Answer
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Hint: The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. In this process only pure primary amines are formed and not secondary or tertiary amines.
Complete step by step answer: This synthesis is used to get primary amines from primary alkyl halides. The reaction has been generalized for applications in the alkylation of sulfonamides and imides & their deprotection in order to obtain amines. Alkylation of ammonia is quite inefficient, therefore it is substituted with phthalimide anion in the Gabriel synthesis.In this reaction N-ethylphthalimide reacts with $N{H}_{2}N{H}_{2}$ in order to form phthalhydrazide and N-ethylamine.
The following is the reaction that is included in this process.
N-ethylphthalimide is made to react with hydrazine $N{H}_{2}N{H}_{2}$. In this N-ethylphthalimide is heated with hydrazine in order to form phthalhydrazide and n-ethylamine.
In this, the hydrazine gets attached to the carbonyl group and through a sequence of steps, the amine ends up leaving the group.
Note: N-ethyphthlamide can also be produced through a different approach also. That is with the use of an alkali KOH. Phthalimide is made to react with KOH. A good nucleophile in the form of an imide ion is formed when potassium hydroxide reacts with the phthalimide. The imide ion executes a nucleophilic substitution reaction on the alkyl halide and creates an intermediate – N-alkyl phthalimide. Then further hydrolysis of this phthalimide yields a primary alkyl amine. However, aryl amines cannot be prepared via Gabriel synthesis as aryl halides don’t undergo simple nucleophilic substitution.
Complete step by step answer: This synthesis is used to get primary amines from primary alkyl halides. The reaction has been generalized for applications in the alkylation of sulfonamides and imides & their deprotection in order to obtain amines. Alkylation of ammonia is quite inefficient, therefore it is substituted with phthalimide anion in the Gabriel synthesis.In this reaction N-ethylphthalimide reacts with $N{H}_{2}N{H}_{2}$ in order to form phthalhydrazide and N-ethylamine.
The following is the reaction that is included in this process.
N-ethylphthalimide is made to react with hydrazine $N{H}_{2}N{H}_{2}$. In this N-ethylphthalimide is heated with hydrazine in order to form phthalhydrazide and n-ethylamine.
![](https://www.vedantu.com/question-sets/ed5d7a62-988e-40c2-af7c-24212b566f984353738959940724688.png)
In this, the hydrazine gets attached to the carbonyl group and through a sequence of steps, the amine ends up leaving the group.
Note: N-ethyphthlamide can also be produced through a different approach also. That is with the use of an alkali KOH. Phthalimide is made to react with KOH. A good nucleophile in the form of an imide ion is formed when potassium hydroxide reacts with the phthalimide. The imide ion executes a nucleophilic substitution reaction on the alkyl halide and creates an intermediate – N-alkyl phthalimide. Then further hydrolysis of this phthalimide yields a primary alkyl amine. However, aryl amines cannot be prepared via Gabriel synthesis as aryl halides don’t undergo simple nucleophilic substitution.
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