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I am asked to consider the reduction of acetaldehyde from the re and si faces. I am also asked whether the added hydrogen (labeled H* in my picture) is pro-R or pro-S in both cases. In my attached picture I replaced the added hydrogen (H*) with D to determine the absolute configuration.

If I did this correctly, addition of hydrogen from the si face results in a pro-S product with respect to the added hydrogen (H*). And addition of hydrogen from the re face results in a pro-R product with respect to the added hydrogen. Am I correct?

enter image description here

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    $\begingroup$ Re is Rhenium, Si is silicon. I think re and si might be what you mean $\endgroup$
    – long
    Oct 21 '14 at 5:50
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Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):

Reduction of deuterated acetaldehyde from both the re and si face.

When we add from the si face, we get an (R) center, and if we add from the re face, it's the other way around (as it should be!) and we have (S) configuration.

Now, what if we reduce "normal" acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-R or pro-S. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.

From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.

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  • $\begingroup$ So it appears I did it right? $\endgroup$
    – Dissenter
    Oct 21 '14 at 18:30
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    $\begingroup$ @Dissenter Yes, it appears so. $\endgroup$
    – tschoppi
    Oct 22 '14 at 10:18

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