Question

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{ }}\xrightarrow[{0^\circ C}]{{{\text{NaN}}{{\text{O}}_2} + {\text{HCl}}}}{\text{X}}\xrightarrow[{warm}]{{water}}{\text{Y }}$
In the above reaction, Y is

{\text{B ) }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}} \\\ {\text{C ) }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH }} \\\ {\text{D ) }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{CHO }} \\\\$$

Answer 1

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$ is aniline. In the given reaction aniline reaction with the diazonium salt and there is formation of product X which further reacts with water and form Y. We have to find out the Y product in the given reaction. Let's discuss it step by step.

Complete step by step answer:
Aniline is an aromatic primary amine. In aniline, the amino group is directly attached to the phenyl ring. Aromatic primary amines react with nitrous acid at low temperature to form diazonium salts.

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{ }}\xrightarrow[{0^\circ C}]{{{\text{NaN}}{{\text{O}}_2} + {\text{HCl}}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{N}}_{\text{2}}}{\text{Cl}}\xrightarrow[{warm}]{{water}}{\text{Y }} \\\\$
Hence, in the above reaction, X is benzene diazonium chloride. The reaction is known as diazotization reaction.
Nitrous acid is produced in situ by using a mixture of nitrous acid and hydrochloric acid. Here, in situ means that the reagent is generated as such in the reaction flask.

${\text{NaN}}{{\text{O}}_{\text{2}}}{\text{ + HCl}} \to {\text{HN}}{{\text{O}}_2}{\text{ + NaCl}}$
The diazonium group of benzene diazonium chloride is highly unstable and using appropriate reagents, the diazonium group can be replaced with a variety of functional groups. Thus, different compounds can be prepared from primary aromatic amines using diazotization reaction.

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{ }}\xrightarrow[{0^\circ C}]{{{\text{NaN}}{{\text{O}}_2} + {\text{HCl}}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{N}}_{\text{2}}}{\text{Cl}}\xrightarrow[{warm}]{{water}}{\text{Y }} \\\\$
When benzene diazonium chloride is mixed with warm water, phenol is obtained. The diazonium group is replaced with phenolic hydroxyl group.

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}{\text{ }}\xrightarrow[{0^\circ C}]{{{\text{NaN}}{{\text{O}}_2} + {\text{HCl}}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{N}}_{\text{2}}}{\text{Cl}}\xrightarrow[{warm}]{{water}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \\\\$
Hence, in the above reaction, the end product Y is phenol.

Thus, option C ) is the correct answer.

Note: Diazonium salt is converted to chlorobenzene via Sandmeyer reaction. The reagent is cuprous chloride in presence of hydrogen chloride. Diazonium salt is converted to benzene by reaction with hypophosphorous acid. Diazonium salt cannot be directly converted to benzaldehyde into single step. Several steps are necessary for this conversion.