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Amines

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Chapter Notes

Overview
This chapter explores amines as derivatives of ammonia, where one or more hydrogen atoms are replaced by alkyl or aryl groups. It covers their classification, nomenclature, structure, methods of preparation, physical and chemical properties, and the importance of diazonium salts in organic synthesis. The one big idea a student must walk away with is that the unshared pair of electrons on the nitrogen atom governs the basicity, nucleophilicity, and reactivity of amines, making them key intermediates in the synthesis of medicines, dyes, and polymers.
Key Concepts (15)
Amines
Organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s).
e.g. Methylamine (CH3NH2) is an amine derived from ammonia by replacing one hydrogen with a methyl group.
Primary (1°) amine
An amine in which one hydrogen atom of ammonia is replaced by an alkyl or aryl group, having the general formula RNH2 or ArNH2.
e.g. Aniline (C6H5NH2) is a primary aromatic amine.
Secondary (2°) amine
An amine in which two hydrogen atoms of ammonia are replaced by alkyl or aryl groups, having the general formula R2NH or R-NHR'.
e.g. Dimethylamine (CH3NHCH3) is a secondary amine.
Tertiary (3°) amine
An amine in which all three hydrogen atoms of ammonia are replaced by alkyl or aryl groups, having the general formula R3N.
e.g. Trimethylamine ((CH3)3N) is a tertiary amine.
Quaternary ammonium salt
A salt formed when a tertiary amine reacts with an alkyl halide, containing a positively charged nitrogen atom bonded to four alkyl/aryl groups and an anion.
e.g. Tetramethylammonium chloride, used as a surfactant.
Ammonolysis
The process of cleavage of the C–X bond by ammonia molecule, where an alkyl or benzyl halide reacts with an ethanolic solution of ammonia to form amines.
e.g. CH3CH2Cl + NH3 (excess) → CH3CH2NH2 + HCl.
Gabriel phthalimide synthesis
A method for the preparation of primary amines by heating phthalimide with ethanolic KOH to form its potassium salt, which on reaction with an alkyl halide followed by alkaline hydrolysis yields the primary amine.
e.g. Phthalimide → Potassium phthalimide → N-alkylphthalimide → Primary amine.
Hoffmann bromamide degradation reaction
A method for preparing primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide, resulting in an amine with one carbon less than the amide.
e.g. Butanamide (CH3CH2CH2CONH2) → Propanamine (CH3CH2CH2NH2).
Carbylamine reaction
A test for primary amines where they react with chloroform and ethanolic potassium hydroxide to form foul-smelling isocyanides or carbylamines.
e.g. CH3NH2 + CHCl3 + 3KOH → CH3NC + 3KCl + 3H2O.
Hinsberg's reagent
Benzenesulphonyl chloride (C6H5SO2Cl) or p-toluenesulphonyl chloride, used to distinguish primary, secondary, and tertiary amines based on the solubility of the sulphonamide formed.
e.g. Primary amine forms a sulphonamide soluble in alkali; secondary amine forms one insoluble in alkali; tertiary amine does not react.
Diazonium salt
A salt with the general formula R–N2+X–, where R is an aryl group and X– is an anion like Cl–, Br–, HSO4–, or BF4–, formed by diazotisation of primary aromatic amines.
e.g. Benzenediazonium chloride (C6H5N2+Cl–).
Diazotisation
The conversion of primary aromatic amines into diazonium salts by reaction with nitrous acid at low temperatures (273-278 K).
e.g. C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl– + NaCl + 2H2O.
Sandmeyer reaction
A reaction where the diazonium group is replaced by Cl–, Br–, or CN– in the presence of Cu(I) ion to form aryl halides or cyanides.
e.g. C6H5N2+Cl– + CuCl → C6H5Cl + N2.
Coupling reaction
A reaction where a diazonium salt couples with phenol or aniline at the para position to form azo dyes, which are coloured compounds with an extended conjugated system.
e.g. C6H5N2+Cl– + C6H5OH → p-hydroxyazobenzene (orange dye).
pKb
The negative logarithm of the base dissociation constant (Kb); smaller pKb indicates a stronger base.
e.g. pKb of methanamine is 3.38, while that of aniline is 9.38, showing aniline is a much weaker base.
Formulas & Equations (10)
RNH2 + H2O ⇌ RNH3+ + OH–
Reads as: Primary amine reacts with water to form substituted ammonium ion and hydroxide ion.
Variables: RNH2 = amine; RNH3+ = substituted ammonium ion; OH– = hydroxide ion.
Use when: Calculating Kb or comparing basic strength of amines.
⚠ Students forget that the equilibrium involves water and write the reaction as RNH2 → RNH2+ + H+ instead of the correct base dissociation.
Kb = [RNH3+][OH–] / [RNH2]
Reads as: Base dissociation constant equals concentration of substituted ammonium ion times concentration of hydroxide ion divided by concentration of amine.
Variables: Kb = base dissociation constant (no unit, but usually expressed in mol/L); [RNH3+] = concentration of substituted ammonium ion; [OH–] = concentration of hydroxide ion; [RNH2] = concentration of amine.
Use when: Calculating the strength of a base from its equilibrium concentrations.
⚠ Students omit water concentration from the expression but incorrectly include it; water is solvent and its concentration is constant, so it is not included in Kb.
pKb = –log Kb
