NCERT Class 12 Chemistry Part 2 Chapter 9 Amines

Question 9.1

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i) \((CH_3)_2CHNH_2\): Propan-2-amine (Primary).
(ii) \(CH_3(CH_2)_2NH_2\): Propan-1-amine (Primary).
(iii) \(CH_3NHCH(CH_3)_2\): N-Methylpropan-2-amine (Secondary).
(iv) \((CH_3)_3CNH_2\): 2-Methylpropan-2-amine (Primary).
(v) \(C_6H_5NHCH_3\): N-Methylbenzenamine (Secondary).
(vi) \((CH_3CH_2)_2NCH_3\): N-Ethyl-N-methylethanamine (Tertiary).
(vii) \(m-BrC_6H_4NH_2\): 3-Bromobenzenamine (Primary).

Question 9.2

Give chemical tests to distinguish between the following pairs of compounds:
(i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline.

(i) Methylamine (\(1^\circ\)) & Dimethylamine (\(2^\circ\))

Carbylamine Test: Methylamine gives a foul smell of isocyanide with \(CHCl_3/KOH\). Dimethylamine does not.

(ii) Secondary & Tertiary Amines

Hinsberg Test: \(2^\circ\) amines react with Benzenesulphonyl chloride to give a solid insoluble in alkali. \(3^\circ\) amines do not react.

(iii) Ethylamine & Aniline

Azo Dye Test: Aniline forms a yellow/orange dye with benzene diazonium chloride at low temp. Ethylamine does not.
Or: Ethylamine liberates \(N_2\) gas rapidly with \(HNO_2\). Aniline forms stable diazonium salt at 0-5°C.

(iv) Aniline & Benzylamine

Nitrous Acid Test: Aniline forms stable diazonium salt. Benzylamine gives Benzyl alcohol and \(N_2\) gas bubbles.

(v) Aniline (\(1^\circ\)) & N-methylaniline (\(2^\circ\))

Carbylamine Test: Aniline gives positive test (foul smell). N-methylaniline does not.

Question 9.3

Account for the following:
(i) \(pK_b\) of aniline is more than methylamine.
(ii) Ethylamine is soluble in water, aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Aniline does not undergo Friedel-Crafts reaction.
(v) Diazonium salts of aromatic amines are more stable than aliphatic amines.

(i): In aniline, the lone pair on N is delocalized over the benzene ring via resonance, making it less available for protonation. In methylamine, the +I effect of the methyl group increases electron density on N. Thus, aniline is a weaker base (higher \(pK_b\)).
(ii): Ethylamine forms intermolecular hydrogen bonds with water. Aniline has a large hydrophobic phenyl group that hinders H-bonding, reducing solubility.
(iii): Methylamine is a base. In water, it produces \(OH^-\) ions: \(CH_3NH_2 + H_2O \to CH_3NH_3^+ + OH^-\). These \(OH^-\) ions react with \(Fe^{3+}\) to precipitate \(Fe(OH)_3\) (Hydrated Ferric Oxide).
(iv): Aniline reacts with the Lewis Acid catalyst (\(AlCl_3\)) to form a salt (\(C_6H_5NH_2^+AlCl_3^-\)). This puts a positive charge on N, making the ring highly deactivated for further electrophilic substitution.
(v): Aromatic diazonium ions are stabilized by resonance (dispersal of positive charge over the ring). Aliphatic diazonium ions are unstable and decompose immediately to carbocations and nitrogen gas.

Question 9.4

Arrange the following in increasing order of their basic strength:
(i) \(C_2H_5NH_2, C_6H_5NH_2, NH_3, C_6H_5CH_2NH_2, (C_2H_5)_2NH\)
(ii) \(C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N, C_6H_5NH_2\)
(iii) \(CH_3NH_2, (CH_3)_2NH, (CH_3)_3N, C_6H_5NH_2, C_6H_5CH_2NH_2\)

Key Concept: Aliphatic amines > Ammonia > Aromatic amines. Secondary aliphatic amines are usually strongest due to combined inductive, solvation, and steric factors.

