Write the following reactions with chemical equation- (i) Carbyl-amine reaction (ii) Hoffmann Bromamide reaction

(i) Carbylamine reaction: Aliphatic or aromatic 1 0 amines react with CHCl 3 and alcoholic potassium hydroxide to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test is used for the identification of 1 0 amines. C 2 H 5 NH 2 Ethylamine + CHCl 3 Chloroform + 3 KOH → C 2 H 5 NC Ethylisocyanide + 3 KCl + 3 H 2 O (ii) Hoffmann Bromamide reaction: When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. R - CO - NH 2 Amide + Br 2 + 4 NaOH → R - NH 2 1 ° Amine + 2 NaBr + Na 2 CO 3 + 2 H 2 O

Write short notes on the reduction of amides.

Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.

Write short notes on Diazotisation

Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273 - 278 K . Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The conversion of primary aromatic amines into diazonium salts is known as diazotisation. Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation. C 6 H 5 NH 2 + NaNO 2 + HCl → 273 - 278 K C 6 H 5 N 2 Cl + NaCl + 2 H 2 O

How would you convert Aniline into following? Write with chemical equation- (a) Chlorobenzene (b) Fluorobenzene

(a) Conversion of aniline to chlorobenzene: Step 1: Sodium nitrite and hydrochloric acid is added to aniline at very low temperatures ( 0 – 5 º C ) to form a benzene diazonium salt. Step 2: This diazonium salt in presence of leftover hydrochloric acid is further treated with Cu 2 Cl 2 to form chlorobenzene. (b) Conversion of aniline to fluorobenzene: Step 1: Sodium nitrite and hydrochloric acid is added to aniline at very low temperatures ( 0 – 5 º C ) to form a benzene diazonium salt. Step 2: This diazonium salt on heating with fluoroboric acid forms fluorobenzene.

Arrange the following in decreasing order of their basic strength- NH 3 , C 2 H 5 NH 2 , C 2 H 5 2 NH , C 6 H 5 NH 2

As we know that, Basic character ∝ + I effect Due to + I effect of - C 2 H 5 , the basic character should be ( C 2 H 5 ) 2 N H > C 2 H 5 N H 2 in the gaseous phase. But in aqueous state hydrogen bonding is also taking place and extent of hydrogen bonding is 1 ° > 2 ° > 3 ° amine. So, overall basic character in aqueous medium is the result of inductive effect and hydrogen bonding. So, order is 2 o > 3 o > 1 o amine (Aliphatic). Considering the inductive effect and the steric hindrance of the alkyl groups in C 2 H 5 N H 2 a n d C 2 H 5 2 N H , their basic strengths, can be arranged as follows: C 2 H 5 NH 2 < C 2 H 5 2 NH Again, due to the - R effect of C 6 H 5 group, the electron density on the N -atom in C 6 H 5 N H 2 is lower than that on the N -atom in C 2 H 5 N H 2 . Therefore, the basicity of C 6 H 5 N H 2 is lower than that of C 2 H 5 N H 2 i.e., aromatic amine is less basic than aliphatic amine. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows: ( C 2 H 5 ) 2 NH > C 2 H 5 NH 2 > C 6 H 5 NH 2 > NH 3 (overall order)

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.

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What do you mean by Diazotisation?

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