What I had in mind was to form but-2-ene with $\ce{H2}$ and Lindlar's catalyst and then simply add bromine ($\ce{Br2}$). But this addition, if I'm not mistaken, is a trans-addition. So I would end up with (2R,3S)-2,3-dibromobutane instead of (2R,3R).

Note that this is (2R,3S)-2,3-dibromobutane:


Have I made a mistake, or is there something wrong in my way of thinking? If my solution is wrong, how could I get (2R,3R)?


1 Answer 1


I think I figured it out. Lindlar is a cis-addition so one would not obtain a (2R,3S), but rather (2S,3S) and (2R,3R) which happen to be enantiomers.

  • 3
    $\begingroup$ Yes, pretty much. And if you wanted the other diastereomer (the (2R,3S) i.e. meso diastereomer), you would just need to use Na, NH3 to reduce the alkyne first. That gives the trans alkene which adds Br2 in an anti fashion (via the bromonium ion) to give the meso compound. $\endgroup$
    – orthocresol
    Feb 11, 2017 at 22:28

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