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enter image description here

I'm not able to figure out the mechanism for this reaction. I don't see how sodium iodide in acetone could lead to rearrangement, and surprisingly ring opening!

What is this reaction called, if it has a special name? How does it occur?

P.S. In general, as far as I know, sodium iodide in acetone promotes excellent SN2 substitution, so all bromine atoms should be replaced by iodine. After this, because 1,2-diodo compounds (vicinal di-iodides) are unstable, elimination takes place (removal of iodine) and formation of alkene occurs (that's what I've seen happen, usually).

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  • $\begingroup$ Do you have a reference to the book / paper you got the reaction from? Looks pretty interesting $\endgroup$ – NotEvans. Sep 1 '18 at 16:15
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    $\begingroup$ @NotEvans. For once the reaction actually has some precedence that looks similar (the starting material is not exactly the same), but the reference is in German: Angew. Chem. 1977, 89 (12), 909–910. My first instinct is a series of electrocyclic reactions, but I'm not sure. I think the SM drawn in the question is too unsaturated. $\endgroup$ – orthocresol Sep 1 '18 at 16:37
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    $\begingroup$ A paper I found discusses a similar reaction: an "exotic" dehalogenation of vicinal dihalides. Actually, it should be noted that the solvent used here is methanol, which might imply a different reaction path. pubs.acs.org/doi/abs/10.1021/ja01607a037?journalCode=jacsat $\endgroup$ – The_Vinz Sep 1 '18 at 22:14
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    $\begingroup$ And I agree with @orthocresol about the starting molecule being too unsaturated $\endgroup$ – The_Vinz Sep 1 '18 at 23:20
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    $\begingroup$ If we take the starting material as the correct one, and perform the reaction, I think only four Bromine atoms can be eliminated out of eight, and the remaining four Bromine atoms will be very difficult to eliminate. So, I think the final product should be logically (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene . $\endgroup$ – Soumik Das Sep 7 '18 at 13:07
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I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture and I don't think the final product given in the picture can ever be got from the given starting material. I will be happy if someone else can find flaws in the logic given below.

$\ce{NaI}$ in acetone is the reagent used for Finkelstein reaction which is nothing but a standard substitution reaction with $\ce{I-}$ ions. But in presence of vicinal dibromides, or any special structure where there is a chance of elimination of hallide ions even further, the iodide ions can actually make the molecule undergo preferably an $E_2$ elimination (in most of the cases).

In this starting compound also, there are substitutable $\ce{Br}$ atoms present. So, the iodide ions first performs a substitution reaction, and then undergo an elimination via bond migration. This same procedure occurs for two times. and finally we are left with a product which has $\ce{Br}$ atoms attached to $sp^2$ carbon atoms, and therefore they are highly reluctant to undergo any substitution or elimination reaction. So, that's why, logically the end product should be, $\text{ (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene}$. Here is the structure of the possible end product.

enter image description here

$\textbf{Mechanism:-}$
enter image description here

As far as I see, there can not be any more further reaction to the final compound. According to me, the given compound in the picture should not be the final product if we take the starting material to be the correct one. Instead, the end product in the mechanism should be the one produced with higher yield at the end.

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  • $\begingroup$ At least two more bromine halogene atoms can be eliminate to form a double bond in the annealed part. It is the fate of two last halogene atoms that is surprising. It seems that NaI is working as a reducing agent for those two. $\endgroup$ – permeakra Sep 9 '18 at 12:42

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