Chapter 9 Amines

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

NCERT Solutions for Class 12 Chemistry Chapter 9 (Amines) play a crucial role in helping students build a strong conceptual foundation for both CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This chapter introduces the mathematical and chemical foundations for understanding nitrogen-containing organic compounds. Understanding these concepts is essential for students as they lead to many of the systems that utilize this technology, including the synthesis of polymers (like Nylon), dyes, and life-saving medicines such as adrenaline and novocaine.

The NCERT Solutions for Class 12 Chemistry Chapter 9 will help students learn how to use electronic effects and solvation energy to explain the complex basicity trends of amines in different phases. Well-explained NCERT solutions enable students to strengthen fundamentals, practice important conversion questions, and perform better in term exams, board exams, and national-level competitive tests.

1.0Class 12 Chemistry Chapter 9 Amines: Key Concepts

Class 12 Chemistry Chapter 9 Amines covers the structure, preparation, properties, basicity, and chemical reactions of amines. It also includes important tests, diazonium salts, and coupling reactions.

Key Concepts

Structure and Classification: Understand the structure of amines and classify them as primary, secondary, and tertiary amines.
Nomenclature: Learn the IUPAC naming of aliphatic and aromatic amines such as aniline.
Preparation of Amines: Study Gabriel Phthalimide Synthesis, Hoffmann Bromamide Degradation, and the reduction of nitriles and amides.
Basicity of Amines: Understand the factors affecting the basic strength of amines, including the +I effect, steric effects, and solvation.
Chemical Reactions: Learn important reactions such as acylation, Carbylamine Reaction, and reactions with nitrous acid.
Hinsberg’s Test: Understand how primary, secondary, and tertiary amines are distinguished using Hinsberg’s test.
Diazonium Salts: Study the preparation and reactions of benzene diazonium chloride.
Coupling Reactions: Learn how diazonium salts undergo coupling reactions to form coloured azo compounds.

2.0NCERT Solutions Class 12 Chemistry Chapter 9 Amines: Detailed Solutions

INTEXT QUESTIONS

Classify the following amines as primary, secondary or tertiary :

Ans. Primary : (i) and (iii) Secondary : (iv) Tertiary : (ii)

(i) Write structures of different isomeric amines corresponding to the molecular formula, $C_{4} H_{11} N$ (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines ? Ans. (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, $C_{4} H_{11} N$ are given below:

(iii) The pairs (a) and (b), (e) and (g) exhibit position isomerism. The pairs (a) and (c); (b) and (d) exhibit chain isomerism The pairs (e) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomersim with secondary and tertiary amines and vice- versa.

How will you convert (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) $Cl - (CH_{2})_{4} - Cl$ into hexan-1, 6-diamine?

Ans.

Arrange the following in increasing order of their basic strength : (i) $C_{2} H_{5} NH_{2}, C_{6} H_{5} NH_{2}, NH_{3}, C_{6} H_{5} CH_{2} NH_{2}$ and $(C_{2} H_{5})_{2} NH$ (ii) $C_{2} H_{5} NH_{2}, (C_{2} H_{5})_{2} NH, (C_{2} H_{5})_{3} N, C_{6} H_{5} NH_{2}$ (iii) $CH_{3} NH_{2}, (CH_{3})_{2} NH, (CH_{3})_{3} N, C_{6} H_{5} NH_{2}$ , $C_{6} H_{5} CH_{2} NH_{2}$

Ans.

(i) $C_{6} H_{5} NH_{2} < NH_{3} < C_{6} H_{5} CH_{2} NH_{2} < C_{2} H_{5} NH_{2}$ $< (C_{2} H_{5})_{2} NH$ (ii) $C_{6} H_{5} NH_{2} < C_{2} H_{5} NH_{2} < (C_{2} H_{5})_{3} N < (C_{2} H_{5})_{2}$ NH (iii) $C_{6} H_{5} NH_{2} < C_{6} H_{5} CH_{2} NH_{2} < (CH_{3})_{3}$ $N < CH_{3} NH_{2} < (CH_{3})_{2} NH$

Complete the following acid-base reactions and name the products : (i) $CH_{3} CH_{2} CH_{2} NH_{2} + HCl ⟶$ (ii) $(C_{2} H_{5})_{3} N + HCl ⟶$

Ans.

(i) $n-Propylamine CH_{2} CH_{2} CH_{2} NH_{2} + HCl$ $⟶$ $n-Propylammoniumchloride CH_{3} CH_{2} CH_{2} N + H_{3} C - l$ (ii) $Triethylamine (C_{2} H_{5})_{3} N + HCl ⟶ Triethylammoniumchloride (C_{2} H_{5})_{3} N + H C - 1$

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

Ans. Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of $Na_{2} CO_{3}$ solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.

