Amines Class 12 Chemistry Important Questions

Students can read the important questions given below for Amines Class 12 Chemistry. All Amines Class 12 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 12 Chemistry Important Questions provid ed for all chapters to get better marks in examinations. Chemistry Question Bank Class 12 is available on our website for free download in PDF.

Important Questions of Amines Class 12

Very Short Answer Questions

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

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

Question. Give the IUPAC name of
H₂N — CH₂ — CH₂ — CH = CH₂.
Answer. But-3-en-1-amine

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. Arrange the following in the increasing order of their boiling point :
C₂H₅NH₂, C₂H₅OH, (CH₃)₃N
Answer. 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.

Question. Arrange the following in increasing order of basic strength :
Aniline, p-nitroaniline, and p-toluidine
Answer.

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

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. 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. 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. 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.

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

Question. 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.

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. 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.

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. 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.

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

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

Question. Arrange the following in increasing order of basic strength :
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
Answer. C₆H₅NH₂ < 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 C_6H5NHCH₃, it is more basic than C₆H₅NH₂.

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

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. 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 structure for N, N-ethylmethylamine.
Answer.

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

Short Answer Questions

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

Question. Give reasons :
(i) Aniline is a weaker base than cyclohexyl amine.
(ii) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
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) The ammonolysis of alkyl halides with ammonia is a nucleophilic substitution reaction in which ammonia acts as a nucleophile by donating the electron pair on nitrogen atom to form primary amine as the initial product. Now, the primary amine can act as a nucleophile and combine with alkyl halide (if available) to give secondary amine and the reaction continues in the same way to form tertiary amine and finally quaternary ammonium salt. Thus, a mixture of products is formed and it is not possible to separate individual amines from the mixture

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 pK_b 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 pK_b value hence, the decreasing order of pK_b values :
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

Question. Complete the following reactions :
(i) CH₃CH₂NH₂ + CHCl₃ + alc. KOH →

Answer. (i) CH₃ — CH₂ — NH₂ + CHCl₃ + 3KOH →
CH₃ — CH₂ — NC + 3KCl + 3H₂O

Question. Give the chemical tests to distinguish between the following pairs of compounds :
(i) Methylamine and dimethylamine
(ii) Aniline and N-methylaniline
Answer. (i) Methyl amine gives carbylamine test, i.e., on treatment with alc. KOH and chloroform, followed by heating it gives offensive odour of methyl isocyanide. Dimethyl amine 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.

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

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.

Long Answer Questions

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 :