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Topic: NBS Reaction & Radicals  (Read 2952 times)

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Offline Cooper

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NBS Reaction & Radicals
« on: January 22, 2013, 12:58:23 PM »
Hi,

I'm doing a synthesis on paper for my class involving a 1,3-diene. I briefly researched a reaction using NBS and noticed that it used radicals to add a Br to an allylic carbon. My question is the Br will add to the more highly substituted carbon, right? Because that would be the most stable radical. I am asking this because I have a secondary and tertiary allylic carbon and I need the Br to go on the tertiary one. I've attached my synthesis below.

Thanks!
~Cooper :)

Offline discodermolide

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Re: NBS Reaction & Radicals
« Reply #1 on: January 22, 2013, 01:07:46 PM »
I think the NBS product will be the compound I've attached.
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Offline Cooper

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Re: NBS Reaction & Radicals
« Reply #2 on: January 22, 2013, 01:10:57 PM »
Hm, maybe they are showing us the basic way it would happen without complicating it. This is the snip-it they gave us about NBS.
~Cooper :)

Offline discodermolide

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Re: NBS Reaction & Radicals
« Reply #3 on: January 22, 2013, 01:15:36 PM »
In that case stick with what they gave you. ;D
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Offline Cooper

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Re: NBS Reaction & Radicals
« Reply #4 on: January 22, 2013, 01:25:16 PM »
Thanks again :)
~Cooper :)

Offline Babcock_Hall

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Re: NBS Reaction & Radicals
« Reply #5 on: January 23, 2013, 08:58:32 AM »
Why would there be an isomerization?

Offline discodermolide

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Re: NBS Reaction & Radicals
« Reply #6 on: January 23, 2013, 09:36:20 AM »
Why not?
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Offline orgopete

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Re: NBS Reaction & Radicals
« Reply #7 on: January 23, 2013, 11:37:57 AM »
Why would there be an isomerization?

More stable alkene and less reactive bromide.
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