In the below-given condition- intermediate -X- and reagent-condition -A- are-Aniline in the presence of NaNO2- HCl at 273-278 K forms -X- which further reacts with -A- to form phenol-

Explanation for the correct option:The correct option is (B)Case 1: When aniline reacts with NaNO2, HCl it will form diazonium salt which further on reaction wi ...

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Standard XII

Chemistry

Reactions of Aryl Amines (Diazo Salts)

In the below-...

Question

In the below-given condition, intermediate “X” and reagent/condition “A” are:
Aniline in the presence of NaNO₂, HCl at 273−278 K forms “X” which further reacts with “A” to form phenol.

“X”: Diazonium salt, “A”: H₂O/NaOH
“X”: Nitrobenzene, “A”: H₂O/NaOH
“X”: Diazonium salt, “A”: H₂O/heat
“X”: Nitrobenzene, “A”: H₂O/heat

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Solution

The correct option is C
“X”: Diazonium salt, “A”: H₂O/heat

Explanation for the correct option:
The correct option is (B)
Case 1: When aniline reacts with NaNO₂, HCl it will form diazonium salt which further on reaction with H₂O/NaOH forms phenylhydrazone.
Case 2: When aniline reacts with NaNO₂, HCl it will form diazonium salt which further on reaction with H₂O/heat it forms phenol with the evolution of nitrogen gas.
aniline diazonium salt phenol
To form nitrobenzene aniline should be treated with sodamide and ${HBF}_{4}$ , which is not mentioned in the above data.
Final Answer: So, the correct answer is option C.

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