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Topic: Help with mechanism (E2/E1)  (Read 3456 times)

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Offline cseil

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Help with mechanism (E2/E1)
« on: December 29, 2014, 10:13:53 AM »
I have to propose a mechanism for the following reaction

CC(C)(Cl)C1CCC1>CO>CC=1CCCC=1C

I started writing the mechanism of the E1.
The product is:

C/C(C)=C1/CCC1

But I don't know how to increase the number of carbons in the cycle.
Could you help me please?
Thank you

Offline critzz

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Re: Help with mechanism (E2/E1)
« Reply #1 on: December 29, 2014, 11:16:27 AM »
Okay, what is the very first step of an E-1-elimination? (also why is it not an E-2-elimination?)

Offline cseil

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Re: Help with mechanism (E2/E1)
« Reply #2 on: December 29, 2014, 11:39:41 AM »
E1 is favoured by a weak base and a protic solvent.
It is the case.

The first step is the formation of the carbocation. The C-Cl bond is broken. The second step is the elimination of the proton.
In this case, the β-carbon is the tertiary carbon of the cycle.

Offline critzz

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Re: Help with mechanism (E2/E1)
« Reply #3 on: December 29, 2014, 12:10:16 PM »
Yes, you are right. Normally the tertiary carbocation would be the most stable one.
However, it has a strained cyclobutyl group next to it that could relieve some strain by rearranging.
Could you draw the initial product?

Offline cseil

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Re: Help with mechanism (E2/E1)
« Reply #4 on: December 29, 2014, 01:49:40 PM »
Yes, you are right. Normally the tertiary carbocation would be the most stable one.
However, it has a strained cyclobutyl group next to it that could relieve some strain by rearranging.
Could you draw the initial product?

Maybe I figured it out.
Is it correct?
Thank you!


Offline Altered State

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Re: Help with mechanism (E2/E1)
« Reply #5 on: December 29, 2014, 02:36:24 PM »
Seems good to me. First rearrangement is favored by the decrease on ring strain and the second one because a more stable carbocation is formed.

Offline critzz

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Re: Help with mechanism (E2/E1)
« Reply #6 on: December 29, 2014, 03:43:35 PM »
Maybe I figured it out.
Is it correct?
Thank you!

Yes, this is what I had in mind.
You're welcome!

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