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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: subro on July 03, 2015, 02:42:43 PM

Title: Help with epoxide opening with MgBr2
Post by: subro on July 03, 2015, 02:42:43 PM
I have to propose a proper mechanism for this reaction

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This is my attempt. I'm not sure if it's right, so i prefer to ask for your wisdom. What do you think?

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Title: Re: Help with epoxide opening with MgBr2
Post by: orgopete on July 03, 2015, 07:41:29 PM
Hint: MgBr2 is a Lewis acid and the intermediate can be written as a carbocation.
Title: Re: Help with epoxide opening with MgBr2
Post by: subro on July 03, 2015, 08:16:14 PM
Hint: MgBr2 is a Lewis acid and the intermediate can be written as a carbocation.

Aah

And now?

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Title: Re: Help with epoxide opening with MgBr2
Post by: orgopete on July 03, 2015, 09:43:06 PM
If the bromide reacted, the product would be the bromo-alcohol. It isn't.
Title: Re: Help with epoxide opening with MgBr2
Post by: subro on July 04, 2015, 05:41:57 AM
Aah, ok. And now?

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Title: Re: Help with epoxide opening with MgBr2
Post by: orgopete on July 04, 2015, 11:01:06 AM
Either something like that or a hydride migration (pinacol-like). The oxygen of ether is probably more basic than bromide. Even though bromide has a net negative charge, it is a very weak base. Bromide cannot deprotonate hydronium ions.
Title: Re: Help with epoxide opening with MgBr2
Post by: pgk on July 04, 2015, 01:10:13 PM
Thanks to Orgopete's suggestions, the last mechanism is correct. Except that the carbocation is a stronger elctrophil than proton cation and therefore, deprotonation can occur without bromide attack.
Title: Re: Help with epoxide opening with MgBr2
Post by: subro on July 04, 2015, 02:53:08 PM
Thanks to Orgopete's suggestions, the last mechanism is correct. Except that the carbocation is a stronger elctrophil than proton cation and therefore, deprotonation can occur without bromide attack.

So, what is the proper mechanism? I don't get how can the H be eliminated without using the bromide ion. Maybe the ether's oxygen abstracts them? Or does the carbocation migrate to the carbonyl carbon?
Title: Re: Help with epoxide opening with MgBr2
Post by: pgk on July 04, 2015, 03:18:28 PM
H-C-C(+)-O-Mg(-)Br2 →  H(+)  +  C=C-O-Mg(-)Br2  →  C=C-OH + MgBr2  →  H-C-C=O  + MgBr2
Is it more clear, now?
What was missing before, is that oxygen is a Bronsted (Lewis) base and MgBr2 is a Bronsted (Lewis) acid and thus, a negative charge must be drawn on the Mg atom of the -O-MgBr2 adduct. There is no leaving of a bromide from MgBr2.
-O- + MgBr2 → =O(+)-Mg(-)Br2
Title: Re: Help with epoxide opening with MgBr2
Post by: orgopete on July 04, 2015, 04:51:32 PM
Thanks to Orgopete's suggestions, the last mechanism is correct. Except that the carbocation is a stronger elctrophil than proton cation and therefore, deprotonation can occur without bromide attack.

So, what is the proper mechanism? I don't get how can the H be eliminated without using the bromide ion. Maybe the ether's oxygen abstracts them? Or does the carbocation migrate to the carbonyl carbon?

In order to complete the mechanism, I wanted a hydride migration, similar to a pinacol rearrangement. No elimination.

If we want to be technical about this problem, look at the following:
http://pubs.acs.org/doi/abs/10.1021/ja01580a046
Title: Re: Help with epoxide opening with MgBr2
Post by: pgk on July 05, 2015, 03:56:51 AM
Not a migration but a hydrogen cation elimination, followed by double bond formation and enol –ketone isomerization.
Title: Re: Help with epoxide opening with MgBr2
Post by: subro on July 05, 2015, 05:46:09 AM
Thanks to Orgopete's suggestions, the last mechanism is correct. Except that the carbocation is a stronger elctrophil than proton cation and therefore, deprotonation can occur without bromide attack.

So, what is the proper mechanism? I don't get how can the H be eliminated without using the bromide ion. Maybe the ether's oxygen abstracts them? Or does the carbocation migrate to the carbonyl carbon?

In order to complete the mechanism, I wanted a hydride migration, similar to a pinacol rearrangement. No elimination.

If we want to be technical about this problem, look at the following:
http://pubs.acs.org/doi/abs/10.1021/ja01580a046

I get it now. Many thanks to both of you ^^