Question

Write the following in order:
(i) Decreasing order of \( pK_b \) value: \( \text{C}_2\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{NHCH}_3, \text{C}_2\text{H}_5\text{NH}_2 \).
(ii) Decreasing order of basic strength: \( \text{C}_6\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{N(CH}_3\text{)}_2, \text{C}_6\text{H}_5\text{NH}_2 \).
(iii) Increasing order of basic strength: Aniline; p-nitroaniline; and paratoludine.
(iv) Solubility order in water: \( \text{C}_6\text{H}_5\text{NH}_2, (\text{C}_2\text{H}_5)_2\text{NH}, \text{C}_2\text{H}_5\text{NH}_2 \).
(v) Increasing order of boiling point: \( \text{C}_2\text{H}_5\text{OH}, \text{CH}_3\text{NH}_2, \text{C}_2\text{H}_5\text{NH}_2 \).

Hint:

For the comparison of basic strengths, remember that electron-donating groups (like methyl) increase basicity, while electron-withdrawing groups (like nitro and phenyl) decrease basicity.

Answer 1

(i) Decreasing order of \( pK_b \) value:
\( pK_b \) is inversely related to the basicity. A higher \( pK_b \) corresponds to a weaker base. The basicity of amines is influenced by the electron-donating ability of the substituents. The order from weakest to strongest base is:
\[ \text{C}_6\text{H}_5\text{NHCH}_3 < \text{C}_2\text{H}_5\text{NH}_2 < \text{C}_6\text{H}_5\text{NH}_2 \] - C\(_6\)H\(_5\)NHCH\(_3\) has the electron-donating methyl group attached to the nitrogen, slightly increasing basicity.
- C\(_2\)H\(_5\)NH\(_2\) has an ethyl group attached, which is more electron-donating than a phenyl group.
- C\(_6\)H\(_5\)NH\(_2\) has a phenyl group attached, which is electron-withdrawing, making it a weaker base.
(ii) Decreasing order of basic strength:
Basic strength is determined by the availability of the lone pair on nitrogen for protonation. The order from strongest to weakest base is:
\[ \text{C}_6\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{N(CH}_3\text{)}_2 > \text{C}_6\text{H}_5\text{NH}_2 \] - C\(_6\)H\(_5\)NH\(_2\) is the strongest base as the lone pair on nitrogen is readily available for protonation.
- C\(_6\)H\(_5\)N(CH\(_3\))\(_2\) has two methyl groups attached, making it more electron-donating, increasing basicity.
- C\(_6\)H\(_5\)NH\(_2\) has an electron-withdrawing phenyl group attached, making it the weakest base.
(iii) Increasing order of basic strength:
In the case of substituted anilines, electron-withdrawing substituents decrease basic strength. The order from weakest to strongest is:
\[ \text{p-nitroaniline} < \text{aniline} < \text{paratoludine} \] - p-Nitroaniline has a nitro group (\(NO_2\)), which is electron-withdrawing, making it the weakest base.
- Aniline has no substituents, so it has a moderate basic strength.
- Paratoludine has a methyl group (\(CH_3\)), which is electron-donating, making it the strongest base.
(iv) Solubility order in water:
Amines with smaller alkyl groups are more soluble in water due to hydrogen bonding. The solubility order is:
\[ \text{C}_6\text{H}_5\text{NH}_2 > (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 \] - C\(_6\)H\(_5\)NH\(_2\) (aniline) is the most soluble in water due to the ability to form hydrogen bonds.
- (\(C_2H_5\))\(_2\)NH (diethylamine) is less soluble due to the increased hydrophobic character of the ethyl groups.
- C\(_2\)H\(_5\)NH\(_2\) (ethylamine) has moderate solubility.
(v) Increasing order of boiling point:
The boiling points of amines increase with the size of the alkyl group and the ability to form hydrogen bonds. The order from lowest to highest boiling point is:
\[ \text{C}_2\text{H}_5\text{OH} < \text{CH}_3\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 \] - C\(_2\)H\(_5\)OH (ethanol) has the lowest boiling point because it is a small molecule and only forms hydrogen bonds weakly.
- CH\(_3\)NH\(_2\) (methylamine) has a higher boiling point due to stronger hydrogen bonding.
- C\(_2\)H\(_5\)NH\(_2\) (ethylamine) has the highest boiling point because it has a larger molecular size and stronger hydrogen bonding.