E1 vs E2: Comparing the E1 and E2 Reactions
Last updated: January 16th, 2020 |
E1 versus E2 : Comparing The E1 and E2 Reactions
Table of Contents
- Comparing The Mechanism Of The E1 and E2 Reactions
- What Do The E1 and E2 Reactions Have In Common?
- How Are The E1 and E2 Reactions Different?
- E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not?
- The Key Requirements Of Stereochemistry In The E2 Reaction
Here’s how each of them work:
Here’s what each of these two reactions has in common:
- in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond
- in both reactions, a species acts as a base to remove a proton, forming the new π bond
- both reactions follow Zaitsev’s rule (where possible)
- both reactions are favored by heat.
Now, let’s also look at how these two mechanisms are different. Let’s look at this handy dandy chart:
The rate of the E1 reaction depends only on the substrate, since the rate limiting step is the formation of a carbocation. Hence, the more stable that carbocation is, the faster the reaction will be. Forming the carbocation is the “slow step”; a strong base is not required to form the alkene, since there is no leaving group that will need to be displaced (more on that in a second). Finally there is no requirement for the stereochemistry of the starting material; the hydrogen can be at any orientation to the leaving group in the starting material [although we’ll see in a sec that we do require that the C-H bond be able to rotate so that it’s in the same plane as the empty p orbital on the carbocation when the new π bond is formed].
The rate of the E2 reaction depends on both substrate and base, since the rate-determining step is bimolecular (concerted). A strong base is generally required, one that will allow for displacement of a polar leaving group. The stereochemistry of the hydrogen to be removed must be anti to that of the leaving group; the pair of electrons from the breaking C-H bond donate into the antibonding orbital of the C-(leaving group) bond, leading to its loss as a leaving group.
4. E1 vs E2: Why Does One Elimination Give The “Zaitsev” Product, And The Other Elimination Does Not?
Now we’re in a position to answer a puzzle that came up when we first looked at elimination reactions. Remember this reaction – where one elimination gave the “Zaitsev” product, whereas the other one did not. Can you see why now?
So what’s going on here?
- The first case is an E2 reaction. The leaving group must be anti to the hydrogen that is removed.
- The second case is an E1 reaction.
- In our cyclohexane ring here, the hydrogen has to be axial. That’s the only way we can form a π bond between these two carbons; we need the p orbital of the carbocation to line up with the pair of electrons from the C-H bond that we’re breaking in the deprotonation step. We can always do a ring flip to make this H axial, so we can form the Zaitsev product.
- Here’s that deprotonation step:
As you can see, cyclohexane rings can cause some interesting complications with elimination reactions! In the next post we’ll take a detour and talk specifically about E2 reactions in cyclohexane rings.