CBSE Class 12 Chemistry Amines Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Write chemical equations for the following conversion:
Benzyl chloride to 2-phenylethanamine.
Answer:
C₆H₅CH₂–Cl (benzyl chloride) KCN
⟶ C₆H₅CH₂–CN LiAlH₄
⟶ C₆H₅CH₂–CH₂–NH₂ (2-phenylethanamine)

Teacher's Note:
1. The product has one carbon more than benzyl chloride, so a step that adds a carbon (cyanide) is needed before reduction.

Question 2. What is carbylamine reaction?
Answer:
Carbylamine reaction is the reaction in which 1° amines produce a bad smelling compound when treated with chloroform in the presence of alkali.
RNH₂ + CHCl₃ + 3KOH Heat
⟶ R–NC + 3KCl + 3H₂O
It is the test for primary amines.

Teacher's Note:
1. Write 'alcoholic KOH' and 'heat' in the equation; examiners look for the conditions as well as the foul-smelling isocyanide.

Question 3. Rearrange the following in an increasing order of their basic strengths:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₆H₅)₂NH and CH₃NH₂
Answer:
(C₆H₅)₂NH < C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂

Teacher's Note:
1. More phenyl groups on N mean more delocalisation of the lone pair (weaker base); methyl groups on N increase basicity.

Question 4. Write IUPAC name of the following compound: (CH₃CH₂)₂NCH₃
Answer:
(CH₃CH₂)₂N–CH₃
IUPAC name: N-Ethyl-N-methylethanamine

Teacher's Note:
1. For a 3° amine, the longest alkyl group gives the parent name and the other groups are written as N-prefixes in alphabetical order.

Question 5. Propanamine and N,N-dimethylmethanamine contain the same number of carbon atoms, even though propanamine has higher boiling point than N,N-dimethylmethanamine. Why?
Answer:
Primary amines (R–NH₂) have two hydrogen atoms on nitrogen which can undergo intermolecular hydrogen bonding whereas no such hydrogen bonding is present in tertiary amines (R3N). So, primary amines boil at a higher temperature than tertiary amines.

Teacher's Note:
1. The two compounds are isomers (C₃H₉N), so molecular mass is the same; only hydrogen bonding makes the difference.

Question 6. Arrange the following compounds in increasing order of solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂
1° amines are more soluble in water than 2° amines. Aniline due to large hydrophobic benzene ring is least soluble.

Teacher's Note:
1. Solubility in water depends on hydrogen bonding with water and falls as the hydrocarbon part becomes larger.

Question 7. Write the structure for N-ethylmethylamine.
Answer:
CH₃–NH–C₂H₅ (N-ethylmethylamine)

Teacher's Note:
1. The 'N-' prefix means the ethyl group is attached to the nitrogen, not to the carbon chain.

Question 8. How will you convert the following:
Aniline into N-phenylethanamide
(Write the chemical equations involved.)
Answer:
C₆H₅NH₂ (aniline) (CH₃CO)₂O
⟶
Pyridine C₆H₅NH–COCH₃ (N-phenylethanamide, acetanilide)

Teacher's Note:
1. Pyridine is added to remove the acid formed and shift the equilibrium to the right.

Question 9. Why primary aromatic amines cannot be prepared by Gabriel phthalimide synthesis?
Answer:
Aromatic amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Teacher's Note:
1. The C–X bond of aryl halides has partial double bond character due to resonance, which is why it resists nucleophilic substitution.

Question 10. Draw the structure of N,N-diethylethanamine.
Answer:
C₂H₅–N(C₂H₅)–C₂H₅, i.e., (C₂H₅)₃N (N,N-diethylethanamine, triethylamine)

Teacher's Note:
1. This is a tertiary amine – all three H atoms of NH₃ are replaced by ethyl groups.

Question 11. Write the IUPAC name of the given compound:
CH₃—CH₂—CH₂—CO—N(CH₃)₂ [CBSE (F) 2016]
Answer:
N, N – Dimethylbutanamide

Teacher's Note:
1. The four-carbon chain including the C=O carbon is the butanamide parent; the two methyls on N are written N,N-.

