Effect of Ring Substituents on Aniline Basicity (p-Nitro vs p-Toluidine)
Chemistry · Amines · NEET
A group on the aniline ring changes how easily the nitrogen lone pair is available. An electron-donating group like -CH3 (in p-toluidine) pushes electron density toward nitrogen, so basicity goes UP. An electron-withdrawing group like -NO2 (in p-nitroaniline) pulls electron density away, so basicity goes DOWN. Memory hook: "Donors Dial UP, Withdrawers Weaken." Order of basicity: p-nitroaniline < aniline < p-toluidine.
Basicity order p-nitroaniline < aniline < p-toluidine: electron-withdrawing -NO2 lowers it, electron-donating -CH3 raises it relative to plain aniline.
Your doubts, answered
Is p-nitroaniline more basic or less basic than aniline?
Less basic. The -NO2 group is strongly electron-withdrawing (both -I and -R effects). It pulls the nitrogen lone pair even more into the ring, so the lone pair is less free to grab a proton. p-Nitroaniline is one of the weakest common anilines.
Why is p-toluidine (p-methylaniline) more basic than aniline?
The -CH3 group is electron-donating (+I effect). It pushes electron density toward the ring and toward nitrogen, making the lone pair more available to accept a proton. So p-toluidine is more basic than plain aniline.
What is the correct order of basic strength?
p-Nitroaniline < Aniline < p-Toluidine. Withdrawing group weakens (below aniline), donating group strengthens (above aniline). This exact order was asked in NEET 2017.
Does the position (ortho, meta, para) of the substituent matter?
Yes for exact values, but for NEET the para position is the clean case. At para, a -R (resonance) group like -NO2 shows its full electron-withdrawing power through the ring. Ortho groups add steric and hydrogen-bonding complications, so NEET usually sticks to para comparisons.
Why does an electron-withdrawing group lower basicity but a donating group raise it?
Basicity depends on how freely the nitrogen lone pair can bond to H+. A withdrawing group ties up that lone pair (pulls it away), so less is available - lower basicity. A donating group frees up the lone pair (adds density), so more is available - higher basicity.
⚠️ The NEET trap ✗ p-Nitroaniline is more basic than aniline because it has an extra polar -NO2 group. ✓ p-Nitroaniline is LESS basic than aniline. The -NO2 group withdraws electrons and locks up the nitrogen lone pair, so it is harder to protonate. 🧠 'Polar' is not 'basic'. Ask only ONE thing: does the group ADD or REMOVE electron density from nitrogen? -NO2 removes, so it weakens.
Real NEET questions
2017
The correct increasing order of basic strength for the following compounds is: (I) Aniline, C6H5NH2 (II) p-Nitroaniline, NH2-C6H4-NO2 (III) p-Toluidine, NH2-C6H4-CH3
A · II < III < I
B · III < I < II
C · III < II < I
D · II < I < III ✓
Solution: The para substituent changes how available the nitrogen lone pair is. In p-nitroaniline (II) the -NO2 group is strongly electron-withdrawing (-I and -R), so it lowers basicity below aniline. In p-toluidine (III) the -CH3 group is electron-donating (+I), so it raises basicity above aniline. Aniline (I) sits in the middle. Increasing order: p-nitroaniline (II) < aniline (I) < p-toluidine (III), which is option (D) II < I < III.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
p-Toluidine is 4-methylaniline: an aniline with a -CH3 group at the para position. The methyl group donates electrons, making it more basic than aniline.
What is p-nitroaniline?
p-Nitroaniline is 4-nitroaniline: an aniline with a -NO2 group at the para position. The nitro group withdraws electrons, making it less basic than aniline.
Which is the strongest base: aniline, p-nitroaniline, or p-toluidine?
p-Toluidine is the strongest of the three because its electron-donating -CH3 group frees up the nitrogen lone pair the most.
How do I decide quickly in an exam?
Look at the group on the ring. Electron-donating (-CH3, -OCH3, -NH2, -OH) raises basicity above aniline. Electron-withdrawing (-NO2, -CN, -X, -COOH) lowers it below aniline.
Why is this concept important for NEET?
Basicity ordering of substituted anilines is a repeated NEET question type (2016, 2017). One clear rule - donors up, withdrawers down - lets you rank any given set quickly.
