Amines

What to remember

• Amines are derivatives of NH₃ and are basic because the nitrogen lone pair accepts H⁺. In water the order for methylamines is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃; in the gas phase it is 3° > 2° > 1° > NH₃.
• Aniline is much weaker as a base than aliphatic amines, because its lone pair is delocalised into the benzene ring.
• Only primary amines give the carbylamine test; Hinsberg's reagent (C₆H₅SO₂Cl) separates 1°, 2° and 3° amines; Hoffmann bromamide degradation gives a primary amine with one carbon less than the amide.
• Arenediazonium salts (from ArNH₂ + NaNO₂/HCl at 273–278 K) lose N₂ with CuCl, CuBr, CuCN, KI, HBF₄, H₂O, H₃PO₂ or C₂H₅OH, and couple with phenol or aniline to give azo dyes.

Structure, classification and nomenclature

In an amine the nitrogen atom is sp³ hybridised and has a lone pair. The shape around N is pyramidal, and the C–N–C angle in trimethylamine is about 108°.

Amines are classified by the number of alkyl or aryl groups on nitrogen:

• Primary (1°): R–NH₂
• Secondary (2°): R₂NH
• Tertiary (3°): R₃N

Common names write the alkyl groups followed by "amine". In IUPAC names the –e of the alkane is replaced by –amine. Substituents on nitrogen are written with the locant N. The simplest aromatic amine, C₆H₅NH₂, is called benzenamine (aniline).

Preparation of amines

Limits of Gabriel synthesis: it gives only primary aliphatic amines. Aromatic primary amines cannot be made, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. It also fails with tertiary halides.

Examples of Hoffmann degradation: CH₃CONH₂ → CH₃NH₂ and C₆H₅CONH₂ → C₆H₅NH₂. Each loses one carbon (as carbonate). Reduction of a nitrile, in contrast, keeps all the carbons: CH₃CN → CH₃CH₂NH₂.

Physical properties

Methylamine, dimethylamine and ethylamine are gases at room temperature, with a smell like fish. Higher amines are liquids or solids. Primary and secondary amines form hydrogen bonds with each other, so their boiling points are higher than those of alkanes and ethers of similar mass. The N–H···N bond is weaker than the O–H···O bond, so amines boil lower than alcohols of similar mass.

Order of boiling point for isomeric amines: primary > secondary > tertiary. A primary amine has two N–H bonds, a secondary has one, and a tertiary has none. Lower amines are soluble in water because they form hydrogen bonds with water. Solubility falls as the size of the alkyl part grows. Aniline is colourless when pure but turns brown on exposure to air because of oxidation.

Basic character of amines

An amine takes H⁺ from water: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻. A smaller pKb means a stronger base.

Aliphatic amines are stronger bases than NH₃, because alkyl groups push electrons (+I effect) and make the lone pair more available.

In the gas phase only the +I effect acts, so the order is 3° > 2° > 1° > NH₃. In water the conjugate acid (R₃NH⁺) must be stabilised by hydrogen bonding with water. The more N–H bonds in the ion, the more solvation. Steric hindrance also matters. These factors compete, giving:

• Methylamines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
• Ethylamines: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃

Aromatic amines: in aniline the nitrogen lone pair is in resonance with the benzene ring and is less available to take H⁺. The anilinium ion has fewer resonance structures than aniline, so it is less stabilised. Hence aniline is a far weaker base than methylamine or ammonia.

Effect of ring substituents on aniline:

• Electron-donating groups (–CH₃, –OCH₃, –NH₂) increase basicity.
• Electron-withdrawing groups (–NO₂, –CN, –Cl) decrease basicity.

So p-toluidine > aniline > p-nitroaniline in basic strength. Order of basic strength in water, from the pKb data above: (C₂H₅)₂NH > (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ > C₆H₅NH₂.

Chemical reactions of amines

Alkylation: amines react with alkyl halides by nucleophilic substitution (RNH₂ → R₂NH → R₃N → R₄N⁺X⁻), giving a mixture.

