Question

Three compounds are given below:
 
The correct decreasing order of their basic strength is:

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

Hint:

Electron-donating groups increase basicity in aniline; electron-withdrawing groups decrease it.

Answer 1

The Correct Option is A

Step 1: Basic strength of aniline depends on the availability of the lone pair on the nitrogen atom.
Step 2: Electron-donating groups (\(+I\) or \(+M\)) increase basic strength by increasing electron density on nitrogen.
Electron-withdrawing groups (\(-I\) or \(-M\)) decrease basic strength by pulling electron density away.
Step 3: In compound I, the \(-NO_2\) group is a strong electron-withdrawing group, so it decreases the basic strength significantly.
In compound II, the \(CH_3\) group is an electron-donating group, increasing the basic strength relative to unsubstituted aniline (III).
Compound III is aniline with no substituents.
Step 4: Therefore, the correct decreasing order is:
II (methyl aniline)>III (aniline)>I (nitroaniline)