CBSE Class 12 Chemistry Amines Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. Write the IUPAC name of the given compound:
A benzene ring with –NH₂ at C-1 and Br atoms at C-2, C-4 and C-6. [CBSE Delhi 2016]
Answer:
2,4,6-Tribromoaniline / 2,4,6-Tribromobenzenamine.

Teacher's Note:
1. This is the white precipitate formed when aniline reacts with bromine water.

Question 2. Give the IUPAC name and structure of the amine obtained when 3-chlorobutanamide undergoes Hoffmann-bromamide reaction. [CBSE SQP 2016]
Answer:
2-Chloropropan-1-amine, CH₃CH(Cl)CH₂NH₂.

Teacher's Note:
1. Hoffmann degradation removes the C=O carbon, so C-3 of the amide becomes C-2 of the amine; renumber the new chain from the –NH₂ carbon.

Question 3. Arrange the following in increasing order of basic strength: Aniline, p-Nitroaniline and p-Toluidine [CBSE Comptt. OD 2015]
Answer:
p-Nitroaniline < Aniline < p-Toluidine.

Teacher's Note:
1. Electron-releasing –CH₃ raises and electron-withdrawing –NO₂ lowers the basicity of aniline.

Question 4. Arrange the following in the decreasing order of their basic strength in aqueous solutions:
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃ [CBSE Delhi 2012]
Answer:
(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.

Teacher's Note:
1. In water, basicity depends on +I effect, solvation (H-bonding) and steric hindrance together; that is why (CH₃)₃N falls below CH₃NH₂.

Question 5. Complete the following reaction equation:
C₆H₅N₂Cl + H₃PO₂ + H₂O → ________ [CBSE Comptt. Delhi 2015]
Answer:
C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ (Benzene) + N₂ + H₃PO₃ + HCl

Teacher's Note:
1. Write all products: N₂, H₃PO₃ and HCl are formed along with benzene.

Question 6. How many of the following amines have a pK_b value less than that of C₆H₅NH₂?
p-CH₃C₆H₄NH₂, o-CH₃C₆H₄NH₂, m-CH₃C₆H₄NH₂, C₆H₅N(CH₃)₂, C₆H₅NHCH₃, p-NO₂C₆H₄NH₂, p-ClC₆H₄NH₂, C₆H₅CH₂NH₂, C₆H₅CH₂CH₂NH₂
Answer:
6
Stronger bases than aniline (lower pK_b): p-toluidine, m-toluidine, C₆H₅N(CH₃)₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂ and C₆H₅CH₂CH₂NH₂.
o-Toluidine (ortho effect), p-nitroaniline and p-chloroaniline are weaker bases than aniline.

Teacher's Note:
1. Benzylamine and 2-phenylethylamine are aliphatic amines – the lone pair on N is not conjugated with the ring, so they are much stronger bases than aniline.

Question 7. Convert m-dinitrobenzene to m-nitroaniline.
Answer:
m-C₆H₄(NO₂)₂ (NH₄)₂S
⟶
[H] m-NO₂C₆H₄NH₂
Ammonium sulphide reduces only one of the two nitro groups (selective reduction).

Teacher's Note:
1. Sn/HCl would reduce both nitro groups; (NH₄)₂S or Na₂S is the reagent for partial reduction.

Question 8. Draw the structure of TNT, an explosive.

Answer:
TNT is 2,4,6-trinitrotoluene:

Teacher's Note:
1. The three NO₂ groups are at positions 2, 4 and 6 – ortho, para and ortho to the CH₃ group.

Question 9. Write the IUPAC name of CH₃–N(CH₃)–C(CH₃)(C₂H₅)–CH₂–CH₃.
Answer:
N,N,3-Trimethylpentan-3-amine

Teacher's Note:
1. Choose the longest carbon chain through the carbon bearing nitrogen (five carbons here); groups on N get the locant N.

Question 10. Give one use of quaternary ammonium salts.
Answer:
It is used as detergents, e.g., [CH₃(CH₂)₁₅N(CH₃)₃]⁺Cl⁻.

Teacher's Note:
1. A quaternary ammonium ion has four groups on nitrogen; a long alkyl chain makes it a cationic detergent.

Question 11. What is Hinsberg’s reagent?
Answer:
Benzene sulphonyl chloride, C₆H₅SO₂Cl.

