Elimination of bromine with sodium iodide and electrocyclization
P&#246;yt&#228;: 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 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. ;)