Question

Which of the following compounds react with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ and ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$ to give alcohol/phenol?
A. ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$
B. ${{\text{C}}_2}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$
C. ${\text{C}}{{\text{H}}_3}{\text{NHC}}{{\text{H}}_3}$
D. ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{NHC}}{{\text{H}}_{\text{3}}}$

Answer 1

The reagent ${\text{NaN}}{{\text{O}}_{\text{2}}}$ and ${\text{HCl}}$ used to distinguish primary, secondary and tertiary amines. Only primary aliphatic amines react with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ and ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$ . The product of the reaction is alcohol.

Complete Step by step answer: The reagent given to us is ${\text{NaN}}{{\text{O}}_{\text{2}}}$ and ${\text{HCl}}$ and the reaction condition is ${\text{0 - 4}}^\circ {\text{C}}$. Only primary aliphatic amines react with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ and ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$ and give alcohol as the product.
The amine given in option A is ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$. Its structure is as follows:

It is aromatic amine so it will not give phenol after reacting with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$.
Thus, option (A) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$ is incorrect.
The amine given in option B is${{\text{C}}_2}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$. Its structure is as follows:

It is a primary aliphatic amine so it will give alcohol after reacting with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$.
So, option (B) ${{\text{C}}_2}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$is correct.
The amine given in option C is ${\text{C}}{{\text{H}}_3}{\text{NHC}}{{\text{H}}_3}$. Its structure is as follows:

It is secondary aliphatic amine so it will not give alcohol after reacting with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$.
Thus, option (C) ${\text{C}}{{\text{H}}_3}{\text{NHC}}{{\text{H}}_3}$ is incorrect.
The amine given in option D is ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{NHC}}{{\text{H}}_{\text{3}}}$. Its structure is as follows:

It is aromatic secondary amine so it will not give phenol after reacting with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$.
Thus, option (D) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{NHC}}{{\text{H}}_{\text{3}}}$ is incorrect.

Hence, option (B) ${{\text{C}}_2}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$is the correct answer.

Note: Aliphatic primary amines react with${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ at ${\text{0 - 4}}^\circ {\text{C}}$and undergo diazotization to form alkane diazonium salt, which however being unstable decomposes to form a mixture of alcohols, alkene with the liberation of ${{\text{N}}_{\text{2}}}$ gas. Secondary amines react with ${\text{NaN}}{{\text{O}}_{\text{2}}}$ in ${\text{HCl}}$ to form N-nitrosamines.