Mechanism[ edit ] The reaction of bromine with sodium hydroxide forms sodium hypobromite in situ , which transforms the primary amide into an intermediate isocyanate. The formation of an intermediate nitrene is not possible because it implies also the formation of a hydroxamic acid as a byproduct, which has never been observed. The intermediate isocyanate is hydrolyzed to a primary amine, giving off carbon dioxide. Base abstraction of the remaining amide proton gives a bromoamide anion. The bromoamide anion rearranges as the R group attached to the carbonyl carbon migrates to nitrogen at the same time the bromide ion leaves, giving an isocyanate.
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Jump to navigation Jump to search Hofmann elimination, also known as exhaustive methylation. In this reaction, least stable alkene is formed, i. A process where a quaternary ammonium reacts to create a tertiary amine and an alkene by treatment with excess methyl iodide followed by treatment with silver oxide , water , and heat. When a quaternary ammonium hydroxide is decomposed on heating then the hydrogen is preferentially eliminated from that beta carbon which is joined to largest number of hydrogen atoms.
Nitrogen in quaternary ammonium hydroxide has positive charge and strong -I effect Electron attracting. It indicates positive character in nearby atoms including beta carbon atom. Which reduces positive character of beta hydrogen atom. So the release of this hydrogen is restricted. Thus, when positively charged nitrogen is linked with ethyl group and n-propyl group then hydrogen is lost from ethyl group and not from propyl group After the first step, a quaternary ammonium iodide salt is created.
After replacement of iodine by a hydroxyl anion, an elimination reaction takes place to form the alkene. With asymmetrical amines, the major alkene product is the least substituted and generally the least stable, an observation known as the Hofmann rule.
The reaction is named after its discoverer, August Wilhelm von Hofmann.
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Hoffmann Bromamide Degradation