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#Post#: 44--------------------------------------------------
Rearrangement of Amides
DIR By: Michel
Date: March 18, 2012, 7:59 pm
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Treating an unsubstituted amide with an aqueous alkaline
solution of bromine. This reaction is called the Hofmann
rearrangement. The carbonyl group is lost as CO3(-2)
eg
CH3(CH2)4CONH2 + 4OH- +Br2--> CH3(CH2)4NH2+CO3(-2)+2H2O+3Br-
Give the mechanism of this reaction.
#Post#: 985--------------------------------------------------
Re: Rearrangement of Amides
DIR By: Michel
Date: October 25, 2012, 10:00 am
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Any takers?
#Post#: 988--------------------------------------------------
Re: Rearrangement of Amides
DIR By: Potla
Date: October 25, 2012, 12:41 pm
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The mechanism is somewhat well-known I guess... it goes through
an isocyanate intermediate.
R-CO-NH2 + Br2+ OH- --> R-CO-NHBr (bromamide) + Br-
R-CO-NHBr + OH- --> R-CO-N--Br (bromamide anion) + H2O
R-CO-N--Br --> Br-+ R-CO-N: (nitrene)
Now this nitrene goes through a C to N R- shift to give the
isocyanate:
R-CO-N: --> R-N--C+=O <--> R-N=C=O (isocyanate)
R-N=C=O +H2O --> R-NH2 + CO2
CO2+2OH- --> CO3-2+H2O.
The reaction is intramolecular, no cross-over products, and the
configuration of R(the migrating alkyl/aromatic group) is
retained.
Hope I wrote it down correctly. :)
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