Question 12. Write the IUPAC name of the given compound:
C₆H₅—NH—COCH₃ [CBSE East 2016]
Answer:
N- Phenylethanamide

Teacher's Note:
1. The common name of this compound is acetanilide – learn both names, as either may be asked.

Question 13. Write the IUPAC name of the given compound: [CBSE South 2016]

Answer:
2,4-Dibromobenzenamine or 2,4-Dibromoaniline

Teacher's Note:
1. Number the ring from the carbon carrying –NH₂ (C-1) so that the bromine atoms get the lowest locants.

Short Answer Questions (2 Marks)

Question 14. How are the following conversions carried out?
(i) CH₃CH₂Cl to CH₃CH₂CH₂NH₂
(ii) Benzene to aniline
Answer:
(i) CH₃CH₂Cl (chloroethane) NaCN
⟶ CH₃CH₂C≡N (propane nitrile) Reduction
⟶
Ni/H₂ CH₃CH₂CH₂NH₂ (1-propanamine)
(ii) C₆H₆ (benzene) Nitration
⟶ C₆H₅NO₂ (nitrobenzene) Fe/HCl
⟶ C₆H₅NH₂ (aniline)

Teacher's Note:
1. In (i) the chain grows by one carbon, which is the signal to use the cyanide route rather than ammonolysis.

Question 15. Account for the following:
(i) Tertiary amines do not undergo acylation reaction.
(ii) Amines are more basic than comparable alcohols.
Answer:
(i) In tertiary amines there are no acidic hydrogen due to which they do not undergo acylation reaction.
(ii) N being less electronegative than O gives lone pair of electron more easily than O atom. Therefore amines are more basic than alcohols.

Teacher's Note:
1. Acylation replaces an H on nitrogen by an acyl group, so an amine with no N–H cannot be acylated.

Question 16. Complete the following reactions:
(i) CH₃CH₂NH₂ + CHCl₃ + alc. KOH Heat
⟶
(ii) C₆H₅NH₂ + HCl_(aq) →
Answer:
(i) CH₃–CH₂–NH₂ + CHCl₃ + alc. 3KOH Heat
⟶ CH₃–CH₂–NC + 3KCl + 3H₂O
(ii) C₆H₅–NH₂ + HCl H₂O
⟶ C₆H₅–NH₃⁺Cl⁻ (anilinium chloride)

Teacher's Note:
1. Balance the carbylamine equation with 3KOH: three HCl molecules are removed from the amine and chloroform.

Question 17. Acetamide is less basic than ethanamine. Why?
Answer:
Ethanamide (acetamide): CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺ ;   Ethanamine: CH₃CH₂NH₂ (lone pair on N is free)
In acetamide, lone pair on nitrogen atom is involved in resonance and hence is not free for donation as in the case of ethanamine.

Teacher's Note:
1. The C=O group in an amide pulls the nitrogen lone pair towards oxygen, so amides are almost neutral.

Question 18. Write the structures of the main products of the following reactions:
(i) C₆H₅NH₂ (CH₃CO)₂O
⟶
Pyridine
(ii) C₆H₅SO₂Cl (CH₃)₂NH
⟶
Answer:
(i) C₆H₅NH₂ (aniline) (CH₃CO)₂O
⟶
Pyridine C₆H₅–NH–CO–CH₃ (N-phenylethanamide, acetanilide)
(ii) C₆H₅SO₂Cl (benzene sulphonyl chloride) + H–N(CH₃)₂ −HCl
⟶ C₆H₅–SO₂–N(CH₃)₂ (N, N-dimethylbenzene sulphonamide)

Teacher's Note:
1. The product in (ii) has no H on nitrogen, so it is insoluble in alkali – the Hinsberg result for a 2° amine.

