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Question: Cyclohexylamine and aniline can be distinguished by: A. Hinsberg test B. Carbylamine test C. L...

Cyclohexylamine and aniline can be distinguished by:
A. Hinsberg test
B. Carbylamine test
C. Lassaigne test
D. Azo dye test

Explanation

Solution

We know that, the test used for the detection of aromatic primary amine is done by dissolving primary amine in dilute hydrochloric acid and cool it to 273278  K273 - 278\;{\rm{K}} and then we will treat it with an ice cold solution of HNO2{\rm{HN}}{{\rm{O}}_2}. Then we will add the resulting solution in a cold alkaline solution of phenol.

Step by step answer: We will get to see the clear observation of the appearance of an orange or red dye. Therefore, we can now confirm that there is the presence of primary amine. This test is known as the Azo dye test.
The given two compounds to us in the question are cyclohexylamine and aniline. We can classify then as primary aliphatic amine and primary aromatic amine respectively.
We can show the chemical equation for the formation of azo dye from aniline as follows.
C6H5NH2+HNO2+HClC6H5N2Cl+2H2O{{\rm{C}}_6}{{\rm{H}}_5}{\rm{N}}{{\rm{H}}_2} + {\rm{HN}}{{\rm{O}}_2} + {\rm{HCl}} \to {{\rm{C}}_6}{{\rm{H}}_5}{{\rm{N}}_2}{\rm{C}}{{\rm{l}}^ - } + 2{{\rm{H}}_2}{\rm{O}}
Now, the second step is the reaction of C6H5N2Cl{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{N}}_2}{\rm{C}}{{\rm{l}}^ - } with phenol which we can show as follows.
C6H5N2Cl+C6H5OHC6H5N=NC6H4OH+HCl{{\rm{C}}_6}{{\rm{H}}_5}{{\rm{N}}_2}{\rm{C}}{{\rm{l}}^ - } + {{\rm{C}}_6}{{\rm{H}}_5}{\rm{OH}} \to {{\rm{C}}_6}{{\rm{H}}_5} - {\rm{N}} = {\rm{N}} - {{\rm{C}}_6}{{\rm{H}}_4} - {\rm{OH}} + {\rm{HCl}}
Cyclohexylamine being an aliphatic amine cannot form azo dye. When we treat cyclohexylamine in the same reaction conditions we will observe that it will result in a brisk evolution of nitrogen gas along with the formation of primary alcohol. We will also observe that the solution will remain clear.
We can show the chemical equation for the reaction of cyclohexylamine with HNO2{\rm{HN}}{{\rm{O}}_2} as follows.
C6H11NH2+HNO2C6H11OH+N2+H2O{{\rm{C}}_6}{{\rm{H}}_{11}}{\rm{N}}{{\rm{H}}_2} + {\rm{HN}}{{\rm{O}}_2} \to {{\rm{C}}_6}{{\rm{H}}_{11}}{\rm{OH}} + {{\rm{N}}_2} \uparrow + {{\rm{H}}_2}{\rm{O}}

Hence, we can conclude that the correct option is D.

Note: We cannot use the carbylamine test for the comparison between cyclohexylamine and aniline because they both are primary amines. When we heat both these amine with an alcoholic solution of KOH{\rm{KOH}} and CHCl3{\rm{CHC}}{{\rm{l}}_{\rm{3}}}, they both gives offensive smell of isocyanide so formed. That’s why we cannot use this test for both the given compounds to us.