orgo destroyer #99

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akimhaneul

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2-iodobutane + (Ph)3P: with DMSO solvent -> C6H5Li with THF solvent -> acetone with ether solvent -> ?


The answer is B (2,3,-methyl-pent-2-ene)

Can anyone explain this problem?

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2-iodobutane + (Ph)3P: with DMSO solvent -> C6H5Li with THF solvent -> acetone with ether solvent -> ?


The answer is B (2,3,-methyl-pent-2-ene)

Can anyone explain this problem?
this is a reaction of alkylhalide with yield with bulky base DMSO and THF which are good leaving groups. the yield a nucleophile will complex with the iodide forming a carbocation which further is attacked by another nucleophile organolithium and acetone both nucleophile's giving the final product alkene. all reactions taking place on carbon 2.
 
2-iodobutane + (Ph)3P: with DMSO solvent -> C6H5Li with THF solvent -> acetone with ether solvent -> ?


The answer is B (2,3,-methyl-pent-2-ene)

Can anyone explain this problem?
Not a chance. NO carbocation is involved with this at all. This is the classic Wittig reaction. Step #1 was a simple SN2 attack. Then.....I removed the acidic proton with the C6H5Li to create an ylide. This ylide reacted with the carbonyl of acetone to form the product. You need to consult any text......Wade, McMurray, Klein, Bruice, Carey.....and familiarize yourself with this very very important reaction. In this reaction we created carbon-carbon bonds and it ranks as one of the most important reactions in all of synthetic organic chemistry. If after you consult your text, and you are still flummoxed, I would be more than happy to draw out the mechanistic details for the entire process for you.

Hope this helps.

Dr. Romano
 
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