Teacher's Note:
1. Hinsberg’s reagent distinguishes 1°, 2° and 3° amines by the solubility of the sulphonamides formed in alkali.

Question 12. Why does aniline dissolve in HCl?
Answer:
C₆H₅NH₂ + HCl → [C₆H₅NH₃]⁺Cl⁻.
It dissolves due to its basic nature.

Teacher's Note:
1. The anilinium chloride formed is an ionic salt, which is why it is soluble in water.

Question 13. How will you test the presence of a primary amine?
Answer:
By carbylamine test.
RNH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O
An offensive smell of isocyanide confirms a primary amine.

Teacher's Note:
1. Secondary and tertiary amines do not give this test – use it only for 1° amines.

Question 14. Write one use each of dopamine and atropine alkaloid.
Answer:
Dopamine: Neurotransmitter.
Atropine alkaloid: 0.5–1.0% solution is used in ophthalmic (eye) examination.

Teacher's Note:
1. Atropine dilates the pupil, which is why eye doctors use it.

Short Answer Questions (2 Marks)

Question 15. Arrange the following:
(i) In increasing order of their basic strength
C₆H₅—NH₂, CH₃—CH₂—NH₂, CH₃—NH—CH₃
(ii) In increasing order of solubility in water
CH₃—NH₂, (CH₃)₃N, CH₃—NH—CH₃ [CBSE Guwahati 2015]
Answer:
(i) C₆H₅NH₂ < CH₃—CH₂—NH₂ < CH₃NHCH₃
(ii) (CH₃)₃N < (CH₃)₂NH < CH₃NH₂

Teacher's Note:
1. Solubility follows the number of N–H bonds available for hydrogen bonding with water: 1° > 2° > 3°.

Question 16. Account for the following:
(i) Aniline gets coloured on standing in air for a long time.
(ii) MeNH₂ is stronger base than MeOH. [NCERT Exemplar]
Answer:
(i) Due to electron-donating effect (+R-effect) of —NH₂ group, the electron density on the benzene ring increases. As a result, aniline is easily oxidised on standing in air for a long time to form coloured products.
(ii) Nitrogen is less electronegative than oxygen therefore lone pair of electrons on nitrogen is readily available for donation. Hence, MeNH₂ is more basic than MeOH.

Teacher's Note:
1. Freshly distilled aniline is colourless; the brown colour on storage is due to atmospheric oxidation.

Question 17. Account for the following:
(i) pK_b of aniline is more than that of methylamine.
(ii) Although trimethylamine and n-propylamine have the same molecular weight, but the former boils at a lower temperature (276 K) than the latter (322 K). Explain. [CBSE (AI) 2014]
Answer:
(i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring due to resonance, so it is less available for protonation. In methylamine, the +I effect of the CH₃ group increases the electron density on nitrogen. Hence aniline is a weaker base than methylamine, its K_b is smaller and its pK_b is higher.
(ii) n-Propylamine has two H-atoms on the N-atom and hence undergoes intermolecular H-bonding, thereby raising its boiling point. Trimethylamine, (CH₃)₃N, being a tertiary amine does not have any H-atom on the N-atom. As a result, it does not undergo H-bonding and hence its boiling point is low.

Teacher's Note:
1. Higher pK_b always means a weaker base – state this link clearly in the answer.

Question 18. Write structures of compounds A and B in each of the following reactions:
(i) C₆H₅NH₂ (aniline) →[conc. H₂SO₄] A →[heat, 453–473 K] B
(ii) C₆H₅CONH₂ →[Br₂/NaOH] A →[(CH₃CO)₂O/pyridine] B [CBSE 2019 (56/3/2)]
Answer:
(i) A = C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogensulphate); B = p-H₂N—C₆H₄—SO₃H (sulphanilic acid) ⇄ p-H₃N⁺—C₆H₄—SO₃⁻ (Zwitter ion)
(ii) A = C₆H₅NH₂ (Aniline); B = C₆H₅NH—CO—CH₃ (Acetanilide)

Teacher's Note:
1. Sulphanilic acid exists mainly as a zwitter ion, which is why it has a high melting point and is insoluble in organic solvents.

