Basicity: Reasoning & Comparison

CHEBasicity Of Amines

Amine basicity is set by how readily the lone pair on nitrogen is donated and how stable the resulting protonated cation is. ISC tests this through comparisons that balance three factors: the $+I$ (electron-releasing) effect of alkyl groups, resonance/conjugation that ties up the lone pair, and solvation of the substituted ammonium ion in water.

Base ionisation equilibrium

RNH_2 + H_2O \rightleftharpoons RNH_3^+ + OH^-

the position of equilibrium (lying further right) measures basic strength

Base dissociation constant

K_b = \dfrac{[RNH_3^+][OH^-]}{[RNH_2]} \qquad pK_b = -\log K_b

larger

K_b

(smaller

pK_b

) means a stronger base

Conjugate-pair relation

pK_a + pK_b = 14

pK_a

is for the conjugate acid

RNH_3^+

298\ K

; a weaker conjugate acid means a stronger base

Basicity correlates with the availability of N's lone pair and the stability of the conjugate acid (the ammonium ion); a more stable cation means a stronger base.
Aliphatic amines are stronger bases than $NH_3$ because the $+I$ effect of alkyl groups increases electron density on N, making the lone pair more available for donation.
Aniline is a weaker base than $NH_3$ : the N lone pair is delocalised into the benzene ring by resonance, so it is less available and the anilinium ion lacks that stabilisation.
Quantitatively: aniline $pK_b \approx 9.4$ , ammonia $pK_b \approx 4.7$ , methylamine $pK_b \approx 3.4$ — smaller $pK_b$ means stronger base.
Gas phase (no solvent) order of methylamines follows $+I$ only: $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ .
In water the order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ because solvation (H-bonding stabilisation of the cation) falls as more alkyl groups crowd the N; $+I$ and solvation oppose each other.
Three opposing factors decide aqueous basicity: $+I$ effect (raises it), steric crowding around N (lowers it for $3^\circ$ ), and solvation/H-bonding of the cation (more N-H bonds give more stabilisation).
Electron-withdrawing groups (e.g. $\text{-}NO_2$ ) weaken aniline's basicity; electron-releasing groups (e.g. $\text{-}CH_3$ ) strengthen it: increasing basic strength $p\text{-nitroaniline} < \text{aniline} < p\text{-toluidine} < NH_3$ .
Among nitroanilines the $ortho$ isomer is weakest (combined $\text{-}I$ plus steric/ortho effect), then $para$ , then $meta$ : order $o < p < m$ in basic strength.
Diphenylamine is far weaker than aniline because the lone pair is delocalised into two rings; two phenyl groups withdraw more electron density than one.
Overall aqueous basicity ranking to remember: $2^\circ$ alkyl amine $>$ $1^\circ$ alkyl amine $>$ $3^\circ$ alkyl amine $>$ $NH_3$ $>$ arylamine $>$ diarylamine $>$ triarylamine.
An amine is a stronger base when it has a higher $K_b$ , a lower $pK_b$ , and (equivalently) a conjugate acid $RNH_3^+$ with a higher $pK_a$ .

Where the marks go

Mixing up $pK_a$ and $pK_b$ : a stronger base has a LOWER $pK_b$ but a HIGHER $pK_a$ for its conjugate acid — quoting the wrong one inverts the answer.
Stating that $3^\circ$ amine is always the strongest base because of maximum $+I$ — that is only true in the gas phase; in water solvation drops it below $1^\circ$ and $2^\circ$ .
Saying aniline's resonance 'donates' the lone pair into the ring to explain weakness — it is delocalisation that makes the lone pair LESS available, not a $+M$ donation that increases basicity.
Getting the nitroaniline order wrong: in basic strength it is $o < p < m$ , but it is easy to flip the $meta$ and $para$ positions — $meta$ is the strongest because $\text{-}NO_2$ cannot withdraw by resonance from the $meta$ position.
Confusing basicity with nucleophilicity, or comparing aromatic with aliphatic amines as if only one factor (the $+I$ effect) operates.

How the board asks it

Give reasonsresonance delocalisation of the N lone pair versus the $+I$ effect of alkyl groups
Account for the fact that aniline is a weaker base than ethylamine, in terms of the availability of the lone pair on nitrogen.
Distinguishaqueous basicity of $1^\circ$ , $2^\circ$ , $3^\circ$ methylamines and $NH_3$
Arrange the following in increasing order of basic strength in aqueous solution and justify your answer: $CH_3NH_2$ , $(CH_3)_2NH$ , $(CH_3)_3N$ , $NH_3$ .
Give reasonselectron-withdrawing $\text{-}NO_2$ versus electron-releasing $\text{-}CH_3$ substituents on aniline
Give reasons: $p$ -nitroaniline is a weaker base than aniline, whereas $p$ -toluidine is a stronger base than aniline.
Distinguishopposing roles of the $+I$ effect and solvation of the ammonium cation
Explain why the basicity of methylamines in the gas phase follows $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ , but in aqueous solution the order becomes $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ .
Assertion–Reasonrelation between $pK_b$ of a base and $pK_a$ of its conjugate acid
Assertion: A stronger base has a lower $pK_b$ value. Reason: A stronger base forms a weaker conjugate acid $RNH_3^+$ , which has a higher $pK_a$ . Choose the correct option regarding these statements.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.