(i): \(C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH\).
(ii): \(C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH\).
(iii): \(C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH\) (In aqueous phase, methyl amines order is \(2^\circ > 1^\circ > 3^\circ\)).

Question 9.5

Complete the following acid-base reactions and name the products:
(i) \(CH_3CH_2CH_2NH_2 + HCl\)
(ii) \((C_2H_5)_3N + HCl\)

(i) CH3CH2CH2NH2 + HCl -> [CH3CH2CH2NH3]+Cl- Product: n-Propylammonium chloride (ii) (C2H5)3N + HCl -> [(C2H5)3NH]+Cl- Product: Triethylammonium chloride

Question 9.6

Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Aniline reacts with excess methyl iodide to undergo exhaustive methylation, forming a quaternary ammonium salt.

C6H5NH2 + 3CH3I –(Na2CO3)–> [C6H5N(CH3)3]+I-

Product: N,N,N-Trimethylanilinium iodide.

Question 9.7

Write chemical reaction of Aniline with benzoyl chloride and write the name of the product obtained.

Schotten-Baumann Reaction:

C6H5NH2 + C6H5COCl –(Pyridine/NaOH)–> C6H5NHCOC6H5 + HCl

Product: N-Phenylbenzamide (or Benzanilide).

Question 9.8

Write structures of different isomers corresponding to the molecular formula \(C_3H_9N\). Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Isomers

\(CH_3CH_2CH_2NH_2\) (Propan-1-amine) – \(1^\circ\)
\(CH_3CH(NH_2)CH_3\) (Propan-2-amine) – \(1^\circ\)
\(CH_3CH_2NHCH_3\) (N-Methylethanamine) – \(2^\circ\)
\((CH_3)_3N\) (N,N-Dimethylmethanamine) – \(3^\circ\)

Reaction with Nitrous Acid

Only primary aliphatic amines liberate nitrogen gas with \(HNO_2\).
So, Propan-1-amine and Propan-2-amine will liberate nitrogen gas.

Question 9.9

Convert:
(i) 3-Methylaniline to 3-nitrotoluene.
(ii) Aniline to 1,3,5-tribromobenzene.

(i) 3-Methylaniline to 3-nitrotoluene

m-Toluidine –(NaNO2/HBF4)–> Diazonium Tetrafluoroborate Salt –(NaNO2/Cu, Heat)–> m-Nitrotoluene

(ii) Aniline to 1,3,5-tribromobenzene

Aniline + 3Br2(aq) -> 2,4,6-Tribromoaniline 2,4,6-Tribromoaniline –(NaNO2/HCl)–> Diazonium Salt Diazonium Salt –(H3PO2/H2O)–> 1,3,5-Tribromobenzene

Question 9.10

Word Problem: An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with \(Br_2\) and \(KOH\) forms a compound ‘C’ of molecular formula \(C_6H_7N\). Write the structures and IUPAC names of compounds A, B and C.

Analysis

Product C is \(C_6H_7N\). This fits Aniline (\(C_6H_5NH_2\)).
C is formed from B by \(Br_2/KOH\) (Hoffmann Bromamide Degradation). This means B is a primary amide with one more carbon than C. B is Benzamide (\(C_6H_5CONH_2\)).
B is formed from A by heating with ammonia. This implies A is a carboxylic acid. A is Benzoic Acid (\(C_6H_5COOH\)).

A: Benzoic Acid, B: Benzamide, C: Aniline.

