Class 12 Chemistry Chapter 13 Question Answer | Amines | English Medium

Class 12 Chemistry Chapter 13 Question Answer | Amines | English Medium | ASSEB

Amines

Welcome to HSLC Guru! In this chapter we study Amines — organic derivatives of ammonia (NH₃) in which one, two or three hydrogen atoms are replaced by alkyl or aryl groups. Amines are widely distributed in nature as alkaloids, vitamins, hormones (e.g., adrenaline, histamine) and building blocks of proteins (amino acids). The chapter covers their classification, structure, nomenclature, methods of preparation, physical and chemical properties, basicity trends, distinction tests (Hinsberg test) and the chemistry of diazonium salts, which are key intermediates in dye and drug synthesis. This material follows the ASSEB Class 12 Chemistry syllabus.

Summary

Classification & Structure: Amines are classified by the number of H of NH₃ replaced by R/Ar groups: primary (1°) R-NH₂, secondary (2°) R₂NH, tertiary (3°) R₃N, and quaternary ammonium salts R₄N⁺X⁻. Nitrogen in amines is sp³ hybridised, carrying a lone pair, giving a pyramidal (trigonal pyramidal) shape similar to NH₃ but with C-N-C angle ≈ 108°. In nomenclature, common names use the suffix “-amine” attached to the alkyl group (e.g., methylamine), while IUPAC names use “alkanamine” (e.g., methanamine). Aromatic amines are named as derivatives of aniline (benzenamine).

Preparation: Major routes include — (i) Hoffmann ammonolysis: alkyl halides + excess NH₃ → mixture of 1°, 2°, 3° amines and quaternary salt; (ii) Reduction of nitro compounds (Sn/HCl, H₂/Ni, Fe/HCl) → 1° amines; (iii) Reduction of nitriles (LiAlH₄ or H₂/Ni) → 1° amines with one extra carbon; (iv) Reduction of amides (LiAlH₄) → 1° amines; (v) Hofmann bromamide degradation: R-CONH₂ + Br₂ + 4NaOH → R-NH₂ (one carbon less); (vi) Gabriel phthalimide synthesis: best for pure 1° aliphatic amines (not used for aromatic amines as aryl halides do not undergo SN₂). Aniline is prepared industrially by reducing nitrobenzene with Sn/HCl or Fe + HCl.

Physical Properties: Lower amines are gases or volatile liquids with a fishy smell. 1° and 2° amines form intermolecular H-bonds (N-H···N), so they have higher boiling points than corresponding alkanes but lower than alcohols (since N is less electronegative than O). 3° amines have no N-H and cannot form H-bonds among themselves, hence have the lowest BP within isomers. Lower amines are water-soluble due to H-bonding with water; solubility decreases as the alkyl group becomes larger.

Chemical Reactions: Amines are basic due to the lone pair on N. In gas phase, basicity follows electron-donation: 3° > 2° > 1° > NH₃. In aqueous solution, basicity depends on +I, steric and solvation effects; for alkyl amines the order is generally 2° > 1° > 3° > NH₃ (Et₂NH > EtNH₂ > Et₃N for ethyl; for methyl: Me₂NH > MeNH₂ > Me₃N > NH₃). Aromatic amines (aniline) are weaker bases than NH₃ because the lone pair is delocalised into the ring. Reactions include — alkylation (with R-X), acylation (with acid chlorides/anhydrides → amides), reaction with nitrous acid (HONO): 1° aliphatic → unstable diazonium → alcohol + N₂; 1° aromatic → stable arene-diazonium salt at 0-5 °C; 2° amines → N-nitroso amines (yellow oil); 3° aliphatic amines → no reaction (form salts), 3° aromatic → p-nitroso compound. Carbylamine test: R-NH₂ + CHCl₃ + KOH → R-NC (foul smelling isocyanide) — only 1° amines respond. Electrophilic aromatic substitution in aniline: -NH₂ is strongly o,p-directing and activating; with conc. HNO₃/H₂SO₄ aniline gives mainly m-nitroaniline (because most aniline is protonated to anilinium, a m-director); for clean p-product, the -NH₂ is first acetylated. Diazonium salts (Ar-N₂⁺X⁻) are prepared by diazotisation (NaNO₂ + HCl, 0-5 °C) and undergo Sandmeyer (CuCl/CuBr/CuCN), Gattermann (Cu/HCl, HBr), replacement by I (KI), F (HBF₄, Balz-Schiemann), OH (warm water), H (H₃PO₂), and coupling with phenol/aniline in alkaline/weakly acidic medium to give brightly coloured azo dyes (e.g., p-hydroxyazobenzene). The Hinsberg test distinguishes 1°, 2°, 3° amines: with PhSO₂Cl + NaOH — 1° gives clear soluble salt of N-alkyl benzenesulphonamide, 2° gives KOH-insoluble solid, 3° does not react.