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

Write structures of different isomers corresponding to the molecular formula, $C_{3} H_{9} N$ . Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid ? Ans. The structures of different isomers corresponding to the molecular formula, $C_{3} H_{9} N$ are given below : (a) $CH_{3} - CH_{2} - CH_{2} - NH_{2}$ Propan-1-amine (1°) (b) (c) $CH_{3} - NH - C_{2} H_{5}$ N-Methylethanamine (2°) (d)

1° amines, i.e. (a) propan -1-amine, and (b) Propan - 2-amine will liberate nitrogen gas on treatment with nitrous acid.'

9. Convert

(i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1, 3, 5- tribromobenzene

Ans.

NCERT EXERCISE

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) $(CH_{3})_{2} CH NH_{2}$ (ii) $CH_{3} (CH_{2})_{2} NH_{2}$ (iii) $CH_{3} NH CH (CH_{3})_{2}$ (iv) $(CH_{3})_{3} C NH_{2}$ (v) $C_{6} H_{5} NHCH_{3}$ (vi) $(CH_{3} CH_{2})_{2} NCH_{3}$ (vii) $m - Br C_{6} H_{4} NH_{2}$

Ans.

(i) Propan - 2 - amine (1° amine) (ii) Propan - 1 - amine (1° amine) (iii) N - Methyl - 2 - propanamine (2° amine) (iv) 2 - Methyl propan - 2 - amine (3° amine) (v) N - Methyl aniline (2° amine) (vi) N - Ethyl - N - methyl ethanamine (3° amine) (vii) 3 - Bromo benzenamine or 3 - Bromoaniline (1° amine)

Give one chemical test 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

Ans.

(i) Methylamine gives carbylamine reaction on heating with $CHCl_{3}$ and alcoholic KOH . Gives foul smelling (Carbyl amine)

$Heat Carbylamine CH_{3} - NH_{2} + CHCl_{3} + 3 KCl + 3 H_{2} O + 3 KOH$

where as Dimethylamine does not give this reaction. (ii) Secondary amines given Hinsberg's test. Its gives insoluble substance with Hinsberg's reagent $(C_{6} H_{5} SO_{2} Cl)$ which is not affected by base.

$Sec. aminin R_{2} NH + Benzens sulphonyl C_{6} H_{5} SO_{2} Cl \Rightarrow chloride R_{2} NSO_{4} SO_{4} C_{6} H_{5}$

(N, N-Dialkyl benzene sulphonamide) Where as tertiary amine does not react with Hinsberg's reagent.

(iii) Aniline gives azo dye test : Dissolve Aniline in conc. HCl and add ice - cold solution of $HNO_{2}$ $(NaNO_{2} + dil . HCl)$ at 273 K and then treat it with an alkaline solution of 2-napthol. Appearance of brilliant orange or red dye indicates aromatic amine (ie Aniline). Where as aliphatic amine (i.e. ethylamine) does not form dye. It will give brisk effervescence due to the evolution of $N_{2}$ but solution remain clear.

(iv) Nitrous acid test : Benzylamine reacts with nitrous acid $(HNO_{2})$ to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of $N_{2}$ gas.

Aniline reacts with $HNO_{2}$ to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve $N_{2}$ gas.

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(v) Carbylamine test : Aniline being a primary amine gives carbylamine test, i.e, when heated with an alcoholic solution of KOH and CHCl 3 , it gives an offensive smell of phenyl isocyanide. In contrast, N-methylaniline, being secondary amine does not give this test.

$Aniline (1°amine) C_{6} H_{5} NH_{2} + CHCl_{3} + 3 (alc.) 3 KOH Δ Phenyliscocyanide (offensive smell) C_{6} H_{5} NC$

$N -methylaniline (2^{\circ} amine) C_{6} H_{5} - NH - CH_{3} CHCl_{3} / KOH (alc) No reaction$

Account for the following : (i) $pK_{b}$ of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabrial phthalimide synthesis is preferred for synthesising primary amines.

Ans.

(i) In aniline, the lonepair of electrons on the N -atom are delocalized over the benzene ring. Resulting, electron density on the nitrogen decreases. On the other hand, in $CH_{3} NH_{2}, + I$ effect of $CH_{3}$ increases the electron density on the N -atom. Thus, aniline is a weaker base than methylamine and hence its $pK_{b}$ value is higher than that of methylamine. (ii) Ethylamine dissolves in water because it forms hydrogen bonds with water molecules.

