Question

\(pK_b\) value of aniline is more in comparison to that of methyl amine. State the reason.

Hint:

Substituent effects play a significant role in determining the basicity of compounds. Electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

Answer 1

Step 1: Understand the structure of aniline and methylamine.
- Aniline (C\(_6\)H\(_5\)NH\(_2\)) has a phenyl group (C\(_6\)H\(_5\)) attached to the amino group (NH\(_2\)). - Methylamine (CH\(_3\)NH\(_2\)) has a methyl group (CH\(_3\)) attached to the amino group.
Step 2: Effect of substituents.
- The phenyl group in aniline is electron-withdrawing through induction and resonance, which decreases the electron density on the nitrogen atom. This makes aniline less basic, increasing its \(pK_b\). - The methyl group in methylamine is electron-donating through induction, which increases the electron density on nitrogen, making it more basic and lowering its \(pK_b\).
Step 3: Conclusion.
Since aniline is less basic than methylamine, its \(pK_b\) is higher, meaning aniline has a weaker tendency to accept a proton compared to methylamine. Final Answer: \[ \boxed{\text{Aniline's } pK_b \text{ is higher than methylamine because the phenyl group is electron-withdrawing, reducing the basicity of aniline.}} \]