
From which of the following does Nitrobenzene give N-phenylhydroxylamine?
A) \[Sn\]/\[HCl\]
B) ${H_2}$/$Pd - C$
C) $Zn$/$NaOH$
D) $Zn$/$N{H_4}Cl$
Answer
484.2k+ views
Hint:As we know that aromatic nitro compounds undergo reduction under different conditions to provide a variety of products. Their reduction can be carried out in four mediums which are acidic, basic, neutral and electrolytic and phenylhydroxylamine is obtained in a neutral medium only.
Complete answer
As we know that aromatic nitro compounds react with different reducing agents to provide a variety of products under various conditions of solutions. In the given question nitrobenzene is giving N-phenylhydroxylamine so we should have an amine, hydroxyl and phenyl groups in our product.
In acidic medium, when \[Sn\]/\[HCl\] is present as a reagent and we know that it is a strong reducing agent, so it will reduce the nitro group in nitrobenzene into aniline and release two molecules of water as well, thus it is not a correct answer.
Next we have ${H_2}$/$Pd - C$ as a reagent and it is also a strong reducing agent acting as a catalyst so it will also reduce the nitro group to amine and results in aniline formation.
Then comes the reagent $Zn$/$NaOH$ which in alkaline medium is again a reducing agent and it will not be able to provide the hydroxyl group to the product rather under different conditions it will form different products. Like when $Zn$ dust and aqueous $NaOH$ is present as a reagent the product formed will be Hydrazobenzene. When $Zn$/$NaOH$ is present along with $C{H_3}OH$ it will form Azobenzene as a product.
Lastly the reagent is $Zn$/$N{H_4}Cl$ which in neutral medium that means $Zn$ dust and aqueous$N{H_4}Cl$ are present in solution, so they will reduces the nitrobenzene and results in the formation of N-phenyl hydroxylamine.
Therefore the correct answer is option (D).
Note:Most of the aromatic nitro compounds are yellow crystalline solids whereas the nitrobenzene is a liquid with a characteristic odour of bitter almonds. Zinc dust and aqueous ammonium chloride in solution acts as a mild reducing agent thereby reducing the nitro group to amine.
Complete answer
As we know that aromatic nitro compounds react with different reducing agents to provide a variety of products under various conditions of solutions. In the given question nitrobenzene is giving N-phenylhydroxylamine so we should have an amine, hydroxyl and phenyl groups in our product.
In acidic medium, when \[Sn\]/\[HCl\] is present as a reagent and we know that it is a strong reducing agent, so it will reduce the nitro group in nitrobenzene into aniline and release two molecules of water as well, thus it is not a correct answer.
Next we have ${H_2}$/$Pd - C$ as a reagent and it is also a strong reducing agent acting as a catalyst so it will also reduce the nitro group to amine and results in aniline formation.
Then comes the reagent $Zn$/$NaOH$ which in alkaline medium is again a reducing agent and it will not be able to provide the hydroxyl group to the product rather under different conditions it will form different products. Like when $Zn$ dust and aqueous $NaOH$ is present as a reagent the product formed will be Hydrazobenzene. When $Zn$/$NaOH$ is present along with $C{H_3}OH$ it will form Azobenzene as a product.
Lastly the reagent is $Zn$/$N{H_4}Cl$ which in neutral medium that means $Zn$ dust and aqueous$N{H_4}Cl$ are present in solution, so they will reduces the nitrobenzene and results in the formation of N-phenyl hydroxylamine.
Therefore the correct answer is option (D).
Note:Most of the aromatic nitro compounds are yellow crystalline solids whereas the nitrobenzene is a liquid with a characteristic odour of bitter almonds. Zinc dust and aqueous ammonium chloride in solution acts as a mild reducing agent thereby reducing the nitro group to amine.
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