Identify A and B in the following reaction. C 6 H 5 N 2 Cl → ∆ HOH A → Br 2 B

The diazonium ions undergoes hydrolysis and on strong heating gives phenol as main product and liberate N 2 gas. Phenol on reacting with Br 2 undergoes electrophilic substitution mechanism giving tribromo substitution product. C 6 H 5 N 2 Cl → ∆ H - OH C 6 H 5 OH + N 2

Identify A , B and C in the following reactions. C 6 H 5 N 2 Cl → HCN CuCN A → H 2 O / H + B

Benzene diazonium chloride on reaction cuprous cyanide gives cyano benzene which on hydrolysis in acidic medium give benzoic acid. The first step of the reaction is electrophilic substitution reaction, and second step is hydrolysis.

Arrange the following in the increasing order of their basic strength: C 2 H 5 NH 2 , C 6 H 5 NH 2 , ( C 2 H 5 ) 2 NH .

Basicity of amines depends on the electron density on nitrogen atom. It means that those group which will have donating nature will increase the electron density. C 2 H 5 NH 2 - This amine has one alkyl group which shows +I effect therefore it will increase the electron density on nitrogen atom. C 2 H 5 2 NH - This amine contain two alkyl group which show +I effect therefore it will increase the electron density on nitrogen atom. But there is more electron density on the nitrogen atom due to two +I groups. Hence, it is more stronger base than C 2 H 5 NH 2 . C 6 H 5 NH 2 - In aniline there is one benzene group. Benzene shows resonance effect due to which it takes the electrons form the nitrogen and hence there is less electron density on the nitrogen atom. Therefore, it is the weakest base among the three. The increasing order of basic strength given organic compounds: C 6 H 5 NH 2 < C 2 H 5 NH 2 < ( C 2 H 5 ) 2 NH

Describe a method for the identification of primary, secondary and tertiary amines. Also, write chemical equations of the reactions involved.

Hinsberg’s test is used to identify and distinguish primary, secondary and tertiary amines. In this test, the amines react with Hinsberg’s reagent, benzene sulphonyl chloride. Different types of amines react differently with the Hinsberg’s reagent. Primary amines react with benzene sulphonyl chloride to form N -alkyl benzene sulphonyl amide which is soluble in alkali. C 6 H 5 SO 2 Cl Benzene sulphonyl chloride + H 2 N - C 3 H 7 Propanamine → - HCl C 6 H 5 SO 2 NHC 3 H 7 N - Propylbenzene sulphonamide . Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide, the H -atom attached to nitrogen can be easily released as a proton. Therefore, it is acidic and dissolves in an alkali. Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali. C 6 H 5 SO 2 Cl Benzene sulphonyl chloride + HN ( CH 3 ) 2 Dimethylamine → - HCl C 6 H 5 SO 2 N ( CH 3 ) 2 N , N - Dimethylbenzene sulphonamide . There is no H -atom attached to the N -atom in the sulphonamide. Therefore, it is not acidic and is insoluble in an alkali. Tertiary amines do not react with Hinsberg’s reagent.