E1 vs E2 Mechanisms: Understanding Catalysts

[MathewsMD]

Hi,

I'm just reviewing E1 and E2 mechanisms and I had one question regarding the catalysts. I've attached the image for the 2 reactions I have a question about.

I understand E1 mechanisms are more common for reactions with tertiary carbocation intermediates while E2 is more common for primary, and that E1 or E2 would occur for secondary (though it would depend on the intermediate stabilities, correct?). In the image posted, there are 2 reactions (E1 and E2) with the same original reagent. I just don't quite understand why OEt-, the catalyst for the E2 mechanism, cannot be used for the E1 mechanism (i.e. why is water used as opposed to OEt-?). They both seem to be LB, though OEt- stronger, and I would presume OEt- can be used as a catalyst for the E1 mechanism, but it appears not. Is there any particular reason why using OEt- (or any other molecule like OCH3-) does not lead to the E1 mechanism but instead to E2 only? Does it have to do with water being a smaller molecule and allowing the reaction to not be stereoselective? Why cannot the stronger LB like OEt- lead to the E1 mechanism?

 

[TeethWhitener]

Take a look at your final (not alkene) products: for E2, you get neutral ethanol, whereas for E1, you get H₃O⁺, a charged species. The answer to your question has to do with the acidity of the alpha proton (the hydrogen on the carbon next to the leaving group). In an E2 reaction, a Lewis base abstracts the alpha proton and the leaving group leaves in a concerted manner ("concerted" just means "at the same time"). Since the alpha proton isn't particularly acidic in this case, you need a pretty strong base to pull it off. In the case of an E1 mechanism, though, the leaving group is abstracted first, and what you're left with is a hydrogen which is alpha to a carbocation. In this case, the carbocation is highly electron withdrawing (as a result of its being positively charged), and the alpha proton becomes extremely acidic. Therefore, the base you need to pull off the remaining proton doesn't have to be nearly as strong (in this case, H₂O is a much weaker base than OEt^-).

 

[MathewsMD]

Take a look at your final (not alkene) products: for E2, you get neutral ethanol, whereas for E1, you get H₃O⁺, a charged species. The answer to your question has to do with the acidity of the alpha proton (the hydrogen on the carbon next to the leaving group). In an E2 reaction, a Lewis base abstracts the alpha proton and the leaving group leaves in a concerted manner ("concerted" just means "at the same time"). Since the alpha proton isn't particularly acidic in this case, you need a pretty strong base to pull it off. In the case of an E1 mechanism, though, the leaving group is abstracted first, and what you're left with is a hydrogen which is alpha to a carbocation. In this case, the carbocation is highly electron withdrawing (as a result of its being positively charged), and the alpha proton becomes extremely acidic. Therefore, the base you need to pull off the remaining proton doesn't have to be nearly as strong (in this case, H₂O is a much weaker base than OEt^-).

Yes. Thank you. Just to clarify: the base you need to pull of the remaining proton therefore doesn't have to be nearly as strong, but can't it still be strong and still allow the mechanism to proceed? Why can't EtO- be used for both reactions? (i.e. for the E2 mechanism, it uses only EtO- and has only one product, but why can't both products, like formed in the E1 pathway, be made if you already have a strong enough base?)

 

[TeethWhitener]

You can use a stronger base. You just don't have to. But I have to caution you that this isn't always true. Think about acid-catalyzed dehydration proceeding through an E1 mechanism. In this case, the first step is to protonate a hydroxyl to -OH₂⁺, which then eliminates as water, at which point a weak base (usually water in this case) can remove the remaining alpha proton. If you try to use a strong base in a reaction like this, you'll just end up deprotonating the protonated hydroxyl -OH₂⁺, turning it back into an -OH group. In this case, the use of a strong base won't push the reaction forward.