Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Amines. This page offers a step-by-step solution to the specific question from Exercise 1, Question 4: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
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.
(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
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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Read MoreWelcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 4: Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H....
Comments
The answers of (I)st and (ii)nd are wrong apart from the (iii)rd one. Kindly update the answers with the correct order
Recheck your answers plz...