Reads as: pKb equals negative logarithm of base dissociation constant.
Variables: pKb = a measure of basic strength (dimensionless); Kb = base dissociation constant.
Use when: Comparing basic strengths of different amines; smaller pKb means stronger base.
⚠ Students think larger pKb means stronger base; the opposite is true.
R–NO2 + 3H2 → R–NH2 + 2H2O (catalytic hydrogenation)
Reads as: Nitro compound reacts with hydrogen gas in presence of catalyst to form amine and water.
Variables: R–NO2 = nitro compound; H2 = hydrogen gas; R–NH2 = primary amine; catalyst = Ni, Pd, or Pt.
Use when: Reducing nitro compounds to amines in board exam questions on preparation methods.
⚠ Students forget that water is a product and write only the amine.
R–X + 2NH3 → R–NH2 + NH4X
Reads as: Alkyl halide reacts with excess ammonia to form primary amine and ammonium halide.
Variables: R–X = alkyl halide; NH3 = ammonia; R–NH2 = primary amine; NH4X = ammonium halide.
Use when: Ammonolysis of alkyl halides to prepare primary amines (with excess ammonia).
⚠ Students forget that excess ammonia is needed to get primary amine as major product; otherwise, secondary and tertiary amines also form.
R–CN + 4[H] → R–CH2–NH2 (LiAlH4 or H2/catalyst)
Reads as: Nitrile is reduced to primary amine using lithium aluminium hydride or catalytic hydrogenation.
Variables: R–CN = nitrile; [H] = reducing agent (LiAlH4 or H2 with catalyst); R–CH2–NH2 = primary amine with one more carbon.
Use when: Ascent of amine series (preparing amines with one more carbon than starting material).
⚠ Students think the product has the same number of carbons as the nitrile; actually, the –CN carbon becomes –CH2NH2, adding one carbon.
R–CONH2 + Br2 + 4NaOH → R–NH2 + 2NaBr + Na2CO3 + 2H2O
Reads as: Amide reacts with bromine and sodium hydroxide to form primary amine with one carbon less, sodium bromide, sodium carbonate, and water.
Variables: R–CONH2 = amide; Br2 = bromine; NaOH = sodium hydroxide; R–NH2 = primary amine (one carbon less); NaBr = sodium bromide; Na2CO3 = sodium carbonate; H2O = water.
Use when: Hoffmann bromamide degradation to prepare primary amines from amides.
⚠ Students forget that the amine has one carbon less than the amide, leading to incorrect product structures.
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl– + NaCl + 2H2O (273-278 K)
Reads as: Aniline reacts with sodium nitrite and hydrochloric acid at low temperature to form benzenediazonium chloride, sodium chloride, and water.
Variables: C6H5NH2 = aniline; NaNO2 = sodium nitrite; HCl = hydrochloric acid; C6H5N2+Cl– = benzenediazonium chloride; NaCl = sodium chloride; H2O = water.
Use when: Diazotisation of primary aromatic amines.
⚠ Students write the reaction at room temperature; the temperature must be 273-278 K (0-5°C) to prevent decomposition of the diazonium salt.
C6H5N2+Cl– + H2O → C6H5OH + N2 + HCl (at 283 K)
Reads as: Benzenediazonium chloride hydrolyses to phenol, nitrogen gas, and hydrochloric acid when warmed.
Variables: C6H5N2+Cl– = benzenediazonium chloride; H2O = water; C6H5OH = phenol; N2 = nitrogen gas; HCl = hydrochloric acid.
Use when: Replacement of diazonium group by hydroxyl group to prepare phenols.
⚠ Students forget that nitrogen gas is evolved and write only phenol.
C6H5N2+Cl– + C6H5OH → p-HO–C6H4–N=N–C6H5 + HCl
Reads as: Benzenediazonium chloride couples with phenol to form p-hydroxyazobenzene and hydrochloric acid.
Variables: C6H5N2+Cl– = benzenediazonium chloride; C6H5OH = phenol; p-HO–C6H4–N=N–C6H5 = p-hydroxyazobenzene (azo dye); HCl = hydrochloric acid.
Use when: Coupling reaction to prepare azo dyes.
⚠ Students write the coupling at the ortho position instead of the para position; para is the major product.
How It Works
Ammonolysis of alkyl halides
  1. Step 1: Take an alkyl or benzyl halide (e.g., CH3CH2Cl) and react with an ethanolic solution of ammonia in a sealed tube at 373 K.
  2. Step 2: The halogen atom is replaced by the –NH2 group via nucleophilic substitution, forming a primary amine salt.
  3. Step 3: The primary amine formed can further react with more alkyl halide to produce secondary, tertiary amines, and finally a quaternary ammonium salt.
  4. Step 4: To obtain the free primary amine as the major product, use a large excess of ammonia.
  5. Step 5: Treat the ammonium salt with a strong base (e.g., NaOH) to liberate the free amine.
💡 The order of reactivity of halides is RI > RBr > RCl. Using excess ammonia is crucial to favour primary amine formation; otherwise, a mixture of all three amines and quaternary salt is obtained.
Gabriel phthalimide synthesis
  1. Step 1: Treat phthalimide with ethanolic potassium hydroxide to form the potassium salt of phthalimide.
  2. Step 2: Heat this salt with an alkyl halide to form N-alkylphthalimide.
  3. Step 3: Perform alkaline hydrolysis (e.g., with NaOH) of N-alkylphthalimide to produce the corresponding primary amine and phthalic acid.
💡 This method gives only primary amines and avoids mixtures. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
Hoffmann bromamide degradation reaction
  1. Step 1: Take an amide (R–CONH2) and treat it with bromine (Br2) in an aqueous or ethanolic solution of sodium hydroxide (NaOH).
