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.
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!
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.