Acylation: 1° and 2° amines react with acyl chlorides, anhydrides or esters to give amides (N–H replaced by N–COR). A base such as pyridine is used to remove the HCl formed and push the reaction forward. Example: C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ (acetanilide) + HCl. Tertiary amines do not acylate because they have no N–H.

Carbylamine reaction (isocyanide test): an aliphatic or aromatic primary amine, warmed with CHCl₃ and alcoholic KOH, gives an isocyanide (carbylamine) with a very bad smell. RNH₂ + CHCl₃ + 3 KOH → RNC + 3 KCl + 3 H₂O. Secondary and tertiary amines do not respond.

Reaction with nitrous acid (HNO₂, made from NaNO₂ + HCl):

• Aliphatic primary amine: gives an unstable diazonium salt that decomposes to alcohol and N₂ (bubbles of N₂): RNH₂ + HNO₂ → ROH + N₂ + H₂O.
• Aromatic primary amine at 273–278 K: gives a stable arenediazonium salt, C₆H₅N₂⁺Cl⁻.
• Secondary amines (aliphatic or aromatic): give N-nitrosoamines, yellow oily liquids.
• Tertiary aliphatic amines dissolve to give nitrite salts.

Hinsberg's test (benzenesulphonyl chloride, C₆H₅SO₂Cl, then aqueous KOH):

Electrophilic substitution in aniline

The –NH₂ group is a strong activating group that directs ortho and para.

Bromination: aniline with bromine water gives a white precipitate of 2,4,6-tribromoaniline at once. It cannot be stopped at the mono stage. To get the monobromo product, first protect the group by acetylation: aniline → acetanilide, then Br₂ in ethanoic acid gives mainly p-bromoacetanilide, and hydrolysis gives p-bromoaniline. Acetylation lowers the activating power because the lone pair now also delocalises into the C=O group.

Nitration: direct nitration with HNO₃ + H₂SO₄ gives a mixture of ortho, meta and para products, including a large amount of meta isomer, because in strong acid aniline is converted to the anilinium ion (–NH₃⁺), which is meta-directing. Direct nitration also causes oxidation and tarry products. Protection by acetylation gives mainly the para product: acetanilide → p-nitroacetanilide → (hydrolysis) → p-nitroaniline.

Sulphonation: aniline + conc. H₂SO₄ gives anilinium hydrogensulphate. On heating at 453–473 K this gives p-aminobenzenesulphonic acid, called sulphanilic acid. It exists as a dipolar ion (zwitterion).

Friedel–Crafts alkylation and acylation do not work with aniline. The basic –NH₂ group forms a salt with the Lewis acid AlCl₃. The nitrogen then carries a positive charge and strongly deactivates the ring.

Diazonium salts

Preparation (diazotisation): C₆H₅NH₂ + NaNO₂ + 2 HCl, at 273–278 K → C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride) + NaCl + 2 H₂O. HNO₂ is made in the flask from NaNO₂ and HCl. The mixture must be kept cold, because the salt decomposes on warming.

Stability: arenediazonium salts are more stable than alkanediazonium salts, because the positive charge is spread by resonance with the benzene ring. Benzenediazonium chloride is unstable and explosive when dry, so it is used in solution. Benzenediazonium fluoroborate (the BF₄⁻ salt) is stable at room temperature and can be isolated.

Reactions with displacement of nitrogen:

Reactions with retention of nitrogen: coupling. Benzenediazonium chloride reacts with electron-rich aromatic rings, at the para position, to give azo compounds (–N=N–), which are brightly coloured dyes.

• With phenol (mildly alkaline, pH 9–10): p-hydroxyazobenzene, an orange dye.
• With aniline (mildly acidic, pH 4–5): p-aminoazobenzene, a yellow dye.

