Please refer to Amines Class 12 Chemistry Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Chemistry based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Chemistry for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.
Class 12 Chemistry Important Questions Amines
Very Short Answer Questions
Question. Write the IUPAC name of the given compound :
Answer. 2, 4, 6-Tribromoaniline
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. How do you convert the following :
Ethanenitrile to ethanamine
Answer.
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. State reasons for the following :
pKb 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 pKb value is more than that for methylamine.
Question. How will you differentiate between aniline and ethylamine?
Answer. Aniline being an aromatic primary amine on treatment with HNO2[NaNO2 + HCl (dil.)] at 273–278 K followed by treatment with an alkaline solution of b-naphthol gives an orange coloured azo dye. Ethylamine does not give this test.
Question. Assign reason for the following :
The pKb 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 pKb value is higher than that of methylamine.
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. 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. Complete the following reaction equation :
Answer. C6H5N2Cl + H3PO2 + H2O →
C6H6 + N2 + H3PO3 + HCl
Question. Why do amines act as nucleophiles?
Answer. Because the electron pair of nitrogen can coordinate with the electron deficient electrophiles.
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. The conversion of primary aromatic amines into diazonium salts is known as __________.
Answer. Diazotisation reaction.
Question. Complete the following reactions :
Answer.
Question. State and illustrate the following :
Coupling reaction.
Answer. Diazonium salts react with aromatic amines in weakly acidic medium and phenols in weakly alkaline medium to form coloured compounds called azo dyes by coupling at p-position of amines or phenols.
Question. How is the following conversion carried out :
Aniline to p-hydroxyazobenzene.
Answer.
Question. How will you bring about the following conversion :
Nitrobenzene to phenol
Answer.
Question. How will you bring about the following conversion :
Aniline to chlorobenzene Write the chemical equation involved.
Answer.
Question. How will you bring about the following conversion :
Aniline to benzonitrile.
Answer.
Question. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.
Answer.
Question. How would you achieve the following conversion :
Aniline to benzonitrile.
Write the chemical equation with reaction conditions in each case.
Answer.
Short Answer Questions
Question. Give IUPAC names of the following compounds :
Answer. (a) But-3-en-2-amine
(b) N-phenylethanamide
Question. How do you convert the following :
(i) C6H5CONH2 to C6H5NH2
(ii) Aniline to phenol
Answer.
Question. Give the structures of A, B and C in the following reactions :
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. Assign reason for the following :
(i) Amines are less acidic than alcohols of comparable molecular masses.
(ii) Aliphatic amines are stronger bases than aromatic amines.
Answer. (i) Loss of proton from amines gives RNH– ion whereas loss of proton from alcohol forms alkoxide ion. Since, O is more electronegative than N therefore, RO– can accommodate the negative charge more easily than RN–H. Further, O H bond is more polar than N H bond. Hence, amines are less acidic than alcohols.
(ii) In aromatic amines, the lone pair of electrons present on nitrogen takes part in resonance and hence, not available for donation. However, in aliphatic amines, the lone pair is available for donation. That’s why aliphatic amines are more basic than aromatic amines.
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 :
C2H5NH2, (C2H5)2NH, (C2H5)3N and CH3NH2.
Answer.(i) e 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. Write chemical equations for the following conversions :
(i) Nitrobenzene to benzoic acid.
(ii) Aniline to benzyl alcohol.
Answer.
Long Answer Questions
Question. Write the structures of A, B and C in the following :
Answer.
Question. Write the structures of main products when aniline reacts with the following reagents :
(i) Br2 water
(ii) HCl
(iii) (CH3CO)2O/pyridine
Answer.
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.
