6
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ 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, 2019 at 11:52
  • 1
    $\begingroup$ 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, 2019 at 16:41
  • 1
    $\begingroup$ Sure will do that in the future. $\endgroup$
    – user600016
    Jan 16, 2019 at 9:49

1 Answer 1

Reset to default
12
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ This answer would be slightly better if you did not use wedge and dashed bonds in the products. $\endgroup$
    – Zhe
    Jan 15, 2019 at 14:27
  • 1
    $\begingroup$ 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, 2019 at 16:38
  • 1
    $\begingroup$ I will take account of that from further on. $\endgroup$
    – Soumik Das
    Jan 15, 2019 at 16:48
  • 1
    $\begingroup$ 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, 2019 at 21:14
  • $\begingroup$ Oh yeah !! My bad.. Sorry.. $\endgroup$
    – Soumik Das
    Jan 16, 2019 at 1:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.