  2. Step 2: The alkyl or aryl group migrates from the carbonyl carbon to the nitrogen atom, resulting in a primary amine.
  3. Step 3: The amine formed contains one carbon less than the starting amide.
💡 The amine has one carbon less than the amide. This is a degradation reaction, so the carbon chain length decreases by one.
Diazotisation
  1. Step 1: Dissolve a primary aromatic amine (e.g., aniline) in a mineral acid (e.g., HCl) and cool the solution to 273-278 K.
  2. Step 2: Add an aqueous solution of sodium nitrite (NaNO2) slowly to the cooled solution to generate nitrous acid in situ.
  3. Step 3: The nitrous acid reacts with the amine to form the diazonium salt (e.g., benzenediazonium chloride).
  4. Step 4: The diazonium salt is unstable and is used immediately without storage.
💡 The temperature must be strictly maintained at 273-278 K (0-5°C). At higher temperatures, the diazonium salt decomposes to form phenol and nitrogen gas.
Sandmeyer reaction
  1. Step 1: Prepare a diazonium salt solution from a primary aromatic amine.
  2. Step 2: Treat the diazonium salt with a solution of Cu(I) halide (e.g., CuCl, CuBr) or CuCN in the corresponding halogen acid.
  3. Step 3: The diazonium group is replaced by the halide or cyanide ion, and nitrogen gas is evolved.
  4. Step 4: The product is an aryl halide or aryl cyanide.
💡 The yield in Sandmeyer reaction is better than in Gattermann reaction. Cu(I) ion is essential for this substitution.
Coupling reaction
  1. Step 1: Take a diazonium salt solution (e.g., benzenediazonium chloride) and add it to a solution of phenol or aniline.
  2. Step 2: The diazonium ion acts as an electrophile and attacks the para position of the phenol or aniline ring.
  3. Step 3: The product is an azo compound (e.g., p-hydroxyazobenzene or p-aminoazobenzene), which is coloured and used as a dye.
💡 Coupling occurs at the para position. If the para position is blocked, ortho coupling may occur, but para is the major product. The reaction is an electrophilic substitution.
Hinsberg test (distinction of primary, secondary, and tertiary amines)
  1. Step 1: Add benzenesulphonyl chloride (Hinsberg's reagent) to the amine in the presence of aqueous KOH or NaOH.
  2. Step 2: If the amine is primary, it forms a sulphonamide that has a hydrogen atom attached to nitrogen, which is acidic and dissolves in alkali (soluble).
  3. Step 3: If the amine is secondary, it forms a sulphonamide with no hydrogen on nitrogen, which is not acidic and remains insoluble in alkali.
  4. Step 4: If the amine is tertiary, it does not react with benzenesulphonyl chloride.
💡 Primary amine sulphonamide is soluble in alkali; secondary amine sulphonamide is insoluble; tertiary amine does not react. This test is also used for separation of a mixture of amines.
Exam Traps — Don't Make These Mistakes
⚠ Confusing order of basic strength in aqueous vs. gas phase
✗ Students write that tertiary amines are always stronger bases than secondary and primary amines in all conditions.
→ In the gas phase, tertiary amines are strongest due to maximum +I effect. However, in aqueous solution, solvation effects and steric hindrance alter the order. For methyl-substituted amines, the order is (CH3)2NH > CH3NH2 > (CH3)3N > NH3. For ethyl-substituted amines, it is (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3.
✓ State the order explicitly for the given condition (aqueous or gas phase) and mention the interplay of inductive effect, solvation effect, and steric hindrance.
⚠ Forgetting that Hoffmann bromamide degradation reduces carbon chain by one
✗ Students write the product amine with the same number of carbon atoms as the starting amide.
→ In Hoffmann bromamide degradation, the alkyl or aryl group migrates from the carbonyl carbon to the nitrogen, and the carbonyl carbon is lost as CO2 (or carbonate). The amine formed has one carbon less than the amide.
✓ Always count the carbons: if the amide has n carbons, the amine has n-1 carbons. For example, butanamide (4 carbons) gives propanamine (3 carbons).
⚠ Writing the product of carbylamine reaction incorrectly
✗ Students write that the product is a cyanide or nitrile.
→ The carbylamine reaction produces isocyanides (R–NC), not cyanides (R–CN). Isocyanides have the functional group –NC, while cyanides have –CN. They are isomers but different compounds.
✓ Write the product as an isocyanide (carbylamine), e.g., CH3NC for methyl isocyanide. The reaction is a test for primary amines.
⚠ Confusing Sandmeyer and Gattermann reactions
✗ Students write that both reactions use Cu(I) ion or that both give the same yield.
→ Sandmeyer reaction uses Cu(I) halide (e.g., CuCl, CuBr, CuCN) in the corresponding halogen acid. Gattermann reaction uses copper powder (Cu) and the corresponding halogen acid. The yield in Sandmeyer reaction is better than in Gattermann reaction.
✓ Clearly specify the reagent: Sandmeyer uses Cu(I) salt; Gattermann uses Cu powder. Mention that Sandmeyer gives better yields.
⚠ Ignoring the effect of protecting the –NH2 group in electrophilic substitution
✗ Students write that aniline undergoes nitration to give only ortho and para products.
