The E2 Reaction and Cyclohexane Rings

by James

in Alkenes, Alkyl Halides, Organic Chemistry 1

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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{ 9 comments… read them below or add one }

mamid

2nd scheme, 2nd reaction, 3rd structure: it would be really helpful if you showed the C2-attached hydrogen, as it is directly involved in the reaction. The same for the 1st reaction probably wouldn’t hurt, either.
Otherwise excellent work, as usual.

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james

Thank you again, I very much appreciate your keen eye.

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Anum Khan

Hi James. For the 4 examples you gave us to try, I worked out the 2nd example (1-bromo-1,2-dimethylcyclohexane) and I get my major product as 1,2-dimethylcyclohex-1-ene (where the double bond is between c1-c2). You have the major product having the double bond between c1-c6. Can you PLEASE explain why that is the major product? Doesn’t Zaitsev’s rule tell us to have a MORE SUBSTITUTED alkene (my answer) versus least substituted (your answer)? I would really appreciate a response!

P.S. Your website is the reason I’m doing well in Organic and I cannot thank you enough. :)
-Anum

Reply

James Ashenhurst

For the second example, which is E2, the bromine needs to be on the opposite side of the ring from hydrogen for the elimination to occur. This isn’t possible with C2, because the hydrogen is on the SAME side of the ring as the bromine. Therefore we can’t form an alkene here – it forms between C1 and C6.

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Parakh

Hello James
Thank you for the wonderful clarification
Two questions (apologies if these sound too naive):
1. To my knowledge anti periplanar conf simply means that dihedral angle between two groups is 180 degrees, and that this is required for e2. Now why can’t this be true if both, the leaving group and alpha-hydrogen are equatorial?
2. In trying to acquire this conf so that e2 can take place, can we convert from cis to trans, and vice versa?

Reply

James

1. There’s no substitute for making a model and seeing this for yourself.
2. Ring flips can never convert cis to trans or vice versa. Ring flips only switch axial to equatorial. Again, make a model and show this to yourself.

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Tim

Excellent work James. I thought I was decent at organic chemistry but these lessons make everything even more simpler. I discovered that I still have a lot to learn.

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Alyx Vance

Dear James:
I want to ask why in the last example, why both reactions give enantiomers? For example, the cyclohexane ring with both methyl and bromine on dashed lines. After elimination, shouldn’t the methyl group stay on the dashed line? How can it be enantiomers?
thanks a lot!!

Reply

Idan

The double bond can form in both carbons adjacent to the carbon with the leaving group (beta carbons). If you draw both products and flip one over you’d see that they are enantiomers: both would have the double bond between the same carbons but the methyl group would be either a dash or a wedge.

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