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NCERT Class 12 Chemistry Chapter 13 Amines
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Amines
Chapter: 13
| Part – II |
1. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3)2CHNH2
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(ii) CH3(CH2)2NH2
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(iii) CH3NHCH(CH3)2
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(iv) (CH3)3CNH2
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(v) C6H5NHCH3
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(vi) (CH3CH2)2NCH3
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(vii) m–BrC6H4NH2
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2. Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine.
Ans: These can be distinguished by carbylamine test. When heated with an alcoholic solution of KOH and chloroform, methyl-amine gives a foul smell of methyl isocyanide.
(ii) Secondary and tertiary amines.
Ans: Secondary and Tertiary Amines: Shake the given amines separately with Hinsberg’s reagent (benzene sulphonyl chloride) in the presence of excess aqueous KOH solution.
A secondary amine reacts to form N,N-dialkyl benzene sulphonamide, which is insoluble in aqueous solutions due to its hydrophobic nature.
(iii) Ethylamine and aniline.
Ans: To ice-cold solutions of ethylamine and aniline in excess dilute HCl, add ice-cold sodium nitrite solution and 2-naphthol prepared in dilute sodium hydroxide, then cool the mixture further.The mixture which forms a brilliant orange dye (azodye) contains aniline while the one in which no dye is formed has ethylamine present in it.
(iv) Aniline and benzylamine.
Ans: Aniline on diazotization (ice cold nitrous acid solution) followed by coupling with 2-naphtol (in alkaline solution) forms brilliant orange or red dye. Benzylamine will not give such test.
(v) Aniline and N-methylaniline.
Ans: Aniline and N-Methylaniline Carbylamine Test: Heat both the compounds separately with chloroform and alcoholic KOH. Aniline produces an unpleasant or offensive odour, while N-methylaniline does not emit any noticeable smell, making it more tolerable in terms of scent.
3. Account for the following:
(i) pKb of aniline is more than that of methylamine.
Ans: pKb of aniline is more than that of methylamine, Because aniline is less basic than methylamine due to stabilisation lone pair of N by resonance effect.
(ii) Ethylamine is soluble in water whereas aniline is not.
Ans: Ethylamine is soluble in water whereas aniline is not. With increase in the molecular weight, the solubility decreases. Aniline has higher molecular weight than ethylamine.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Ans: Methylamine is more basic than water due to the +I effect of the −CH3 group. It increases the OH⁻ concentration by accepting H⁺ from water. FeCl₃ dissociates into Fe³⁺ and Cl⁻ ions, and Fe³⁺ reacts with OH⁻ to form Fe(OH)₃, precipitating as hydrated ferric oxide.
(iv) Although the amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
Ans: −NO2 group that is nitro group is an electron withdrawing group. so it does not undergo friedel craft reaction an ortho and para position as it deactivates benzene ring mainly at ortho and para position for electrophilic substitution reaction.there fore meta product is formed so ans is true.
(v) Aniline does not undergo Friedel-Crafts reaction.
Ans: Aniline does not under Friedel Craft reaction (alkylation and acetylation) due to the salt formation with aluminium chloride, the Lewis acid which is used as a catalyst. This positive charge deactivates the benzene ring towards further electrophilic substitution.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Ans: Primary aliphatic amines produce highly unstable alkyl diazonium salts, while primary aromatic amines form aryl diazonium salts, which remain stable for a short time in solution at 273-278 K. The stability of diazonium ions of aromatic amines is explained on the basis of resonance.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: Using excess of ammonia in ammonolysis of alkyl halide is not a very practicable method for the synthesis of primary amines. Alternatively, phthalimide is alkylated with alkyl or benzyl halide, followed by hydrolysis or hydrazinolysis to yield pure primary amines. Phthalic acid formed can be recycled into phthalimide for reuse.
4. Arrange the following:
(i) In decreasing order of the pKb values:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
Ans: (C2H5)2NH > C2H5NH2 > C6H5NHCH3 > C6H5NH2
(ii) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
Ans: C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine.
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(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.
Ans: C6H5NH2 < C6H5CH2NH2 < C6H5NHCH3.
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
Ans: (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
(v) In increasing order of boiling point:
C2H5OH, (CH3)2NH, C2H5NH2
Ans: (CH3)2NH < C2H5NH2 < C2H5OH2
(vi) In increasing order of solubility in water:
C6H5NH2 , (C2H5 )2NH, C2H5NH2 .
Ans: C6H5NH2 < C2H5NH2 < (C2H5)2NH
5. How will you convert:
(i) Ethanoic acid into methanamine.
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(ii) Hexanenitrile into 1-aminopentane.
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(iii) Methanol to ethanoic acid.
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(iv) Ethanamine to methanamine.
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(v) Ethanoic acid into propanoic acid.
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(vi) Methanamine into ethanamine.
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(vii) Nitromethane into dimethylamine.
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(viii) Propanoic acid into ethanoic acid?
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6. Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Ans: Primary amine reacts with benzene sulphonyl chloride with primary amine produces N-alkyl benzene sulphonamide. The hydrogen attached to nitrogen in the sulfonamide product of primary amines is acidic, making it soluble in alkali.
Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.
Tertiary amines does not react with benzene sulphonyl chloride.
7. Write short notes on the following:
(i) Carbylamine reaction.
