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Free radical bromination at benzylic positon
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Leslie Coduti
Free radical bromination at benzylic positon
The benzylic radical and the benzylic cation also are quite stable, and with the correct substituents they can be isolated. The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
The benzylic radical and the benzylic cation also are quite stable, and with the correct substituents they can be isolated. The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
You need something that can form a radical. If it's like t-Butylbenzene then you won't get benzylic bromination. But you may form other radicals which can re-arrange to give a multitude of products.
You need something that can form a radical. If it's like t-Butylbenzene then you won't get benzylic bromination. But you may form other radicals which can re-arrange to give a multitude of products.
The benzylic radical and the benzylic cation also are quite stable, and with the correct substituents they can be isolated. The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
Thank you for your confirmation!!!
Just to ask another quick question: the bromination won't occur if the benzylic position doesn't have any proton, correct?
The benzylic radical and the benzylic cation also are quite stable, and with the correct substituents they can be isolated. The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
Thank you for your confirmation!!!
Just to ask another quick question: the bromination won't occur if the benzylic position doesn't have any proton, correct?
You need something that can form a radical. If it's like t-Butylbenzene then you won't get benzylic bromination. But you may form other radicals which can re-arrange to give a multitude of products.
You need something that can form a radical. If it's like t-Butylbenzene then you won't get benzylic bromination. But you may form other radicals which can re-arrange to give a multitude of products.
The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
The resonance contribution, as you have shown, from the other radicals is small as you correctly say addition there will initially destroy the aromatic system. If you do this reaction on a large enough scale (kg) you can sometimes see the ring brominated products in very small quantities (<0.5%).
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But you may form other radicals which can re-arrange to give a multitude of products.
But you may form other radicals which can re-arrange to give a multitude of products.
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Thank you for your confirmation!!!
Just to ask another quick question: the bromination won't occur if the benzylic position doesn't have any proton, correct?
Thank you for your confirmation!!!
Just to ask another quick question: the bromination won't occur if the benzylic position doesn't have any proton, correct?
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Thank you very much
Thank you very much
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