Give the structures of A, B and C in the following reactions: CH 3 CH 2 I → NaCN A → Partial hydrolysis OH - B → NaOH + Br 2 C .

CH 3 CH 2 I → NaCN CH 3 CH 2 CN Propanenitrile A → Partial hydrolysis OH - CH 3 - C O ∥ - N Ethanamide B H 2 → NaOH + Br 2 CH 3 - NH 2 Methanamine C In the above reaction, alkyl halide is first converted into alkyl cyanide by nucleophilic substitution reaction then on partial alkaline hydrolysis acetamide is formed and then by Hoffmann's bromamide degradation reaction methanamine is formed as a final product. The amine formed which contains one carbon less than that parent amide.

Give the structures of A, B and C in the following reactions: CH 3 CH 2 Br → KCN A → LiAlH 4 B → 0 ∘ C HNO 2 C

CH 3 CH 2 Br → KCN CH 3 CH 2 CN Cyanoethane A → LiAlH 4 CH 3 CH 2 CH 2 NH 2 Propan - 1 - amine B → 0 ∘ C HNO 2 CH 3 CH 2 CH 2 OH Propan - 1 - ol C Ethyl bromide on reaction with potassium cyanide gives cyanoethane which on reduction gives Propan- 1 -amine. The resulting product on reaction with cold nitrous acid gives Propan- 1 -ol.

Give the structures of A, B and C in the following reactions: C 6 H 5 NO 2 → Fe / HCl A → 273 K NaNO 2 + HCl B → ∆ H 2 O / H + C

In the above reactions, the reduction of nitrobenzene gives aniline and then on diazotisation, benzene diazonium chloride formed then after acidic hydrolysis of benzene diazonium chloride gives phenol.

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.

E M B I B E

Chemistry Part 2 Textbook for Class 12>Chapter 4 - Amines>Exercises>Q 13.8

HARD

12th CBSE

IMPORTANT

Earn 100

Accomplish the following conversion:
Chlorobenzene to $p$ -chloroaniline

Important Questions on Amines

EASY

12th CBSE

IMPORTANT

Chemistry Part 2 Textbook for Class 12>Chapter 4 - Amines>Exercises>Q 13.8

Accomplish the following conversion: Aniline to

p

-bromoaniline.

EASY

12th CBSE

IMPORTANT

Chemistry Part 2 Textbook for Class 12>Chapter 4 - Amines>Exercises>Q 13.8

Accomplish the following conversion: Benzamide to toluene.

HARD

12th CBSE

IMPORTANT

Chemistry Part 2 Textbook for Class 12>Chapter 4 - Amines>Exercises>Q 13.8

Accomplish the following conversion:
Aniline to benzyl alcohol.

EASY

12th CBSE

IMPORTANT

Chemistry Part 2 Textbook for Class 12>Chapter 4 - Amines>Exercises>Q 13.9

Give the structures of A, B and C in the following reactions: ${CH}_{3} {CH}_{2} I \overset{NaCN}{\to} A \to_{Partial hydrolysis}^{{OH}^{-}} B \overset{NaOH + {Br}_{2}}{\to} C$ .