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I have included some snapshots from clayden organic chemistry: enter image description here

enter image description here

The photos show a discussion of the addition of an electrophile to the bottom of the vinyl silane. However, what if the electrophile attacks from the top? Why does it still lead to the same product as seen in the photo? I seem to get the opposite product (e.g. E isomer becomes Z isomer instead of remaining as E)

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  • $\begingroup$ You are starting with an achiral vinylsilane and forming an achiral product with retention of stereochemistry (Si --> E). Addition of the electrophile to either face of the double bond is equally likely. It forms a pair of mirror image intermediates. The collapse of each one leads to the same product. Make sure your intermediates are enantiomeric. If both mirror image transition states had to be drawn, Clayden would cost a lot more!!! Only one transition state is normally shown. $\endgroup$
    – user55119
    Sep 14, 2021 at 20:14
  • $\begingroup$ I think the stereochem of the final compound depends on which way the Si group is--up or down. Is there a way to predict this? $\endgroup$
    – Sloth123
    Sep 15, 2021 at 0:58
  • $\begingroup$ BTW, this reaction is stereospecific, not stereoselective. See: chemistry.stackexchange.com/questions/77061/… $\endgroup$
    – user55119
    Sep 15, 2021 at 19:57
  • $\begingroup$ Clayden mentioned that the rxn is stereoselective not specific though... I suspect they are just trying to explain why a particular path is favoured $\endgroup$
    – Sloth123
    Sep 16, 2021 at 7:43
  • $\begingroup$ The terms have specific meanings. You can google them. If you have 2-phenylpropanal and add CH3MgBr and get a lot more of one diastereomer than the other, then the reaction is stereoselective. Sometimes synthetic organic chemists, who complete a total synthesis, use the term stereospecific to describe the synthesis when they should say stereoselective. Stereospecific reactions are invariably mechanism controlled. The bromination of 2-butene is another example. $\endgroup$
    – user55119
    Sep 16, 2021 at 13:09

1 Answer 1

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I have used the achiral (Z)-vinylsilane 1 as an example with one of the p-orbitals of the double bond colored red. When the electrophile reacts on the "top" of the black p-orbital the carbon rehybridizes from sp2 to sp3 producing the chiral carbocation conformation 2. The same operation on the "bottom" lobe of the black p-orbital affords carbocation conformation 3. These two carbocations are mirror images of one another. In these Newman projections the polar silicon-carbon bond (Si$\rightarrow$C) makes a 60o angle with the vacant p-orbital. There is already partial overlap between the Si-C σ-bond and the vacant orbital. Rotation about the central C-C σ-bond of the sp3 carbon by 60o in a clockwise direction in 2 brings the E-substituent proximate to the phenyl group. Cleavage of the silyl group leads to (Z)-4. The same operation with 3, but in a counterclockwise direction, also leads to (Z)-4. There is no overlap between the Si-C bond and the other lobe of the vacant p-orbital.

Addendum: There can be no rapid rotation about the central C-C bond in 2 or 3 because that would lead to the same distribution of products as afforded by both (Z)- and (E)-1. If so, the reaction would no longer be stereospecific.

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