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Question
1) Arrange the following:
A) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3,(C2H5)2NH and C6H5NH2
2) Arrange the following:
B) In increasing order of basic strength: C6H5NH2,C6H5N(CH3)2,(C2H5)2NH and CH3NH2
3)(A) Arrange the following:
C) In increasing order of basic strength: Aniline, p-nitroaniline and p-toluidine
3) (B) Arrange the following:
D) In increasing order of basic strength: C6H5NH2,C6H5NHCH3,C6H5CH2NH2.
4) Arrange the following:
E) In decreasing order of basic strength in gas phase: C2H5NH2,(C2H5)2NH,(C2H5)3NandNH3
5) Arrange the following:
F) In increasing order of boiling point: C2H5OH,(CH3)2NH,C2H5NH2
6) Arrange the following:
G) In increasing order of solubility in water: C6H5NH2,(C2H5)2NH,C2H5NH2
A) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3,(C2H5)2NH and C6H5NH2
2) Arrange the following:
B) In increasing order of basic strength: C6H5NH2,C6H5N(CH3)2,(C2H5)2NH and CH3NH2
3)(A) Arrange the following:
C) In increasing order of basic strength: Aniline, p-nitroaniline and p-toluidine
3) (B) Arrange the following:
D) In increasing order of basic strength: C6H5NH2,C6H5NHCH3,C6H5CH2NH2.
4) Arrange the following:
E) In decreasing order of basic strength in gas phase: C2H5NH2,(C2H5)2NH,(C2H5)3NandNH3
5) Arrange the following:
F) In increasing order of boiling point: C2H5OH,(CH3)2NH,C2H5NH2
6) Arrange the following:
G) In increasing order of solubility in water: C6H5NH2,(C2H5)2NH,C2H5NH2
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Solution
Decreasing order of pKb values
In C2H5NH2, one −C2H5 group is present and in (C2H5)2NH, two −C2H5 groups are present. Therefore, the +I effect of ethyl is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.
Also, both C6H5NHCH3 & C6H5NH2 are less basic than (C2H5)2NH & C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 & C6H5NH2,C6H5NHCH3 will be more basic due to extra +I effect of −CH3 group which is absent in C6H5NH2.
Hence, the order of increasing basicity of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<C2H5NH2<(C2H5)2NH
We know that the higher the basic strength, the lower is the pKb value.
So, decreasing order of the pKb values is:
C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH.
Increasing order of basic strength
In CH3NH2, one −CH3 group is present and in (C2H5)2NH, two −C2H5 groups are present. Therefore, the +I effect of ethyl group is more in (C2H5)2NH than +I effect of methyl group in CH3NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in CH3NH2. Hence, (C2H5)2NH is more basic than CH3NH2.
Also, both C6H5N(CH3)2 & C6H5NH2 are less basic than CH3NH2 & (C2H5)2NH due to the delocalization of the lone pair in the former two. Further, among C6H5N(CH3)2 & C6H5NH2, C6H5N(CH3)2 will be more basic due to extra +I effect of −CH3 groups which is absent in C6H5NH2.
Hence, the increasing order of the basic strengths of the given compounds is as follows: C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH
Increasing order of basic strength
In p - toluidine, the presence of electron-donating −CH3 group increases the electron density on the N-atom. Thus, p - toluidine is more basic than aniline.
In the other case, the presence of electron-withdrawing −NO2 group decreases the electron density over the N atom in p - nitroaniline.
Thus, p-Nitroaniline is less basic than aniline.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
p−Nitroaniline<Aniline<p–Toluidine
Increasing order of basic strength
C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.
Again, in C6H5NHCH3,−C6H5 group is directly attached to the N - atom. However, it is not so in C6H5CH2NH2.
Thus, in C6H5NHCH3, lone pair of electrons is delocalised on benzene ring which decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.
Hence, the increasing order of the basic strength of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<C6H5CH2NH2
Decreasing order of basic strength of amines in gas phase
In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:
(C2H5)3N>(C2H5)2NH>C2H5NH2>NH3
Increasing order of boiling point
The extent of H−bonding existing in some compounds can decide the boiling point of compounds.
Stronger the H−bonding in the compound, the higher will be the boiling point.
(CH3)2NH has only one H−atom while C2H5NH2 has two H−atoms. Subsequently, C2H5NH2 undergoes more extensive H−bonding than (CH3)2NH. Therefore, the boiling point of C2H5NH2 is more than that of (CH3)2NH.
Also, oxygen is more electronegative than nitrogen. Thus, C2H5OH forms stronger H-bonds than C2H5NH2.
As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 & (CH3)2NH.
