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Elimination Reactions

By James Ashenhurst

The E2 Mechanism

Last updated: March 27th, 2019

Having gone through the E1 mechanism for elimination reactions, we’ve accounted for one way in which elimination reactions can occur. However, there’s still another set of data that describes some elimination reactions that we haven’t adequately explained yet.

Here’s an example of the reaction I’m talking about:

What’s interesting about this reaction is that it doesn’t follow the same rules that we saw for the E1 reaction. We’ll talk about two key differences here.

Clue #1 – The Rate Law Is Bimolecular

Remember that the E1 reaction has a “unimolecular” rate determining step (that is, the rate only depends on the concentration of the substrate?)

Well, when we look at the rate law for this reaction, we find that it depends on two factors. It’s dependent on the concentration of both substrate and the base.

That means that whatever mechanism we propose for this reaction has to explain this data.

By the way, see how useful chemical kinetics can be? They’re such simple experiments – measure reaction rate versus concentration – and you get these nice graphs out of it.  I can’t even begin to stress how important this data can be in understanding reaction mechanisms. So simple, so elegant, and so useful.

Another note – you might notice that the base here (CH3O) is a stronger base than we see for the E1 reaction (more on that later).

Clue #2: Stereochemistry Of The C–H Bond And The Leaving Group Is Crucial

Here’s the second key piece of information – and we didn’t talk about this for the E1. The reaction below  is very dependent on the stereochemistry of the starting material. 

When we treat this alkyl halide with the strong base, CH3ONa, look at this interesting result. What’s weird about this? Well, this seems to fly in the face of Zaitsev’s rule, right? Why don’t we get the tetrasubstituted alkene here?

The mystery gets a little deeper. If, instead of starting with the alkyl halide above, we “label” it with deuterium – that is, we replace one of the hydrogens with its heavy-isotope cousin that has essentially identical chemical properties – we see this interesting pattern:

Note how the group that is on the opposite face of the cyclohexane ring to the leaving group (Br) is always broken.

In fact, if we use the molecule above and make just one modification, now we actually do get the Zaitsev product!

See what’s going on? The hydrogen that is broken is always opposite, or “anti” to the leaving group.

So how do we explain these two factors? 

The Mechanism Of The E2 Reaction

Here’s a hypothesis for how this elimination reaction works. It accounts for all the bonds that form and break, as well as the rate law, and – crucially – the stereochemistry.

In this mechanism, the base removes the proton from the alkyl halide that is oriented anti to the leaving group, and the leaving group leaves – all in one concerted step.

Since it’s an elimination reaction, and the rate law is “bimolecular”, we call this mechanism the E2.

In the next post, we’ll directly compare the E1 and E2 reactions.

Next Post: Comparing the E1 and E2 Reactions

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Comment section

14 thoughts on “The E2 Mechanism

  1. I’m not sure I see what the “one modification” was to make the molecule follow Zaitsev’s rule? The substrate still looks the same as before?

    1. In the first molecule, the wedge was referring to a methyl group, while in the second molecule (the modification), the wedge showed a hydrogen, which was available to be broken off by OCH3

  2. Oh I see now. But, how can you just pus that H downward? Is that allowed? If I had a problem like this on the test, should I just redraw the cyclohexane with the H pointing downward and making it 180 from Br?

    1. Those are actually two different molecules. In the second case, the hydrogen on the more substituted beta carbon is anti. Thus, this molecule can give a Zaitsev product.a

  3. I was just wondering: for E2 and E1 we don’t have to worry about inversion at any chiral center or a racemic mixture of products the way we would have for SN2 and SN1 reactions, right?

  4. I have a question regarding the bonds – how come the wedges and dashes become flat bonds? Is it possible that some of them could stay wedges and dashes and others could become flat – in other words, what’s the criteria for when a bond stays a wedge/dash or when it becomes planar?

    1. I assume you mean the bottom mechanism? I forgot to mention one thing – that we rotated our “view” from a “side-view” to a “top view” . When we look at the alkene from the top, it’s flat, so it shouldn’t have dashes or wedges. When we look at the alkene from the side, some groups will point towards us and some will point away. I should update this figure.

    1. @Varun this is because the transition state has an alkene-like structure. I am sure of the fact that 3>2>1, but I cannot illustrate the diagram here. You should look it up in a reference book if you still doubt it

  5. If the Morse Potential allows bond breaking, by spectroscopic stretching and rotation; could this allow the disruption of inter and intra molecular attachments?
    In turn, could selection of an appropriate bond, application of correct downfield absorption, and suitable reduced mass point, allow complex bond breaking?
    Also, would the molecular resonance be tolerable and penetrable in organisms?
    If you follow my drift?

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