Importance in synthesis: diazonium salts let us introduce –F, –I, –CN, –OH and –H into the benzene ring, which cannot be done by direct substitution on benzene. They are also the route to azo dyes. Converting –NO₂ to –NH₂ to –N₂⁺ to –X is a standard route to many substituted benzenes.

NEET traps

• 1. Gas phase: 3° > 2° > 1° > NH₃. Water: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N, so tertiary methylamine is weaker than primary in water.
• 2. In aqueous ethylamines, the 2° amine is the strongest and 3° is stronger than 1°; do not copy the methylamine order.
• 3. Hoffmann degradation loses one carbon; reduction of nitrile adds one carbon relative to the halide.
• 4. Gabriel synthesis makes primary aliphatic amines only, never aniline.
• 5. Carbylamine reaction is for primary amines only, aliphatic and aromatic.
• 6. In Hinsberg's test, the primary sulphonamide dissolves in alkali; the secondary one does not.
• 7. Aliphatic primary amines give N₂ and alcohol with HNO₂; aromatic ones give a diazonium salt at 273–278 K.
• 8. Aniline cannot be Friedel–Crafts alkylated or acylated; direct nitration also gives meta product because of the anilinium ion.
• 9. Sandmeyer needs a copper(I) salt; Gattermann uses copper powder; fluorides come from HBF₄ (Balz–Schiemann) and iodides from KI.
• 10. Coupling with phenol is in mildly alkaline medium and with aniline in mildly acidic medium.

One-liners

• 1. In amines nitrogen is sp³ hybridised with a lone pair.
• 2. 1°, 2° and 3° amines have two, one and no N–H bonds, respectively.
• 3. C₆H₅NH₂ is called benzenamine (aniline).
• 4. Fe/HCl reduction of nitro compounds is preferred because only a little HCl is needed.
• 5. Ammonolysis of alkyl halides gives a mixture of amines and a quaternary salt.
• 6. Gabriel synthesis uses potassium phthalimide and an alkyl halide.
• 7. Hoffmann bromamide reaction: amide → amine with one carbon less.
• 8. The pKb of aniline (9.38) is far larger than that of methylamine (3.38).
• 9. Electron-donating groups on aniline raise basicity; electron-withdrawing groups lower it.
• 10. The carbylamine test gives an isocyanide with a foul smell.
• 11. Acetylation of aniline gives acetanilide and reduces the activating effect of –NH₂.
• 12. Aniline with bromine water gives white 2,4,6-tribromoaniline.
• 13. Sulphonation of aniline gives sulphanilic acid, a zwitterion.
• 14. Benzenediazonium chloride is made at 273–278 K from aniline, NaNO₂ and HCl.
• 15. Diazonium coupling gives azo dyes containing –N=N–.

Practice questions

1. The nitrogen atom in an amine is hybridised as

   1. sp²
   2. sp³d
   3. sp³
   4. sp
   Answer

   C. sp³

   Nitrogen has three bonds and one lone pair, so it is sp³ and the shape is pyramidal.

2. What is the IUPAC name of CH₃NHCH₃?

   1. ethanamine
   2. N-methylmethanamine
   3. N-methylethanamine
   4. dimethylmethanamine
   Answer

   B. N-methylmethanamine

   Two methyl groups on N: methanamine with an N-methyl substituent.

3. Propanamide is treated with Br₂ and NaOH. The product is

   1. propan-1-amine
   2. methanamine
   3. propan-2-amine
   4. ethanamine
   Answer

   D. ethanamine

   Hoffmann bromamide degradation removes the carbonyl carbon: CH₃CH₂CONH₂ → CH₃CH₂NH₂.

4. Gabriel phthalimide synthesis cannot be used to prepare

   1. aniline
   2. methanamine
   3. ethanamine
   4. benzylamine
   Answer

   A. aniline

   The phthalimide anion does not displace halide from an aryl halide, so aromatic primary amines are not formed.

