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Chapter 13- Amines Interview Questions Answers

Question 1 :

Classify the following amines as primary, secondary ortertiary:

Answer 1 :

(i)


Answer

Primary: (i) and (iii)

Secondary: (iv)

Tertiary: (ii)

Question 2 :

(i) Writestructures of different isomeric amines corresponding to the molecular formula,C4H11N


Answer 2 :

(ii) Write IUPAC names of all the isomers.

(iii) What type of isomerism is exhibited by different pairs of amines?


Answer

(i), (ii) Thestructures and their IUPAC names of different isomeric amines corresponding tothe molecular formula, C4H11N are given below:

(a) CH3-CH2-CH2-CH2-NH2

Butanamine (10)

(b) 

Butan-2-amine (10)

(c)

2-Methylpropanamine (10)

(d)

2-Methylpropan-2-amine (10)

(e) CH3-CH2-CH2-NH-CH3

N-Methylpropanamine (20)

(f) CH3-CH2-NH-CH2-CH3

N-Ethylethanamine (20)

(g)

N-Methylpropan-2-amine (20)

(h)

N,N-Dimethylethanamine (3°)

(iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism.

The pairs (a) and (c); (a) and (d); (b) and (c); (b) and(d) exhibit chain isomerism.

The pairs (e) and (f) and (f) and (g) exhibit metamerism.

All primary amines exhibit functional isomerism withsecondary and tertiary amines and vice-versa.

Question 3 : How will you convert?

Answer 3 :

(i) Benzeneinto aniline    (ii) Benzeneinto N, N-dimethylaniline

(iii) Cl−(CH2)4−Clinto hexan-1, 6-diamine?


Answer

(i)

Question 4 :

Arrange the following in increasing order of their basicstrength:


Answer 4 :

(i) C_2H_5NH_2, C_6H_5NH_2, NH_3, C_6H_5CH_2NH_2 and (C_2H_5)_2NH

(ii) C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N, C_6H_5NH_2

(iii) CH_3NH_2, (CH_3)_2NH, (CH_3)_3N, C_6H_5NH_2, C_6H_5CH_2NH_2.

Answer

(i) Consideringthe inductive effect of alkyl groups, NH3, C2H5NH2, and(C2H5)2NH can be arranged in the increasing order of their basicstrengths as:

Again, C6H5NH2 hasproton acceptability less than NH3. Thus, we have:

Due to the −I effect of C6H5 group,the electron density on the N-atom in C6H5CH2NH2 islower than that on the N-atom in C2H5NH2, butmore than that in NH3. Therefore, the given compounds can be arranged in theorder of their basic strengths as:

(ii) Consideringthe inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (CH5)2NH2, and their basic strengths as follows:

Again, due to the −R effect of C6H5 group,the electron density on the N atom in C6HNH2 is lower than that on the N atom in C2H5NH2.Therefore, the basicity of C6H5NH2 islower than that of C2H5NH2. Hence, the given compounds can be arranged in theincreasing order of their basic strengths as follows:

(iii) Consideringthe inductive effect and the steric hindrance of alkyl groups, CH3NH2,(CH3)2NH, and (CH3)3Ncan be arranged in the increasing order of their basic strengths as:

In C6H5NH2, Nis directly attached to the benzene ring. Thus, the lone pair of electrons onthe N−atom is delocalized over the benzene ring. In C6H5CH2NH2, Nis not directly attached to the benzene ring. Thus, its lone pair is not delocalizedover the benzene ring. Therefore, the electrons on the N atom are more easilyavailable for protonation in C6H5CH2NH2 thanin C6H5NH2 i.e., C6H5CHNH2 ismore basic than C6H5NH2.

Again, due to the −I effect of C6H5 group,the electron density on the N−atom in C6H5CH2NH2 islower than that on the N−atom in (CH3)3N.Therefore, (CH3)3N is more basic than C6H5CH2NH2.Thus, the given compounds can be arranged in the increasing order of theirbasic strengths as follows.

Question 5 :

Complete the following acid-base reactions and name theproducts:


Answer 5 : (i) CH3CH2CH2NH2 +HCl 

(ii) (C2H5)3N +HCl 

Answer
(i) 

(ii) 

C2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride

Question 6 :

Write reactions of the final alkylation product of anilinewith excess of methyl iodide in the presence of sodium carbonate solution.

Answer 6 :

Aniline reacts with methyl iodide to produce N,N-dimethylaniline.

With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate.

Question 7 :

Write chemical reaction of aniline with benzoyl chlorideand write the name of the product obtained.

Answer 7 :

Question 8 :

Write structures of different isomers corresponding to themolecular formula, C3H9N. 

Answer 8 : Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.


Answer

The structures of different isomerscorresponding to the molecular formula, C3H9Nare given below:

(a) 

Propan-1-amine (10)

(b)

Propan-2-amine (10)

(c)

(d)

N,N-Dimethylmethanamine (30)

10amines, (a)propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatmentwith nitrous acid.

Question 9 :

Convert:-(i) 3-Methylanilineinto 3-nitrotoluene.    (ii) Anilineinto 1,3,5-tribromobenzene.

Answer 9 :

(i)

(ii)

Question 10 :

Write IUPAC names of the following compounds and classifythem into primary, secondary and tertiary amines.


Answer 10 :

(i) (CH_3)_CHNH_(ii) CH_3(CH_2)_2NH_2

(iii) CH_3NHCH(CH_3)_(iv) (CH_3)_3CNH_2

(v) C_6H_5NHCH_3 (vi) (CH_3CH_2)_2NCH_3

(vii) m−BrC_6H_4NH_2

_

_Answer


(i) 1-Methylethanamine(10 amine)

(ii) Propan-1-amine(10 amine)

(iii) N−Methyl-2-methylethanamine(20 amine)

(iv) 2-Methylpropan-2-amine(10 amine)

(v) N−Methylbenzamineor N-methylaniline (20 amine)

(vi) N-Ethyl-N-methylethanamine(30 amine)

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(vii) 3-Bromobenzenamineor 3-bromoaniline (10 amine)


Selected

 

Chapter 13- Amines Contributors

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