[robert_l]

can someone explain whether this is correct?

what function does n-BuLi play?

[sjb]

Under basic conditions, yes, you'll get the (E)-olefin. http://en.wikipedia.org/w/index.php?title=Peterson_olefination&oldid=256578418

BuLi deprotonates the nitrile compound.

S

[robert_l]

Thanks.

Is my mechanism right?

Where is deprotonation needed?

[sjb]

Thanks.

Is my mechanism right?

Where is deprotonation needed?

I'm not 100% happy with the bond angles you've drawn for the intermediates, but in essence the mechanism is right. Consider the TMS-acetonitrile compound you've drawn as a reagent, which has then added to the aldehyde.

[robert_l]

thanks sjb.

i think a part of of the compound (CN-??) attacks the carbonyl c of the aldehyde but not sure how that works.

[sjb]

no, not the CN part. remember you have a C-C bond to break if you want free cyanide or similar.

Look at the difference between the TMS-acetonitrile and the first intermediate that you've drawn, focussing on the bottom half of it.

[robert_l]

SiMe3 attacks the carbonyl?

[sjb]

Here's a copy of your starting materials, with certain atoms numbered. Can you take the initial product, that I've drawn and put the numbers on that? Can you then see where you deprotonate?

[robert_l]

i think it deprotonates at 6. there are 3 Hs starting and only 2 in the product.

thanks a lot for helping me.

i am learning.

[sjb]

That's fine, I think you've got it

S

[lutesium]

How can that SiMe3 group be eliminated???

[sjb]

It's a sort of retro-Brook rearrangement, or similar (if you want a formalism), imo

[lutesium]

It's a sort of retro-Brook rearrangement, or similar (if you want a formalism), imo

But at retro-Brook rearrangement the SiMe3 group is not eliminated. Only the molecue changes his angle.
I'm asking for the last step where SiMe3 is eliminated but how?

Lutesium...