Question 19. Give the structures of A, B and C in the following reactions:
(i) C₆H₅NO₂ Sn + HCl
⟶ A NaNO₂ + HCl
⟶
273 K B H₂O
⟶ C
(ii) CH₃CN H₂O/H⁺
⟶ A NH₃
⟶
Δ B Br₂ + KOH
⟶ C
Answer:
(i) C₆H₅NO₂ Sn + HCl
⟶ C₆H₅NH₂ (A) NaNO₂ + HCl
⟶
273 K C₆H₅N₂⁺Cl⁻ (B) H₂O
⟶ C₆H₅OH (C)
(ii) CH₃CN H₂O/H⁺
⟶ CH₃COOH (A) NH₃
⟶ CH₃CONH₂ (B) Br₂ + KOH
⟶ CH₃NH₂ (C)

Teacher's Note:
1. Warming a diazonium salt solution with water replaces –N₂⁺ by –OH, which is the usual route from aniline to phenol.

Question 20. Give the chemical tests to distinguish between the following pairs of compounds:
(i) Methylamine and dimethylamine
(ii) Aniline and N-methylaniline
Answer:
(i) Methylamine gives carbylamine test, i.e., on treatment with alc. KOH and chloroform, followed by heating it gives offensive odour of methyl isocyanide. Dimethylamine does not give this test.
(ii) Aniline gives carbylamine test, i.e., on treatment with alc. KOH and chloroform followed by heating it gives offensive odour of phenylisocyanide but N-methylaniline being secondary amine, does not show this test.

Teacher's Note:
1. In a distinguishing test always state what each compound does – the positive result for one and the absence of it for the other.

Question 21. How would you account for the following:
(i) Aniline is a weaker base than cyclohexylamine.
(ii) Methylamine in aqueous medium gives reddish-brown precipitate with FeCl₃.
Answer:
(i) Aniline is weaker base than cyclohexylamine because of resonance. Due to electromeric effect, the lone pair on nitrogen is attracted by benzene ring. Hence, donor tendency of –NH₂ group decreases. There is no resonance in cyclohexylamine. Electron repelling nature of cyclohexyl group further increases the donor property of NH₂ group. So, cyclohexylamine is a stronger base.
Aniline: C₆H₅–NH₂ ↔ resonance structures with =NH₂⁺ and negative charge at o- and p-positions; cyclohexylamine, C₆H₁₁–NH₂: no resonance.
(ii) Methylamine forms hydroxide ions when dissolved in water due to the following acid - base equilibrium.
CH₃–NH₂ + H₂O ⇌ CH₃–NH₃⁺ + OH⁻
These OH⁻ ions react with Fe³⁺ ions to form ferric hydroxide.
Fe³⁺ + 3OH⁻ → Fe(OH)₃ (reddish-brown ppt.)

Teacher's Note:
1. The brown precipitate is Fe(OH)₃, the same one NaOH would give – it simply shows that aqueous methylamine is basic.

Question 22. How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
Answer:
(i) C₆H₅NO₂ (nitrobenzene) Fe/HCl
⟶ C₆H₅NH₂ (aniline)
(ii) CH₃COOH (ethanoic acid) PCl₅
⟶ CH₃–CO–Cl NH₃
⟶ CH₃–CO–NH₂ Br₂/KOH
⟶ CH₃–NH₂ (methanamine)

Teacher's Note:
1. Ethanoic acid (C₂) to methanamine (C₁) loses a carbon, which points to Hoffmann bromamide degradation as the last step.

Question 23. Give the structure of products A and B in the given sequence of reactions.
(i) R–C≡N H₂/Ni
⟶
Na(Hg)/C₂H₅OH A
(ii) R–CO–NH₂ (i) LiAlH₄
⟶
(ii) H₂O B
Answer:
(i) R–C≡N H₂/Ni
⟶
Na(Hg)/C₂H₅OH R–CH₂–NH₂ (A)
(ii) R–CO–NH₂ (i) LiAlH₄
⟶
(ii) H₂O R–CH₂–NH₂ (B)

Teacher's Note:
1. Reduction keeps all the carbon atoms, unlike Hoffmann bromamide degradation which removes one.

