What is the correct order of pK_b values of the following amines?

Benzenamine > Methanamine > Ethanamine

The correct answer is option 3. Benzenamine > Methanamine > Ethanamine.

To determine the correct order of \( pK_b \) values for ethanamine, methanamine, and benzenamine, let's consider the factors affecting their basicity:

Methanamine (Methylamine, \(CH_3NH_2\)):

Methylamine has a single methyl group attached to the nitrogen. The methyl group is an electron-donating group, which increases the electron density on the nitrogen, making methylamine more basic. Thus, methylamine has a relatively low \( pK_b \), indicating higher basicity.

Ethanamine (Ethylamine, \(C_2H_5NH_2\)):

Ethylamine has an ethyl group attached to the nitrogen. The ethyl group is also an electron-donating group, similar to the methyl group but slightly stronger due to the additional carbon. Therefore, ethylamine is slightly more basic than methylamine, giving it a lower \( pK_b \) value.

Benzenamine (Aniline, \(C_6H_5NH_2\)):

Aniline has a phenyl group attached to the nitrogen. The phenyl group has a resonance effect that delocalizes the lone pair of electrons on the nitrogen into the aromatic ring, significantly decreasing the electron density on the nitrogen. This makes aniline much less basic than alkylamines, resulting in a higher \( pK_b \) value.

Relationship of \( pK_b \) Values:

Lower \( pK_b \) value corresponds to higher basicity. Ethylamine \((C_2H_5NH_2)\) is more basic than methylamine \((CH_3NH_2)\), which is more basic than aniline \((C_6H_5NH_2)\).

Thus, the correct order of \( pK_b \) values is:
\[ \text{Aniline (Benzenamine)} > \text{Methylamine (Methanamine)} > \text{Ethylamine (Ethanamine)} \]

This order indicates that benzenamine has the highest \( pK_b \) value (least basic), followed by methanamine, and then ethanamine with the lowest \( pK_b \) value (most basic).

Therefore, the correct answer is Benzenamine > Methanamine > Ethanamine.