Question 9.11

Complete the following reactions:
(i) \(C_6H_5NH_2 + CHCl_3 + 3KOH \to\)
(ii) \(C_6H_5N_2Cl + H_3PO_2 + H_2O \to\)
(iii) \(C_6H_5NH_2 + H_2SO_4 (\text{conc}) \to\)
(iv) \(C_6H_5N_2Cl + C_2H_5OH \to\)
(v) \(C_6H_5NH_2 + Br_2(aq) \to\)
(vi) \(C_6H_5NH_2 + (CH_3CO)_2O \to\)

(i) C6H5NC (Phenyl isocyanide) + 3KCl + 3H2O (Carbylamine reaction) (ii) C6H6 (Benzene) + N2 + H3PO3 + HCl (iii) C6H5NH3+HSO4- –(Heat)–> Sulfanilic acid (p-Aminobenzenesulphonic acid) (iv) C6H6 (Benzene) + N2 + CH3CHO + HCl (v) 2,4,6-Tribromoaniline (White ppt) (vi) C6H5NHCOCH3 (Acetanilide) + CH3COOH

Question 9.12

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Gabriel phthalimide synthesis involves the nucleophilic attack of the phthalimide anion on an alkyl halide (\(S_N2\) reaction).
Aryl halides do not undergo nucleophilic substitution easily due to the partial double bond character of the C-X bond and the instability of the phenyl cation. Therefore, aryl halides cannot react with potassium phthalimide to form N-phenylphthalimide.

Question 9.13

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

(i) Aromatic Primary Amine (Aniline)

Reacts at low temperature (0-5°C) to form stable Diazonium salt.

C6H5NH2 + HNO2 + HCl –(0-5°C)–> [C6H5N2]+Cl- + 2H2O

(ii) Aliphatic Primary Amine

Forms unstable diazonium salt which decomposes immediately to release nitrogen gas and form alcohol.

R-NH2 + HNO2 -> [R-N2]+Cl- -> R-OH + N2 + HCl

Question 9.14

Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?

(i): Acidity depends on the stability of the conjugate base. Alkoxide ion (\(RO^-\)) is formed from alcohol, and amide ion (\(RNH^-\)) from amine. Since Oxygen is more electronegative than Nitrogen, it can accommodate the negative charge better, making \(RO^-\) more stable. Hence, alcohols are more acidic.
(ii): Primary amines (\(RNH_2\)) have two hydrogen atoms on Nitrogen available for intermolecular Hydrogen Bonding. Tertiary amines (\(R_3N\)) have no hydrogen on Nitrogen, so they cannot form H-bonds with each other. Stronger intermolecular forces in primary amines lead to higher boiling points.
(iii): In aromatic amines (like Aniline), the lone pair on Nitrogen is delocalized into the benzene ring via resonance, making it less available for donation. In aliphatic amines, the alkyl groups exert a +I effect, increasing electron density on Nitrogen and making the lone pair more available. Hence, aliphatic amines are stronger bases.

Intext 9.1

Classify the following amines as primary, secondary or tertiary:

(i) \(CH_3NH_2\): Primary
(ii) \((CH_3CH_2)_2NH\): Secondary
(iii) \((CH_3CH_2)_3N\): Tertiary
(iv) \(C_6H_5NHCH_3\): Secondary

Intext 9.2

(i) Write structures of different isomeric amines corresponding to the molecular formula \(C_4H_{11}N\). (ii) Write IUPAC names of all the isomers.

Primary: Butan-1-amine, Butan-2-amine, 2-Methylpropan-1-amine, 2-Methylpropan-2-amine.
Secondary: N-Methylpropan-1-amine, N-Methylpropan-2-amine, N-Ethylethanamine.
Tertiary: N,N-Dimethylethanamine.

Intext 9.3

How will you convert: (i) Benzene into aniline (ii) Benzene into N,N-dimethylaniline (iii) Cl-(CH2)4-Cl into hexan-1,6-diamine?

(i) Benzene -> Nitrobenzene (HNO3/H2SO4) -> Aniline (Sn/HCl) (ii) Benzene -> Aniline -> N,N-Dimethylaniline (2CH3I, Heat) (iii) Cl-(CH2)4-Cl -> NC-(CH2)4-CN (KCN) -> H2N-(CH2)6-NH2 (Reduction H2/Ni)

NCERT Solutions