Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings
Last updated: November 9th, 2020 |
Antiperiplanar Relationships Between C-H And The Leaving Group: The E2 Reaction and Cyclohexane Rings
Here we come to a very testable application of the E2 reaction – how to draw the products of E2 reactions in cyclohexane rings!
Table of Contents
- In The E2 Reaction, The Leaving Group Is Always “Anti-Periplanar” To The Hydrogen That Is Removed On The Adjacent Carbon (“Beta-Carbon”)
- In Cyclohexane Rings, E2 Reactions Can Only Occur When The Leaving Group Is Axial
- In E2 Reactions Of Cyclohexane Rings, The Only Way The Leaving Group And C-H Bond Can Be Anti-Periplanar Is If They Are On Opposite Faces Of The Ring
- Some Examples: What Would Be The Major E2 Product In Each Case?
- Cyclohexane Substituents Can Affect The Rate Of E2 Elimination Reactions In Cyclohexane Rings
- The Higher The Concentration Of The Conformer With An Axial Leaving Group, The Faster The Elimination Will Be
- (Advanced) References and Further Reading
1. In The E2 Reaction, The Leaving Group Is Always “Anti-Periplanar” To The Hydrogen That Is Removed On The Adjacent Carbon (i.e. the “Beta-Carbon”)
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.
3. In E2 Reactions Of Cyclohexane Rings, The Only Way The Leaving Group And C-H Bond Can Be Anti-Periplanar Is If They Are On Opposite Faces Of The Ring
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 comes up in tests, a lot!] 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?
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* [CH3 is 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.
6. The Higher The Concentration Of The Conformer With An Axial Leaving Group, The Faster The Elimination Will Be
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.
Answers for the question above:
- Abspaltungsreaktionen und ihr sterischer Verlauf
Walter Hückel, Werner Tappe, Günter Legutke
Lieb. Ann. Chem. 1940, 543 (1), 191-230
An early study of E2 stereochemistry on menthyl derivatives.
- Mechanism of elimination reactions. Part XVIII. Kinetics and steric course of elimination from isomeric benzene hexachlorides
E. D. Hughes, C. K. Ingold, and R. Pasternak
J. Chem. Soc. 1953, 3832-3839
The name of this paper is a little confusing since ‘benzene hexachloride’ does not refer to hexachlorobenzene, but 1,2,3,4,5,6–hexachlorocyclohexane. Hexachlorocyclohexane has a few isomers depending on the orientation of the Cl atoms, as explained in the paper. Depending on the orientation of the Cl atoms in the isomer, different products are obtained, based on whether an E2 pathway is possible at those position(s).
- Mechanisms of Elimination Reactions. X. Deuterium Exchange in Base-Promoted Dehydrochlorination of β-Benzene Hexachloride
Stanley J. Cristol and Delbert D. Fix
Journal of the American Chemical Society 1953, 75 (11), 2647-2648
Classic study of hexachlorocyclohexane isomers – all chlorine atoms on adjacent carbon atoms are trans to each other, so it is impossible to bring adjacent hydrogen and chlorine into an anti relationship without causing ring strain.
- Mechanism of elimination reactions. Part XIX. Kinetics and steric course of elimination from isomeric menthyl chlorides
E. D. Hughes, C. K. Ingold, and J. B. Rose
J. Chem. Soc. 1953, 3839-3845
This paper illustrates Zaitsev’s rule in a cyclohexane system. Neomenthyl chloride gives 78% 3-menthene and 22% 2-menthene with EtO– in ethanol .
- Mechanism of elimination reactions. Part XXII. Anomalous elimination from the trimethylneomenthylammonium ion
E. D. Hughes and J. Wilby
J. Chem. Soc. 1960, 4094-4101
This paper perfectly illustrated the concepts explained in this section. Menthylammonium ion cannot undergo elimination to give the 3-olefin, since the anti position on C4 is occupied by the isopropyl group. Only the 2-olefin is obtained from elimination with menthyl-LG (LG = leaving group, chloride, trimethylamine, etc.). In neomenthyl-LG, the isopropyl group is not anti, and so the 3-olefin is accessible. However, the yield of the 3-olefin decreases from 88% to 65% when EtO– is used instead of OH–, and it would be likely to decrease further when using bulkier bases (e.g. tBuO–).
- Neighboring Carbon and Hydrogen. XIX. t-Butylcyclohexyl Derivatives. Quantitative Conformational Analysis
S. Winstein and N. J. Holness
Journal of the American Chemical Society 1955, 77 (21), 5562-5578
This is the classic paper by Prof. Winstein introducing the concept of A-values. There is a brief mention of the effect of axial/equatorial orientation of the leaving group on reactivity. When the leaving group is axial, an E2 pathway is favored, but when the leaving group is equatorial, a mix of E1 and SN1 takes place.