5. LiAlH₄ reduces CH₃CN to

   1. CH₃CH₂NH₂
   2. CH₃CH₂OH
   3. CH₃CONH₂
   4. CH₃NH₂
   Answer

   A. CH₃CH₂NH₂

   A nitrile is reduced to a primary amine with the same number of carbon atoms: RCN → RCH₂NH₂.

6. Among methylamine, dimethylamine, trimethylamine and ammonia, the strongest base in aqueous solution is

   1. (CH₃)₃N
   2. NH₃
   3. CH₃NH₂
   4. (CH₃)₂NH
   Answer

   D. (CH₃)₂NH

   Aqueous order: (CH₃)₂NH (pKb 3.27) > CH₃NH₂ (3.38) > (CH₃)₃N (4.22) > NH₃ (4.75).

7. In the gas phase the strongest base among CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃ is

   1. CH₃NH₂
   2. (CH₃)₃N
   3. (CH₃)₂NH
   4. NH₃
   Answer

   B. (CH₃)₃N

   In the gas phase only the +I effect acts, so basicity rises with the number of alkyl groups: 3° > 2° > 1° > NH₃.

8. Which compound gives the carbylamine test?

   1. triethylamine
   2. N,N-dimethylaniline
   3. ethylamine
   4. diethylamine
   Answer

   C. ethylamine

   Only primary amines give the foul-smelling isocyanide with CHCl₃ and alcoholic KOH.

9. Aniline reacts with NaNO₂ and HCl at 273–278 K to give

   1. phenol
   2. nitrobenzene
   3. benzenediazonium chloride
   4. N-nitrosoaniline
   Answer

   C. benzenediazonium chloride

   Diazotisation of a primary aromatic amine at low temperature gives the arenediazonium salt.

10. Which amine does not react with benzenesulphonyl chloride?

    1. aniline
    2. triethylamine
    3. diethylamine
    4. ethylamine
    Answer

    B. triethylamine

    A tertiary amine has no N–H bond, so no sulphonamide can form.

11. Aniline reacts with bromine water to give

    1. 2,4,6-tribromoaniline
    2. 2-bromoaniline
    3. 4-bromoaniline
    4. 3,5-dibromoaniline
    Answer

    A. 2,4,6-tribromoaniline

    The –NH₂ group strongly activates the ortho and para positions, so all three are brominated at once.

12. Aniline heated with conc. H₂SO₄ at 453–473 K gives

    1. benzenesulphonic acid
    2. nitrobenzene
    3. 3-aminobenzenesulphonic acid
    4. sulphanilic acid (4-aminobenzenesulphonic acid)
    Answer

    D. sulphanilic acid (4-aminobenzenesulphonic acid)

    Anilinium hydrogensulphate rearranges on heating to the para sulphonic acid.

13. Benzenediazonium chloride reacts with CuCN/KCN to give

    1. phenyl isocyanide
    2. benzonitrile
    3. benzamide
    4. aniline
    Answer

    B. benzonitrile

    Sandmeyer reaction replaces –N₂⁺ by –CN.

14. The correct order of increasing basic strength is

    1. aniline < p-nitroaniline < p-toluidine
    2. p-nitroaniline < p-toluidine < aniline
    3. p-toluidine < aniline < p-nitroaniline
    4. p-nitroaniline < aniline < p-toluidine
    Answer

    D. p-nitroaniline < aniline < p-toluidine

    –NO₂ withdraws electrons and lowers basicity; –CH₃ donates electrons and raises it.

15. Heating benzenediazonium fluoroborate gives

    1. benzenediazonium fluoride
    2. benzene
    3. fluorobenzene
    4. phenol
    Answer

    C. fluorobenzene

    Balz–Schiemann reaction: ArN₂⁺BF₄⁻ → ArF + BF₃ + N₂.

16. Benzenediazonium chloride couples with phenol in mildly alkaline medium to give

    1. p-hydroxyazobenzene
    2. p-aminoazobenzene
    3. p-nitrophenol
    4. azoxybenzene
    Answer

    A. p-hydroxyazobenzene

    Coupling at the para position of phenoxide gives the orange azo dye p-hydroxyazobenzene.