Based on the above explanation, the compounds given in the question can be arranged in the increasing order of their boiling points, which is given below:
(CH3)2NH<C2H5NH2<C2H5OH
Increasing order of solubility in water
Stronger the H−bonding with water, more will be the solubility in water. C2H5NH2 contain two H− atoms whereas (C2H5)2NH contains only one H−atom.
Thus, C2H5NH2 undergoes stronger H-bonding than (C2H5)2NH with water. Hence, the solubility of C2H5NH2 in water is more than that of (C2H5)2NH in water. Further, the solubility of amines decreases with an increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part, bigger the hydrophobic part, lesser will be the interaction of compound with water, hence less will be the solubility .The molecular mass of C6H5NH2 is greater than that of C2H5NH2 & (C2H5)2NH.
Hence, the increasing order of their solubility in water is as follows:
C6H5NH2<(C2H5)2NH<C2H5NH2
In C2H5NH2, one −C2H5 group is present and in (C2H5)2NH, two −C2H5 groups are present. Therefore, the +I effect of ethyl is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.
Also, both C6H5NHCH3 & C6H5NH2 are less basic than (C2H5)2NH & C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 & C6H5NH2,C6H5NHCH3 will be more basic due to extra +I effect of −CH3 group which is absent in C6H5NH2.
Hence, the order of increasing basicity of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<C2H5NH2<(C2H5)2NH
We know that the higher the basic strength, the lower is the pKb value.
So, decreasing order of the pKb values is:
C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH.
Increasing order of basic strength
In CH3NH2, one −CH3 group is present and in (C2H5)2NH, two −C2H5 groups are present. Therefore, the +I effect of ethyl group is more in (C2H5)2NH than +I effect of methyl group in CH3NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in CH3NH2. Hence, (C2H5)2NH is more basic than CH3NH2.Also, both C6H5N(CH3)2 & C6H5NH2 are less basic than CH3NH2 & (C2H5)2NH due to the delocalization of the lone pair in the former two. Further, among C6H5N(CH3)2 & C6H5NH2, C6H5N(CH3)2 will be more basic due to extra +I effect of −CH3 groups which is absent in C6H5NH2.
Hence, the increasing order of the basic strengths of the given compounds is as follows: C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH
Increasing order of basic strength
In p - toluidine, the presence of electron-donating −CH3 group increases the electron density on the N-atom. Thus, p - toluidine is more basic than aniline.
In the other case, the presence of electron-withdrawing −NO2 group decreases the electron density over the N atom in p - nitroaniline.Thus, p-Nitroaniline is less basic than aniline.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
p−Nitroaniline<Aniline<p–Toluidine
Increasing order of basic strength
C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.
Again, in C6H5NHCH3,−C6H5 group is directly attached to the N - atom. However, it is not so in C6H5CH2NH2.
Thus, in C6H5NHCH3, lone pair of electrons is delocalised on benzene ring which decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.
Hence, the increasing order of the basic strength of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<C6H5CH2NH2
Decreasing order of basic strength of amines in gas phase
In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:
(C2H5)3N>(C2H5)2NH>C2H5NH2>NH3
Increasing order of boiling point
The extent of H−bonding existing in some compounds can decide the boiling point of compounds.
Stronger the H−bonding in the compound, the higher will be the boiling point.
(CH3)2NH has only one H−atom while C2H5NH2 has two H−atoms. Subsequently, C2H5NH2 undergoes more extensive H−bonding than (CH3)2NH. Therefore, the boiling point of C2H5NH2 is more than that of (CH3)2NH.
Also, oxygen is more electronegative than nitrogen. Thus, C2H5OH forms stronger H-bonds than C2H5NH2.
As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 & (CH3)2NH.
Based on the above explanation, the compounds given in the question can be arranged in the increasing order of their boiling points, which is given below:
(CH3)2NH<C2H5NH2<C2H5OH
Increasing order of solubility in water
Stronger the H−bonding with water, more will be the solubility in water. C2H5NH2 contain two H− atoms whereas (C2H5)2NH contains only one H−atom.
Thus, C2H5NH2 undergoes stronger H-bonding than (C2H5)2NH with water. Hence, the solubility of C2H5NH2 in water is more than that of (C2H5)2NH in water. Further, the solubility of amines decreases with an increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part, bigger the hydrophobic part, lesser will be the interaction of compound with water, hence less will be the solubility .The molecular mass of C6H5NH2 is greater than that of C2H5NH2 & (C2H5)2NH.
Hence, the increasing order of their solubility in water is as follows:
C6H5NH2<(C2H5)2NH<C2H5NH2
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