Question 19. How will you carry out the following conversions?
(i) toluene → p-toluidine
(ii) p-toluidine diazonium chloride → p-toluic acid [NCERT Exemplar]
Answer:
(i) C₆H₅CH₃ (Toluene) →[HNO₃/H₂SO₄] p-CH₃C₆H₄NO₂ →[Fe/HCl] p-CH₃C₆H₄NH₂ (p-Toluidine)
(ii) p-CH₃C₆H₄N₂⁺Cl⁻ →[CuCN/KCN] p-CH₃C₆H₄CN →[H₂O/H⁺] p-CH₃C₆H₄COOH (p-Toluic acid)

Teacher's Note:
1. Nitration of toluene gives both o- and p-nitrotoluene; the p-isomer is separated before reduction.

Question 20. Write the following conversions:
(i) nitrobenzene → acetanilide
(ii) acetanilide → p-nitroaniline [NCERT Exemplar]
Answer:
(i) C₆H₅NO₂ (Nitrobenzene) →[Sn/HCl] C₆H₅NH₂ →[(CH₃CO)₂O, Pyridine] C₆H₅NHCOCH₃ (Acetanilide)
(ii) C₆H₅NHCOCH₃ (Acetanilide) →[HNO₃/H₂SO₄] p-O₂N—C₆H₄—NHCOCH₃ →[H⁺/H₂O] p-O₂N—C₆H₄—NH₂ (p-nitroaniline)

Teacher's Note:
1. Acetylation protects –NH₂ from oxidation by HNO₃ and stops the formation of m-product via the anilinium ion.

Question 21. Explain the following reactions:
(i) Gabriel phthalimide reaction
(ii) Coupling reaction [CBSE Comptt. Delhi 2013]
Answer:
(i) Gabriel's phthalimide synthesis: Phthalimide when treated with alcoholic potassium hydroxide is easily converted into potassium phthalimide which on treatment with alkyl halide followed by hydrolysis with acid or alkali yields phthalic acid and a primary amine.
C₆H₄(CO)₂NH (Phthalimide) → [KOH, −H₂O] → C₆H₄(CO)₂N⁻K⁺ (Potassium phthalimide) → [RX] → C₆H₄(CO)₂N–R (N-alkyl phthalimide)
C₆H₄(CO)₂N–R → [(aq.) NaOH] → C₆H₄(COONa)₂ (Salts of phthalic acid) + RNH₂ (Primary amine)
(ii) Coupling reaction: It involves the reaction of benzene diazonium salts with phenols or aryl amines. Coupling of phenol takes place in mild alkaline solution while with aromatic 1° amines in mild acidic medium.
C₆H₅N₂Cl + C₆H₅OH → [OH⁻, 273 K – 278 K] → C₆H₅–N=N–C₆H₄–OH (p-Hydroxyazobenzene, orange dye) + HCl

Teacher's Note:
1. Coupling occurs at the para position of phenol or aniline; the extended conjugation of the –N=N– link is what makes azo compounds coloured.

Question 22. Give chemical tests to distinguish between the following pairs of compounds:
(i) Aniline and Ethylamine
(ii) Ethylamine and Dimethylamine [CBSE Comptt. Delhi 2013]
Answer:
(i) Aniline forms diazonium salt with NaNO₂/HCl which gives orange-red dye with β-naphthol. Ethylamine does not give this test.
C₆H₅NH₂ + NaNO₂ + HCl → [273–278 K] → C₆H₅N₂⁺Cl⁻ + NaCl + H₂O
C₆H₅N₂Cl + β-naphthol → [dil. NaOH, −NaCl, H₂O] → 1-Phenylazo-2-naphthol (orange-red dye) + HCl
(ii) Ethylamine gives foul smelling ethyl isocyanide on heating with chloroform and potassium hydroxide solution. Dimethylamine does not give this test.
C₂H₅NH₂ + CHCl₃ + 3KOH → C₂H₅NC + 3KCl + 3H₂O

Teacher's Note:
1. Choose a test that only one compound of the pair gives, and always write the reagent and the observation, not just the name of the test.

