Explain the reactions involved in the Hoffman exhaustive methylation and degradation of ppyrrole?
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Answer:
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Hofmann elimination is an elimination reaction of an amine where the least stable (least substituted) alkene, the Hofmann product, is formed. This tendency, known as the Hofmann alkene synthesis rule, is in contrast to usual elimination reactions, where Zaitsev's rule predicts the formation of the most stable alkene. It is named after its discoverer, August Wilhelm von Hofmann.[1][2]
The reaction involves the formation of a quaternary ammonium iodide salt by treatment of the amine with excess methyl iodide (exhaustive methylation), followed by treatment with silver oxide and water to form a quaternary ammonium hydroxide. When this salt is decomposed by heat, the Hofmann product is preferentially formed due to the steric bulk of the leaving group causing the hydroxide to abstract the more easily accessible hydrogen.
Explanation:
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