1-Mark Questions

Q1. What is the hybridisation of nitrogen in amines?

Answer: The nitrogen atom in amines is sp³ hybridised, with a lone pair occupying one of the four sp³ orbitals; the geometry is pyramidal.

Q2. Give the IUPAC name of (CH₃)₂NH.

Answer: N-methylmethanamine (common name: dimethylamine).

Q3. Why is aniline a weaker base than methylamine?

Answer: In aniline the lone pair on N is delocalised into the benzene ring (resonance), making it less available for protonation; methylamine has the lone pair fully available and is enhanced by +I of methyl.

Q4. Name the reagent used in the Hinsberg test.

Answer: Benzenesulphonyl chloride (C₆H₅SO₂Cl) in the presence of aqueous KOH.

Q5. What is a quaternary ammonium salt?

Answer: A salt of the form R₄N⁺ X⁻ in which all four hydrogens of NH₄⁺ are replaced by alkyl/aryl groups (e.g., (CH₃)₄N⁺Cl⁻).

Q6. Which amine cannot be prepared by Gabriel phthalimide synthesis?

Answer: Aromatic primary amines (e.g., aniline) cannot be prepared because aryl halides do not undergo nucleophilic substitution with potassium phthalimide.

Q7. What is the temperature for diazotisation?

Answer: 0 °C to 5 °C (ice-cold conditions); above 5 °C the diazonium salt decomposes to phenol and N₂.

Q8. Give one use of azo dyes.

Answer: Azo dyes (e.g., methyl orange, p-hydroxyazobenzene) are widely used as colouring agents in textiles and as acid-base indicators.

Q9. Which amine gives a foul smelling carbylamine?

Answer: Only primary amines (R-NH₂) give the carbylamine reaction with CHCl₃ + alc. KOH to form R-NC (isocyanide).

Q10. Why are 3° amines insoluble in HCl-free water but soluble in HCl?

Answer: 3° amines have no N-H bond, so cannot H-bond with water effectively; with HCl they form water-soluble ammonium chloride salts (R₃NH⁺Cl⁻).

2 / 3-Mark Questions

Q1. Classify amines with one example each.

Answer: (i) Primary (1°): R-NH₂, e.g., CH₃NH₂ (methylamine); (ii) Secondary (2°): R₂NH, e.g., (CH₃)₂NH (dimethylamine); (iii) Tertiary (3°): R₃N, e.g., (CH₃)₃N (trimethylamine); (iv) Quaternary ammonium salts: R₄N⁺X⁻, e.g., (CH₃)₄N⁺Cl⁻ (tetramethylammonium chloride).

Q2. Compare the boiling points of n-butylamine, di-n-propylamine and tri-ethylamine of similar molar mass.

Answer: 1° (n-butylamine) has two N-H bonds → strongest H-bonding → highest BP. 2° (di-n-propylamine) has one N-H → moderate H-bonding → intermediate BP. 3° (triethylamine) has no N-H → no H-bonding among themselves → lowest BP. Order: 1° > 2° > 3°.