Question 23. A primary amine, RNH₂ can be reacted with CH₃–X to get secondary amine, R–NHCH₃ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH₂ forms only 2° amine? [NCERT Exemplar]
Answer:
R–NH₂ + CHCl₃ + 3KOH → R–NC (Alkyl isocyanide) → [H₂/Pd (Reduction)] → R–NH–CH₃ (2° Amine)
Carbylamine reaction is shown by 1° amine only which results in the replacement of two hydrogen atoms attached to nitrogen atom of NH₂ group by one carbon atom. On catalytic reduction, the isocyanide will give a secondary amine with one methyl group.

Teacher's Note:
1. Catalytic reduction of R–NC adds the new carbon to nitrogen (R–NH–CH₃), whereas reduction of R–CN gives R–CH₂–NH₂.

Question 24. Complete the following acid-base reactions and name the products:
(i) CH₃CH₂CH₂NH₂ + HCl →
(ii) (CH₃CH₂)₃N + HCl →
Answer:
(i) CH₃CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂NH₃⁺Cl⁻ (n-propylammonium chloride)
(ii) (CH₃CH₂)₃N + HCl → (CH₃CH₂)₃NH⁺Cl⁻ (triethylammonium chloride)

Teacher's Note:
1. Amines are bases, so with acids they form ammonium salts; name them by replacing ‘amine’ with ‘ammonium’ plus the anion.

Question 25. Write the chemical reaction of C₆H₅NH₂ with C₆H₅COCl and name the product obtained.
Answer:
C₆H₅COCl + H₂N–C₆H₅ → C₆H₅CONH–C₆H₅ + HCl
The product is N-phenylbenzamide (benzanilide).

Teacher's Note:
1. This is benzoylation (Schotten–Baumann reaction); a base such as pyridine or NaOH removes the HCl and drives the reaction forward.

Question 26. How will you convert:
(i) 3-methylaniline → 3-nitrotoluene
(ii) Aniline → 1,3,5-tribromobenzene
Answer:
(i) 3-Methylaniline Diazo
⟶
NaNO₂/HBF₄ 3-methylbenzenediazonium fluoroborate (m-CH₃C₆H₄N₂⁺BF₄⁻) NaNO₂/Cu
⟶
Δ 3-nitrotoluene + N₂ + NaBF₄
(ii) Aniline Br₂/H₂O
⟶ 2,4,6-tribromoaniline Diazo
⟶
NaNO₂/HCl 2,4,6-tribromobenzenediazonium chloride H₃PO₂ + H₂O
⟶
(Cu⁺) 1,3,5-tribromobenzene

Teacher's Note:
1. In (ii) the –NH₂ group is used only to direct three Br atoms to alternate positions and is then removed by H₃PO₂ – a classic exam trick.

Question 27. How will you convert:
(i) Propanoic acid → Ethanoic acid
(ii) Nitromethane → Dimethylamine
Answer:
(i) C₂H₅COOH NH₃
⟶
Δ C₂H₅CONH₂ Br₂/KOH
⟶ C₂H₅NH₂ HNO₂
⟶ CH₃CH₂OH [O]
⟶ CH₃CHO [O]
⟶ CH₃COOH
(ii) CH₃NO₂ H₂/Ni
⟶ CH₃NH₂ CHCl₃/KOH
⟶ CH₃N≡C H₂/Ni
⟶ CH₃–NH–CH₃

Teacher's Note:
1. Hoffmann bromamide degradation is the standard step for ‘one carbon less’ conversions; reduction of an isocyanide gives a secondary amine.

Short Answer Questions (3 Marks)

Question 28. Give the structure of A, B and C in the following reactions:
(i) CH₃Br \( \xrightarrow{\text{KCN}} \) A \( \xrightarrow{\text{LiAlH}_{4}} \) B \( \xrightarrow[\text{273 K}]{\text{HNO}_{2}} \) C
(ii) CH₃COOH \( \xrightarrow[\Delta]{\text{NH}_{3}} \) A \( \xrightarrow{\text{Br}_{2}\text{ + KOH}} \) B \( \xrightarrow{\text{CHCl}_{3}\text{ + NaOH}} \) C
Answer:
(i) CH₃Br \( \xrightarrow{\text{KCN}} \) CH₃CN (A) \( \xrightarrow{\text{LiAlH}_{4}} \) CH₃CH₂NH₂ (B) \( \xrightarrow[\text{273 K}]{\text{HNO}_{2}} \) CH₃CH₂OH (C)
(ii) CH₃COOH \( \xrightarrow[\Delta]{\text{NH}_{3}} \) CH₃CONH₂ (A) \( \xrightarrow{\text{Br}_{2}\text{ + KOH}} \) CH₃NH₂ (B) \( \xrightarrow{\text{CHCl}_{3}\text{ + NaOH}} \) CH₃NC (C)

Teacher's Note:
1. Primary aliphatic amines with HNO₂ give alcohols (with N₂), unlike aniline, which gives a stable diazonium salt at 273 K.

