The E2 Mechanism
Last updated: September 26th, 2019 |
E2 Mechanism – How The E2 (Elimination, Biomolecular) Reaction Works
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.
Table of Contents
- Example Of An “E2” Reaction: How Do We Explain What Happens In This Reaction?
- Clue #1 About The Mechanism Of The E2 Reaction: The Rate Depends on Concentration of Both Substrate and Base
- Clue #2 About The Mechanism Of The E2 Reaction: Stereochemistry Of The C–H Bond And The Leaving Group Is Always “Anti”
- Putting It Together: The Mechanism Of The E2 Reaction
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.
2. Clue #1 About The Mechanism Of The E2 Reaction: The Rate Depends on Concentration of Both Substrate and Base
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).
3. Clue #2 About The Mechanism Of The E2 Reaction: Stereochemistry Of The C–H Bond And The Leaving Group Is “Anti”
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?
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.