(i) A Friedel Crafts reaction is carried out in the presence of AlCl₃. But AlCl₃ used as catalyst and is acidic in nature i.e., Lewis acid whereas aniline is a strong Lewis base. Thus, aniline reacts with AlCl₃ to form a salt.
Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo
Friedel-Crafts reaction.
(ii) (CH₃)₂NH is more basic than (CH₃)₃ in an aqueous solution. + I effect will increase in alkyl group that results in increasing the case of donation of lone pair electron. Amine accepts a proton and form cation which will be stabilised in water by solvation. Higher the solvation by hydrogen bonding, higher will be the basic strength.
Therefore, with increase in methyl group, hydrogen bonding and stabilisation by solvation decreases. This net effect results in decrease of basic strength from secondary to tertiary amine.
(iii) In tertiary amine, there are no H-atoms whereas, in primary amines, two H-atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding.
As a result, extra energy is required to separate the molecules of primary amine. Therefore, primary amines have higher boiling point than tertiary amine.

(i) \({ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }\overset { Sn/HCI }{ \longrightarrow } { { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }\overset { NaNO_{ 2 }+HCI }{ \underset { 273K }{ \longrightarrow } } }{ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }^{ + }{ CI }^{ - }\overset { { H }_{ 2 }O }{ \longrightarrow } { C }_{ 6 }{ H }_{ 5 }OH+{ N }_{ 2 }+HCI\)
(ii)\({ C }{ H }_{ 3 }CN\overset { { H }_{ 2 }O/{ H }^{ + } }{ \longrightarrow } { C }{ H }_{ 3 }-\overset { \begin{matrix} O \\ \parallel \end{matrix} }{ C } -OH\overset { NH_{ 3 } }{ \underset { \triangle }{ \longrightarrow } } { CH }_{ 3 }{ CONH }_{ 2 }\overset { { Br }_{ 2 }+KOH }{ \longrightarrow } { CH }_{ 3 }{ NH }_{ 2 }\)

Hinsberg's test is used for the identification of primary, secondary, and tertiary amines.
Hinsberg's reagent is benzenesulphonyl chloride (C_6​H₅​SO_2​Cl).
It reacts differently with primary, secondary, and tertiary amines.
(i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.
Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.
(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.
Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.
(iii) Hinsberg's reagent does not react with tertiary amines.