Question 29. (a) Write the reactions involved in the following:
(i) Hoffmann bromamide degradation reaction
(ii) Gabriel phthalimide synthesis
(b) Give reason: (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. [CBSE Delhi/Outside Delhi 2018]
Answer:
(a) (i) Ar/R–CONH₂ + Br₂ + 4NaOH → Ar/R–NH₂ + 2NaBr + Na₂CO₃ + 2H₂O
(ii) Phthalimide \( \xrightarrow{\text{KOH}} \) Potassium phthalimide \( \xrightarrow{\text{R–X}} \) N-Alkylphthalimide
N-Alkylphthalimide \( \xrightarrow{\text{NaOH(aq)}} \) Sodium phthalate, C₆H₄(COO⁻Na⁺)₂ + R–NH₂
(b) Because of the combined factors of inductive effect and solvation or hydration effect.
(CH₃)₂NH + H⁺ → (CH₃)₂NH₂⁺; (CH₃)₃N + H⁺ → (CH₃)₃NH⁺
2° amine salt form is more stable than 3° amine due to inductive effect and higher degree of hydration. Therefore, higher the stability of salt, greater will be the basic strength of corresponding amine.

Teacher's Note:
1. In Gabriel synthesis write all three steps (salt formation, alkylation, hydrolysis); the last step releasing R–NH₂ is often missed.

Question 30. (i) Arrange the following compounds in increasing order of dipole moment.
CH₃CH₂CH₃, CH₃CH₂NH₂, CH₃CH₂OH [NCERT Exemplar]
(ii) Give possible explanation for each of the following:
(a) The presence of a base is needed in the ammonolysis of alkyl halides.
(b) Amides are more acidic than amines.
Answer:
(i) CH₃CH₂CH₃ < CH₃CH₂NH₂ < CH₃CH₂OH
(ii) (a) To remove HX formed so that the reaction shifts in the forward direction.
(b) R—C(=O)—NH₂ ↔ R—C(—O⁻)=N⁺H₂
Due to +R effect, availability of lone pair of electron on N of —NH₂ group decreases. As a result, acid amide is much weaker base than amines. Because of the positive charge on N, as a result of resonance, N can easily lose a proton and behaves, as a weak acid.

Teacher's Note:
1. Dipole moment follows the electronegativity of the heteroatom: O > N > C (propane is almost non-polar).

Question 31. Write the structures of A, B and C in the following:
(i) C₆H₅—CONH₂ →[Br₂/aq. KOH] A →[NaNO₂ + HCl, 0–5°C] B →[KI] C
(ii) CH₃—Cl →[KCN] A →[LiAlH₄] B →[CHCl₃ + alc. KOH, Δ] C [CBSE Delhi 2016]
Answer:
(i) A = C₆H₅—NH₂ (Aniline), B = C₆H₅—N⁺≡NCl⁻ (Benzene diazonium chloride), C = C₆H₅—I (Iodobenzene)
(ii) A = CH₃CN (Methyl cyanide), B = CH₃—CH₂—NH₂ (Ethyl amine), C = CH₃—CH₂—N⇄C (Ethyl isocyanide)

Teacher's Note:
1. Write the reagent conditions over each arrow in the exam – especially 0–5°C for diazotisation.

Question 32. Complete the following reactions:
(i) C₆H₅—COOH →[NH₃] A →[Heat] B →[Br₂/KOH] C
(ii) C₆H₅NO₂ →[Fe/HCl] A →[NaNO₂ + HCl, 273–278 K] B →[CuCN] C [CBSE East 2016]
Answer:
(i) A = C₆H₅COO⁻NH₄⁺ (Ammonium benzoate), B = C₆H₅CONH₂ (Benzamide), C = C₆H₅NH₂ (Aniline)
(ii) A = C₆H₅NH₂ (Aniline), B = C₆H₅N⁺≡NCl⁻ (Benzene diazonium chloride), C = C₆H₅CN (Benzonitrile)

Teacher's Note:
1. An ammonium salt of a carboxylic acid loses water on heating to give the amide – a common first step before Hoffmann degradation.