Q3. Explain Hofmann bromamide degradation with an example.

Answer: An aliphatic or aromatic amide is treated with Br₂ and aq. NaOH (or KOH) to give a 1° amine with one carbon less than the amide.

R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.

Example: CH₃CONH₂ → CH₃NH₂ (acetamide → methylamine).

Q4. Why does aniline not undergo Friedel-Crafts reaction?

Answer: Aniline forms a Lewis acid–base complex with AlCl₃ (the Lewis acid catalyst) at the lone pair on nitrogen. This makes the aromatic ring strongly electron-deficient (acts like -NH₂AlCl₃⁻), deactivating the ring towards electrophilic substitution. Hence Friedel-Crafts alkylation/acylation does not work on aniline.

Q5. Give the order of basic strength of methylamine, dimethylamine, trimethylamine and ammonia (i) in gas phase, (ii) in water.

Answer: (i) Gas phase (only +I effect): (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃. (ii) In water (steric + solvation effects): (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.

Q6. Write the preparation of aniline from nitrobenzene.

Answer: Nitrobenzene is reduced with Sn (or Fe) and conc. HCl, then made alkaline with NaOH:

C₆H₅NO₂ + 6[H] → (Sn/HCl) → C₆H₅NH₂ + 2H₂O.

Steam distillation gives pure aniline. Industrially Fe/HCl is preferred (cheaper, recyclable).

5 / 7-Mark Questions

Q1. Describe the Hinsberg test for distinguishing 1°, 2° and 3° amines.

Answer: The amine is shaken with benzenesulphonyl chloride (C₆H₅SO₂Cl) in aqueous KOH/NaOH.

(i) 1° amine: R-NH₂ + PhSO₂Cl → PhSO₂NHR (N-alkylbenzenesulphonamide). The N-H is acidic (electron-withdrawing SO₂) and dissolves in alkali to give a clear solution. On acidification a solid reappears.

(ii) 2° amine: R₂NH + PhSO₂Cl → PhSO₂NR₂ (N,N-dialkylbenzenesulphonamide). No N-H is present, so the solid is insoluble in KOH.

(iii) 3° amine: R₃N has no N-H to react; it does not form a sulphonamide. The 3° amine remains as a separate liquid layer; on adding HCl it dissolves as R₃NH⁺Cl⁻.

Conclusion: Soluble in KOH → 1°; insoluble solid → 2°; no reaction → 3°.

Q2. How are diazonium salts prepared? Give five reactions of benzenediazonium chloride.

Answer: Preparation (Diazotisation): Aniline is dissolved in dilute HCl and treated with sodium nitrite at 0-5 °C: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O.

Reactions of C₆H₅N₂⁺Cl⁻:

(i) Sandmeyer reaction: with CuCl/HCl → C₆H₅Cl (chlorobenzene); with CuBr/HBr → C₆H₅Br; with CuCN → C₆H₅CN.

(ii) Gattermann reaction: with Cu powder + HCl/HBr → C₆H₅Cl / C₆H₅Br (lower yield but cheaper).

(iii) Replacement by -I: with KI → C₆H₅I (iodobenzene); replacement by -F: with HBF₄ followed by heat (Balz-Schiemann) → C₆H₅F.

(iv) Replacement by -OH: warm with H₂O → C₆H₅OH (phenol) + N₂ + HCl.

(v) Coupling with phenol (in alkaline medium) → p-hydroxyazobenzene (orange azo dye); with aniline (weakly acidic) → p-aminoazobenzene (yellow dye). Also, with H₃PO₂ → benzene (C₆H₆) + N₂ + HCl + H₃PO₃ (replacement by -H).

Q3. Discuss the basicity of amines. Why is diethylamine more basic than ethylamine in water but less basic than triethylamine in gas phase?

