[nj_bartel]

 

In step 3, where the electron pair from the carbonyl bond transfers to the carbonyl oxygen to form the negative charge, why don't you get a positive charge on the carbonyl carbon?  I feel like I'm just missing something obvious.

Thanks.

[macman104]

Because you form a C=C bond in the process, which replaces the C=O double bond.

[Yggdrasil]

The carbon still has the hydrogen on it even though it is not shown in compound 4.

[nj_bartel]

Because you form a C=C bond in the process, which replaces the C=O double bond.

But the C=C bond isn't using the carbon in question.

The carbon still has the hydrogen on it even though it is not shown in compound 4.

Thanks, that would make sense.  The fact that they showed one of the hydrogens for the proton transfer had my head in a mix.

[nj_bartel]

Am I correct in assuming that the attached is one possible stereoisomer of the product?  I'm attempting to come up with a way to control the stereochemistry at the alcohol carbon - just want to make sure that I'm very clear first.

[nj_bartel]

'Modify' already disappeared, sorry.  And the other forms would just be the R-group, alcohol, and hydrogen changing places, yes?

[azmanam]

sounds good to me

[macman104]

Sorry, I completely missed the step 3 text.

[nj_bartel]

Thank you   

[MrDax]

In step 3, where the electron pair from the carbonyl bond transfers to the carbonyl oxygen to form the negative charge, why don't you get a positive charge on the carbonyl carbon?

The carbon in question receives electrons from enolate 3