→ Direct nitration of aniline in strongly acidic medium protonates the –NH2 group to –NH3+, which is meta-directing and deactivating. This leads to a mixture including significant meta-nitroaniline, along with tarry oxidation products. To get para-nitroaniline as the major product, the –NH2 group must be protected by acetylation (forming acetanilide) before nitration.
✓ Explain that direct nitration gives a mixture including meta product due to protonation. For controlled nitration, acetylate the amine first, then nitrate, then hydrolyse to get the desired substituted aniline.
⚠ Writing the coupling reaction at the wrong position
✗ Students write the azo coupling at the ortho position of phenol or aniline.
→ The coupling reaction is an electrophilic substitution that occurs preferentially at the para position because it is less sterically hindered and the intermediate is more stable. Ortho coupling is possible only if the para position is blocked.
✓ Write the product with the azo group (–N=N–) attached at the para position of the phenol or aniline ring. For example, benzenediazonium chloride with phenol gives p-hydroxyazobenzene.
Compare & Contrast
Reaction with Hinsberg's reagent (benzenesulphonyl chloride)
Primary amine
Secondary amine
Primary amine forms a sulphonamide that is soluble in alkali (due to acidic H on N), while secondary amine forms a sulphonamide that is insoluble in alkali (no H on N).
Basicity (pKb values)
Aliphatic amine
Aromatic amine
Aliphatic amines are stronger bases than ammonia (pKb 3-4) due to +I effect of alkyl groups. Aromatic amines are weaker bases than ammonia (pKb ~9.38 for aniline) due to resonance delocalisation of the lone pair on nitrogen into the benzene ring.
Reagent used for replacement of diazonium group by halogen
Sandmeyer reaction
Gattermann reaction
Sandmeyer reaction uses Cu(I) halide (e.g., CuCl) in the corresponding halogen acid; Gattermann reaction uses copper powder (Cu) and the halogen acid. Sandmeyer gives better yields.
Stability of diazonium salt formed with nitrous acid
Primary aliphatic amine
Primary aromatic amine
Primary aliphatic amines form highly unstable alkyldiazonium salts that immediately liberate nitrogen gas and form alcohols. Primary aromatic amines form relatively stable arenediazonium salts at low temperatures (273-278 K), which can be used for further synthesis.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11-12 marks.
Most asked: Basicity of amines (order and factors affecting it) · Hoffmann bromamide degradation (mechanism and product prediction) · Gabriel phthalimide synthesis (why it gives only primary amines) · Hinsberg test (distinction of primary, secondary, tertiary amines) · Diazotisation and Sandmeyer reaction · Coupling reaction (azo dyes) · Carbylamine reaction (test for primary amines) · Electrophilic substitution in aniline (nitration, bromination, effect of protecting group)
Don't skip: Conversion reactions (e.g., aniline to p-bromoaniline, nitrobenzene to benzoic acid) — often 3-5 marks; Reasoning questions on basic strength order (e.g., why aniline is weaker base than methylamine) — 2-3 marks; Distinction tests (Hinsberg, carbylamine) — 2-3 marks; Mechanism of diazotisation and coupling — 3 marks
One-Line Revision (read aloud before the exam)
  1. Amines are derivatives of ammonia where one or more hydrogen atoms are replaced by alkyl or aryl groups.
  2. Primary amines have the formula RNH2, secondary amines R2NH, and tertiary amines R3N.
  3. The nitrogen atom in amines is sp3 hybridised with a pyramidal geometry and an unshared pair of electrons.
  4. Aliphatic amines are stronger bases than ammonia due to the +I effect of alkyl groups.
  5. Aromatic amines are weaker bases than ammonia because the lone pair on nitrogen is delocalised into the benzene ring.
  6. Gabriel phthalimide synthesis gives only primary amines and cannot be used for aromatic primary amines.
  7. Hoffmann bromamide degradation reduces the carbon chain by one, converting an amide to a primary amine.
  8. Carbylamine reaction is a test for primary amines, producing foul-smelling isocyanides.
  9. Hinsberg's reagent (benzenesulphonyl chloride) distinguishes primary, secondary, and tertiary amines.
  10. Diazotisation converts primary aromatic amines to diazonium salts at 273-278 K using NaNO2 and HCl.
  11. Sandmeyer reaction replaces the diazonium group with Cl, Br, or CN using Cu(I) salts.
  12. Coupling reaction of diazonium salts with phenols or anilines produces coloured azo dyes.
  13. Direct nitration of aniline gives a mixture including meta-nitroaniline due to protonation of the amino group.
  14. Acetylation of aniline protects the amino group and allows controlled electrophilic substitution.
  15. Quaternary ammonium salts are used as surfactants.
  16. Amines are soluble in water due to hydrogen bonding, but solubility decreases with increasing molar mass.
  17. Boiling point order of isomeric amines: primary > secondary > tertiary due to intermolecular hydrogen bonding.
  18. The order of reactivity of alkyl halides in ammonolysis is RI > RBr > RCl.
  19. Benzenediazonium chloride is stable only in cold solution and decomposes on warming to give phenol.
  20. Aryl fluorides and iodides cannot be prepared by direct halogenation but are easily obtained from diazonium salts.