Answer: Basicity of amines depends on three factors:

(a) +I effect of alkyl groups: increases electron density on N → increases basicity. By this factor alone, 3° > 2° > 1° > NH₃ (operates in gas phase).

(b) Steric effect: bulky groups around N hinder protonation; hence 3° amines are less basic than expected.

(c) Solvation: in water, the conjugate acid R-NH₃⁺ / R₂NH₂⁺ / R₃NH⁺ is stabilised by H-bonding with water. R-NH₃⁺ has 3 N-H bonds → most solvated; R₃NH⁺ has only 1 N-H → least solvated. Solvation favours 1° > 2° > 3°.

In water the net order from balance of these effects is: (C₂H₅)₂NH > C₂H₅NH₂ > (C₂H₅)₃N > NH₃. Diethylamine has optimum balance: enough +I, moderate steric bulk and good solvation. In gas phase (no solvent), only +I matters → triethylamine > diethylamine > ethylamine > NH₃.

Aromatic amines (aniline) are far weaker bases (Kb ≈ 4.0 × 10⁻¹⁰) than aliphatic amines because the lone pair on N is delocalised into the benzene ring through resonance, lowering its availability.

Q4. Explain Gabriel phthalimide synthesis with mechanism and limitations.

Answer: The Gabriel synthesis is used to prepare pure 1° aliphatic amines free from 2° and 3° contaminants.

Steps:

(1) Phthalimide is treated with KOH to form potassium phthalimide (K-N(CO)₂C₆H₄), a strong nucleophile because the negative charge on N is delocalised into two carbonyl groups.

(2) Potassium phthalimide is heated with an alkyl halide R-X (SN₂) to give N-alkyl phthalimide.

(3) N-alkyl phthalimide is hydrolysed with aq. NaOH (or NH₂NH₂, hydrazine, in modern procedures) to release the 1° amine R-NH₂ along with phthalic acid (or phthalhydrazide).

Limitations: Aromatic 1° amines (aniline, etc.) cannot be prepared because aryl halides do not undergo SN₂ with potassium phthalimide (sp² C, no backside attack). Only 1° R-NH₂ is obtained — no further alkylation occurs because product amine is not a stronger nucleophile than phthalimide ion.

Q5. Account for the following: (a) Aniline is acetylated before nitration. (b) Methylamine is more basic than NH₃. (c) Aniline does not react with NaNO₂/HCl above 5 °C to give a stable product. (d) Quaternary ammonium salts cannot be reduced. (e) Carbylamine reaction is a test for 1° amines.

Answer: (a) Aniline is highly activated by -NH₂; direct nitration with HNO₃/H₂SO₄ protonates -NH₂ to -NH₃⁺ (m-director) and may oxidise the ring. So -NH₂ is first acetylated to -NHCOCH₃ (less activating, still o,p-directing); after p-nitration, hydrolysis regenerates -NH₂ giving p-nitroaniline cleanly.

(b) The +I effect of -CH₃ pushes electron density on N, making the lone pair more available; also CH₃NH₃⁺ is stabilised better than NH₄⁺ by hyperconjugation/+I.

(c) Above 5 °C the diazonium ion C₆H₅N₂⁺ decomposes to phenol + N₂; hence the reaction must be carried out at 0-5 °C to obtain a stable diazonium salt.

(d) In R₄N⁺X⁻ the N already has four bonds and a positive charge; there is no lone pair and no remaining bond to reduce — they can however undergo Hofmann elimination (β-elimination) on heating with moist Ag₂O.

(e) Only 1° amines have two N-H bonds; with CHCl₃/KOH dichlorocarbene (:CCl₂) attacks N twice, giving R-N=C: (isocyanide), which has a foul smell. 2° and 3° amines lack the required N-H pattern, so the test is specific for 1° amines.

