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Q1 (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Ans: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the 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 with secondary and tertiary amines and vice-versa.
Q2 Classify the following amines as primary, secondary or tertiary:
(i)
(ii)
(iii) (C2H5)2CHNH2
(iv) (C2H5)2NH
Ans: Primary: (i) and (iii)
Secondary: (iv)
Tertiary: (ii)
Q3 How will you convert?
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) Cl-(CH2)4-Cl into hexan-1, 6-diamine?
Ans: (i)
(ii)
(iii)
Q4 Arrange the following in increasing order of their basic strength:
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.
Ans: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as:
NH3 2H5NH2 <(C2H5)2NH
Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have:
C6H5NH2 3 2H5NH2 <(C2H5)2NH
Due to the - I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as:
C6H5NH2 3 2H5NH2 C6H5CH2NH22H5NH2<(C2H5)2NH
(ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows:
C2H5NH2 <(C2 H5)3N <(C2 H5)2NH
Again, due to the - R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows:
C2H5NH2 2 H5NH2 <(C2 H5)3N <(C2 H5)2NH
(iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as:
(CH3)3N 3NH2 (CH3)2NH
In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N - atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2 NH2 is more basic than C6H5NH2.
Again, due to the - I effect of C6H5 group, the electron density on the N - atom in C6H5CH2NH2 is lower 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 their basic strengths as follows.
C6H5NH2 6H5CH2NH2 <(CH3)3N 3NH2 <(CH3)2NH
Q5 Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2 + HCl →
(ii) (C2H5)3N + HCl →
Ans: (i)
(ii) (C2H5)3N Triethylamine+ HCI → (C2H5)3N+HCI - Triethylammoniumchloride
Q6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans: Aniline reacts with methyl iodide to produce N, N-dimethylaniline.
With excess methyl iodide, in the presence of Na2CO3 solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.
Q7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans: Q8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on
treatment with nitrous acid.
Ans: The structures of different isomers corresponding to the molecular formula, C3H9N are given below:
(a) CH3 –CH2 –CH2 –NH2
Propan-1-amine (10)
(b)
Propan-2-amine (10)
(c) CH3 –NH –C2H5
N – Methylethanamine(20)
(d)
N,N-Dimethylmethanamine (30)
10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.
Q9 Convert
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5-tribromobenzene.
Ans: (i)
(ii)