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:
Amines act as Lewis bases because the nitrogen atom possesses a lone pair of electrons that can be donated to bond with a proton (\(\text{H}^+\)). The basic strength of an amine is directly proportional to the availability of this lone pair. Any factor that increases electron density on the nitrogen makes the amine a stronger base, while factors that withdraw or delocalize that density make it weaker.
Step 2: Key Formula or Approach:
To rank the amines, consider two main structural features:
1. Aromaticity: Is the nitrogen atom's lone pair involved in resonance with an aromatic ring? If yes, availability decreases drastically.
2. Substituent Effects: If it's an aromatic amine, look for substituents on the ring. Electron-Donating Groups (EDG) increase basicity, whereas Electron-Withdrawing Groups (EWG) decrease it.
Step 3: Detailed Explanation:
Let's evaluate the structural features of each given amine:
- Aniline (I) - \(\text{C}_6\text{H}_5\text{NH}_2\):
In aniline, the amino group is attached directly to the benzene ring. The lone pair of electrons on the nitrogen atom is conjugated with the \(\pi\)-electron system of the ring. Because these electrons are delocalized into the ring, they are less available to be shared with an incoming proton. This resonance effect makes aniline a notably weak base.
- Benzylamine (II) - \(\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2\):
Here, the amino group is not attached directly to the aromatic ring; it is separated by an \(sp^3\)-hybridized methylene (\(-\text{CH}_2-\)) group. Because of this separation, the lone pair on the nitrogen cannot participate in resonance with the benzene ring. Benzylamine functions as a typical primary aliphatic amine. Since its lone pair is localized and fully available, it is a much stronger base than typical aromatic amines.
- p-toluidine (III) - \(p\text{-CH}_3\text{-C}_6\text{H}_4\text{NH}_2\):
This molecule is an aniline derivative featuring a methyl group (\(-\text{CH}_3\)) at the para position relative to the amino group. The lone pair is still delocalized into the ring as in aniline. However, the methyl group acts as an Electron-Donating Group through hyperconjugation and a weak +I inductive effect. This added electron density partially offsets the resonance withdrawal, increasing the electron density at the nitrogen atom relative to unsubstituted aniline. Therefore, p-toluidine is a stronger base than aniline.
Establishing the Order:
- Benzylamine (II) is an aliphatic amine, making it the strongest base among the three.
- p-toluidine (III) and Aniline (I) are both aromatic amines, which are weaker than aliphatic ones.
- Between the two aromatic amines, p-toluidine (III) has an activating EDG, making it stronger than the unsubstituted Aniline (I).
Thus, the decreasing order of basic strength is: Benzylamine (II) > p-toluidine (III) > Aniline (I).
In Roman numerals, this is II > III > I.
Step 4: Final Answer:
The correct decreasing order of basic strength is II > III > I, which corresponds to option (e).