Hoffmann Bromamide reaction

Started by Pranjal Singh, February 04, 2022, 03:02:02 AM

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Pranjal Singh

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chenbeier

The who reaction  only envolves  the nitrogen and carboxyl C atom. So mainly product 4 will be obtained.

uma

#2
Observe the mechanism and decide about the stereochemistry of migratory R group. It will try to connect to NH2 group with minimum steric hindrance.
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chenbeier

There is no C from the Cyclohexyl Ring called R involved in the reaction. Therefore in my opinion the stereo chemistry is not changed. Which Front side should be blocked. The only movement of the R is to go from carboxy C to nitrogen to form the Isocyanate.

jakecunningham

The major product will be number 4.
In this reaction, an alkali is used as a strong base to attack the amide, which leads to the deprotonation and generation of an anion. The reaction of bromine with sodium hydroxide leads to the formation of sodium hypobromite. NaOBr (sodium hypobromite) transforms the primary amide into an isocyanate intermediate. Now water attacks this isocyanate intermediate, which leads to a series of proton transfer steps. The thermal conditions cause two things. 1: Explosion of carbon dioxide gas and 2: Quenching of the ammonium cation to the required amine product. The reaction is also called Hoffmann degradation of amide.
This reaction is used to convert a primary amide into a primary amine with one less carbon atom.

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