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2-Bromoethyl (2-bromoethyl)carbamate can react through either Path A or Path B to form salt 1 or salt 2. In both instances the carbonyl oxygen functions as the nucleophile in an intramolecular alkylation. What is the preferred reaction path for the transformation in question?

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I would go with A, but I don't know how to explain it.

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    $\begingroup$ Well you must have a reason for choosing A so try to give us some idea of what it is. $\endgroup$
    – bon
    Sep 15, 2015 at 20:51
  • $\begingroup$ The flow of electrons is more linear in the TS of reaction pathway A? $\endgroup$
    – EJC
    Sep 16, 2015 at 8:17
  • $\begingroup$ I would tentatively suggest A based on less ring strain due to not having a double bond inside the ring and having oxygens instead of nitrogens. I'm not particularly knowledgeable in this field though. If you want the question to receive more attention you could consider putting a bounty on it. $\endgroup$
    – bon
    Sep 17, 2015 at 11:04
  • $\begingroup$ But double bonds inside rings are more stable than exocyclic ones. A bounty seems inevitable. $\endgroup$
    – EJC
    Sep 17, 2015 at 13:32
  • $\begingroup$ AFAIK that's only because they usually have more substituent groups (compare 1-methylcyclohexene and methylenecyclohexane) but in this case that's not the case. Since the geometry of the double bond is restricted by the ring, I think that would make it less stable but this is only speculation. $\endgroup$
    – bon
    Sep 17, 2015 at 16:39

1 Answer 1

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Path A resembles a 5-enolexo-exo-tet reaction which is allowed by Baldwin rules while path B represents a 5-enolendo-exo-tet reaction which is disfavored. The orbital interactions of path B suffer severe geometrical constraints (even in the conformation where the C-Br bond is pointing towards the viewer when looking at the attached image). Path A is presented in the attached image.

enter image description here

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