MCQs — 10 Questions with Answers

Q1. Which of the following is NOT a correct IUPAC name?

  • A. CH₃CH₂NH₂ is ethanamine
  • B. CH₃NHCH₂CH₃ is N-methylethanamine
  • C. (CH₃)₂CHNH₂ is propan-1-amine ✓
  • D. H₂N–CH₂–CH₂–NH₂ is ethane-1,2-diamine

Answer: C — (CH₃)₂CHNH₂ is propan-2-amine (isopropylamine), not propan-1-amine, because the –NH₂ group is attached to the 2nd carbon atom of the propane chain.

Q2. In ammonolysis of alkyl halides, the primary amine formed can further react with the alkyl halide. Which condition prevents excessive secondary and tertiary amine formation?

  • A. Low temperature
  • B. High pressure
  • C. Large excess of ammonia ✓
  • D. Presence of catalyst

Answer: C — Excess ammonia increases the concentration of ammonia relative to the primary amine, so ammonia competes more effectively as a nucleophile, favouring the formation of primary amine as the major product.

Q3. The reduction of nitrobenzene to aniline by iron scrap and dilute HCl is preferred over other reducing agents because:

  • A. Iron is cheaper than other metals
  • B. FeCl₂ formed hydrolyses to release HCl, which maintains acidic conditions needed for the reaction ✓
  • C. Iron produces no side products
  • D. The reaction is faster at room temperature

Answer: B — FeCl₂ formed during the reaction undergoes hydrolysis to regenerate HCl, ensuring that acidic conditions are maintained throughout without requiring continuous addition of more HCl.

Q4. Which statement correctly explains the basicity order: aliphatic amines > ammonia > aromatic amines?

  • A. Alkyl groups are electron-withdrawing and stabilise the conjugate acid
  • B. Alkyl groups are electron-donating and stabilise the conjugate acid, while in arylamines the lone pair is delocalised into the benzene ring ✓
  • C. Aromatic amines have sp² hybridised nitrogen, which is more basic
  • D. The lone pair in aromatic amines is more readily available for bonding

Answer: B — Electron-donating alkyl groups in aliphatic amines stabilise the conjugate acid (RNH₃⁺), increasing basicity; in arylamines, resonance delocalisation of the lone pair into the benzene ring reduces basicity.

