Question

Arrange the following:

(i) In decreasing order of the pK_bvalues:

C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

(ii) In increasing order of basic strength:

C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂

(iii) In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine

(b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂.

(iv) In decreasing order of basic strength in gas phase:

C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃

(v) In increasing order of boiling point:

C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂

(vi) In increasing order of solubility in water:

C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂.

Answer 1

(i) In C₂H₅NH₂, only one −C₂H₅ group is present while in (C₂H₅)₂NH, two −C₂H₅ groups are present. Thus, the +I effect 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 and C₆H₅NH₂ are less basic than (C₂H₅)₂NH and C₂H₅NH₂ due to the delocalization of the lone pair in the former two. Further, among C₆H₅NHCH₃ and C₆H₅NH₂, the former will be more basic due to the +T effect of −CH₃ group. 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 pK_b values.

C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

(ii) C₆H₅N(CH₃)₂ is more basic than C₆H₅NH₂ due to the presence of the +I effect of two −CH₃ groups in C₆H₅N(CH₃)₂. Further, CH₃NH₂ contains one −CH₃ group while (C₂H₅)₂NH contains two −C₂H₅ groups. Thus, (C₂H₅)₂ NH is more basic than C₂H₅NH₂.

Now, C₆H₅N(CH₃)₂ is less basic than CH₃NH₂ because of the−R effect of −C₆H₅ group.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

(iii) (a)

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.

On the other hand, 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

(b) 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, the −R effect of −C₆H₅ group decreases the electron density over the N-atom. Therefore, C₆H₅CH₂NH₂ is more basic than C₆H₅NHCH₃.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂.

(iv) 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

(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH₃)₂NH contains only one H−atom whereas C₂H₅NH₂ contains two H-atoms. Then, C₂H₅NH₂ undergoes more extensive H-bonding than (CH₃)₂NH. Hence, the boiling point of C₂H₅NH₂ is higher than that of (CH₃)₂NH.

Further, O is more electronegative than N. Thus, C₂H₅OH forms stronger H−bonds than C₂H₅NH₂. As a result, the boiling point of C₂H₅OH is higher than that of C₂H₅NH₂ and (CH₃)₂NH.

Now, the given compounds can be arranged in the increasing order of their boiling points as follows:

(CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(vi) The more extensive the H−bonding, the higher is the solubility. C₂H₅NH₂ contains two H-atoms whereas (C₂H₅)₂NH contains only one H-atom. Thus, C₂H₅NH₂ undergoes more extensive H−bonding than (C₂H₅)₂NH. Hence, the solubility in water of C₂H₅NH₂ is more than that of (C₂H₅)₂NH.

Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part. The molecular mass of C₆H₅NH₂ is greater than that of C₂H₅NH₂ and (C₂H₅)₂NH.

Hence, the increasing order of their solubility in water is as follows:

C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