Question 24. (i) Write the IUPAC name of C₆H₅–N(CH₃)₂.
(ii) Out of (CH₃)₃N and (CH₃)₂NH, which one is more basic in aqueous solution?
Answer:
(i) N,N-Dimethylbenzenamine.
(ii) (CH₃)₂NH = Secondary amine; (CH₃)₃N = Tertiary amine
Secondary amines are more basic than tertiary amines in aqueous solution due to steric factor. So, (CH₃)₂NH is more basic in aqueous solution.

Teacher's Note:
1. In water, the tertiary amine’s cation is poorly solvated (only one N–H) and crowded, so it falls below the secondary amine despite having more +I groups.

Question 25. (i) Write an isomer of C₃H₉N which gives foul smell of isocyanide when treated with chloroform and ethanolic NaOH.
(ii) Write an isomer of C₃H₉N which does not react with Hinsberg reagent.
Answer:
(i) Isomer of C₃H₉N is CH₃CH₂CH₂NH₂ (Propan-1-amine), gives foul smell of isocyanides (because primary amine gives carbylamine reaction).
(ii) Isomer of C₃H₉N which does not react with Hinsberg reagent is (CH₃)₃N i.e., tertiary amine.
(CH₃)₃N (Trimethylamine) + C₆H₅SO₂Cl (Hinsberg reagent) → No reaction

Teacher's Note:
1. Only 1° amines give the carbylamine test and only 3° amines fail to react with Hinsberg’s reagent; propan-2-amine would also be correct for (i).

Question 26. (i) Arrange the following in increasing order of boiling points.
(CH₃)₃N, C₂H₅OH, C₂H₅NH₂
(ii) Arrange the following in increasing order of base strength in gas phase:
(C₂H₅)₃N, C₂H₅NH₂, (C₂H₅)₂NH
Answer:
(i) (CH₃)₃N < C₂H₅NH₂ < C₂H₅OH
(ii) C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N
Base strength is the ability to donate lone pair. Due to inductive +I effect of C₂H₅, the negative charge density on nitrogen atom increases. Therefore, lone pair is easily available for donation.

Teacher's Note:
1. In the gas phase there is no solvation, so the order follows the +I effect alone: 3° > 2° > 1°. Read the phase in the question carefully.

Short Answer Questions (3 Marks)

Question 27. How are the following reactions carried out? Write the equations and conditions.
(i) Acetic acid to ethylamine
(ii) Bromocyclohexane to cyclohexanamine.
(iii) Methylamine to dimethylamine.
Answer:
(i) CH₃COOH (acetic acid) + NH₃ → CH₃CONH₂ (acetamide) (i) LiAlH₄
⟶
(ii) H₂O CH₃CH₂NH₂ (ethylamine)
(ii) C₆H₁₁–Br (bromocyclohexane) + KCN → C₆H₁₁–CN H₃O⁺
⟶ C₆H₁₁–COOH NH₃
⟶ C₆H₁₁–CONH₂ Br₂
⟶
NaOH C₆H₁₁–NH₂ (cyclohexanamine)
(iii) CH₃NH₂ + CHCl₃ + KOH Δ
⟶ CH₃NC H₂
⟶
Na/C₂H₅OH CH₃NHCH₃

Teacher's Note:
1. In (i) the carbon count stays the same (C₂ to C₂), so the amide is reduced with LiAlH₄ rather than degraded with Br₂/KOH.

Question 28. Which amine in each of the following pairs is a stronger base? Give reason.
(i) CH₃–CH(NH₂)–CH₃ and CH₃–CH(NH₂)–COOCH₃
(ii) CH₃CH₂CH₂NH₂ and CH₃NHCH₂CH₃
Answer:
(i) CH₃–CH(NH₂)–CH₃ is more basic than CH₃–CH(NH₂)–COOCH₃ because –COOCH₃ is an electron withdrawing group which decrease the electron density on nitrogen atom.
(ii) 2° amines are more basic than 1° amines, because in 2° amine there are two electron releasing groups and in 1° amine only one electron releasing group is present, so, CH₃NHCH₂CH₃ is more basic than CH₃CH₂CH₂NH₂.

Teacher's Note:
1. When comparing two bases, look first at the groups near nitrogen: electron-releasing groups strengthen, electron-withdrawing groups weaken.

