
Give an example for each describing the following reactions.
Hoffmann’s bromamide reaction.
Answer
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Hint: Hoffmann bromamide reaction uses an amide, bromine, along with a base, to obtain aliphatic or aromatic primary amines. This reaction is also known as Hoffmann bromamide degradation. The degradation is such that amine has one carbon atom less from the amide.
Complete answer:
The reaction when an amide reacts with bromine, in the presence of sodium hydroxide, NaOH or potassium hydroxide, KOH, which is a base, along with an ethanolic solution, then aliphatic or aromatic amines are formed, and this reaction is called as Hoffmann’s bromamide reaction or Hoffmann bromamide degradation. The strong base NaOH is used to attack the amide, and deprotonate it and form an anion. The amide is degraded in this reaction that forms an amine which has one carbon atom less than the reactant amide. The reaction is as follows:
$C{{H}_{3}}CON{{H}_{2}}+B{{r}_{2}}+4NaOH\to C{{H}_{3}}N{{H}_{2}}+2NaBr+N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}O$
This reaction has ethanamide as an amide, that reacts with bromine and sodium hydroxide. The products formed are, ethanamine, sodium bromide, sodium carbonate, and water.
Hence, Hoffmann bromamide reaction results in the formation of amines, and has amide, bromine and sodium hydroxide as reactants.
Note:
Hoffmann degradation reaction is different from Hofmann elimination reaction. Hofmann elimination involves formation of tertiary amines and alkenes. The reactants are quaternary ammonium salt with excess methyl iodide. The resulting compound is quaternary ammonium iodide salt, which is treated with silver oxide and water to form an alkene and a tertiary amine.
Complete answer:
The reaction when an amide reacts with bromine, in the presence of sodium hydroxide, NaOH or potassium hydroxide, KOH, which is a base, along with an ethanolic solution, then aliphatic or aromatic amines are formed, and this reaction is called as Hoffmann’s bromamide reaction or Hoffmann bromamide degradation. The strong base NaOH is used to attack the amide, and deprotonate it and form an anion. The amide is degraded in this reaction that forms an amine which has one carbon atom less than the reactant amide. The reaction is as follows:
$C{{H}_{3}}CON{{H}_{2}}+B{{r}_{2}}+4NaOH\to C{{H}_{3}}N{{H}_{2}}+2NaBr+N{{a}_{2}}C{{O}_{3}}+2{{H}_{2}}O$
This reaction has ethanamide as an amide, that reacts with bromine and sodium hydroxide. The products formed are, ethanamine, sodium bromide, sodium carbonate, and water.
Hence, Hoffmann bromamide reaction results in the formation of amines, and has amide, bromine and sodium hydroxide as reactants.
Note:
Hoffmann degradation reaction is different from Hofmann elimination reaction. Hofmann elimination involves formation of tertiary amines and alkenes. The reactants are quaternary ammonium salt with excess methyl iodide. The resulting compound is quaternary ammonium iodide salt, which is treated with silver oxide and water to form an alkene and a tertiary amine.
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