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?

enter image description here

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

  • 3
    $\begingroup$ Well you must have a reason for choosing A so try to give us some idea of what it is. $\endgroup$ – bon Sep 15 '15 at 20:51
  • $\begingroup$ The flow of electrons is more linear in the TS of reaction pathway A? $\endgroup$ – RBW Sep 16 '15 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 '15 at 11:04
  • $\begingroup$ But double bonds inside rings are more stable than exocyclic ones. A bounty seems inevitable. $\endgroup$ – RBW Sep 17 '15 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 '15 at 16:39

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

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.