Basicity of amines

Dissolve ammonia gas in water and the solution turns litmus blue — a lone pair on nitrogen has grabbed a proton. Amines, built by replacing one or more hydrogens of $NH_3$ with alkyl or aryl groups, do the same trick, and how eagerly they do it is one of the most exam-tested comparisons in ISC organic chemistry. Basicity is not a fixed label attached to "amine" as a class — it is a tug-of-war between how much electron density piles up on nitrogen and how easily the resulting ammonium ion can be stabilised in water. Three effects run this contest: the electron-donating push of alkyl groups, the steric crowding that blocks solvation of the ammonium ion, and — for aniline — the drain of the lone pair into the benzene ring by resonance. This article compresses those three effects into a single ranking rule, shows how a measured $pK_b$ converts a "which is more basic" question into an actual number, and works a substituted-aniline example end to end.

The core insight

Amine basicity is set by how available the nitrogen lone pair is for protonation, which alkyl groups increase (+I push) but bulky alkyl groups and ring resonance (as in aniline) decrease, giving the aqueous-phase order 2° > 1° > 3° > $NH_3$ > aniline, quantified by $K_b$ or $pK_b$ .

The core idea

The proton-transfer equilibrium. Every amine's basicity is measured by the same reaction, an amine pulling a proton from water:

R\text{-}NH_2 + H_2O \rightleftharpoons R\text{-}NH_3^+ + OH^-

where: the equilibrium constant for this reaction is $K_b$ , the base dissociation constant; a larger $K_b$ (equivalently, a smaller $pK_b = -\log_{10}K_b$ ) means the amine holds onto protons from water more readily and is the stronger base. Ranking amines by basicity is exactly ranking them by $K_b$ (or, in reverse, by $pK_b$ ).

Three factors set the lone pair's availability. Inductive push: alkyl groups are weakly electron-donating (+I), so each one bonded to nitrogen pushes extra electron density onto the lone pair, making it more willing to bond to $H^+$ — alkylamines are therefore more basic than $NH_3$ itself. Steric hindrance and solvation: the ammonium ion $R\text{-}NH_3^+$ is stabilised in water by hydrogen bonding, and bulky alkyl groups crowd around nitrogen and block that solvation shell, which weakens the effective basicity — this penalty grows with each extra alkyl group. Resonance in aniline: in $C_6H_5NH_2$ , the nitrogen lone pair delocalises into the aromatic ring (spread over several resonance structures placing negative charge on ring carbons), so it is far less available to accept a proton, making aniline much weaker than ammonia or any alkylamine. Balancing the +I push against the solvation penalty gives the aqueous-phase order for simple methylamines that ISC exams test directly:

(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3 > C_6H_5NH_2

Ring substituents further tune aniline: electron-withdrawing groups ( $-NO_2$ , $-Cl$ ) pull still more density off the already-delocalised lone pair and make aniline an even weaker base; electron-donating groups ( $-CH_3$ , $-OCH_3$ ) partly compensate and raise its basicity back up.

Figure

Bar chart comparing pKb values of dimethylamine methylamine ammonia and aniline showing basicity falling from dimethylamine to aniline

Derivation AFrom a difference in pKb to how many times stronger a base is

The goal is to turn a table of measured $pK_b$ values into an actual multiplying factor for base strength, so that "aniline is a much weaker base than dimethylamine" becomes a specific number rather than a vague comparison — using nothing beyond the definition $pK_b = -\log_{10}K_b$ and one rule of logarithms.

Step 1. Write $K_b$ in terms of $pK_b$ for each amine. By definition $pK_b = -\log_{10}K_b$ , so inverting gives $K_b = 10^{-pK_b}$ . Using literature values $pK_b\big[(CH_3)_2NH\big] = 3.27$ and $pK_b\big[C_6H_5NH_2\big] = 9.38$ , the two base constants are $K_b\big[(CH_3)_2NH\big] = 10^{-3.27}$ and $K_b\big[C_6H_5NH_2\big] = 10^{-9.38}$ .

Step 2. Form the ratio of the two $K_b$ values, since that ratio is exactly "how many times stronger" one base is than the other. $\dfrac{K_b\big[(CH_3)_2NH\big]}{K_b\big[C_6H_5NH_2\big]} = \dfrac{10^{-3.27}}{10^{-9.38}} = 10^{(-3.27) - (-9.38)} = 10^{6.11}$ .

Step 3. Evaluate the power of ten. $10^{6.11} = 10^{6}\times10^{0.11} \approx 1.0\times10^{6}\times1.29 \approx 1.3\times10^{6}$ .

Step 4. Read the physical meaning. Dimethylamine's $K_b$ is about $1.3\times10^{6}$ times larger than aniline's, so dimethylamine is roughly 1.3 million times the stronger base. This is the general rule worth remembering: because $pK_b$ is a logarithmic scale, every one unit of $pK_b$ difference is a factor of ten in $K_b$ , so a $pK_b$ gap of about $6$ units (as here) corresponds to a million-fold difference in base strength — the qualitative "aniline is much weaker" statement, made numeric.