Question 29. Write the structures of main products when aniline reacts with the following reagents:
(i) Br₂ water
(ii) HCl
(iii) (CH₃CO)₂O/pyridine
Answer:
(i) C₆H₅NH₂ + 3Br_2(aq) → 2,4,6-Br₃C₆H₂NH₂ (2,4,6-tribromoaniline) + 3HBr
(ii) C₆H₅–NH₂ + HCl H₂O
⟶ C₆H₅–NH₃⁺Cl⁻ (anilinium chloride)
(iii) C₆H₅NH₂ (aniline) (CH₃CO)₂O
⟶
Pyridine C₆H₅–NH–CO–CH₃ (N-phenylethanamide, acetanilide)

Teacher's Note:
1. With bromine water the white precipitate of 2,4,6-tribromoaniline forms at once; no catalyst is needed.

Question 30. A compound X (C₇H₇Br) reacts with KCN to give Y (C₈H₇N). Reduction of Y with LiAlH₄ yields Z (C₈H₁₁N). Z gives carbylamine reaction, reacts with Hinsberg’s reagent in the presence of aq. KOH to give a clear solution. With NaNO₂ and HCl at 0°C (Z) gives a neutral compound which gives red colour with ammonium cerric nitrate. What are X, Y and Z?
Answer:
C₇H₇Br (X) KCN
⟶
−KBr C₈H₇N (Y) LiAlH₄
⟶ C₈H₁₁N (Z)
or C₆H₅CH₂Br (X) KCN
⟶
−KBr C₆H₅CH₂CN (Y) LiAlH₄
⟶ C₆H₅CH₂CH₂NH₂ (Z)
X = benzyl bromide, Y = phenylacetonitrile (benzyl cyanide), Z = 2-phenylethanamine.

Teacher's Note:
1. Every test on Z points to an aliphatic 1° amine: carbylamine test, soluble Hinsberg product, and an alcohol (red with ceric ammonium nitrate) with HNO₂.

Question 31. (a) Give one chemical test to distinguish between the compounds of the following pairs:
(i) CH₃NH₂ and (CH₃)₂NH
(ii) (C₂H₅)₂NH and (C₂H₅)₃N
(b) Why aniline does not undergo Friedel–Crafts reaction?
Answer:
(a) (i) Methylamine gives carbylamine test, i.e., on treatment with alc. KOH and chloroform, followed by heating it gives offensive odour of methyl isocyanide. Dimethylamine does not give this test.
(ii) (C₂H₅)₂NH and (C₂H₅)₃N can be distinguished by Hinsberg’s reagent. (C₂H₅)₂NH gives N,N-diethylbenzenesulphonamide, a solid insoluble in alkali, while (C₂H₅)₃N does not react.
(b) In Friedel Crafts reaction, AlCl₃ is added as a catalyst which is a Lewis acid. It forms a salt with aniline due to which the nitrogen of aniline acquires positive charge. This positively charged nitrogen acts as a strong deactivating group, hence aniline does not undergo Friedel Crafts reaction.
C₆H₅–NH₂ + AlCl₃ → C₆H₅–NH₂⁺–AlCl₃⁻

Teacher's Note:
1. For Hinsberg's test, state what each amine actually does with the reagent, not just the name of the test.

Question 32. Write the structures of A, B and C in the following sequence of reactions:
(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
Answer:
(i) C₆H₅CONH₂ Br₂
⟶
aq. KOH C₆H₅NH₂ (A) 0-5°C
⟶
NaNO₂ + HCl C₆H₅N₂⁺Cl⁻ (B) KI
⟶ C₆H₅I (C)
(ii) CH₃–Cl KCN
⟶ CH₃CN (A) LiAlH₄
⟶ CH₃CH₂NH₂ (B) CHCl₃ + alc. KOH
⟶
Δ CH₃CH₂NC (C)

Teacher's Note:
1. Iodobenzene is made from the diazonium salt with KI directly; no copper(I) catalyst is needed, unlike for Cl and Br.

