This is what I have read/know about the dehalogenation of vicinal dihalides by NaI/acetone: The two halides which are leaving must be on opposite sides (from the mechanism). In case they are present in the same side, we must rotate the C-C bond first and then proceed with the reaction. In both cases we end by with a trans alkene.
But now I came across this:
meso-2,3-dibromobutane on reaction with NaI/acetone gives trans-2-butene.
d/l-2,3-dibromobutane on reaction with NaI/acetone gives cis-2-butene.
I have no problem regarding the first statement. But regarding the second, from what I have learnt, the product again must be trans-2-butene.
I drew the Fisher projection and tried to figure out the reaction. For the 1st reaction, meso-2,3-dibromobutane must undergo C-C bond rotation, (performing two "swaps") and then we would end up with d/l-2,3-dibromobutane leading both 1st and 2nd reactions to trans-2-butene.
Where is my mistake? Is my understanding of the mechanism of the dehalogenation of NaI/acetone wrong? Any good sources to read regarding the same (I couldn't find it in any of my books) are appreciated, too.