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I am considering the hydrogenation of 2-butanone with respect to its Re and Si faces.

So if we look at 2-butanone as I've drawn below from above we are looking at its rectus face.

And from below we are looking at its sinister face.

Now, time to consider hydrogenation.

We'll keep the wedge and dash groups as they are and the carbon and oxygen both on lines. In other words we keep two points as is in the originally trigonal planar 2-butanone and when the hydrogen inserts itself, the other two points (a hydrogen and an alcohol group) pop up out the plane to form the classic tetrahedral shape of an $\ce{sp3}$ carbon atom.

So it appears that hydrogenation from the sinister face adds an hydrogen to the central carbon and forms an alcohol, and the alcohol is above the added hydrogen spatially, and this gives us the R isomer. Hydrogenation from the rectal face gives the opposite S isomer.

How's the reasoning? Am I right? I am fairly sure. But not completely sure.

enter image description here

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    $\begingroup$ I really do not understand what you are asking about. Do you just want to hear a Yay or Nay? $\endgroup$ – Martin - マーチン Oct 15 '14 at 3:29
  • $\begingroup$ I need constructive criticism and a yay or nay would be nice too. $\endgroup$ – Dissenter Oct 15 '14 at 3:51
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When looking at the corresponding Wikipedia article for prochirality you will find out, that your reasoning is correct for the hydrogenation. Of course if you have a different nucleophile, then the situation would change, as the Cahn-Ingold-Prelog identifiers would change.

re si faces and products of hydrogenation

See also prochirality in the IUPAC goldbook.

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  • $\begingroup$ The Re/Si enantiotopic faces of 2-butanone are independent of the nucleophile. As your first link notes, the priority of the 3 CIP groups are arranged clockwise. (I point my thumb in the direction of an attacking nucleophile and follow my fingers from highest to lowest priority.) It is true that R/S for the product will be nucleophile dependent, i.e., hydrogen vs. sulfur. Your last sentence is a bit confusing. $\endgroup$ – user55119 Nov 20 '19 at 23:21

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