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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.

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  • $\begingroup$ Build a 3d model of the two compounds: it is far more obvious what is happening. Fisher projections don't help nearly as much as a 3D model. $\endgroup$ – matt_black Jan 15 at 11:52
  • $\begingroup$ I just noticed that you give your questions very, very non-descriptive titles. We would appreciate it very much if you could avoid 'buzzwords' like question, doubt, problem, ... , and give your question a titel that summarises the actual question. $\endgroup$ – Martin - マーチン Jan 15 at 16:41
  • $\begingroup$ Sure will do that in the future. $\endgroup$ – thewitness Jan 16 at 9:49
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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:

mechanism with the meso-compound as starting material

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

enter image description here

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

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    $\begingroup$ This answer would be slightly better if you did not use wedge and dashed bonds in the products. $\endgroup$ – Zhe Jan 15 at 14:27
  • $\begingroup$ I believe the wedge/dash bonds are wrong according to IUPAC guidelines. // Please do not abuse MathJax for styling elements. $\endgroup$ – Martin - マーチン Jan 15 at 16:38
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    $\begingroup$ I will take account of that from further on. $\endgroup$ – Soumik Das Jan 15 at 16:48
  • $\begingroup$ Your stereo configuration of the structure of starting material in last scheme is actually $(\mathrm{2}R,\mathrm{3}R)$-, not $(\mathrm{2}R,\mathrm{3}S)$- as indicated. It is worth note that $(\mathrm{2}R,\mathrm{3}S)$- is the stereo configuration of meso- compound in first scheme. $\endgroup$ – Mathew Mahindaratne Jan 15 at 21:14
  • $\begingroup$ Oh yeah !! My bad.. Sorry.. $\endgroup$ – Soumik Das Jan 16 at 1:54

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