2 Improved Formatting edited Jan 15 at 16:35 Martin - マーチン♦ 34.9k99 gold badges118118 silver badges245245 bronze badges 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 $$S_N2$$ SN2) by $$\ce{I-}$$ on the two carbon atoms containing bromine. As Iodideiodide is a better nucleophile than Bromidebromide, 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 Iodineiodine atoms orient themselves in an anti-periplanar condition for $$E_2$$a E2 elimination. So, essentially not only elimination, but substitution also occurs prior to elimination, and thus $$\text{meso-2,3 dibromobutane}$$meso-2,3-dibromobutane never actually gets converted into $$\text{(d/l) 2,3-dibromobutane}$$(d/l)-2,3-dibromobutane during the reaction.Also 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 $$\text{2,3-dibromobutane}$$2,3-dibromobutane as your initial compound,: Now you can easily see how the corresponding alkenes are formed depending on the substrates. 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 $$S_N2$$ ) 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 in an anti-periplanar condition for $$E_2$$ elimination. So, essentially not only elimination, but substitution also occurs prior to elimination, and thus $$\text{meso-2,3 dibromobutane}$$ never actually gets converted into $$\text{(d/l) 2,3-dibromobutane}$$ 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 $$\text{2,3-dibromobutane}$$ as your initial compound, Now you can easily see how the corresponding alkenes are formed depending on the substrates. 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. 1 answered Jan 15 at 13:47 Soumik Das 4,42911 gold badge55 silver badges2323 bronze badges 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 $$S_N2$$ ) 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 in an anti-periplanar condition for $$E_2$$ elimination. So, essentially not only elimination, but substitution also occurs prior to elimination, and thus $$\text{meso-2,3 dibromobutane}$$ never actually gets converted into $$\text{(d/l) 2,3-dibromobutane}$$ 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 $$\text{2,3-dibromobutane}$$ as your initial compound, Now you can easily see how the corresponding alkenes are formed depending on the substrates.