In aniline due to large, hydro carbon part, the extent of H - bonding decreases considerably and hence aniline is insoluble in water. (iii) Methylamine being more basic than water, accepts a proton from water liberating $OH^{Θ}$ ions.

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These $OH^{-}$ ions combine with $Fe^{+ 3}$ ions present in $H_{2} O$ to form brown precipitate of hydrated ferric oxide.

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$2 Fe^{+ 3} + 6 OH^{-} \to 2 Fe (OH)_{3}$ or $Fe_{2} O_{3} .3 H_{2} O$ Hydrated ferric oxide (Brown ppt.) (iv) Nitration is usually carried out with a mixture of conc. $HNO_{3}$ and conc. $H_{2} SO_{4}$ . In presence of these acids, most of aniline gets protonated to form anilinium ion. Thus in presence of acids, the reaction mixture consists of aniline and anilinium ion. Now $- NH_{2}$ groups in aniline is o, p-directing and activating while the $NH_{3}^{\oplus}$ group in anilinium ion in m-directing and deactivating.

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(v) Aniline being a Lewis base, reacts with lewis acid $AlCl_{3}$ to from a salt.

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As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction, consequently, aniline does not undergo Friedel - Crafts reaction (vi) The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to dispersal of the positive charge on the benzene ring as shown below. (vii) Gabriel phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore, It is preferred for synthesising (aliphatic) primary amines.

Arrange the following : (i) In decreasing order of the $pK_{b}$ values : $C_{2} H_{5} NH_{2}, C_{6} H_{5} NHCH_{3}, (C_{2} H_{5})_{2} NH$ and $C_{6} H_{5} NH_{2}$ (ii) In increasing order of basic strength: $C_{6} H_{5} NH_{2}, C_{6} H_{5} N (CH_{3})_{2}, (C_{2} H_{5})_{2} NH$ and $CH_{3} NH_{2}$ (iii) In increasing order of basic strength : (a) Aniline, p-nitroaniline and p-toluidine (b) $C_{6} H_{5} NH_{2}, C_{6} H_{5} NHCH_{3}, C_{6} H_{5} CH_{2} NH_{2}$ (iv) In decreasing order of basic strength in gas phase : $C_{2} H_{5} NH_{2}, (C_{2} H_{5})_{2} NH, (C_{2} H_{5})_{3} N$ and $NH_{3}$ (v) In increasing order of boiling point : $C_{2} H_{5} OH, (CH_{3})_{2} NH, C_{2} H_{5} NH_{2}$ (vi) In increasing order of solubility in water : $C_{6} H_{5} NH_{2}, (C_{2} H_{5})_{2} NH, C_{2} H_{5} NH_{2}$

Ans.

(i) $C_{6} H_{5} NH_{2} > C_{6} H_{5} NH - CH_{3} > C_{2} H_{5} NH_{2} >$ $(C_{2} H_{5})_{2} NH$ (ii) $C_{6} H_{5} NH_{2} < C_{6} H_{5} N (CH_{3})_{2} < CH_{3} NH_{2} < (C_{2} H_{5})_{2}$ NH (iii) (a) p - nitroaniline < aniline < p - toluidine (b) $C_{6} H_{5} NH_{2} < C_{6} H_{5} NHCH_{3} < C_{6} H_{5} CH_{2} NH_{2}$

(iv) $(C_{2} H_{5})_{3} N > (C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2} > NH_{3}$ (v) $(CH_{3})_{2} NH < C_{2} H_{5} NH_{2} < C_{2} H_{5} OH$ (vi) $C_{6} H_{5} NH_{2} < (C_{2} H_{5})_{2} NH < C_{2} H_{5} NH_{2}$

How will you convert : (i) Ethanoic acid into methanamine (ii) Hexanentirile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid

Ans.

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Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equation of the reactions involved.

Ans. Benzene sulphonyl chloride $(C_{6} H_{5} SO_{2} Cl)$ , Which is also known as Hinsberg's reagent, reacts with primary and secondary amine to form sulphonamides.

The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence it is soluble in alkali.

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(iii) Tertiary amines do not react with benzene sulphonyl chloride. This property of amines reacting with benzene sulphonylchloride in a different manner is used for the distinction of primary, secondary and tertiary amines.

Write short notes on the following : (i) Carbylamine reaction (ii) Diazotisation (iii) Hoffmann's bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis

Ans.

(i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanide or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amine.

