CBSE · NCERT · Class 12 Chemistry · Chapter 9

NCERT Solutions: Class 12 Chemistry Chapter 9 - Amines

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Chapter-wise NCERT intext questions and exercise answers for Amines, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Q.9.1Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2CHNH2(ii) CH3(CH2)2NH2(iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2(v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii) m–BrC6H4NH2v
Solution

Primary amines have one carbon group on nitrogen, secondary amines have two, and tertiary amines have three. In IUPAC names, N-substituents are indicated with the locant N; substituted anilines are named with –NH2 on benzene as the parent.

Answer:

(i) propan-2-amine, primary. (ii) propan-1-amine, primary. (iii) N-methylpropan-2-amine, secondary. (iv) 2-methylpropan-2-amine, primary. (v) N-methylaniline, secondary. (vi) N-ethyl-N-methylethanamine, tertiary. (vii) 3-bromoaniline, primary.

Q.9.2Give 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.v
Solution

(i) Methylamine gives carbylamine test with CHCl3/alc. KOH; dimethylamine does not. (ii) In Hinsberg test, secondary amine gives an insoluble sulphonamide, while tertiary amine does not form sulphonamide and dissolves in acid. (iii) Aniline gives diazonium salt at 273-278 K and azo dye on coupling; ethylamine gives nitrogen gas and alcohol with nitrous acid. (iv) Aniline forms a stable diazonium salt and couples with phenol; benzylamine evolves N2 with nitrous acid. (v) Aniline gives carbylamine test; N-methylaniline does not.

Answer:

Use carbylamine, Hinsberg and nitrous acid/diazotisation tests depending on the pair.

Q.9.3Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.v
Solution

(i) Aniline is less basic because the nitrogen lone pair is delocalised into the benzene ring; methylamine has +I donation. (ii) Ethylamine forms hydrogen bonds with water and has a small hydrophobic part; aniline has a large phenyl group and is poorly soluble. (iii) Methylamine produces OH- in water, precipitating Fe(OH)3 from FeCl3. (iv) In nitrating mixture aniline is protonated to anilinium ion, which is meta directing. (v) Aniline forms a salt/complex with AlCl3, strongly deactivating the ring. (vi) Arenediazonium ions are resonance stabilised; alkyldiazonium ions are not. (vii) Gabriel synthesis gives primary amines without secondary/tertiary amine mixtures.

Answer:

The reasons involve resonance in aniline, hydrogen bonding/size, basicity, protonation in acid, salt formation with Lewis acids, resonance stabilisation of arenediazonium ions and selectivity of Gabriel synthesis.

Q.9.4Arrange the following: (i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2. (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.v
Solution

Lower pKb means stronger base. Alkyl groups increase basicity by +I effect; aryl groups reduce it by delocalising the nitrogen lone pair. In gas phase, solvation is absent, so more alkyl substitution increases basicity. Boiling point follows hydrogen bonding strength, and water solubility decreases with larger hydrophobic groups.

Answer:

(i) C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH. (ii) C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH. (iii)(a) p-nitroaniline < aniline < p-toluidine. (iii)(b) C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2. (iv) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3. (v) (CH3)2NH < C2H5NH2 < C2H5OH. (vi) C6H5NH2 < (C2H5)2NH < C2H5NH2.

Q.9.6Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.v
Solution

Primary amine: RNH2 + C6H5SO2Cl → C6H5SO2NHR + HCl; the product has acidic N-H and dissolves in alkali. Secondary amine: R2NH + C6H5SO2Cl → C6H5SO2NR2 + HCl; the product has no N-H and is insoluble in alkali. Tertiary amines do not react with benzenesulphonyl chloride but dissolve in dilute acid as ammonium salts.

Answer:

Hinsberg test distinguishes primary, secondary and tertiary amines using benzenesulphonyl chloride.

Q.9.7Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.v
Solution

(i) Primary amines + CHCl3 + alc. KOH → foul-smelling isocyanides. (ii) Diazotisation: ArNH2 + NaNO2 + HCl → ArN2+Cl- at 273-278 K. (iii) Hofmann bromamide reaction: RCONH2 + Br2/NaOH → RNH2 with one carbon less. (iv) Coupling reaction joins diazonium salts with phenols/anilines to form azo compounds. (v) Ammonolysis is cleavage of C-X by NH3 to form amines. (vi) Acetylation replaces N-H hydrogen by CH3CO- using acetyl chloride or acetic anhydride. (vii) Gabriel synthesis uses potassium phthalimide and alkyl halides followed by hydrolysis to give primary amines.

Answer:

These are named reactions/processes used for identification, preparation and transformations of amines and diazonium salts.

Q.9.10An 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.v
Solution

Benzoic acid reacts with ammonia to form ammonium benzoate, which on heating gives benzamide. Benzamide undergoes Hofmann bromamide degradation with Br2/KOH to give aniline, C6H5NH2, whose formula is C6H7N. IUPAC names: benzoic acid, benzamide and benzenamine.

Answer:

A is benzoic acid, B is benzamide, and C is aniline.

Q.9.12Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?v
Solution

Gabriel synthesis needs an alkyl halide that can undergo SN2 attack by potassium phthalimide. In aryl halides, the C–X bond has partial double-bond character and the sp2 carbon does not undergo ordinary SN2 substitution, so aromatic primary amines cannot be prepared by this method.

Answer:

Aryl halides do not undergo the required nucleophilic substitution with phthalimide anion.

Q.9.13Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.v
Solution

(i) C6H5NH2 + NaNO2 + 2HCl →(273-278 K) C6H5N2+Cl- + NaCl + 2H2O. (ii) RNH2 + HNO2 → ROH + N2 + H2O; the aliphatic diazonium salt is too unstable to isolate.

Answer:

Aromatic primary amines form relatively stable diazonium salts at 273-278 K; aliphatic primary amines form unstable diazonium salts that decompose to alcohols with nitrogen gas.

Q.9.14Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines?v
Solution

Oxygen is more electronegative than nitrogen, so alcohols release H+ more readily than amines. Primary amines contain N–H bonds and associate through hydrogen bonding, increasing boiling point; tertiary amines lack N–H bonds. In aliphatic amines, +I effect of alkyl groups increases electron density on nitrogen, whereas in aromatic amines the lone pair is partly delocalised into the benzene ring and is less available for protonation.

Answer:

(i) N–H bonds are less polar than O–H bonds and amide ions are less stable than alkoxide ions. (ii) Primary amines form intermolecular hydrogen bonds; tertiary amines cannot donate N–H hydrogen bonds. (iii) Alkyl groups donate electron density to nitrogen, while aryl groups delocalise the nitrogen lone pair.