Please refer to Amines Class 12 Chemistry Important Questions with answers below. These solved questions for Chapter 13 Amines in NCERT Book for Class 12 Chemistry have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 12 Chemistry for all chapters in your textbooks.

Important Questions Class 12 Chemistry Chapter 13 Amines

All Amines Class 12 Chemistry Important Questions provided below have been prepared by expert teachers of Standard 12 Chemistry. Please learn them and let us know if you have any questions.

Short Answer Questions :

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

Answer : (i) 

(ii)

Question. Complete the following reactions :
C₆H₅N₂ +Cl– → ^H₂O_{(Room temp.)}  
Answer : 

Question. How will you bring about the following conversion :
Aniline to chlorobenzene Write the chemical equation involved. 
Answer : 

Question. Describe the following giving the relevant chemical equation :
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 → R—N ≡C + 3KCl + 3H₂O
It is the test for primary amines.

Question. How are the following conversions carried out?
(i) CH₃CH₂Cl to CH₃CH₂CH₂NH₂
(ii) Benzene to aniline. 
Answer : 

Question. Give IUPAC names of the following compounds :

Answer : (a) But-3-en-2-amine
(b) N-phenylethanamide

Question. Why is an alkylamine more basic than ammonia?
Answer : Electron density of N-atom increases due to the +I effect of the alkyl group. Hence, alkylamines are stronger bases than ammonia.
                                       ^¨NH₃, R → ^¨NH₂

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

Question. Account for the following :
Ethylamine is soluble in water whereas aniline is not. 
Answer : Ethylamine is soluble in water due to formation of intermolecular hydrogen bonds with water molecules. However, in aniline due to large hydrophobic aryl group the extent of hydrogen bonding decreases considerably and hence aniline is insoluble in water.

Question. An aromatic compound ‘A’ of molecular formula C₇H₇ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions :

Answer : 

Question. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.  
Answer : Formula of the compound ‘C’ indicates it to be an amine. Since it is obtained by the reaction of Br₂ and KOH with the compound ‘B’ so compound ‘B’ can be an amide. As ‘B’ is obtained from compound ‘A’ by reaction with ammonia followed by heating so, compound ‘A’ could be an aromatic acid. Formula of compound ‘C’ shows it to be aniline, then ‘B’ is benzamide and compound ‘A’ is benzoic acid. The sequence of reactions can be written as follows :

Question. (i) Arrange the following in an increasing order of basic strength in water :
C₆H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃.
(ii) Arrange the following in increasing order or basic strength in gas phase :
C₆H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and CH₃NH₂. 
Answer : (i) The increasing order of basic strength in water of the given amines and ammonia follows the following order :
C₆H₅NH₂ < NH₃ < (C₂H₅)₃N < (C₂H₅)₂NH
(ii) The increasing order of basic strength in gas phase of the given amines follows the order :
CH₃NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH < (C₂H₅)₃N

Question. Account for the following observations :
(i) pK_b for aniline is more than that for methylamine.
(ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide.
(iii) Aniline does not undergo Friedel-Crafts reaction.  
Answer : (i) In aniline, the lone pair of electrons of N-atom are delocalised over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in CH₃—NH₂, +I effect of —CH₃ group increases the electron density on the N-atom.
Therefore, aniline is a weaker base than methylamine and hence, its pK_b value is higher than that of methylamine.
(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.
        2Fe + 6OH^– → 2Fe(OH)₃
(iii)

Question. How will you convert the following :
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
Answer : 

Question. Which of the two is more basic and why?

Answer : CH₃(C₆H₄)NH₂ is more basic than C₆H₅NH₂ due to electron releasing nature of methyl group which pushes electrons towards nitrogen.

Question. How would you achieve the following conversions :
(i) Nitrobenzene to aniline.
(ii) An alkyl halide to a quaternary ammonium salt.
Write the chemical equation with reaction conditions in each case. 
Answer :

(ii) Alkyl halides when treated with ethanolic solution of ammonia give a mixture of primary, secondary, tertiary amines and quaternary ammonium salt.

Question. How do you convert the following :
Ethanenitrile to ethanamine 
Answer : CH₃CN + 4[H] →^{Na + C₂H₅OH} CH₃CH₂NH₂
              Ethanenitrile                         Ethanamine

Question. Give reasons for the following :
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. 
Answer : (i) 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.

(ii) In aqueous solution 2° amine is more basic than 3° amine due to the combination of inductive effect, solvation effect and steric reasons.

Question. Why cannot primary aromatic amines 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.

Question. Assign reason for the following :
The pK_b of aniline is higher than that of methylamine.
Answer : In aniline, the lone pair of electrons of N-atom are delocalised over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in CH₃—NH₂, +I effect of —CH₃ group increases the electron density on the N-atom.
Therefore, aniline is a weaker base than methylamine and hence, its pK_b value is higher than that of methylamine.

Question. Write chemical equations for the following conversion :
Benzyl chloride to 2-phenylethanamine.
Answer : 

Question. Arrange the following in increasing order of their basic strength :

Answer : (i) C₆H₅CH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂
C₆H₅NH₂ and C₆H₅NHCH₃ are less basic than aliphatic amine C₆H₅CH₂NH₂ due to lone pair of nitrogen is in conjugation with benzene ring. But due to +I effect of —CH₃ group in C₆H₅NHCH₃, it is more basic than C₆H₅NH₂.
(ii)

Electron withdrawing group (–NO₂) on benzene ring decreases the basicity and electron donating group (–CH₃) on benzene ring increases the basicity of compound.

Question. Arrange the following in increasing order of basic strength :
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅N(CH₃)₂
Answer : Increasing order of basic strength in gaseous state is as follows :
C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅N(CH₃)₂
As the number of —CH₃ groups (+I effect) attached to nitrogen increases, its basicity its basicity will increases.

Question. 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

Question. Illustrate the following reactions giving suitable example in each case :
(i) Ammonolysis
(ii) Acetylation of amines 
Answer : (i) Alkyl halides when treated with ethanolic solution of ammonia give a mixture of primary, secondary, tertiary amines and quaternary ammonium salt.

(ii) Acetylation of amines : The process of introducing an acetyl group 

into a molecule is called acetylation.

Question. 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.

(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.
        2Fe + 6OH^– → 2Fe(OH)₃

Question. Give one chemical test each to distinguish between the following pairs of compounds :
(i) Ethylamine and aniline
(ii) Aniline and N-methylaniline  
Answer : (i) Aniline gives white or brown precipitate with bromine water.

Ethylamine does not react with bromine water.
(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.

Question. (i) Arrange the following compounds in an increasing order of basic strength :
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(ii) Arrange the following compounds in a decreasing order of pKb values :
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
Answer : (i) Increasing order of basic strength is
C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH
(ii) Stronger the base lower will be its pKb value hence, the decreasing order of pKb values :
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

Question. Give the chemical tests to distinguish between the following pairs of compounds :
(i) Ethylamine and aniline
(ii) Aniline and benzylamine  
Answer : (i) Aniline gives white or brown precipitate with bromine water.

Ethylamine does not react with bromine water.
(ii) Aniline and benzylamine : Benzylamine reacts with nitrous acid to form a diazonium salt which is unstable at low temperature and decomposes with evolution of N₂ gas.

Question. Complete the following reactions :

Answer :