Why does d/l-2,3-dibromobutane on reaction with NaI/acetone give cis-2-butene? 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 &quot;swaps&quot;) 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.
The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the formation of vicinal di-iodide, the compound now undergoes bond rotations if required and make the two iodine atoms orient themselves anti-periplanar for a E2 elimination. 

So, essentially not only elimination, but substitution also occurs prior to elimination, and thus meso-2,3-dibromobutane never actually gets converted into (d/l)-2,3-dibromobutane during the reaction. Also, going by Sawhorse Projection formula is particularly useful in this case rather than Fisher projection. The former gives you a better visualisation of the syn or anti-periplanarity. 

Here is the mechanism of the reaction with the meso-compound as starting material: 



Similarly, you can draw the mechanism with the optically active 2,3-dibromobutane as your initial compound:



Now you can easily see how the corresponding alkenes are formed depending on the substrates.