Q5. The Gabriel phthalimide synthesis is preferred over direct ammonolysis for preparing primary amines because:

  • A. It uses a cheaper reagent than ammonia
  • B. It prevents the formation of secondary and tertiary amines, yielding only pure primary amine ✓
  • C. It works at room temperature, unlike ammonolysis which requires heating
  • D. It does not require any acidic conditions

Answer: B — Phthalimide reacts with alkyl halides once, preventing over-alkylation; hydrolysis then yields pure primary amine, unlike ammonolysis which gives a mixture of 1°, 2°, and 3° amines.

Q6. In the Carbylamine test, a primary amine reacts with CHCl₃ and KOH to produce a foul-smelling isocyanide. Which statement is NOT correct about this reaction?

  • A. Only primary amines give a positive test
  • B. Secondary and tertiary amines do not give the foul smell characteristic of isocyanides
  • C. The product is R–N≡C, where R is the alkyl group from the amine
  • D. The test is specific to aromatic amines only ✓

Answer: D — The Carbylamine test is specific to primary amines (both aliphatic and aromatic), not aromatic amines only; both R–NH₂ and Ar–NH₂ produce the foul-smelling isocyanide (R–N≡C or Ar–N≡C).

Q7. Reduction of nitriles using LiAlH₄ is described as 'ascent of amine series.' This is because:

  • A. The process converts secondary amines to tertiary amines
  • B. The nitrile (R–C≡N) is converted to a primary amine (R–CH₂–NH₂) with one additional carbon atom ✓
  • C. It increases the boiling point of the amine formed
  • D. It converts aromatic nitriles to aliphatic amines

Answer: B — Reduction of R–C≡N by LiAlH₄ produces R–CH₂–NH₂, which has one more carbon atom than the starting material, effectively moving up one step in the homologous series of amines.

Q8. Assertion (A): In amines, the C–N–C bond angle is approximately 108°, which is less than the tetrahedral angle of 109.5°. Reason (R): The lone pair on nitrogen occupies more space than a bonding pair, causing greater repulsion and compression of bond angles.

  • A. Both A and R are correct, and R is the correct explanation of A ✓
  • B. Both A and R are correct, but R is not the correct explanation of A
  • C. A is correct, but R is incorrect
  • D. A is incorrect, but R is correct

Answer: A — The assertion is correct: the C–N–C angle is ~108°. The reason correctly explains this: the lone pair on sp³-hybridised nitrogen exerts greater repulsion than bonding pairs, compressing the bond angles below 109.5°.

Q9. Which of the following conversions is NOT possible using the reduction methods described in this unit?

  • A. C₆H₅–NO₂ → C₆H₅–NH₂ (reduction by Fe/HCl)
  • B. CH₃–CN → CH₃–CH₂–NH₂ (reduction by LiAlH₄)
  • C. CH₃–CH₂–Br → CH₃–CH₂–NH₂ (ammonolysis with excess NH₃)
  • D. CH₃–CH₂–NH₂ → CH₃–CH₂–N≡C (treatment with CHCl₃ and KOH) ✓

Answer: D — The Carbylamine test (CHCl₃/KOH) does not work on primary aliphatic amines; it works on primary amines to produce isocyanides (R–N≡C), but the reverse conversion shown is not a reduction method and produces a different product type.

Q10. The coupling reaction of diazonium salts with phenols or anilines produces azo dyes. If 0.5 mol of aniline is coupled with 0.5 mol of benzene diazonium salt, how many moles of azo dye product are formed (assuming 100% conversion)?

  • A. 0.25 mol
  • B. 0.5 mol ✓
  • C. 1.0 mol
  • D. 0.75 mol

Answer: B — The coupling reaction is Ar–N₂⁺ + Ar'–OH (or Ar'–NH₂) → Ar–N=N–Ar'–OH (or Ar–N=N–Ar'–NH₂); 1 mole of diazonium salt reacts with 1 mole of phenol/aniline to form 1 mole of azo dye, so 0.5 mol each gives 0.5 mol product.

Flashcards

What is the hybridisation state of nitrogen in amines and why is the C–N–C angle not 109.5°?

Nitrogen is sp³ hybridised but the C–N–C angle is ~108° (less than 109.5°) because the lone pair occupies more space than a bonding pair, compressing the bond angles.

How are primary, secondary and tertiary amines classified?

Primary amines (RNH₂) have one alkyl/aryl group; secondary amines (R₂NH) have two; tertiary amines (R₃N) have three alkyl/aryl groups attached to nitrogen.

Write the IUPAC name of CH₃CH₂NHCH₃.

The IUPAC name is N-methylethanamine because the longest carbon chain has 2 carbons (ethanamine) and the methyl substituent on nitrogen is prefixed with 'N-'.

