Question
Compare the basic strength of aniline and cyclohexylamine. Explain why one is a stronger base than the other, even though both have an —NH₂ group attached to a ring structure.
Solution — Step by Step
An amine acts as a base by donating the lone pair on nitrogen to a proton (H⁺). So the availability of that lone pair is everything — the more freely available it is, the stronger the base.
In cyclohexylamine, the —NH₂ is attached to a sp³ carbon (saturated ring). The lone pair on nitrogen is purely localised on N — no resonance, no conjugation. It’s fully available to attack H⁺.
pKb ≈ 3.34 — a reasonably strong amine base.
In aniline, the —NH₂ is attached directly to a benzene ring (sp² carbons). The lone pair on nitrogen overlaps with the π-system of the ring. This is called conjugation or resonance delocalisation.
The lone pair gets pulled into the ring, making it far less available to donate to H⁺. Aniline is a much weaker base.
pKb ≈ 9.40 — orders of magnitude weaker than cyclohexylamine.
We can draw four resonance contributors for aniline:
In these structures, nitrogen carries a positive charge — meaning it has already “donated” its lone pair into the ring. This is why it cannot easily donate to H⁺.
Cyclohexylamine is a stronger base than aniline.
The lone pair in cyclohexylamine is localised on N and freely available. In aniline, the lone pair is delocalised into the benzene ring via resonance, reducing its basicity dramatically.
Why This Works
The key concept here is resonance vs. localisation of the lone pair. Whenever a nitrogen’s lone pair is conjugated with a π-system (aromatic ring, carbonyl, etc.), basicity drops. This is because donation into the ring and donation to H⁺ are competing — and the ring usually wins due to the stability gained from delocalisation.
Cyclohexylamine has no such competition. The sp³ ring is inert to the lone pair. Nitrogen just sits there with full electron density, ready to grab a proton.
A quick heuristic: sp³ N > sp² N > sp N in basicity. Amines attached to saturated carbons are stronger bases than those attached to aromatic rings, which in turn are stronger than amides (where lone pair is next to C=O). This trend appears repeatedly in JEE and NEET MCQs.
Alternative Method — Using pKb Values
Instead of the lone-pair argument, we can use the pKb comparison directly:
| Amine | pKb | Conclusion |
|---|---|---|
| Cyclohexylamine | 3.34 | Stronger base |
| Aniline | 9.40 | Weaker base |
Lower pKb = stronger base (just like lower pKa = stronger acid). The difference is ~6 units, meaning cyclohexylamine is roughly 10⁶ times more basic than aniline. That’s not a small difference — it’s a consequence of full resonance delocalisation in aniline.
This numerical approach works well when the question gives you pKb data directly, as CBSE sometimes does in 2-mark questions.
Common Mistake
Many students write: “Aniline is weaker because the benzene ring withdraws electrons by induction.” This is incorrect reasoning. Benzene is not strongly electron-withdrawing by induction — the correct reason is resonance delocalisation of the lone pair into the π-system. Induction and resonance are different effects; here resonance dominates completely. Writing “induction” in a board exam will likely cost you the explanation mark.