# Rearrangement with sodium iodide (ring opening)

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).

• Do you have a reference to the book / paper you got the reaction from? Looks pretty interesting – NotEvans. Sep 1 '18 at 16:15
• @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. – orthocresol Sep 1 '18 at 16:37
• 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 – The_Vinz Sep 1 '18 at 22:14
• And I agree with @orthocresol about the starting molecule being too unsaturated – The_Vinz Sep 1 '18 at 23:20
• 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 . – Soumik Das Sep 7 '18 at 13:07

$\ce{NaI}$ 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.
$\textbf{Mechanism:-}$