An unsubstituted amide is converted to primary amine in hofmann degradation. Carbonyl group of amide is completely eliminated and primary amine is given as the product. Due to remove of one carbonyl group, one carbon atom is eliminated from the amide compound.
This reaction should be done in alkali medium. sodium hypobromite (NaOBr) is the most used reagent for this reaction. Sodium hypobromite can be prepared by NaOH and Br2 reaction. KOH and Br2 are also used as reagents. . This reaction occurs in a special rearrangement mechanism.
In this reaction, number of carbon atoms are reduced by one. Hence this reaction is a very important in organic conversion questions to produce a compound which have a less carbon atom than previous compound.
The R group may be alkyl or aryl.
R = -CH3 , -CH2CH3 , -C6H5 (phenyl)
but if it is an alkyl group of more than about six or seven carbons, low yields are obtained. To gey higher yields Br2 and NaOMe are used instead of Br2 and NaOH.
Hofmann degradation of ethanamide will give methanamine. Here, ethanamide reacts with Br2 / KOH and give methamine (primary amine), KBr, K2CO3 and water as products.
Hofmann degradation of propanamide will give ethanamine as the primary amine. If Br2 / NaOH is usedfor hofmann degradation, Na2CO3, NaBr and water are given as other products.
Hofmann degradation of benzanamide will give aniline. This is one of the reaction aniline can be prepared without benzene.
Benzene to aniline is a two step reaction and benzanamide to aniline is only a single step in hofmann degradation.
Isocyanate, is given as the actual product. But due to reaction condition, this isocyanate product hydrolysis to give primary amine.
MARCH’S ADVANCED ORGANIC CHEMISTRY, REACTIONS, MECHANISMS,AND STRUCTURE 6th edition, Michael B. Smith, Jerry March, 18-13 The Hofmann Rearrangement (p,1607)
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