Korialstrasz69
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Selecticity in Free Radical Halogenation
I really am enjoying myself through discovering these reactions
As i know,as the number of carbon substitutions to a carbon atom increase,reactivity of the C-H bonds of that atom decreases, And thus halogination is
prefered. But how does a functional group change that ? Specificly Alcohol group.
For example is i am brominating n-Butanon, may i assume that C-H bonds get weaker getting away from the OH group because of the effect of the
electronegative oxygen atom ? And so the weakest would be the last one and so the main product will be 4-bromo butanon ?
That makes perfect sense to me as a rout to synthesise THF, but i want to make sure
Thank ahead
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halogen
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You will end up oxidizing the alcohol itself.
The carbon next to an -OR group (ether or alcohol, called "alpha" position, that is right next to something, followed by beta, etc.) is activated to
radicals. You would find butyraldehyde, alpha bromobutyraldehyde, and other things depending on conditions. There are reasons for this.
F. de Lalande and M. Prud'homme showed that a mixture of boric oxide and sodium chloride is decomposed in a stream of dry air or oxygen at a red heat
with the evolution of chlorine.
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Darkstar
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Electron-withdrawing groups decrease radical stability UNLESS they have lone pairs that can be donated to the electron-deficient radical center. So
with n-butanol, a radical at the 1-carbon would actually be more stable than a radical at the 4-carbon.
By the way, free-radical bromination is a lot more selective than chlorination. You will likely get little if any 4-bromobutan-1-ol. Also, as
mentioned above, there's a good chance that you'll just end up oxidizing the alcohol to an aldehyde. If bromine adds to the 1-carbon, the unstable
geminal halohydrin can then undergo an elimination reaction as oxygen's lone pair comes down to form a new pi-bond with the now highly
electron-deficient alpha-carbon.
[Edited on 3-18-2016 by Darkstar]
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