Question

In diazotization reaction of aniline with \[{\rm{NaN}}{{\rm{O}}_2}\] and HCl, an excess of HCl is used to:
A) suppress the formation of acetanilide
B) suppress the concentration of free aniline available for coupling
C) maintain the pH of the reaction
D) prevent the formation of anilinium ion

Answer 1

Hint: Benzene diazonium chloride (the diazonium salt) is obtained by the action of nitrous acid on aniline at low temperature. Benzene diazonium chloride (the diazonium salt) is used in the preparation of azo dyes.

 Complete step by step answer:
Aniline reacts with nitrous acid to form benzene diazonium chloride (the diazonium salt) which is highly reactive (unstable). This reaction is carried out at low temperature of \[0^\circ {\rm{C}} - 5^\circ {\rm{C}}\]. Nitrous acid is prepared in situ by reaction of \[{\rm{NaN}}{{\rm{O}}_2}\] and HCl.

\[{{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}}{{\rm{H}}_2}{\rm{ }} \xrightarrow [0^oC-5^oC]{NaNO_2/HCl}
  {\rm{ }}{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{N}}_2}{\rm{Cl}}\]

In the above reaction, the aromatic amino group is converted into a diazo group.
Azo dyes are prepared by coupling of benzene diazonium chloride with either phenol or aromatic amine in presence of sodium hydroxide.

\[{{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}}{{\rm{H}}_2}{\rm{ + }}{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{N}}_2}{\rm{Cl }} \xrightarrow {NaOH}
 {\rm{ }}{{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}} = {\rm{N}} - {{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}}{{\rm{H}}_2}\]
In the above reaction, aniline reacts with benzene diazonium chloride, in presence of sodium hydroxide, to form azo dye. This reaction is known as azo coupling.

To avoid azo coupling, the diazotization reaction is carried out in presence of excess HCl. Free aniline reacts with excess HCl to form hydrochloride salt and is not available for azo coupling. In this way, the concentration of free aniline is suppressed.

\[{{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}}{{\rm{H}}_2}{\rm{ + HCl }} \to {\rm{ }}{{\rm{C}}_6}{{\rm{H}}_5}{\rm{NH}}_3^ + {\rm{C}}{{\rm{l}}^ - }\]

For azo coupling, aniline should be present as a free base. Hydrochloride salt of aniline does not participate in azo coupling. Thus, conversion of aniline into hydrochloride salt can easily avoid the side reaction of azo coupling.

Hence, option B is the correct answer.

Note:
Acetanilide \[\left( {{{\rm{C}}_6}{{\rm{H}}_5} - {\rm{NHCOC}}{{\rm{H}}_3}} \right)\] is not formed in this reaction. If excess HCl is used then formation of anilinium ion \[\left( {{{\rm{C}}_6}{{\rm{H}}_5}{\rm{NH}}_3^ + } \right)\] cannot be prevented, as aniline (base) readily reacts with protons (from hydrochloric acid) to form anilinium ion.