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The E2 Reaction and Cyclohexane Rings

Last time we compared the E1 and E2 reactions and mentioned one of the key differences was the stereochemistry of the E2 reaction. Remember that in the E2, the leaving group is always “anti” to the hydrogen that is removed on the adjacent carbon. [That means that they’re directly opposed to each other, or 180°; kind of like the minute hand and the hour hand when a clock reads 6:00].

This is an extremely important detail to be able to apply in reactions. One way this often comes up is in discussions of cyclohexane rings. If you’ll recall, in the cyclohexane chair conformation, groups can either be axial (pointing straight up or down) or equatorial (pointing “somewhat up” or “somewhat down”).

In order for a hydrogen to be “anti” to a leaving group, it’s required that both groups be axial. Look closely at the cyclohexane ring on the left, where the leaving group is equatorial – see how the group that is “anti” is the C-C bond [highlighed in red]? That E2 is never gonna work.


So if you draw the leaving group equatorial in a cyclohexane chair, you’ll have to do a chair flip so that the leaving group is axial. That’s shown in the right hand example, where an E2 can actually happen.

This brings us to the second point. If the leaving group is, let’s say, on the “top” face of the cyclohexane, you can only form an alkene to adjacent carbons where the hydrogen is on the opposite face. 
You might remember  the example from last time where we couldn’t form the “Zaitsev” alkene because the Br was a wedge and there was an alkyl group on the carbon next door that was on the opposite face. In this case we can only form the less substituted alkene. If the methyl group is switched, however, then the E2 to give the Zaitsev product becomes possible:

The bottom line here [and trust me, this is very testable!] is that you always want to pay attention to what side of the ring your leaving group is on, and make sure that the E2 you draw is indeed possible.

Here are some more examples to think about. What would be the major E2 product in each case?

Now, let’s talk about a very interesting application of what we just discussed. This is a little more advanced, but see if you can follow it through. It ties together what we’ve discussed about the E2 with what you’ve previously learned about cyclohexane chair flips.

Imagine you’ve got two alkyl halides, and they’ve got slightly different structures. We make the following observation: E2 with the second starting material is significantly faster than E2 with the first product. Question: why might this be?

 In order to understand what’s going on, it would help to draw the cyclohexane chair forms of both of these molecules. So let’s do that and then have a closer look.

What’s going on? Each molecule will have an equilibrium between two chair forms.

In the top molecule, the left-hand conformation is favored, because the bulky methyl group* [CHis actually bulkier than Br] is equatorial. So equilibrium will favor the left hand molecule.

In the bottom molecule, the rightmost conformation is favored, because the bulky methyl group is equatorial. So equilibrium will favor the right-hand molecule.

Notice something interesting? Remember that in order for E2 to occur, the leaving group must be axial. So there’s only one conformation where this will be possible for each ring. However, in the top example, Br is axial only in the least stable conformation, whereas in the bottom example, Br is axial in the most stable conformation. Since the bottom example will have a higher concentration where Br is axial, it will be faster. 

Isn’t it interesting how it all ties together? Concepts you learn in one chapter can come back and be applied in later chapters!

In the next post we’ll talk about another example where Zaitsev’s rule doesn’t apply.

P.S. Answers for the question above:

Next Post: Bulky Bases in Elimination Reactions

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