$R - NH_{2} + CHCl_{3} + 3 KOH Heat R - N ≅ C + 3 KCl + 3 H_{2} O$

(ii) Diazotisation reaction : When a cold solution of a primary aromatic amine in a dilute mineral acid (HCl or $H_{2} SO_{4}$ ) is treated with a cold solution of nitrous acid at 273-278 K. arenediazonium salt is formed. This reaction is called diazotisation reaction.

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(iii) Hoffmann's Bromamide Reactions : The conversions of a primary amide to a primary amine containing one carbon atom less than the original amide on heating with a mixture of $Br_{2}$ in presence of NaOH or KOH is called Haffmann's bromamide reaction. For example,

$CH_{3} CH_{2} CONH_{2} + Br_{2} + 4 KOH Δ Ethylamine C CH_{3} CH_{2} NH_{2} + 2 KBr + K_{2} CO_{3} + 2 H_{2} O$

$C_{6} H_{5} CONH_{2} + Br_{2} + 4 KOH Benzamide Δ Aniline C_{6} H_{5} NH_{2} + 2 KBr + K_{2} CO_{3} + 2 H_{2} O$

This reaction is extremely useful for converting a higher homologue to the next lower homologue. (iv) Coupling Reaction : The reaction of diazonium salts with phenols and aromatic amines to form azo compounds of the general formula $Ar - N = N - Ar$ is called coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in mildly alkaline medium while with amines it occurs under faintly acidic conditions.

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(v) Ammonolysis : The process of cleavage of the C-X bond by ammonia molecule is known as ammonolysis.

(vi) Acetylation : The process of introducing an acetyl group

$(C H_{3} - C (= O) -)$

into a molecule is called acetylation. Common acetylating agents used are acetyl chloride and acetic anhydride.

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(vii) Gabriel Phthalimide Synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potoassium phthalimide is heated with an alkylhalide to yield an N-alkylphthalimide which is hydrolysed to phthalic acid and primary amine by heating with HCl or KOH solution.

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Aromatic primary amines cannot be prepared by this method.

Accomplish the following conversions :

(i) Nirobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-chloroaniline (vii) Aniline to p-bromoaniline (viii) Benzamide to toluene (ix) Aniline to benzyl alcohol

Ans.

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Give the structures of A, B and C in the following reactions :

Ans :

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 $C_{6} H_{7} N$ . Write the structures and IUPAC names of compounds A, B and C. Ans.

Complete the following reactions : (i) $C_{6} H_{5} NH_{2} + CHCl_{3} +$ alc. $KOH ⟶$ (ii) $C_{6} H_{5} N_{2} Cl + H_{3} PO_{2} + H_{2} O ⟶$ (iii) $C_{6} H_{5} NH_{2} + H_{2} SO_{4}$ (conc.) ⟶ (iv) $C_{6} H_{5} N_{2} Cl + C_{2} H_{5} OH ⟶$ (v) $C_{6} H_{5} NH_{2} + Br_{2}$ (aq.) ⟶ (vi) $C_{6} H_{5} NH_{2} + (CH_{3} CO)_{2} O ⟶$ (vii) $C_{6} H_{5} N_{2} Cl (i) HBF_{4} (ii) NaNO_{2} / Cuu, Δ$ Ans.

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Why aromatic primary amines can not be prepared by Gabriel phthalimide synthesis ? Ans. The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.

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As arylhalides do not undergo nucleophilic substitution reaction easily, aromatic primary amines cannot be prepared by Gabriel phthalimide reaction

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

Ans.

Give possible 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 aliphatic amines are stronger bases than aromatic amines ?

Ans.

(i) Loss of a proton from an amine gives amide - ion $(N H ⊖)$ while loss of a proton from alcohol gives an alkoxide ion $(O R ⊖)$ as shown below.

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As O is more electronegative than N, So negative charge on more electronegative atom is more stable, As $R - O ⊖$ is more stable than $RN H - \cdot$ Therefore amines are less acidic than alcohols.

(ii) Due to the presence of two H-atoms on N-atoms of 1° amines, they undergo extensive intermolecular H-bonding while 3° amines due to the absence of H-atoms on N-atom there is no hydrogen bonding takes place. So primary amines have higher b.p. than tertiary amines of comparable molecular mass.

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(iii) Aliphatic amines are stronger bases than aromatic amines because : (a) Due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring thus is less available for protonation. (b) The aryl amine ions, have lower stability than the corresponding aliphatic amines i.e., protonation of aromatic amines is not favoured.

