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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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For product D I got the following, but I'm unsure about compound E, which should be a reactive intermediate.

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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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All in all, I am wondering what products E and F will be and how to figure them out.

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  • $\begingroup$ There are no charges in E. Calculate the degree of unsaturation of E. There is no benzene ring in E. $\endgroup$
    – user55119
    Apr 17, 2020 at 16:19
  • $\begingroup$ I know the IHD is 5, but how can I determine where the double/triple bonds will be? $\endgroup$
    – Pöytä Laatikko
    Apr 17, 2020 at 21:03
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    $\begingroup$ For E you have a degree of unsaturation of 5, one of which is a ring. So how do you introduce 4 double bonds (forget triple bonds) into the carbon framework? Put the two bromine atoms in their most stable orientation. $\endgroup$
    – user55119
    Apr 17, 2020 at 21:54
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    $\begingroup$ F arises as a racemate from a conrotatory electrocyclization of E. $\endgroup$
    – user55119
    Apr 18, 2020 at 2:28
  • $\begingroup$ Thanks for the tips, this really helped! $\endgroup$
    – Pöytä Laatikko
    Apr 18, 2020 at 7:39

1 Answer 1

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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 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. ;)

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  • $\begingroup$ Thanks! Seems like I did it correctly after your tip about the degree of saturation. $\endgroup$
    – Pöytä Laatikko
    Apr 18, 2020 at 16:48
  • $\begingroup$ Just a technical note: H should be in squire parenthesis since it is unstable. Also, it would ne nicer to include a reference for relevant reactions, e.g., doi.org/10.1021/ja01583a071. $\endgroup$
    – Mathew Mahindaratne
    Apr 19, 2020 at 14:51
  • $\begingroup$ This would be better reference: doi.org/10.1021/cr60266a002. $\endgroup$
    – Mathew Mahindaratne
    Apr 19, 2020 at 15:07
  • $\begingroup$ The 50 year old review covers the early steps of the sequence of reactions and it provides an historical perspective. I don't recall seeing any examples of electrocyclizations. Good find nonetheless. $\endgroup$
    – user55119
    Apr 19, 2020 at 16:16
  • $\begingroup$ Typo alert, H is actually benzocyclobutadiene not benzocyclobutene. $\endgroup$
    – ron
    Apr 19, 2020 at 23:40

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