17. Given below are two statements: Statement I: Aniline is a weaker base than methylamine. Statement II: In aniline the lone pair on nitrogen is delocalised into the benzene ring. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Delocalisation makes the lone pair less available (pKb 9.38 against 3.38). Both are correct.

18. Given below are two statements: Statement I: Gabriel synthesis gives pure primary aliphatic amines. Statement II: Gabriel synthesis can be used to prepare aniline. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Gabriel synthesis works with alkyl halides only; aniline cannot be made this way.

19. Given below are two statements: Statement I: In aqueous solution, trimethylamine is a stronger base than methylamine. Statement II: In the gas phase, trimethylamine is a stronger base than methylamine. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    In water CH₃NH₂ (pKb 3.38) is stronger than (CH₃)₃N (4.22); in the gas phase the order is reversed.

20. Which of the following statements are correct? 1. Primary aromatic amines give the carbylamine test. 2. Secondary amines give the carbylamine test. 3. Aliphatic primary amines release N₂ with HNO₂.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    C. 1 and 3 only

    Only primary amines give the carbylamine test, and aliphatic primary amines give N₂ with HNO₂. So 1 and 3.

21. Given below are two statements: Statement I: Direct nitration of aniline with HNO₃ and H₂SO₄ gives a considerable amount of meta product. Statement II: In strongly acidic medium aniline is converted to the anilinium ion, which is meta-directing. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    –NH₃⁺ is electron-withdrawing and meta-directing. Both are correct.

22. Which of the following statements are correct? 1. Hoffmann bromamide degradation gives an amine with one carbon less than the amide. 2. Reduction of RCN gives an amine with the same number of carbon atoms as RCN. 3. Ammonolysis of an alkyl halide gives only the primary amine.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    Ammonolysis gives a mixture of 1°, 2°, 3° amines and a quaternary salt. So statements 1 and 2 only.

23. Given below are two statements: Statement I: The benzenesulphonamide formed from a primary amine dissolves in aqueous KOH. Statement II: The benzenesulphonamide formed from a secondary amine dissolves in aqueous KOH. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Only the primary sulphonamide has an acidic N–H and dissolves in alkali; the secondary one is insoluble.

24. Given below are two statements: Statement I: Arenediazonium salts are more stable than alkanediazonium salts. Statement II: Dry benzenediazonium chloride is stable and is stored as a solid. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Resonance gives arenediazonium ions extra stability, but dry benzenediazonium chloride is explosive and is used in solution.

25. Given below are two statements: Statement I: Acetylation of aniline reduces the activating effect of the –NH₂ group. Statement II: In acetanilide the lone pair on nitrogen is also delocalised towards the C=O group. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    The lone pair is shared with the carbonyl group, so it activates the ring less. Both are correct.

26. Given below are two statements: Statement I: Coupling of benzenediazonium chloride with phenol is done in mildly acidic medium. Statement II: Coupling with aniline is done in mildly alkaline medium. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Phenol couples at pH 9–10 and aniline at pH 4–5; both statements are incorrect.

27. Assertion (A): Methylamine is a stronger base than aniline. Reason (R): The lone pair of nitrogen in aniline is delocalised into the benzene ring.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Delocalisation reduces the availability of the lone pair, so R explains A.

28. Assertion (A): Aniline gives 2,4,6-tribromoaniline with bromine water. Reason (R): Aniline is a weaker base than ammonia.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true, but the tribromo product results from the strongly activating –NH₂ group, not from basicity.

29. Assertion (A): Diazotisation of aniline is carried out at 273–278 K. Reason (R): Benzenediazonium chloride is stable at high temperature.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    The salt decomposes on warming, so R is false.

30. Assertion (A): Hoffmann bromamide degradation gives an amine with one more carbon than the amide. Reason (R): The carbonyl carbon is lost as carbonate.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A is false: the amine has one carbon fewer. R is true.