3.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

Get chapter-wise NCERT Solutions for Class 12 Chemistry with well-explained answers, important concepts, key formulae and detailed solutions for the numerical problems.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Solutions
Chapter 2	Electrochemistry
Chapter 3	Chemical Kinetics
Chapter 4	d- and f-Block Elements
Chapter 5	Coordination Compounds
Chapter 6	Haloalkanes and Haloarenes
Chapter 7	Alcohols, Phenols and Ethers
Chapter 8	Aldehydes, Ketones and Carboxylic Acids
Chapter 10	Biomolecules

4.0Key Features of NCERT Solutions for Class 12 Chemistry Chapter 9

Basicity Concepts: Clear explanations of the basic strength of primary, secondary, and tertiary amines in different conditions.
Important Name Reactions: Step-by-step explanations of Gabriel Phthalimide and Hoffmann Bromamide reactions.
Distinction Tests: Simple procedures and observations for Carbylamine and Hinsberg’s tests.
Organic Conversions: Easy methods for conversions involving aniline and diazonium salts.

NCERT Solutions Class 12 Chemistry Chapter 9 - Amines

1.0Class 12 Chemistry Chapter 9 Amines: Key Concepts

Key Concepts

Structure and Classification: Understand the structure of amines and classify them as primary, secondary, and tertiary amines.
Nomenclature: Learn the IUPAC naming of aliphatic and aromatic amines such as aniline.
Preparation of Amines: Study Gabriel Phthalimide Synthesis, Hoffmann Bromamide Degradation, and the reduction of nitriles and amides.
Basicity of Amines: Understand the factors affecting the basic strength of amines, including the +I effect, steric effects, and solvation.
Chemical Reactions: Learn important reactions such as acylation, Carbylamine Reaction, and reactions with nitrous acid.
Hinsberg’s Test: Understand how primary, secondary, and tertiary amines are distinguished using Hinsberg’s test.
Diazonium Salts: Study the preparation and reactions of benzene diazonium chloride.
Coupling Reactions: Learn how diazonium salts undergo coupling reactions to form coloured azo compounds.

2.0NCERT Solutions Class 12 Chemistry Chapter 9 Amines: Detailed Solutions

INTEXT QUESTIONS

Classify the following amines as primary, secondary or tertiary :

Ans. Primary : (i) and (iii) Secondary : (iv) Tertiary : (ii)

(i) Write structures of different isomeric amines corresponding to the molecular formula, $C_{4} H_{11} N$ (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines ? Ans. (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, $C_{4} H_{11} N$ are given below:

How will you convert (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) $Cl - (CH_{2})_{4} - Cl$ into hexan-1, 6-diamine?

Ans.

Arrange the following in increasing order of their basic strength : (i) $C_{2} H_{5} NH_{2}, C_{6} H_{5} NH_{2}, NH_{3}, C_{6} H_{5} CH_{2} NH_{2}$ and $(C_{2} H_{5})_{2} NH$ (ii) $C_{2} H_{5} NH_{2}, (C_{2} H_{5})_{2} NH, (C_{2} H_{5})_{3} N, C_{6} H_{5} NH_{2}$ (iii) $CH_{3} NH_{2}, (CH_{3})_{2} NH, (CH_{3})_{3} N, C_{6} H_{5} NH_{2}$ , $C_{6} H_{5} CH_{2} NH_{2}$

Ans.

Complete the following acid-base reactions and name the products : (i) $CH_{3} CH_{2} CH_{2} NH_{2} + HCl ⟶$ (ii) $(C_{2} H_{5})_{3} N + HCl ⟶$

Ans.

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

Ans. Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of $Na_{2} CO_{3}$ solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.

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

Write structures of different isomers corresponding to the molecular formula, $C_{3} H_{9} N$ . Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid ? Ans. The structures of different isomers corresponding to the molecular formula, $C_{3} H_{9} N$ are given below : (a) $CH_{3} - CH_{2} - CH_{2} - NH_{2}$ Propan-1-amine (1°) (b) (c) $CH_{3} - NH - C_{2} H_{5}$ N-Methylethanamine (2°) (d)

1° amines, i.e. (a) propan -1-amine, and (b) Propan - 2-amine will liberate nitrogen gas on treatment with nitrous acid.'

9. Convert

(i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1, 3, 5- tribromobenzene

Ans.

NCERT EXERCISE

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) $(CH_{3})_{2} CH NH_{2}$ (ii) $CH_{3} (CH_{2})_{2} NH_{2}$ (iii) $CH_{3} NH CH (CH_{3})_{2}$ (iv) $(CH_{3})_{3} C NH_{2}$ (v) $C_{6} H_{5} NHCH_{3}$ (vi) $(CH_{3} CH_{2})_{2} NCH_{3}$ (vii) $m - Br C_{6} H_{4} NH_{2}$

Ans.

Give one chemical test 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