Distinction Between Pairs of Amines - Class XII

Aniline is a primary amine, while N-methylaniline is a secondary amine. The Carbylamine test (C) is specific to primary amines, and the Hinsberg reagent (B) reacts differently with primary and secondary amines, making both useful for distinction.

Aniline and diphenylamine may be distinguished by carbylamine reaction.
Aniline is a primary amine and diphenylamine is a secondary amine.
carbylamine reaction is given by primary amines. It is not given by secondary or tertiary amines.
In Hoffmann's carbylamine test (also known as Isocyanide test), aliphatic or aromatic primary amine is heated with chloroform  in presence of alcoholic potassium hydroxide to give foul smelling alkyl/aryl  isocyanide or carbylamine.
$\underset {} {C _6H _5-NH _2} + CHCl _3  + 3KOH  \xrightarrow {heat} C _6H _5-NC  + 3KCl + 3H _2O$

Azo dye are a large class of synthetic organic dyes, that contain nitrogen as the azo group or $−N=N−$ as a part of their molecular structures. Orange-1 is an example of azo dye.

Separation of $1^o,, 2^o$ and $3^o$ amines:

Ethylamine (aliphatic) and aniline (aromatic) can be distinguished using the azo dye test. Aniline reacts with nitrous acid to form a diazonium salt, which couples with beta-naphthol to form an orange-red dye, whereas ethylamine does not form a stable diazonium salt under these conditions.

Pyridine is an aromatic amine and aromatic amine gives azo dye test which implies pyridine would also give azo-dye test.

p-chloroaniline and anilinium hydrochloride can be distinguished by by using $AgNO _3$ and $NaHCO _3$, With $AgNO _3 $ anilinium hydrochloride gives white precipitate of $AgCl$. 

As only aryl systems undergo azo dye test. Therefore, propylamine gives negative azo dye test.

Aniline and benzylamine both are aryl systems as only highly activated benzene system gives positive azo dye test below room temperature. Therefore, aniline and benzylamine cannot be distinguished by azo dye test as both give positive azo dye test.

Since the compound reacts with benzenesulphonyl chloride to give a product which is insoluble in alkali, it shows there is no $H$ attached to $N$ in the product . Hence, the compound $X$ is a secondary amine.
$CH _3 - NH - C-2H _5 + C _6H _5SO _2Cl \rightarrow \underset{N - Ethyl - N - methylbenzene sulphonamide}{CH _3 - {\underset{C _2H _5}{\underset{|}{N}}} - SO _2C _6H _5}$

Aniline is the colourless liquid A having boiling point $184 ^oC$.
On treatment with $NaNO _{2}:and:dil:HCl$ at low temperature, it forms benzne diazonum salt.
The diazonium salt reacts with napthol in alkaline medium to form azo dye which is n orange yellow precipitate.

The amine is treated with benzenesulphonyl chloride  and shaken with alkali solution when the three amine because in different way$.$

Salt solution such as sodium chloride ($NaCl$) conducts an electric current because it has ions in it that have the freedom to move about in solution. These ions are produced when sodium chloride dissolves in pure water to produce sodium ($Na^+$) and chloride ions ($Cl^–$).