have an extra C-H that forms and an extra C-H that breaks.

This is a sure sign of a rearrangement step!

So what’s going on here?

Well, we start by protonating the alcohol. This allows for water to leave in the next step, which is going to form a carbocation. Here’s the thing: the carbocation is secondary, and we’re adjacent to a tertiary carbon. So if the hydrogen (and its pair of electrons) were to migrate from C3 in our example to C-2, we’d now have a tertiary carbocation, which is more stable. Then, a base (water in this example) could remove C-H, forming the more substituted alkene (the Zaitsev product in this case). And that’s how the alkene ends up there.

OK. So that’s one mystery solved.

Elimination (E1) With Alkyl Shift

You might remember that these types of rearrangements can occur in SN1 reactions too. And if you read that post, you might recall that in addition to shifts of hydrogen (“hydride”, because there’s a pair of electrons attached) we can also have alkyl shifts. Here’s a final example. Note – I’ve also made a video of this, you can watch it here.

This pretty much does it for elimination reactions.

In the next series of posts, let’s go though one of the biggest questions students struggle with. Okay, now that we’ve gone through substitution and elimination reactions, HOW DO WE DECIDE WHICH ONE IS GOING TO OCCUR IN EACH SITUATION?

Great question. That’s next.

Next Series, post 1: SN1/SN2/E1/E2 Decision (1) – The Substrate


(Advanced) References and Further Reading

  1. Check this paper out for some very clean, classic examples of dehydration with alkyl shift. The authors take 1-cyclohexyl-1-methylethanol and treat it with either TsOH/benzene or BF3•OEt2. You might think that they’d get the tetrasubstituted olefin, but the dominant product is the trisubstituted alkene (90:10). Reason is greater acidity of the axial C-H bonds which are aligned with the intermediate carbocation.
    BF3·OEt2 Promotes Fast, Mild, Clean and Regioselective Dehydration of Tertiary Alcohols.

    Posner, G. H.; Shulman-Roskes, E. M.; Oh, C. H.; Carry, J.-C.; Green, J. V.; Clark, A. B.; Dai, H.; Anjeh, T. E. N.
    Tetrahedron Lett. 1991, 32 (45), 6489–6492.
    DOI: 10.1016/0040-4039(91)80200-P

  2. Lanostane to Cucurbitane Transformations.
    Edwards, O. E.; Kolt, R. J. .
    Can. J. Chem. 1987, 65 (3), 595–612.
    DOI: 10.1139/v87-104
    The authors take a very rigid system (the steroid lanostane) containing a tertiary alcohol and observe what happens when it is dehydrated with strong acid (H2SO4 – AcOH – Ac2O, so-called, “Westphalen conditions”). After loss of water, a methyl shift from the adjacent quaternary carbon is observed (NOT a hydride shift, interestingly!) and the authors compare the ratio of alkenes (trisubstituted vs tetrasubstituted). Ratios are greatly affected by subtle electronic effects of remote groups.
  3. A Mild One-Pot Method for Conversion of Various Steroidal Secondary Alcohols into the Corresponding Olefins.
    Kumar, R. R.; Haveli, S. D.; Kagan, H. B.
    Synlett 2011, 2011 (12), 1709–1712.
    DOI: 10.1055/s-0030-1260803
    Slightly different steroid system, giving mixture of rearrangement + elimination products.

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Comments

Comment section

2 thoughts on “Elimination (E1) Reactions With Rearrangements

  1. Hi , thanks for your great post, i am 16 but i love chemistry and i am going to paticipate in IChO.but a question, if both alkyl and hydrid shift is possible, what will happen?

    1. Hydride shifts, generally, will perform 1,2 shifts much faster than alkyl shifts. First, think about the driving force for a 1,2 shift. The driving force is formation of a less substituted carbocation. If you have a secondary carbocation adjacent to a tertiary carbon, and the tertiary C-H migrates, you obtain a (more stable) tertiary carbocation. However if an alkyl group migrates, you obtain an (equally stable) secondary carbocation.

      Exceptions can occur. In the lab, there are cases where 1,2- alkyl shifts will happen preferentially, especially in rigid cyclic systems. In rigid, cyclic systems containing a free carbocation the groups likeliest to migrate are those which have their bonds aligned with the empty p-orbital; axial groups, in other words. See the references at the end of the post for some examples.

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