Question 33. Complete the following chemical equations:
(i) C₆H₅N₂Cl + C₆H₅NH₂ →[OH⁻]
(ii) C₆H₅N₂Cl + CH₃CH₂OH →
(iii) RNH₂ + CHCl₃ + KOH → [CBSE Delhi 2010]
Answer:
(i) C₆H₅—N⁺≡NCl⁻ (Benzene diazonium chloride) + C₆H₅—NH₂ (Aniline) →[273–278 K, H⁺ (pH 4–5)] C₆H₅—N=N—C₆H₄—NH₂(p) (p-Aminoazobenzene, yellow dye) + HCl
(ii) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) + CH₃—CH₂—OH (Ethanol) → C₆H₆ (Benzene) + N₂ + CH₃—CHO + HCl
(iii) R—NH₂ (1° amine) + CHCl₃ + 3KOH (alc.) →[Δ] R—NC (Alkyl isocyanide) + 3KCl + 3H₂O

Teacher's Note:
1. In (ii) ethanol acts as a mild reducing agent and is itself oxidised to ethanal.

Question 34. How are the following conversions carried out:
(i) Aniline to fluorobenzene
(ii) Benzene diazonium chloride to benzene
(iii) Methyl chloride to ethylamine [CBSE (F) 2013]
Answer:
(i) C₆H₅NH₂ →[NaNO₂/HCl, 0–5°C] C₆H₅N₂⁺Cl⁻ →[HBF₄] C₆H₅N₂⁺BF₄⁻ →[Δ] C₆H₅F (Fluoro benzene)
(ii) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) + H₃PO₂ + H₂O → C₆H₆ (Benzene) + N₂ + H₃PO₃ + HCl
(iii) CH₃—Cl (Methyl chloride) →[alc. KCN] CH₃—CN →[LiAlH₄] CH₃CH₂NH₂ (Ethylamine)

Teacher's Note:
1. Going from one carbon (CH₃Cl) to a two-carbon amine signals the cyanide route, which adds one carbon atom.

Question 35. How will you convert the following:
(i) Aniline to chlorobenzene
(ii) Ethanoic acid to methanamine
(iii) Benzene diazonium chloride to phenol [CBSE (F) 2013]
Answer:
(i) C₆H₅NH₂ (Aniline) →[NaNO₂/HCl, 273–278 K] C₆H₅N₂⁺Cl⁻ (Diazonium salt) →[Cu₂Cl₂/HCl] C₆H₅Cl (Chlorobenzene)
(ii) CH₃COOH (Ethanoic acid) →[NH₃, Δ] CH₃CONH₂ (Ethanamide) →[Br₂/KOH (Hoffmann bromamide reaction)] CH₃NH₂ (Methanamine)
(iii) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) + H₂O →[boil] C₆H₅OH (Phenol) + HCl + N₂↑

Teacher's Note:
1. Ethanoic acid has two carbons and methanamine one – the loss of one carbon points to Hoffmann bromamide degradation.

Question 36. Complete the following reactions:
(i) Cyclobutanecarbonitrile (C₄H₇—CN) →[H₂/Ni]
(ii) The diazonium salt shown below →[H₃PO₂ + H₂O]
(iii) C₆H₅CH₂–NH₂ + CHCl₃ →[Ethanolic KOH] [CBSE 2019 (56/2/1)]

Answer:
(i) C₄H₇—CN →[H₂/Ni] C₄H₇—CH₂NH₂ (cyclobutylmethanamine)
(ii) 3-Bromo-4-methylbenzenediazonium chloride →[H₃PO₂ + H₂O] 2-bromotoluene (o-CH₃C₆H₄Br; the –N₂⁺Cl⁻ group is replaced by H)
(iii) C₆H₅CH₂–NH₂ + CHCl₃ →[Ethanolic KOH] C₆H₅CH₂–NC + 3KCl + 3H₂O

Teacher's Note:
1. In (ii) only the diazonium group is replaced by H; the CH₃ and Br stay in their positions, so renumber the product carefully.

