Comments on: Bonus Topic: Allylic Rearrangements

By: Chaminng

Chaminng — Tue, 20 Feb 2018 11:06:31 +0000

Hello james! I thoroughly read your post

but I have a question..

compared to radical structures..

radical structure of A is TERTIARY allyl radical, radical structure of B is SECONDARY allyl radical.

the former is more stable radical structure..

But Major product is B.

Just Zaitsev’s rule(hyperconjugation) has more afffection than Radical Stability?

By: jess

jess — Fri, 21 Apr 2017 15:31:43 +0000

In reply to James. Hi! I was coming here to ask a similar question -- specifically, if one started with 1-methylcyclohexene (the unbrominated version of Product B), would NBS favor bromination of the methyl group? Thanks so much!

By: Farzeen

Farzeen — Thu, 23 Feb 2017 16:39:39 +0000

In reply to Andrea Jurado.

More substituted alkenes are more stable because the inductive effect and hyperconjugative effects satisfies the hunger of the (relatively) electron hungry sp2 carbon atom.
But here, isn’t Bromine more electrong ‘hungry’ than sp2 carbon?
Wouldn’t product A where bromine is on a tertiary carbon atom be more stabilized?

Thanks James and Andrea Jurado.

By: Nandakumar U K

Nandakumar U K — Fri, 10 Feb 2017 03:01:57 +0000

Excellent explanations

By: Andrea Jurado

Andrea Jurado — Tue, 20 Dec 2016 00:07:22 +0000

In reply to Tuneer.

It depends on the temperature!

I think the main point why we covered the thermodynamically product is because it’s conceptually the point of the rearrangement. The most stable alkyl radical would be the major product at low temperatures, because it is kinetically favored due to the faster rate determining step. However at HT, thermodynamics trump kinetics to give you the more substituted alkene/ (mark pdct).

Good point!

By: Freelancer

Freelancer — Wed, 23 Nov 2016 07:51:38 +0000

I Think I got your explanation of more substituted alkene B being more stable than A , but then if we look at the transition state leading to B and A we find that A has a tertiary free radical which is more stable than secondary free radical(for B) , so why shouldn’t A be the major product by this line of thought , even though I agree That the final product B Is more stable than A but shouldn’t we also compare the transition states leading to the two products … I am confused as to why we are not considering this free-Radical stability … Can you please elaborate on this part ?

By: James

James — Mon, 08 Aug 2016 21:13:28 +0000

In reply to Michael. Product B is major because the double bond is more substituted, which is ultimately more stable (similar to why eliminations tend to occur in a way to give the most substituted alkene - Zaitsev's rule).

By: James

James — Mon, 08 Aug 2016 21:12:21 +0000

In reply to Ian.

Hi – if one had an excess of NBS (>1 equivalents) then a second allylic bromination could occur, and that CH3 hanging off the double bond in B is one position where it could happen.

The rate of that reaction will be proportional to the concentration of NBS and the concentration of B, so as the reaction proceeds (and the concentration of B goes up) we should expect to see a little bit of it happening.

By: Ian

Ian — Sat, 16 Jul 2016 15:22:53 +0000

Hey James. Is it possible for the bromine to also bond with the CH3 hanging off the double bond in product B?

By: Michael

Michael — Wed, 20 Apr 2016 17:05:58 +0000

Love the site! Just want put in encouragement to finish this conversation. Both a and b seem like they could be the major product for mutually exclusive reasons. Looking for a way to reconcile all the points of view and get the definitive answer.