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

Hint: 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.