Elimination of bromine with sodium iodide and electrocyclization

I've been trying to solve the following reaction scheme for a while now and I am having trouble especially with the products **E** and **F**.



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> [![enter image description here](https://i.stack.imgur.com/Sh6lr.png)](https://i.stack.imgur.com/Sh6lr.png)
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For product **D** I got the following, but I'm unsure about compound **E**, which should be a reactive intermediate.



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> [![enter image description here](https://i.stack.imgur.com/LN16a.png)](https://i.stack.imgur.com/LN16a.png)
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While trying to figure out the structure of compound **E**, I drew the following structure, but I'm not sure how the rearrangements will happen to get rid of the charges and get product **E**. I do expect that **E** won't be aromatic but that **F** will be.



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> [![enter image description here](https://i.stack.imgur.com/51NLC.png)](https://i.stack.imgur.com/51NLC.png)
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All in all, I am wondering what products **E** and **F** will be and how to figure them out.




Pöytä: Your conclusion that **D** contains four bromine atoms symmetrically distributed is sound. The treatment of **D** with NaI serves to substitute one, if not two bromine atoms with iodine to form "**J**". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.) The stereochemistry of **E** is likely to be *E,E* which arises from a transition state for elimination that avoids steric interactions involving bromine. I am assuming that the "electrocyclization" is thermal since there is no indication that light was involved. Achiral **E** will undergo a conrotatory electrocyclization in two directions to give racemic **F** and, eventually, diiodide **G** via SN2 displacement. (The electrocyclization may be rationalized as involving the HOMO of a butadiene or octatetraene. [Look here](http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/pericyclic/pericyclic.htm) for a discussion.) Reductive elimination of iodine via Zn-Cu couple affords benzocyclobutadiene **H**. (The elimination is akin to the formation of ethylene from 1,2-dibromoethane with magnesium.) The strained, reactive double bond of **H** undergoes a Diels-Alder reaction with cyclopentadiene to form adduct **I**. The stereochemistry shown arises from a so-called "endo" transition state.  
  
I found this question somewhat amusing in that information was provided for all species save the final product. ;)


![](https://i.stack.imgur.com/Y8ppR.jpg)


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