Question 33. Write the structure of 2, 4-dinitrochlorobenzene.

Answer:
Structure of 2,4-dinitrochlorobenzene (1-chloro-2,4-dinitrobenzene) is shown below.

Teacher's Note:
1. Number the ring from the carbon carrying Cl: the NO₂ groups are then at the ortho (2) and para (4) positions.

Question 34. Write the structure of N-methylethanamine.
Answer:
CH₃CH₂NHCH₃

Teacher's Note:
1. The parent chain is ethanamine; the methyl group is on the nitrogen, shown by the 'N-' prefix.

Question 35. Write the structure of 2-aminotoluene.

Answer:
Structure of 2-aminotoluene (o-toluidine) is shown below.

Teacher's Note:
1. In 2-aminotoluene the CH₃ carbon is C-1, so the NH₂ group is on the next (ortho) carbon.

Question 36. Amit wants to manufacture aniline for the synthesis of dye stuff. For this he has selected tin and hydrochloric acid as reducing agent for the reduction of nitrobenzene. But his friend suggested to use iron scrap and hydrochloric acid as the reducing agent.
Now answer the following questions:
(i) Write the chemical equation for the reduction of nitrobenzene to aniline.
(ii) Why Amit’s friends has suggested to use scrap iron and HCl in place of tin and HCl?
Answer:
(i) [Fe + 2HCl → FeCl₂ + 2H] × 3
C₆H₅NO₂ (nitrobenzene) + 6H
⟶ C₆H₅NH₂ (aniline) + 2H₂O
(ii) Amit’s friend has suggested to use scrap iron and hydrochloric acid because in this reaction FeCl₂ formed gets hydrolysed to release HCl during the reaction. Thus only small amount of HCl is required to initiate the reaction and scrap iron is also cheaper.

Teacher's Note:
1. Fe/HCl is the industrial choice; Sn/HCl is mainly a laboratory method.

Question 37. (a) Give a simple chemical test to distinguish between aniline and N, N-dimethylaniline.
(b) Arrange the following in the increasing order of their boiling point:
C₂H₅NH₂, C₂H₅OH, (CH₃)₃N
Answer:
(a) Aniline undergoes isocyanide test (carbylamine reaction) whereas, N, N-dimethylaniline does not.
C₆H₅NH₂ (aniline) + CHCl₃ + 3KOH Δ
⟶ C₆H₅NC (phenyl isocyanide, foul smell) + 3KCl + 3H₂O
(b) Increasing order of boiling points:
(CH₃)₃N < C₂H₅NH₂ < C₂H₅OH
Tertiary amine does not have hydrogen to form hydrogen bonding and hydrogen bonding in alcohol is stronger than that of amines because oxygen is more electronegative than nitrogen.

Teacher's Note:
1. The three compounds have similar molecular masses, so the order is decided by the strength of hydrogen bonding.

Question 38. Write the chemical equations for the following conversions:
(i) Aniline to benzonitrile
(ii) Aniline to p-nitroaniline.
Answer:
(i) C₆H₅–NH₂ NaNO₂ + HCl
⟶
273-278 K C₆H₅–N₂⁺Cl⁻ (benzenediazonium chloride) CuCN
⟶
KCN C₆H₅–CN (benzonitrile)
(ii) C₆H₅–NH₂ (aniline) (CH₃CO)₂O
⟶
Pyridine C₆H₅NHCOCH₃ (acetanilide) HNO₃ + H₂SO₄
⟶
288 K p-O₂N–C₆H₄–NHCOCH₃ OH⁻ or H⁺
⟶
Hydrolysis p-O₂N–C₆H₄–NH₂ (p-nitroaniline)

Teacher's Note:
1. Acetylation is needed in (ii); direct nitration would oxidise aniline and give a lot of m-nitroaniline.

