Amines Class 12 Chemistry Exam Questions

Please refer to the below Amines important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.

Short Answer Questions :

Question. Complete the following reaction equation :
C₆H₅NH₂ + Br_2(aq) → 
Answer :

Question. Arrange the following in increasing order of basic strength :
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
Answer : 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.

Question. State and illustrate the following :
Gabriel synthesis 
Answer : Gabriel phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.

Question. How will you bring about the following conversion :
Ethanamine to ethanoic acid
Answer :

Question. Give reasons for the following :
Primary amines have higher boiling point than tertiary amines. 
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.

Question. Give a simple chemical test to distinguish between the following pair of compounds :
(CH₃)₂NH and (CH₃)₃N 
Answer : When treated with benzenesulphonyl chloride (Hinsberg’s reagent), (CH₃)₂NH forms insoluble N, N-dialkylbenzene sulphonamide which is insoluble in KOH whereas tertiary amine does not react at all.

Question. Write the structure of 2-aminotoluene.
Answer :

Question. State reasons for the following :
pK_b value for aniline is more than that for ethylamine. 
Answer : In aniline, the lone pair of electrons on N-atom is delocalised over benzene ring due to resonance. As a result, electron density on the nitrogen atom decreases. In contrast, in methylamine, +I-effect of CH₃ group increases electron density on the nitrogen atom. Therefore, aniline is a weaker base than methylamine hence, its pK_b value is more than that for methylamine.

Question. Out of CH₃NH₂ and (CH₃)₃N, which one has higher boiling point?
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.

Question. Arrange the following in increasing order of their basic strength in aqueous solution :
CH₃NH₂, (CH₃)₃N, (CH₃)₂NH
Answer : In case of small alkyl groups like CH₃ the order of basicity is secondary amine > primary amine > tertiary amine due to solvation effect and +I effect of —CH₃ group.
    (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
pK_b      3.27        3.38          4.22

Question. Why do amines act as nucleophiles? 
Answer : Because the electron pair of nitrogen can coordinate with the electron deffcient electrophiles.

Question. How do you convert the following :
(i) C₆H₅CONH₂ to C₆H₅NH₂
(ii) Aniline to phenol 
Answer :

Question. Write the structure for N, N-ethylmethylamine.
Answer : CH₃N—C₂H₅ (N, N-ethylmethylamine)
                    l
                   H

Question. Write the chemical reaction to illustrate the following :
Ammonolysis 
Answer : Alkyl halides when treated with ethanolic solution of ammonia give a mixture of primary, secondary, tertiary amines and quaternary ammonium salt.

Question. Write the chemical equation involved in the following reaction :
Hofmann bromamide degradation reaction
Answer : R—CONH₂ + Br₂ + 4NaOH →  R—NH₂ + Na₂CO₃ + 2 NaBr + 2H₂O
              Acid amide                            1° amine

Question. Illustrate the following reactions giving a chemical equation in each case :
(i) Gabriel phthalimide synthesis
(ii) Hofmann’s bromamide reaction.
Answer : (i) Gabriel phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.

(ii) R—CONH₂ + Br₂ + 4NaOH →  R—NH₂ + Na₂CO₃ + 2 NaBr + 2H₂O
     Acid amide                            1° amine

Question. Which of the two is more basic and why?
CH₃NH₂ or NH₃
Answer : Methyl amine is more basic than ammonia because of the presence of electron donating methyl group (+I effect), which increases the electron density on nitrogen atom.

Question. Account for the following :
Primary amines (R—NH₂) have higher boiling point than tertiary amines (R₃N). 
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.

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

Question. State reasons for the following :
(i) Ethylamine is soluble in water whereas aniline is not soluble in water.
(ii) Primary amines have higher boiling points than tertiary amines. 
Answer : (i) 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.
(ii) 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.

Question. How will you convert the following :
Aniline into N-phenylethanamide
(Write the chemical equations involved.)
Answer : 

Question. Account for the following :
Amines are basic substances while amides are neutral.
Answer : In CH₃CONH₂, the lone pair of electrons on nitrogen atom is involved in resonance with the carbonyl group. So, the electron pair of nitrogen is not easily available for protonation. Hence, CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.

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

Answer : CH₃NH₂ is more basic than C₆H₅NH₂ because in aniline the lone pair of electrons on nitrogen are involved in resonance.

Question. Account for the following :
Nitro compounds have higher boiling points than the hydrocarbons having almost the same molecular mass. 
Answer : The nitro compounds are highly polar molecules. Due to this polarity they have strong intermolecular dipole – dipole interactions which causes them to have higher boiling points in comparison to the hydrocarbons having almost same molecular mass.

Question. Write the IUPAC name of the given compound :

Answer : 2, 4, 6-Tribromoaniline

Question. Write the main products of the following reactions :

Answer : (i) 

(ii) CH₃—NH₂
Methanamine

Question. How will you bring about the following conversion :
Nitrobenzene to phenol  
Answer : 

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

Ethylamine does not react with bromine water.
(ii) When heated with an alcoholic solution of KOH and CHCl₃, ethylamine gives foul smelling ethyl isocyanide. Dimethylamine does not give this test.

Question. Write the chemical equations involved when aniline is treated with the following reagents :
(i) Br₂ water
(ii) CHCl₃ + KOH
(iii) HCl  
Answer : (i)

(ii) 

(iii)