Hoffmann degradation requires a halogen molecule to be present....during Hoffmann degradation, some amount of base may add to the carbonyl group, but given the chance, it finds abstracting a proton from the amine group easier. The double bonded amine group then attacks the halogen-halogen bond as a nucleophile.
The part upto which the proton is abstracted is an equilibrium reaction. The proton may be given back also. What drives the reaction after that is the nucleophilicity of the double bonded amine (-NH) group. After it attacks the halogen bond, the reaction cannot "go back". It must proceed.
In the base-catalysed hydrolysis of amides, they undergo nucleophilic addition of OH- and finally form acetate.
In hoffmann degradation, we use OH- (of an alkali) where it extracts a proton from the -NH2 group.
Why is the action of same OH- different in two cases?
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3 Answers
Pritish Chakraborty
·2010-12-09 20:01:19
swordfish
·2010-12-10 03:40:14
Thanks for your answer.
You answered my second part. What about the first one? I mean why it does not abstract the hydrogen in hydrolysis.
Pritish Chakraborty
·2010-12-10 04:16:42
Again it is similar to what I said for the second part. It does abstract the proton, but with nothing to drive the reaction, the proton is given back.