Question 37. How do you convert the following:
(i) N-phenylethanamide to p-bromoaniline
(ii) Benzene diazonium chloride to nitrobenzene
(iii) Benzoic acid to aniline [CBSE 2019 (56/2/1)]
Answer:
(i) C₆H₅NHCOCH₃ (N-phenylethanamide) →[Br₂/CH₃COOH] p-Br—C₆H₄—NHCOCH₃ (p-Bromo acetanilide) →[H₂O/H⁺] p-Br—C₆H₄—NH₂ (p-Bromo aniline)
(ii) C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride) →[HBF₄] C₆H₅N₂⁺BF₄⁻ →[NaNO₂, Cu, Δ] C₆H₅NO₂ (Nitrobenzene)
(iii) C₆H₅COOH (Benzoic acid) →[+ NH₃, Δ] C₆H₅CONH₂ (Benzamide) →[Br₂/KOH] C₆H₅NH₂ (Aniline)

Teacher's Note:
1. Direct bromination of aniline gives the 2,4,6-tribromo product; the acetyl group is used to get only the p-bromo compound.

Question 38. Give the structures of A, B and C in the following reaction:
(i) C₆H₅NO₂ → [Fe/HCl] → A → [HNO₂, 273 K] → B → [C₆H₅OH] → C
(ii) C₆H₅N₂Cl → [CuCN] → A → [H₂O/H⁺] → B → [NH₃, Δ] → C [CBSE SQP 2016]
Answer:
(i) A – C₆H₅NH₂, B – C₆H₅N₂⁺Cl⁻, C – C₆H₅–N=N–C₆H₄–OH (p-hydroxyazobenzene)
(ii) A – C₆H₅CN, B – C₆H₅COOH, C – C₆H₅CONH₂

Teacher's Note:
1. CuCN replaces –N₂⁺ by –CN (Sandmeyer reaction); hydrolysis then gives the acid and heating with NH₃ gives the amide.

Question 39. Give the structure of A, B and C in the following reactions:
(i) CH₃Br → [KCN] → A → [LiAlH₄] → B → [HNO₂, 273 K] → C
(ii) CH₃COOH → [NH₃, Δ] → A → [Br₂ + KOH] → B → [CHCl₃ + NaOH] → C [CBSE Delhi 2014]
Answer:
(i) CH₃Br → [KCN] → CH₃CN (A) → [LiAlH₄] → CH₃CH₂NH₂ (B) → [HNO₂, 273 K] → CH₃CH₂–OH (C)
(ii) CH₃COOH → [NH₃, Δ] → CH₃CONH₂ (A) → [Br₂ + KOH] → CH₃NH₂ (B) → [CHCl₃ + NaOH] → CH₃NC (C)

Teacher's Note:
1. Aliphatic diazonium salts are unstable even at low temperature, so a 1° aliphatic amine with HNO₂ gives the alcohol with evolution of N₂.

Long Answer Questions (5 Marks)

Question 40. Predict the reagent or the product in the following reaction sequence:
p-Nitrotoluene \( \xrightarrow{\text{1}} \) p-Toluidine (4-methylaniline) \( \xrightarrow[\text{Pyridine}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) 4-Methylacetanilide \( \xrightarrow[\text{H}_{2}\text{SO}_{4}]{\text{HNO}_{3}} \) [2] \( \xrightarrow{\text{3}} \) 4-Methyl-2-nitroaniline \( \xrightarrow{\text{NaNO}_{2}\text{/HCl}} \) [4] \( \xrightarrow{\text{5}} \) m-Nitrotoluene (1-methyl-3-nitrobenzene)
Answer:
1. Sn/HCl
2. 4-Methyl-2-nitroacetanilide: CH₃ at C-4 and NO₂ at C-2 with respect to –NHCOCH₃ (nitration)
3. H₂O/H⁺ or OH⁻ (hydrolysis)
4. 4-Methyl-2-nitrobenzenediazonium chloride, CH₃C₆H₃(NO₂)N₂⁺Cl⁻ (diazotisation)
5. H₃PO₂/H₂O (deamination)

Teacher's Note:
1. This sequence uses –NH₂ as a temporary directing group: it puts NO₂ meta to CH₃, and is then removed by diazotisation and H₃PO₂.