Multiple Choice Questions (MCQs)

1. The hybridisation of N in trimethylamine is —
(a) sp (b) sp² (c) sp³ (d) sp³d

2. Which is most basic in water?
(a) NH₃ (b) CH₃NH₂ (c) (CH₃)₂NH (d) (CH₃)₃N

3. Aniline is prepared from nitrobenzene by —
(a) Sn/HCl (b) NaOH (c) HNO₃ (d) Br₂/AlCl₃

4. The carbylamine test is given by —
(a) 3° amine (b) 2° amine (c) 1° amine (d) Quaternary salt

5. Hinsberg’s reagent is —
(a) PCl₅ (b) CH₃COCl (c) C₆H₅SO₂Cl (d) C₆H₅COCl

6. Diazonium salts are prepared at —
(a) 25-30 °C (b) 0-5 °C (c) 50-60 °C (d) 100 °C

7. Sandmeyer reaction uses —
(a) Cu/HCl (b) CuCl/HCl (c) Fe/HCl (d) Sn/HCl

8. Hofmann bromamide reaction converts —
(a) Amine to nitrile (b) Nitrile to amine (c) Amide to amine with one less C (d) Acid to amide

9. Gabriel synthesis is unsuitable for —
(a) CH₃NH₂ (b) C₂H₅NH₂ (c) C₆H₅NH₂ (d) (CH₃)₂CHNH₂

10. Coupling of benzene-diazonium chloride with phenol in alkali gives —
(a) Benzene (b) p-hydroxyazobenzene (c) Phenol (d) Aniline

Fill in the Blanks

1. The geometry around N in amines is __________ . (pyramidal)

2. The order of basicity of methyl amines in water is __________ . ((CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃)

3. Aniline is prepared from __________ by reduction with Sn/HCl. (nitrobenzene)

4. The reagent in carbylamine test is __________ + alc. KOH. (CHCl₃)

5. Replacement of -N₂⁺ by -CN using CuCN is the __________ reaction. (Sandmeyer)

True / False

1. Tertiary amines do not give the Hinsberg test. — True

2. Aniline is more basic than methylamine. — False (aniline is much weaker)

3. Diazotisation is carried out at room temperature. — False (0-5 °C)

4. Gabriel synthesis gives pure primary aliphatic amines. — True

5. Aniline undergoes Friedel-Crafts alkylation easily. — False (lone pair complexes with AlCl₃, deactivating ring)

Glossary

Term	Meaning
Amine	Organic derivative of NH₃ in which one or more H are replaced by R/Ar groups.
Primary amine	R-NH₂ — one H of NH₃ replaced.
Secondary amine	R₂NH — two Hs replaced.
Tertiary amine	R₃N — all three Hs replaced.
Quaternary salt	R₄N⁺X⁻ — four R groups; positively charged N.
Hofmann ammonolysis	R-X + NH₃ (excess) → 1°/2°/3° amines + R₄N⁺X⁻.
Hofmann bromamide	R-CONH₂ + Br₂/4NaOH → R-NH₂ (one C less).
Gabriel synthesis	Phthalimide route to pure 1° aliphatic amines.
Diazotisation	ArNH₂ + NaNO₂ + HCl, 0-5 °C → ArN₂⁺Cl⁻.
Sandmeyer	ArN₂⁺ + CuCl/CuBr/CuCN → ArCl/ArBr/ArCN.
Gattermann	ArN₂⁺ + Cu/HX → ArX (cheaper, lower yield).
Carbylamine test	1° amine + CHCl₃ + KOH → R-NC (foul smell).
Hinsberg test	Distinguishes 1°/2°/3° amines using PhSO₂Cl.
Azo coupling	ArN₂⁺ + phenol/aniline → coloured azo dye Ar-N=N-Ar’.
Pyramidal	Trigonal pyramidal shape of sp³ N with one lone pair.

Continue your ASSEB Class 12 Chemistry preparation with the next chapters on HSLC Guru. Practise distinguishing 1°/2°/3° amines, learn the Hinsberg and carbylamine tests, and master diazonium chemistry — these are recurring favourites in board exams.