SN2 reaction of organocuprates (Gilman reagents) with alkyl halides to give alkanes
Description: Alkyl halides (or tosylates) will react with organocuprates (Gilman reagents) to form alkanes in an SN2 reaction
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I was asking about (3-methyl iodide)cyclopentyl methyl ketone. Why isn’t any inversion shown on the product side? Shouldn’t it be a dash on the product?
Look closely. The C-I bond breaks, and a new C-CH3 bond forms.
The wedged group is CH2I. If this bond were to be inverted, it would mean we are forming C-CH3 and breaking C-CH2I . That would make carbon the leaving group. Not good!
Why isn’t there any inversion in the second last example?
Hi, thanks for your question. In the second-last example OTs is being replaced by CH3. In the product the carbon undergoing attack is attached to two methyl groups, so it’s not a chiral center. That’s why I didn’t show inversion.
In example 2, I believe, there is an extra carbon that shouldn’t be there. You use CH3-CH2-CH2-I and a cyclopentane directly attached to the CuLi, thus you should have 1-propyl-cyclopentane and not a 1-butyl-cyclopentane.
You are correct! I fixed the image. Thanks for spotting!
That is correct. C-2 attached to Cu is nucleophilic (partially negative) and C1 attached to Br is electrophilic (partially positive).
Oh, my mistake. Electronegativity (Cu) = 1,9 < Electronegativity (C) = 2,5. :-) Thx
Hi, please for explaining how is created bond C1 and C2 in Mechanism? C1 is (+) and Cu is (-) —> C2 is (+) I suppose. Thx. P.