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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: ryin0 on August 02, 2021, 07:52:55 PM

Title: Axial Requirement/Hofmann E2 elimination
Post by: ryin0 on August 02, 2021, 07:52:55 PM

(https://i.ibb.co/bBZghPM/Screen-Shot-2021-08-02-at-6-47-32-PM.png) (https://ibb.co/ncvs2bN)
Hey ya'll, problem from Klein, did this question first with a sterically hindered base for E2 elimination to get the hofmann product for step 2, but the answer just says use any strong base. Thought it was because the H at the bridgehead wasn't axial, and therefore isn't reactive, but I'm pretty terrible at drawing Newman projections. Anyone know if this explanation makes sense? Thanks!

Title: Re: Axial Requirement/Hofmann E2 elimination
Post by: kriggy on August 03, 2021, 02:52:54 AM

Yeah makes sense but you could have gotten E1 to the bridgehead as well.
Bridgehead does not have double bonds connected to them unless the ring is sufficiently large for steric reasons

Title: Re: Axial Requirement/Hofmann E2 elimination
Post by: ryin0 on August 04, 2021, 01:21:08 AM

ah poop bredt's, i'm an idiot
Thanks!

Title: Re: Axial Requirement/Hofmann E2 elimination
Post by: OrganicH2O on August 11, 2021, 12:42:59 PM

That specific molecule is very interesting for E2. Bredt's rule is important, but there's another cool thing happening also: anti-elimination isn't possible, so instead it eliminates by a syn E2. This is a weird exception that shows E2 only has a very strong preference for anti-elimination. If the molecule is locked into a shape where a beta-hydrogen and a leaving group are syn, then a syn E2 is possible.