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Topic: Radical addition scheme  (Read 9681 times)

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Offline Rutherford

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Radical addition scheme
« on: February 23, 2014, 05:21:44 AM »
The problem is attached, I need to determine the unknown compounds. I will attach my attempt in the next post.

The whole problem can be found here http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html it is the 23rd.
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Offline Rutherford

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Re: Radical addition scheme
« Reply #1 on: February 23, 2014, 05:22:29 AM »
What transformation will F undergo?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #2 on: February 23, 2014, 05:52:47 AM »
What happens to the primary radical you formed by closing the ring. Is there not a substituent on that methyl group?
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Offline Rutherford

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Re: Radical addition scheme
« Reply #3 on: February 23, 2014, 06:00:08 AM »
What substituent?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #4 on: February 23, 2014, 06:02:28 AM »
Do the radical cyclisation you end up with a primary radical, what happens to it?
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Offline Rutherford

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Re: Radical addition scheme
« Reply #5 on: February 23, 2014, 06:02:42 AM »
I forgot the radical on the methyl group, it will undergo a transformation with the azide, right? But where does the methyl on nitrogen come from?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #6 on: February 23, 2014, 06:08:27 AM »
I don't think it reacts with the azide
« Last Edit: February 23, 2014, 06:40:10 AM by discodermolide »
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Offline Rutherford

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Re: Radical addition scheme
« Reply #7 on: February 23, 2014, 06:22:32 AM »
What else path could afford the final compound?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #8 on: February 23, 2014, 06:26:35 AM »
If your compound F is correct, what could compound G be?, Compound G looses N2 after acid treatment. So F must be some sort of reactive intermediate that transforms to G, your F will not do anything.
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Offline discodermolide

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Re: Radical addition scheme
« Reply #9 on: February 23, 2014, 06:29:10 AM »
OK, you have a primary radical, lets say it reacts with the azide, what compound do you get?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #10 on: February 23, 2014, 06:33:14 AM »
this compound
[C]C1(CCN=[N+]=[N-])C2=CC(OC)=CC=C2N(CC3=CC=CC=C3)C1=O

the thing the smiles had marked C(1) is supposed to be a radical
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Offline Rutherford

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Re: Radical addition scheme
« Reply #11 on: February 23, 2014, 06:51:44 AM »
The primary radical has to react with nitrogen, N2 will be released, so the final compound after treatment with CH3X would be obtained, I think. But I can't think of a mechanism for the reaction.
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Offline discodermolide

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Re: Radical addition scheme
« Reply #12 on: February 23, 2014, 06:53:10 AM »
No, do it stepwise:
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Offline Rutherford

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Re: Radical addition scheme
« Reply #13 on: February 23, 2014, 06:58:54 AM »
Then the lone pair from nitrogen will end up forming a triple bond, and the other N-N bond will be cleaved to form nitrogen molecule and a radical that will abstract a hydrogen to obtain I?
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Offline discodermolide

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Re: Radical addition scheme
« Reply #14 on: February 23, 2014, 07:01:01 AM »
more or less:
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