Chemistry Class 12 NCERT Solutions: Chapter 13 Amines Part 6

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Q: 4. Arrange the following:

(i) In decreasing order of the values:

(ii) In increasing order of basic strength:

(iii) In increasing order of basic strength:

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

(b)

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

(v) In increasing order of boiling point:

(vi) In increasing order of solubility in water:

,

Answer:

(i) In , only one group is present while in , two groups are present. Thus, the effect is more in than in . Therefore, the electron density over the N-atom is more in than in . Hence, is more basic than .

Also, both are less basic than and due to the delocalization of the lone pair in the former two. Further, among and , the former will be more basic due to the effect of group. Hence, the order of increasing basicity of the given compounds is as follows:

We know that the higher the basic strength, the lower is the values.

(ii) is more basic than due to the presence of the effect of two groups in . Further, contains one group while contains two groups. Thus, is more basic than .

Now, is less basic than because of the−R effect of group. Hence, the increasing order of the basic strengths of the given compounds is as follows:

(iii)

(a)

Q 4 iii p-Toluidine and Aniline, p-Nitroaniline

Q 4 Iii P-Toluidine and Aniline, P-Nitroaniline

Q 4 iii p-Toluidine and Aniline, p-Nitroaniline

In p-toluidine, the presence of electron-donating 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

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) is more basic than due to the presence of electron-donating group in .

Again, in , group is directly attached to the N/atom. However, it is not so in . Thus, in , the effect of group decreases the electron density over the N-atom. Therefore, is more basic than . Hence, the increasing order of the basic strengths of the given compounds is as follows: .

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the effect. The higher the effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:

(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. contains only one H−atom whereas contains two H-atoms. Then, undergoes more extensive H-bonding than . Hence, the boiling point of is higher than that of .

Further, O is more electronegative than N. Thus, forms stronger bonds than . As a result, the boiling point of is higher than that of and .

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

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

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 is greater than that of and .

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