Question

Arrange the following amines in the decreasing order of their basic strength : Aniline (I), Benzylamine (II), p-toluidine (III)

• A. I>II>III
• B. III>II>I
• C. II>I>III
• D. III>I>II
• E. II>III>I

Hint:

If Nitrogen is directly attached to the benzene ring, its basicity drops significantly. If there is a "spacer" carbon (like in Benzylamine), it is much more basic. For aromatic amines, \textbf{EDGs} (Electron Donating Groups) increase basicity, and \textbf{EWGs} (Electron Withdrawing Groups) decrease it.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The basic strength of an amine depends on the availability of the lone pair of electrons on the Nitrogen atom for donation to a proton ($H^+$). Factors that increase electron density on Nitrogen (like $+I$ or $+M$ effects) increase basicity, while factors that decrease it (like resonance or $-I$ effects) decrease basicity.
Step 2: Detailed Explanation:
1. Benzylamine (II): In $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$, the lone pair on Nitrogen is localized (it is not in resonance with the benzene ring) because of the $-\text{CH}_2-$ group. It behaves like an aliphatic amine, making it the strongest base among the three.
2. Aniline (I): In $\text{C}_6\text{H}_5\text{NH}_2$, the lone pair on Nitrogen is delocalized into the benzene ring through resonance. This significantly reduces the electron density on Nitrogen, making it a very weak base.
3. p-toluidine (III): This is aniline with a methyl group ($-\text{CH}_3$) at the para position. The $-\text{CH}_3$ group has a $+I$ effect and hyperconjugation, which pushes electron density toward the ring and Nitrogen. While the lone pair is still delocalized (like aniline), it is more available than in pure aniline.
Step 3: Comparing the three:
- II (Localized lone pair) is strongest.
- III (Delocalized lone pair + electron-donating group) is stronger than Aniline.
- I (Delocalized lone pair) is the weakest.
Order: II>III>I.
Step 4: Final Answer:
The decreasing order of basic strength is II>III>I.