Ans: Carbylamine reaction is also known as Hofmann’s isocyanide test. The carbylamine reaction is a chemical test that detects primary amines and produces isocyanides, which have a terrible odour. Carbylamine reaction is used as a test for the identification of primary amines.
(ii) Diazotization.
Ans: Diazotization is a chemical reaction that converts primary aromatic amines into diazonium salts. It’s an important reaction in organic chemistry, especially for synthesising aromatic compounds. The process of conversion of a primary aromatic amino compound into a diazonium salt is known as diazotization. They are used to prepare benzene, phenol,and haloarenes.
(iii) Hofmann’s bromamide reaction.
Ans: Hofmann’s bromamide reaction involves degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. The Hofmann bromamide reaction is an organic reaction that converts a primary amide into a primary amine.
(iv) Coupling reaction.
Ans: A coupling reaction refers to the reaction between a diazonium salt and a phenol or an aromatic amine to form an azo compound, which is typically brightly coloured and used in dyes.
ArN2 + Cl− + Ar′−OH → Ar−N = N − Ar′ − OH
(v) Ammonolysis.
Ans: Ammonolysis is a chemical reaction that involves ammonia and is used in a variety of ways,Ammonolysis is the process where an organic compound reacts with ammonia (NH₃), replacing a functional group such as a halogen or an ester with an amino group. It is commonly used in the synthesis of amines. Ammonolysis is similar to hydrolysis, but in ammonolysis, ammonia ions replace water molecules.
RX + NH3 → RNH2 + HX
(vi) Acetylation.
Ans: Acetylation is a chemical reaction that adds an acetyl group to a chemical compound, replacing a hydrogen atom. When an alkyl or benzyl halide reacts with an ethanolic solution of ammonia, it undergoes a nucleophilic substitution reaction where the halogen atom is replaced by an amino (−NH₂) group. This process of cleavage of the carbon-halogen bond is known as ammonolysis.
(vii) Gabriel phthalimide synthesis.
Ans: Gabriel phthalimide synthesis is a chemical reaction that converts primary alkyl halides into primary amines. gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide.
8. Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid.
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(ii) Benzene to m-bromophenol.
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(iii) Benzoic acid to aniline.
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(iv) Aniline to 2,4,6-tri bromofluorobenzene.
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(v) Benzyl chloride to 2-phenylethanamine.
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(vi) Chlorobenzene to p-chloroaniline.
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(vii) Aniline to p-bromoaniline.
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(viii) Benzamide to toluene.
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(ix) Aniline to benzyl alcohol.
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9. Give the structures of A, B and C in the following reactions:
(i)
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(ii)
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(iii)
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(iv)
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(v)
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(vi)
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10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.
Ans: (i) determine the structures of B and C. Compound ‘B’ on heating with Br, and forms compound C. So, ‘B’ is amide and ‘C’ is amine. The molecular formula C₆H₇N of compound ‘C’ confirms it is aniline, with the structure C₆H₅NH₂, an aromatic amine.
(ii) To determine the structure of compound ‘A’, it reacts with aqueous ammonia to form an amide (‘B’), indicating that ‘A’ is likely an acyl halide, specifically benzoyl chloride (C₆H₅COCl).
(a) (C₆H₅COCl): IUPAC name: Benzoyl chloride.
(b) (C₆H₅CONH₂): IUPAC name: Benzamide.
(c) (C₆H₇N): IUPAC name: Aniline.
11. Complete the following reactions:
(i) C6H5NH2 + CHCl3 + alc.KOH →
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(ii) C5H5N2Cl + H3PO2H2O →
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(iii) C6H5NH2 + H3SO4(conc.) →
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(iv) C6H5N2CI +C2H5OH →
Ans: C6H5N2CI + C2H5OH → C6H5 + N2 + HCI + CH3CHO
(v) C6H5NH2 + Br2(aq) →
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(vi) C6H5N2 + (ch3co)2O →
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(vii)
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12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans: In Gabriel phthalimide reaction, potassium salt of phthalimide is formed. It reacts readily with alkyl halide to form the corresponding alkyl derivative.
Aryl halides don’t react with potassium phthalimide due to the partial double bond character of the C-X bond in haloarenes, making aromatic amines unpreparable via Gabriel phthalimide synthesis.
13. Write the reactions of:
(i) aromatic. and
Ans: Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273- 278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.
(ii) aliphatic primary amines with nitrous acid.
Ans: Amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.
14. Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
Ans: Amines are less acidic than alcohols of comparable molecular masses because the N-H bond is less polar than O-H bond. Hence, amines release H+ ion with more difficulty as compared to alcohol. RO-H5+ as compared to RN-H bond in RNH₂. Alcohols can lose protons to some extent, but amines are proton acceptors.
(ii) Why do primary amines have higher boiling point than tertiary amines?
Ans: In primary amines, N atoms have two H atoms which results in extensive intermolecular H bonding. In tertiary amines, N atoms do not have H atoms and hydrogen bonding is not possible. Hence, primary amines have higher boiling point than tertiary amines.
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans: Aromatic amines exhibit resonance, causing the lone pair of electrons on the nitrogen atom to become delocalized over the benzene ring, making them less available for protonation. The stability of aryl amine ions is lower than the stability of alkyl amines. Protonation of aromatic amines is not favoured.