31. Assertion (A): Aniline does not undergo Friedel–Crafts alkylation. Reason (R): Aniline forms a salt with AlCl₃, which makes the ring strongly deactivated.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The –NH₂ group coordinates to AlCl₃ and becomes positively charged. R explains A.

32. Assertion (A): Sulphanilic acid exists as a zwitterion. Reason (R): Sulphanilic acid is formed on heating anilinium hydrogensulphate.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true; the dipolar form comes from the acidic –SO₃H and basic –NH₂ groups, not from how it is made.

33. Match List I with List II. List I: a. Sandmeyer reaction b. Gattermann reaction c. Balz–Schiemann reaction d. Carbylamine reaction List II: i. HBF₄, then heat ii. Cu powder, HX iii. CuCl/HCl iv. CHCl₃, alcoholic KOH Choose the correct answer.

    1. a-ii, b-iii, c-i, d-iv
    2. a-iii, b-ii, c-iv, d-i
    3. a-iii, b-i, c-ii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    D. a-iii, b-ii, c-i, d-iv

    Sandmeyer: Cu(I) salt; Gattermann: Cu powder; Balz–Schiemann: HBF₄; carbylamine: CHCl₃/KOH.

34. Match List I with List II. List I: a. KI b. H₂O, warm c. H₃PO₂ d. CuCN/KCN List II: i. benzene ii. iodobenzene iii. phenol iv. benzonitrile Choose the correct answer.

    1. a-ii, b-i, c-iii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-ii, b-iii, c-iv, d-i
    4. a-ii, b-iii, c-i, d-iv
    Answer

    D. a-ii, b-iii, c-i, d-iv

    Reagents with benzenediazonium chloride: KI gives iodobenzene, water gives phenol, H₃PO₂ gives benzene, CuCN gives benzonitrile.

35. Match List I with List II. List I: a. C₆H₅NH₂ b. (CH₃)₂NH c. (CH₃)₃N d. CH₃NH₂ List II: i. 3.27 ii. 9.38 iii. 4.22 iv. 3.38 Choose the correct answer.

    1. a-iv, b-i, c-iii, d-ii
    2. a-ii, b-i, c-iii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-i, c-iv, d-iii
    Answer

    B. a-ii, b-i, c-iii, d-iv

    pKb values: aniline 9.38, dimethylamine 3.27, trimethylamine 4.22, methylamine 3.38.

36. Match List I with List II. List I: a. Gabriel synthesis b. Hoffmann bromamide reaction c. Ammonolysis of RX d. Reduction of RCN List II: i. amine with one carbon less than the amide ii. mixture of 1°, 2°, 3° amines and quaternary salt iii. pure primary aliphatic amine iv. amine with the same number of carbons as RCN Choose the correct answer.

    1. a-iii, b-iv, c-ii, d-i
    2. a-iii, b-i, c-ii, d-iv
    3. a-i, b-iii, c-ii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    B. a-iii, b-i, c-ii, d-iv

    Gabriel: pure 1° amine; Hoffmann: one carbon less; ammonolysis: mixture; nitrile reduction: same carbons.

37. Match List I with List II. List I: a. Hinsberg's reagent b. Carbylamine test c. HNO₂ with aromatic primary amine at 273–278 K d. Bromine water with aniline List II: i. foul-smelling isocyanide ii. benzenediazonium salt iii. white precipitate iv. distinguishes 1°, 2° and 3° amines Choose the correct answer.

    1. a-i, b-iv, c-ii, d-iii
    2. a-iii, b-i, c-ii, d-iv
    3. a-iv, b-ii, c-i, d-iii
    4. a-iv, b-i, c-ii, d-iii
    Answer

    D. a-iv, b-i, c-ii, d-iii

    Hinsberg's reagent separates the three classes; carbylamine gives an isocyanide; HNO₂ gives the diazonium salt; Br₂ water gives a white precipitate.