Why are aliphatic amines more basic than aniline?

Aliphatic amines are more basic because the lone pair on nitrogen is freely available; in aniline, the lone pair is delocalised into the benzene ring, reducing its basicity.

In ammonolysis of alkyl halides, why is a large excess of ammonia used?

Excess ammonia favours the formation of primary amine as the major product because the primary amine formed is less likely to react further when ammonia is in large excess.

What is the Gabriel phthalimide synthesis and when is it preferred over ammonolysis?

Gabriel phthalimide synthesis converts phthalimide to phthalamide using alkyl halide, then hydrolyses to give pure primary amines; it is preferred because it avoids the mixture of 1°, 2°, and 3° amines from ammonolysis.

Describe the Carbylamine test for primary amines.

A primary amine reacts with chloroform (CHCl₃) and potassium hydroxide (KOH) to produce a foul-smelling isocyanide (R–N≡C), confirming the presence of a primary amine.

Write the equation for reduction of nitrobenzene by iron scrap and dilute HCl.

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O; iron reduces the nitro group and HCl released from hydrolysis of FeCl₂ maintains acidic conditions.

What is the order of reactivity of alkyl halides in ammonolysis?

The reactivity order is RI > RBr > RCl because the C–I bond is weakest and most easily cleaved by the nucleophilic ammonia molecule.

Why is reduction of nitriles by LiAlH₄ called 'ascent of amine series'?

Reduction of nitriles (R–C≡N) by LiAlH₄ produces primary amines (R–CH₂–NH₂) with one additional carbon atom, effectively ascending the homologous series of amines by one carbon.

Important Board Questions

Write the structure and IUPAC name of the primary amine with molecular formula C₃H₉N. [2 marks]

There are two structural isomers for C₃H₉N as primary amines; identify the straight-chain and branched structures, then apply IUPAC naming rules where the longest carbon chain containing the –NH₂ group determines the base name.

Explain why aliphatic amines are more basic than aromatic amines. Also show the conjugate acid formation for both types and explain how electron effects stabilise or destabilise the conjugate acid. [5 marks]

Compare the lone pair availability on nitrogen: in aliphatic amines, alkyl groups donate electron density (inductive effect) stabilising the conjugate acid RNH₃⁺; in aromatic amines, the lone pair delocalises into the benzene ring via resonance, reducing basicity. Write the conjugate acid structures (RNH₃⁺ and [C₆H₅NH₃]⁺) and explain the stabilisation mechanism in each case.

Starting from nitrobenzene, outline the complete preparation of aniline using iron scrap and hydrochloric acid. Explain why this method is preferred over other reducing agents like tin or zinc. Also describe a chemical test that can be used to confirm the presence of a primary amino group in the product. [6 marks]

Write the reduction equation C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O; explain that FeCl₂ hydrolyses to release HCl continuously (Fe²⁺ + 2H₂O ⇌ Fe(OH)₂ + 2H⁺), maintaining acidic conditions with minimal acid input; then describe the Carbylamine test (RNH₂ + CHCl₃ + KOH → R–N≡C with foul smell) to confirm the primary amine functional group.

Assertion-Reason Questions for Amines

Q1. Assertion (A):

Nitrogen in amines has sp3 hybridisation and the geometry is pyramidal.

Reason (R):

The fourth sp3 orbital contains an unshared pair of electrons which occupies more space, reducing bond angles below 109.5°.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The unshared pair in the fourth sp3 orbital causes lone pair repulsion, which compresses bond angles (e.g., C–N–C = 108° in trimethylamine). The reason correctly explains why the geometry is pyramidal and not tetrahedral.

Q2. Assertion (A):

Primary amines are obtained as the major product in ammonolysis when excess ammonia is used.

Reason (R):

Excess ammonia promotes the formation of quaternary ammonium salts.

Answer: C — A is true but R is false — The assertion is true; excess ammonia suppresses secondary and tertiary amine formation. However, the reason is false—excess ammonia prevents quaternary salt formation by out-competing the primary amine for the alkyl halide. The reason states the opposite of what actually happens.

Q3. Assertion (A):

Reduction of a nitrile RCN with LiAlH4 produces a primary amine with one additional carbon atom compared to the starting material.

Reason (R):

The nitrile carbon is retained as a methylene carbon in the amine product.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and correctly linked. The product is RCH2NH2, where the C≡N becomes CH2NH2, effectively adding one carbon to the chain. This is the basis of the ascent of amine series.

