March 1, 2024

Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 4 on Amines, complete with easy-to-follow step-by-step explanations. Widely favored among Class 12 Science students, these Chemistry Amines Solutions are invaluable for efficiently tackling homework assignments and gearing up for exams. Free of charge, you can access all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry, ensuring a convenient resource for your academic needs. Page No 384: Question 13.1: Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH ANSWER: Primary: (i) and (iii) Secondary: (iv) Tertiary: (ii) Page No 384: Question 13.2: (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? ANSWER: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below: (a) CH3-CH2-CH2-CH2-NH2 Butanamine (10) (b)  Butan-2-amine (10) (c)  2-Methylpropanamine (10) (d) 2-Methylpropan-2-amine (10) (e) CH3-CH2-CH2-NH-CH3 N-Methylpropanamine (20) (f) CH3-CH2-NH-CH2-CH3 N-Ethylethanamine (20) (g) N-Methylpropan-2-amine (20) (h) N,N-Dimethylethanamine (3°) (iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism. The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism. The pairs (e) and (f) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa. Page No 387: Question 13.3: How will you convert? (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine? ANSWER: (i) (ii) (iii) Page No 396: Question 13.4: Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. ANSWER: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as: Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have: Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as: (ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows: Again, due to the −R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows: (iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as: In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2­NH2 is more basic than C6H5NH2. Again, due to the −I effect of C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows. Page No 396: Question 13.5: Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl  (ii) (C2H5)3N + HCl  ANSWER: (i)  (ii) (C2H5)3N Triethylamine+ HCl → (C2H5)3N+HCl−TriethylammoniumchlorideC2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride Page No 396: Question 13.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. ANSWER: Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate. Page No 396: Question 13.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. ANSWER: Page No 396: Question 13.8: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. ANSWER: The structures of different isomers corresponding to the molecular formula, C3H9N are given below: (a)  Propan-1-amine (10) (b) Propan-2-amine (10) (c) (d) N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid. Page No 399: Question 13.9: Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene. ANSWER: (i) (ii) Page No 400: Question 13.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii) m−BrC6H4NH2 ANSWER: (i) 1-Methylethanamine (10 amine) (ii) Propan-1-amine (10 amine) (iii) N−Methyl-2-methylethanamine (20 amine) (iv) 2-Methylpropan-2-amine (10 amine) (v) N−Methylbenzamine or N-methylaniline (20 amine) (vi) N-Ethyl-N-methylethanamine (30 amine) (vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine) Page No 400: Question 13.2: Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. ANSWER: (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not. (ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl). Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent. (iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions. ​ (iv) Aniline …

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