Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: livelaughlove00 on February 03, 2017, 10:49:55 PM
-
I completely understand where the first reagent comes into play. But for the duels-alder reaction to occur, the single double bond needs to positioned differently and made into two double bonds. So, I figured NBS would add bromide to the cyclopentene and then Bu4N+I- would abstract the proton to displace double bond positioning. First of all it's not strong enough of a base to do that AND it does not lead to an I-Br byproduct. Please help for good karma!
-
I don't agree with the formation of bromo-iodide. It is a product of a net reduction (electron donation). Formation of a diene requires a net oxidation (electron loss). A loss of bromide (such as HBr) would accomplish this. This reaction needs a (weak?) base and probably simultaneous with the Diels-Alder reaction.
-
Was the 'hint' just mis-spoken? I can imagine TBAI reacting with NBS to form IBr as a more electrophilic source of 'Br+'. Then just do a C3-bromination of the indole followed by an elimination (presumably the succinimide anion is the base?), remove a proton to turn the N-acyl iminium into the enamine and you're ready to go. Thoughts?
-
could radical allylic bromination followed by elimination be feasible explanation?
-
Was the 'hint' just mis-spoken? I can imagine TBAI reacting with NBS to form IBr as a more electrophilic source of 'Br+'. Then just do a C3-bromination of the indole followed by an elimination (presumably the succinimide anion is the base?), remove a proton to turn the N-acyl iminium into the enamine and you're ready to go. Thoughts?
I agree, this makes sense.