Long Answer Questions (5 Marks)

Question 39. (a) Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC name of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
(b) Classify the following amines as primary, secondary and tertiary:
(i) 1-naphthylamine (C₁₀H₇–NH₂)
(ii) N,N-dimethyl-1-naphthylamine (C₁₀H₇–N(CH₃)₂)
(iii) (C₂H₅)₂CHNH₂
(iv) (C₂H₅)₂NH
Answer:
(a) In all, four structural isomers are possible. These are as follows:
Primary amines: CH₃CH₂CH₂NH₂ (propan-1-amine), CH₃–CH(NH₂)–CH₃ (propan-2-amine)
Secondary amines: CH₃–NH–C₂H₅ (N-methylethanamine)
Tertiary amines: CH₃–N(CH₃)–CH₃ (N,N-dimethylmethanamine)
Only primary amines react with HNO₂ to liberate N₂ gas.
(i) CH₃CH₂CH₂NH₂, propan-1-amine   (ii) (CH₃)₂CH–NH₂, propan-2-amine
(b) (i) 1° amine   (ii) 3° amine   (iii) (C₂H₅)₂–CHNH₂: 1° amine   (iv) (C₂H₅)₂NH: 2° amine

Teacher's Note:
1. Classify an amine by the number of carbon groups on nitrogen, not by the type of carbon; (C₂H₅)₂CHNH₂ is 1° even though N is on a 2° carbon.

Question 40. (a) C₅H₁₃N (X, optically active) NaNO₂/HCl
⟶
−N₂ Y (tertiary alcohol) + other products
Find X and Y. Is Y optically active? Write the intermediate steps.
(b) Which of the following is more acidic and why? C₆H₅NH₃⁺ or p-F–C₆H₄–NH₃⁺
Answer:
(a) CH₃–CH(CH₃)–C*H(NH₂)–CH₃ (X) NaNO₂/HCl
⟶ CH₃–C(OH)(CH₃)–CH₂–CH₃ (Y, rearranged product) + N₂ + NaCl + H₂O
Intermediate steps:
CH₃–CH(CH₃)–CH(NH₂)–CH₃ (X) NaNO₂/HCl
⟶
−N₂ CH₃–CH(CH₃)–C⁺H–CH₃ 1,2-H⁻ shift
⟶ CH₃–C⁺(CH₃)–CH₂–CH₃ H₂O
⟶ CH₃–C(OH)(CH₃)–CH₂–CH₃ (Y)
X = 3-methylbutan-2-amine; Y = 2-methylbutan-2-ol, which is optically inactive (it has no chiral carbon).
(b) p-F–C₆H₄–NH₃⁺ is more acidic due to the strong electron withdrawing power (−I effect) of fluorine.

Teacher's Note:
1. The 1,2-hydride shift converts a 2° carbocation into a more stable 3° carbocation, which is why a tertiary alcohol is obtained.

Question 41. (a) Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
(b) Write the chemical equations for the following conversions
(i) Chlorobenzene to p-chloroaniline
(ii) Aniline to p-bromoaniline
Answer:
(a) (i) Aromatic primary amines react with nitrous acid to form diazonium salts.
C₆H₅NH₂ + HNO₂ HCl
⟶
273 - 278 K C₆H₅N₂⁺Cl⁻
(ii) Aliphatic primary amines also form diazonium salts on reaction with nitrous acid but they are unstable and decompose to give the corresponding alcohols as the major product with the evolution of nitrogen.
R–NH₂ + HNO₂ HCl
⟶
273 - 278 K [R–N₂⁺Cl⁻] H₂O
⟶ R–OH + N₂ + HCl
(b) (i) C₆H₅Cl (chlorobenzene) HNO₃/H₂SO₄
⟶ p-Cl–C₆H₄–NO₂ Sn / HCl
⟶ p-Cl–C₆H₄–NH₂ (p-chloroaniline, separated from the ortho isomer)
(ii) C₆H₅NH₂ (aniline) (CH₃CO)₂O/CH₃COOH
⟶ C₆H₅NHCOCH₃ Br₂/CH₃COOH
⟶ p-Br–C₆H₄–NHCOCH₃ H₂O / OH⁻
⟶ p-Br–C₆H₄–NH₂ (p-bromoaniline)

Teacher's Note:
1. Aromatic diazonium salts are stable only at 273-278 K, so the temperature is an essential part of the answer.