38. Match List I with List II. List I: a. Aniline + (CH₃CO)₂O b. Aniline + conc. H₂SO₄ at 453–473 K c. Aniline + Br₂ water d. C₆H₅N₂⁺Cl⁻ + phenol (alkaline) List II: i. sulphanilic acid ii. p-hydroxyazobenzene iii. 2,4,6-tribromoaniline iv. acetanilide Choose the correct answer.

    1. a-iv, b-iii, c-i, d-ii
    2. a-i, b-iv, c-iii, d-ii
    3. a-iii, b-i, c-iv, d-ii
    4. a-iv, b-i, c-iii, d-ii
    Answer

    D. a-iv, b-i, c-iii, d-ii

    Acetylation gives acetanilide; sulphonation gives sulphanilic acid; bromination gives the tribromo compound; coupling gives the azo dye.

39. Nitrobenzene is reduced to aniline with scrap iron and HCl. Only a small amount of HCl is needed because

    1. nitrobenzene cannot be reduced by HCl in any amount
    2. iron does not react with HCl
    3. FeCl₂ formed is hydrolysed and releases HCl again
    4. aniline neutralises the HCl
    Answer

    C. FeCl₂ formed is hydrolysed and releases HCl again

    Fe/HCl is preferred industrially: hydrolysis of FeCl₂ regenerates some HCl.

40. A compound C₂H₇N gives a foul-smelling substance with CHCl₃ and alcoholic KOH, and gives N₂ with nitrous acid. It is

    1. propan-1-amine
    2. ethanamine
    3. N-methylmethanamine
    4. N,N-dimethylmethanamine
    Answer

    B. ethanamine

    A primary amine gives both tests. C₂H₇N as a primary amine is CH₃CH₂NH₂.

41. Which amine gives a benzenesulphonamide that is insoluble in aqueous KOH?

    1. ethylamine
    2. triethylamine
    3. diethylamine
    4. aniline
    Answer

    C. diethylamine

    A secondary amine gives a sulphonamide with no N–H. Primary gives a soluble one; tertiary does not react.

42. Which sequence converts aniline into 4-bromoaniline?

    1. bromination with Br₂ water, then hydrolysis
    2. bromination, then acetylation
    3. nitration, bromination, reduction
    4. acetylation, bromination, hydrolysis
    Answer

    D. acetylation, bromination, hydrolysis

    Protection by acetylation limits substitution to mono-bromination at the para position; hydrolysis restores –NH₂.

43. Which sequence converts nitrobenzene into iodobenzene?

    1. Sn/HCl; KI; NaNO₂/HCl
    2. Sn/HCl; NaNO₂/HCl at 273–278 K; KI
    3. NaNO₂/HCl; Sn/HCl; KI
    4. KI; Sn/HCl; NaNO₂/HCl
    Answer

    B. Sn/HCl; NaNO₂/HCl at 273–278 K; KI

    Reduce to aniline, diazotise at low temperature, then replace –N₂⁺ with I using KI.

44. Benzenediazonium chloride is coupled with aniline. The product and the medium used are

    1. p-aminoazobenzene; mildly acidic
    2. p-hydroxyazobenzene; mildly alkaline
    3. p-aminoazobenzene; strongly alkaline
    4. azobenzene; strongly acidic
    Answer

    A. p-aminoazobenzene; mildly acidic

    Coupling with aniline gives yellow p-aminoazobenzene at pH 4–5.

45. The correct increasing order of basic strength in aqueous solution is

    1. NH₃ < C₆H₅NH₂ < CH₃NH₂ < (CH₃)₂NH
    2. C₆H₅NH₂ < CH₃NH₂ < NH₃ < (CH₃)₂NH
    3. C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH
    4. C₆H₅NH₂ < NH₃ < (CH₃)₂NH < CH₃NH₂
    Answer

    C. C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH

    pKb: aniline 9.38 > NH₃ 4.75 > CH₃NH₂ 3.38 > (CH₃)₂NH 3.27, so basicity rises in that order.