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Case-Based Questions (CBQ) for Amines

Case Study:

A pharmaceutical company synthesises phenylamine (aniline) derivatives to produce antihistamine drugs. In the laboratory, aniline is prepared by reducing nitrobenzene using iron scrap and dilute hydrochloric acid. The reaction is exothermic and produces a complex mixture that requires careful workup. A researcher observes that when aniline is treated with excess methyl iodide in the presence of base, a quaternary ammonium salt is formed. However, when methylamine (a simple aliphatic amine) undergoes the same reaction, the quaternary salt forms much more readily and rapidly.

Q1. In the reduction of nitrobenzene with iron scrap and HCl, the role of HCl regeneration is to:

  • A. Increase the temperature of the reaction
  • B. Maintain acidic conditions and continuously regenerate the reducing agent ✓
  • C. Protonate the aniline product to form an anilinium salt
  • D. Prevent the formation of secondary amines

Answer: B

Q2. The difference in rate of quaternary salt formation between aniline and methylamine is primarily due to:

  • A. Methylamine is more soluble in the reaction solvent
  • B. Aniline is less basic because the lone pair is delocalised into the benzene ring ✓
  • C. Methylamine forms a more stable quaternary salt
  • D. Aniline undergoes competitive protonation in basic solution

Answer: B

Q3. During the workup of the reduction mixture, aniline is recovered by treating with a strong base. The purpose of this step is to:

  • A. Dehydrate the aniline product
  • B. Precipitate iron salts formed during reduction
  • C. Protonate aniline to form anilinium chloride for easier separation
  • D. Convert anilinium salt back to free aniline base for extraction ✓

Answer: D

Q4. If a mixture of aniline and methylamine is treated with dilute HCl, both form chloride salts. Which order of basicity is correct?

  • A. Methylamine > Aniline > Ammonia ✓
  • B. Aniline > Methylamine > Ammonia
  • C. Ammonia > Methylamine > Aniline
  • D. Methylamine > Ammonia > Aniline

Answer: A

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Previous Year Questions (PYQ) for Amines

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

The reaction of ammonia with a large excess of ethyl chloride will give mainly :

  • Ethylamine
  • Tetraethylammonium chloride
  • Triethylamine
  • Diethylamine

CBSE 2026 · Set 56(B) · 1 mark

Acetamide reacts with Br2 and aqueous KOH to form :

  • Ethanamine
  • Ammonia
  • Methanamine
  • Ethanenitrile

CBSE 2026 · Set 56(B) · 1 mark

Assertion (A) : Boiling point of propylamine is higher than that of trimethylamine. Reason (R) : In propylamine, intermolecular hydrogen bonding occurs, whereas trimethylamine does not undergo intermolecular hydrogen bonding.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56(B) · 3 marks

Explain the following : (a) Carbylamine reaction (b) Hoffmann's bromamide degradation reaction (c) Diazotisation reaction

CBSE 2026 · Set 56/1/1 · 1 mark

Aniline on direct nitration yields

  • 51%-ortho, 47%-para, 2%-meta derivatives
  • 51%-meta, 47%-ortho, 2%-para derivatives
  • 51%-para, 47%-meta, 2%-ortho derivatives
  • 51%-ortho, 47%-meta, 2%-para derivatives

CBSE 2026 · Set 56/1/1 · 1 mark

Which of the following amines has lowest pKb value?

  • C6H5 – N(CH3)2
  • C6H5 – NH(CH3)
  • C6H5 – NH2
  • [Option D - not clearly visible in source]

CBSE 2026 · Set 56/1/1 · 4 marks

The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho-and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.

Why CH3 – NH2 is a stronger base than (CH3)3N in aqueous solution?

CBSE 2026 · Set 56/1/2 · 1 mark

Benzene diazonium chloride on reaction with phenol in weakly basic medium gives

  • Azobenzene
  • Benzene
  • Chlorobenzene
  • p-hydroxyazobenzene

CBSE 2026 · Set 56/1/2 · 4 marks

The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho- and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.

(a) (i) Why is CH3–NH2 a stronger base than (CH3)3N in aqueous solution? (2 marks) (ii) Write structural formulae of the compound A and B: CH3CONH2 →(NaOBr) A →(C6H5COCl Base) B

CBSE 2026 · Set 56/1/3 · 1 mark

CH3 – NH2 on reaction with (CH3CO)2O gives

  • CH 3CONH2
  • CH3COONHCH3
  • CH 3 – NH – CO – CH3
  • CH3 – CO – CH2 – NH2

CBSE 2026 · Set 56/1/3 · 4 marks

The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho-and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.

(a) (i) Why CH 3 – NH2 is a stronger base than (CH3)3N in aqueous solution ?

CBSE 2026 · Set 56/2/1 · 1 mark

Identify the compound produced by the reduction of Ethanenitrile with Lithium aluminium hydride:

  • Ethylamine
  • Ethanal
  • Propylamine
  • Methylamine

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Class 12 Chemistry Formula Sheet

Every formula and named reaction from all 10 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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