6
$\begingroup$

Ultimate question: Is Bredt's rule equally valid for fused bicyclic system (no bridgehead alkenes without a ring 8 carbons or more), but for other reasons (historical?) it is only defined as applying to bridged systems? Or am I missing something else fundamental about bridged vs fused bicyclic systems?

--Background--

Everywhere I can look (online, reference texts), Bredt's rule is specifically in reference to bridged bicyclic systems. However, shouldn't the rationale behind Bredt's rule -- that by having a bicyclic alkene with the double bond at the bridgehead means it has to have a trans alkene and therefore needs 8 carbons or more -- also apply to a fused bicyclic alkene as well?

The only example my book gives students is a bicyclo[4.4.0]dec-1-ene. It says "well this doesn't need Bredt's rule, because it's not a bridged system, but it's still stable because the largest ring has 10 carbons to accommodate the trans bond". That answer is unsatisfying to me, because the basis for them recognizing that the molecule has a trans alkene in it, comes from the discussion about Bredt's rule in the first place.

Bicyclo[3.1.0]hex-1-ene

It never gives an example of something like bicyclo[3.1.0]hex-1-ene (above)...but I assume based upon a trans-alkene with less than 8 carbons this should also not be stable? Would this just be an application of the idea that trans-alkenes need to be in rings 8 or larger to be stable (without being part of Bredt's rule specifically)?

$\endgroup$
  • $\begingroup$ Here's a way to look at it - bicyclo[3.1.0]hex-1-ene is a bridged compound, the bridge just has "0" units as the name indicates. $\endgroup$ – ron Jun 2 '18 at 23:18
  • $\begingroup$ @ron That's fair, and sort of how I planned on presenting this particular topic (if I couldn't find a more complete answer). However, it still leaves me wanting, because Bredt's rule is presented as specifically for bridged (as in non-zero carbon) bicyclic. So...I think my original question still remains :) $\endgroup$ – J M Jun 3 '18 at 0:26

Your Answer

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

Browse other questions tagged or ask your own question.