Amines Class 12 Chemistry Important Questions

Important Questions Class 12

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.

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

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

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 + Br2(aq)

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

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

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

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 :
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.

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

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 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. 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

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 :

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