Question:
Which of the following is less basic than aniline?
Which of the following is less basic than aniline?
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An electron withdrawing group on the ring lowers the availability of the nitrogen lone pair.
Updated On: Nov 12, 2026
- p-nitroaniline
- \(\text{NH}_{3}\)
- \(\text{CH}_{3}\text{NH}_{2}\)
- N-methyl aniline
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Basicity depends on the availability of the lone pair on nitrogen.
In aniline the lone pair is partly delocalised into the benzene ring, so it is a weak base.
Step 2: Compare with the others
\(\text{NH}_3\) and \(\text{CH}_3\text{NH}_2\) have a localised lone pair (and \(\text{CH}_3\) adds +I effect), so they are stronger bases than aniline.
N-methylaniline is slightly more basic because of the +I effect of the methyl group.
Step 3: The weaker base
In p-nitroaniline, the \(-\text{NO}_2\) group withdraws electrons by -M and -I effects.
This pulls the lone pair further from nitrogen, so p-nitroaniline is less basic than aniline.
Final Answer:
p-Nitroaniline is less basic than aniline. \[ \boxed{\text{(A) }\text{p-nitroaniline}} \]
Basicity depends on the availability of the lone pair on nitrogen.
In aniline the lone pair is partly delocalised into the benzene ring, so it is a weak base.
Step 2: Compare with the others
\(\text{NH}_3\) and \(\text{CH}_3\text{NH}_2\) have a localised lone pair (and \(\text{CH}_3\) adds +I effect), so they are stronger bases than aniline.
N-methylaniline is slightly more basic because of the +I effect of the methyl group.
Step 3: The weaker base
In p-nitroaniline, the \(-\text{NO}_2\) group withdraws electrons by -M and -I effects.
This pulls the lone pair further from nitrogen, so p-nitroaniline is less basic than aniline.
Final Answer:
p-Nitroaniline is less basic than aniline. \[ \boxed{\text{(A) }\text{p-nitroaniline}} \]
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