Identify X,Y, AND Z in the given reaction
$ C{}_6H{}_5CONH{}_2\mathop \to \limits_{KOH}^{Br{}_2} X\mathop \to \limits_{HCl}^{HNO{}_2} Y\mathop \to \limits_{Cu{}^ + }^{NaNO{}_2} Z $
Answer
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Hint :The given question can be solved with the help of Hoffmann bromamide reaction which is one of the most important reactions in organic chemistry. This reaction involves degradation of amide and hence its known as Hoffmann bromamide degradation reaction.
Complete Step By Step Answer:
The reaction given above is Hoffmann bromamide degradation which gives aniline. Aniline is diazotized which results in the formation of benzene diazonium chloride. This diazonium salt is further converted to nitrobenzene when treated with $ NaNO{}_2/Cu{}^ + $ the reaction is given by;
$ C{}_6H{}_5CONH{}_2\mathop \to \limits_{KOH}^{Br{}_2} C{}_6H{}_5NH{}_2\mathop \to \limits_{HCl}^{HNO{}_2} C{}_6H{}_5N{}_2Cl\mathop \to \limits_{Cu{}^ + }^{NaNO{}_2} C{}_6H{}_5NO{}_2 $
So the correct answer is;
X= $ C{}_6H{}_5NH{}_2 $
Y= $ C{}_6H{}_5N{}_2Cl $
Z= $ C{}_6H{}_5NO{}_2 $
Additional Information:
When an amide is treated with bromine in aqueous solution of sodium hydroxide which degrades amide which leads to the formation of primary amine. The reaction involves the degradation of amide which is known as Hoffmann bromamide degradation reaction. The primary amine thus formed will contain one carbon less than the number of carbon atoms in that amide.
The mechanism of this reaction is given below,
Deprotonation of amide takes place when the base attacks amide and generates
Anion reacts with bromine which leads to the formation bromoamide.
When deprotonation of bromoamide takes place there is formation of bromoamide anion which undergoes rearrangement and forms isocyanate.
Isocyanate forms carbamic acid when there is addition of water molecules.
Decarboxylation of carbamic acid occurs and forms primary amine.
Note :
The limitations of this reaction are that secondary and tertiary amide cannot be used in Hoffmann bromamide reaction which produces primary amines. The application includes; used in production of primary aromatic and aliphatic amines, used in the preparation of aniline etc.
Complete Step By Step Answer:
The reaction given above is Hoffmann bromamide degradation which gives aniline. Aniline is diazotized which results in the formation of benzene diazonium chloride. This diazonium salt is further converted to nitrobenzene when treated with $ NaNO{}_2/Cu{}^ + $ the reaction is given by;
$ C{}_6H{}_5CONH{}_2\mathop \to \limits_{KOH}^{Br{}_2} C{}_6H{}_5NH{}_2\mathop \to \limits_{HCl}^{HNO{}_2} C{}_6H{}_5N{}_2Cl\mathop \to \limits_{Cu{}^ + }^{NaNO{}_2} C{}_6H{}_5NO{}_2 $
So the correct answer is;
X= $ C{}_6H{}_5NH{}_2 $
Y= $ C{}_6H{}_5N{}_2Cl $
Z= $ C{}_6H{}_5NO{}_2 $
Additional Information:
When an amide is treated with bromine in aqueous solution of sodium hydroxide which degrades amide which leads to the formation of primary amine. The reaction involves the degradation of amide which is known as Hoffmann bromamide degradation reaction. The primary amine thus formed will contain one carbon less than the number of carbon atoms in that amide.
The mechanism of this reaction is given below,
Deprotonation of amide takes place when the base attacks amide and generates
Anion reacts with bromine which leads to the formation bromoamide.
When deprotonation of bromoamide takes place there is formation of bromoamide anion which undergoes rearrangement and forms isocyanate.
Isocyanate forms carbamic acid when there is addition of water molecules.
Decarboxylation of carbamic acid occurs and forms primary amine.
Note :
The limitations of this reaction are that secondary and tertiary amide cannot be used in Hoffmann bromamide reaction which produces primary amines. The application includes; used in production of primary aromatic and aliphatic amines, used in the preparation of aniline etc.
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