Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: ILoveISO on March 04, 2011, 05:23:10 PM
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How would this work?
BrMgCH2CH2CH2CH2MgBr, THF
------------------------>
H, H2O
How does a grignard reaction work when there is two BrMg's present? The compound I start with is a 5 membered ring with a O at one of the points, carbonyl oxygen, and 2 methyl groups at two different carbon corners.
Anyone got any idea? Thanks
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Also once the grignard opens up this lactone, how does the ring reform cause that is what the product ends up to be so I assume it had to have opened up then reformed...
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Draw or explain properly. Preferably both.
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Okay basically it is a Lactone with 2 methyl groups at the top right and bottom left of the ring. (This is a 5 membered ring) Now it undergoes this reaction
BrMgCH2CH2CH2CH2MgBr, THF
----------------------------->
H+, H2O
My question is how does a grignard reagent work when there is TWO MgBr's in there? Where does it attack? The lactone will break apart but how will the ring reform?
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Okay basically it is a Lactone with 2 methyl groups at the top right and bottom left of the ring. (This is a 5 membered ring)
This is exactly why we have nomenclature. Do you mean 3,5-dimethyldihydrofuran-2(3H)-one?
The lactone will break apart but how will the ring reform?
Have you been given the product's structure? What makes you think the "the ring" will reform?
Here's a question for you - if you treat ethyl acetate with two equivalents of MeMgBr, what do you get?