Worked example

Problem: p-Nitroaniline has $pK_b \approx 13.0$ , aniline has $pK_b \approx 9.38$ , and p-toluidine (p-methylaniline) has $pK_b \approx 8.92$ . Rank the three bases and state how many times stronger p-toluidine's $K_b$ is compared with p-nitroaniline's.

Step 1. Identify the electronic effects first, since $pK_b$ ordering must match them as a sanity check. The $-NO_2$ group in p-nitroaniline is strongly electron-withdrawing (by both induction and resonance), pulling still more density off the already-delocalised nitrogen lone pair, so p-nitroaniline should be the weakest base of the three (highest $pK_b$ ). The $-CH_3$ group in p-toluidine is electron-donating, partially compensating for the ring delocalisation, so p-toluidine should be the strongest base of the three (lowest $pK_b$ ).

Step 2. Rank by $pK_b$ — smaller $pK_b$ means stronger base. Reading the given values, $8.92 < 9.38 < 13.0$ , so the basicity order is p-toluidine greater than aniline greater than p-nitroaniline, matching the electronic reasoning from Step 1.

Step 3. Convert the $pK_b$ gap between the two extremes into a $K_b$ ratio, exactly as in the derivation above. $pK_b$ difference $= 13.0 - 8.92 = 4.08$ , so $\dfrac{K_b(\text{p-toluidine})}{K_b(\text{p-nitroaniline})} = 10^{4.08} \approx 1.2\times10^{4}$ .

Step 4. State the final answer. Basicity order: p-toluidine greater than aniline greater than p-nitroaniline, and p-toluidine's $K_b$ is about $1.2\times10^{4}$ (twelve thousand) times larger than p-nitroaniline's — the ring substituent alone swings base strength by four orders of magnitude.

Common traps

Assuming the gas-phase order (3° > 2° > 1° > $NH_3$ , from +I alone) is what ISC wants — in aqueous solution, hindered solvation of the bulkier ammonium ion reverses part of this, giving 2° > 1° > 3° > $NH_3$ for small alkylamines.
Attributing aniline's weak basicity to an inductive effect — it is resonance delocalisation of the lone pair into the ring, not induction, that is responsible.
Forgetting that a smaller $pK_b$ means a stronger base — it is easy to reverse the inequality when converting between $K_b$ and $pK_b$ .
Treating every ring substituent the same way — electron-withdrawing groups ( $-NO_2$ , $-Cl$ ) always decrease aniline's basicity, electron-donating groups ( $-CH_3$ , $-OCH_3$ ) always increase it, and the position on the ring matters for how strongly a resonance donor or acceptor communicates with nitrogen.

Ask the tutor: Ask the AI tutor to walk through the resonance structures of aniline's lone-pair delocalisation, or to rank a fresh mixed set of substituted anilines by predicted $pK_b$ .

Why does adding more alkyl groups not just keep making an amine more and more basic?

Because two competing effects are at work, not one. Each alkyl group does push extra electron density onto nitrogen through the +I effect, which on its own would make trialkylamines the strongest bases. But in water, the resulting ammonium ion needs to be stabilised by hydrogen bonding to solvent, and a crowded tertiary nitrogen with three bulky groups blocks that stabilisation more than a primary amine does. For small alkyl groups the solvation penalty wins out, so the aqueous order ends up 2° > 1° > 3° > $NH_3$ rather than a simple 3° > 2° > 1° > $NH_3$ .

Is basicity the same thing as nucleophilicity for amines?

No, though they often move together. Basicity is an equilibrium property — how favourably an amine's lone pair accepts a proton at equilibrium, measured by $K_b$ . Nucleophilicity is a kinetic property — how fast that lone pair attacks an electrophilic carbon in a reaction like $S_N2$ substitution. Steric bulk can lower nucleophilicity (a crowded nitrogen approaches a carbon atom less easily) without lowering basicity nearly as much (a proton is tiny and far less sensitive to crowding), so the two properties can disagree for bulky amines even though they usually rank the same way for simple ones.

Does every electron-withdrawing group on aniline's ring reduce its basicity by the same amount?

No — both the strength and the mechanism of the substituent matter, along with its position. A strong resonance-withdrawing group like $-NO_2$ at the para or ortho position can conjugate directly with the ring and pull far more density off the nitrogen lone pair than a weaker or purely inductive withdrawing group at the meta position, where resonance communication with nitrogen is cut off. That is why p-nitroaniline ( $pK_b \approx 13$ ) is dramatically weaker than aniline itself ( $pK_b \approx 9.38$ ), while a meta-substituted analogue would show a much smaller shift for a similarly electron-withdrawing group.

Basicity of amines

The core idea

Worked example

Common traps

Before this

Where this leads