Question 41. An organic compound A with molecular formula C₇H₇NO reacts with Br₂/aq. KOH to give compound B, which upon reaction with NaNO₂ and HCl at 0°C gives C. Compound C on heating with CH₃CH₂OH gives a hydrocarbon D. Compound B on further reaction with Br₂ water gives white precipitate of compound E. Identify the compound A, B, C, D and E; also justify your answer by giving relevant chemical equations.
Answer:
C₆H₅CONH₂ (A) \( \xrightarrow[\Delta\text{, KOH(aq)}]{\text{Br}_{2}} \) C₆H₅NH₂, Aniline (B) \( \xrightarrow[\text{0°C}]{\text{NaNO}_{2}\text{, HCl}} \) C₆H₅N₂Cl (C) (diazonium salt)
C₆H₅N₂Cl (C) \( \xrightarrow[\Delta]{\text{CH}_{3}\text{CH}_{2}\text{OH}} \) C₆H₆, Benzene (D) + CH₃CHO + N₂ + HCl
C₆H₅NH₂ (B) \( \xrightarrow{\text{Br}_{2}\text{, H}_{2}\text{O}} \) 2,4,6-Tribromoaniline (E), white ppt.
A = Benzamide, B = Aniline, C = Benzenediazonium chloride, D = Benzene, E = 2,4,6-Tribromoaniline

Teacher's Note:
1. C₇H₇NO plus Br₂/KOH points straight to an amide (Hoffmann); losing one carbon gives C₆H₅NH₂.

Question 42. (i) Write the structures of main products when aniline reacts with the following reagents:
(a) Br₂ water
(b) HCl
(c) (CH₃CO)₂O / pyridine
(ii) Arrange the following in the increasing order of their boiling point:
C₂H₅NH₂, C₂H₅OH, (CH₃)₃N
(iii) Give a simple chemical test to distinguish between the following pair of compounds:
(CH₃)₂NH and (CH₃)₃N [CBSE Delhi 2015]
Answer:
(i) (a) C₆H₅NH₂ \( \xrightarrow{\text{Br}_{2}\text{/H}_{2}\text{O}} \) 2,4,6-Tribromoaniline
(b) C₆H₅NH₂ \( \xrightarrow{\text{HCl}} \) C₆H₅NH₃⁺Cl⁻ (Anilinium chloride)
(c) C₆H₅NH₂ \( \xrightarrow[\text{Pyridine}]{\text{(CH}_{3}\text{CO)}_{2}\text{O}} \) C₆H₅NHCOCH₃ (Acetanilide)
(ii) (CH₃)₃N < C₂H₅NH₂ < C₂H₅OH (increasing order of b.p.). This order is due to H-bonding.
(iii) (CH₃)₂NH reacts with nitrous acid to form an oily layer of N-nitrosamine which is insoluble in aqueous mineral acids.
(CH₃)₂NH + HO–NO → (CH₃)₂N–N=O (N-nitrosamine, yellow oily layer) + H₂O
Whereas (CH₃)₃N reacts with nitrous acid to form soluble nitrite salts with no ppt.
(CH₃)₃N + HONO → (CH₃)₃NH⁺ ONO⁻ (Trimethyl ammonium nitrite)

Teacher's Note:
1. In a distinction test always write the reagent, what each compound shows, and the equation; reagents alone do not earn full marks.

Question 43. (i) Write the structures of main products when benzenediazonium chloride reacts with the following reagents:
(a) H₃PO₂ + H₂O
(b) CuCN/KCN
(c) H₂O
(ii) Arrange the following in the increasing order of their basic character in an aqueous solution:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N
(iii) Give a simple chemical test to distinguish between the following pair of compounds:
C₆H₅–NH₂ and C₆H₅–NH–CH₃ [CBSE Outside Delhi 2015]
Answer:
(i) (a) C₆H₆ (Benzene)
(b) C₆H₅CN (Benzonitrile)
(c) C₆H₅OH (Phenol)
(ii) C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) Add CHCl₃ and alc. KOH, C₆H₅–NH₂ gives foul smell of isocyanide whereas C₆H₅–NH–CH₃ does not (or any other correct test).

Teacher's Note:
1. Remember that the diazonium group is replaced in each case: by H (H₃PO₂), by CN (CuCN, Sandmeyer) and by OH (warm water).