Q.Which of the following compounds is the weakest Brönsted base?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Amine Basicity Order
Amine Basicity Order – From Intuition to Precision
You already know that a base is something that accepts a proton (H+). An amine does this through the lone pair on its nitrogen. The more readily that lone pair grabs a proton, the stronger the base. So the question becomes: what makes that lone pair more or less willing to accept a proton?
The answer depends on three competing effects: inductive effect (electron pushing/pulling from nearby groups), steric hindrance (bulky groups blocking the proton), and solvation (how water molecules stabilise the protonated amine). And critically, the order flips depending on whether you are in the gas phase or in water.
The Intuition – What Would You Expect?
Imagine a nitrogen with three hydrogens: ammonia (NH3). Now replace one hydrogen with a methyl group (CH3). Methyl is an electron-donating group — it pushes electron density toward the nitrogen. That makes the lone pair richer, so it should grab a proton more easily. So methylamine (CH3NH2) should be a stronger base than ammonia.
Replace another hydrogen with a second methyl group. Now you have dimethylamine ((CH3)2NH). Even more electron density on nitrogen — stronger base still. Replace the third hydrogen: trimethylamine ((CH3)3N). Maximum electron donation — so you would expect it to be the strongest base of all.
That is the inductive effect prediction: more alkyl groups → stronger base. In the gas phase, this is exactly what happens. The order is:
Gas phase: NH3<1∘<2∘<3∘
But in water, the observed order is different. Why?
The Complication – Solvation and Steric Hindrance
When an amine accepts a proton, it becomes a positively charged ammonium ion (RNH3+). In water, that positive charge is stabilised by hydrogen bonding with water molecules. The more hydrogens on the nitrogen, the more hydrogen bonds it can form — and the more stable the protonated form becomes.
A primary amine (1∘) has three N–H hydrogens after protonation. A secondary (2∘) has two. A tertiary (3∘) has only one. So solvation stabilises the protonated form in the order: 1∘>2∘>3∘.
At the same time, bulky alkyl groups physically block the approach of a proton to the nitrogen — this is steric hindrance. A tertiary amine has three bulky groups crowding the nitrogen, making it harder for H+ to reach the lone pair.
So in water, two effects oppose the inductive effect: solvation (which favours more N–H bonds) and steric hindrance (which favours less crowding). The result is a compromise.
The Precise Statement – Aqueous Phase Order
In water, the typical order of basic strength for aliphatic amines is:
Aqueous phase: NH3<3∘<1∘<2∘
That is, secondary amines are the strongest bases in water, followed by primary, then tertiary, then ammonia.
The order in water is not simply "more alkyl groups = stronger base". The secondary amine wins because it has a good balance: enough electron donation from two alkyl groups, but still two N–H hydrogens for solvation, and not too much steric hindrance.
The Full Picture – A Table
| Amine | Gas-phase order (inductive only) | Aqueous order (all effects) | Reason |
|---|---|---|---|
| NH3 | Weakest | Weakest | No alkyl donation; only 3 H-bonds |
| 1∘ (RNH2) | Second | Second | One alkyl group donates; 3 H-bonds after protonation |
| 2∘ (R2NH) | Third | Strongest | Two alkyl groups donate; 2 H-bonds; steric hindrance still low |
| 3∘ (R3N) | Strongest | Third | Three alkyl groups donate most, but only 1 H-bond and high steric hindrance |
Amines are far more basic than alcohols/phenols. Among the two oxygen bases, phenol's oxygen lone pair is delocalised into the aromatic ring, so it is the least available — phenol is the weakest base. …
Basicity depends on how readily the heteroatom lone pair accepts a proton. Nitrogen bases (the amines) beat oxygen bases (the alcohol and phenol), and among the oxygen bases phenol is weakest because its lone pair is tied up in the ring — phenol is the weakest Brönsted base.
Concept
A Brönsted base accepts H+ using a lone pair. Nitrogen is less electronegative than oxygen, so it holds its lone pair more loosely and donates it more readily → amines are stronger bases than alcohols and phenols.
Ranking the four
- (ii) cyclohexylamine and (i) aniline are amines → relatively good bases (cyclohexylamine > aniline, but both far more basic than the O compounds). …
Method: Nitrogen Bases vs. Oxygen Bases, and Ranking Within Each
Core Concept
Nitrogen is less electronegative than oxygen and so donates its lone pair more readily, making amines stronger Bronsted bases than alcohols or phenols in general; within the oxygen bases, a lone pair delocalised into an aromatic ring (as in phenol) is far less available than one fully localised on a saturated carbon (as in an alcohol).
Steps
- Sort the candidates into nitrogen bases (amines) and oxygen bases (alcohol, phenol).
- Note that N-bases are inherently more basic than O-bases of comparable structure, since N holds its lone pair more loosely.
- Among the oxygen bases, check for resonance delocalisation of the lone pair: an aromatic -OH (phenol) conjugates its oxygen lone pair into the ring pi-system, while a saturated -OH (alcohol) keeps its lone pair fully localised.
- Conclude that the resonance-delocalised oxygen base (phenol) is weaker than the localised one (alcohol), and both O-bases are weaker than either N-base.
- Select the overall weakest of all four as the answer. …
- Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 markQ.Why is acetanilide (C6H5NHCOCH3) less basic than aniline?
›Reveal solutionSolution
In acetanilide the nitrogen's lone pair is tied up in resonance with the adjacent C=O group (amide resonance) in addition to the ring, so it is even less available for protonation than in aniline.
In aniline (C6H5-NH2), the lone pair on nitrogen is already partly delocalised into the benzene ring by resonance, which reduces (but does not eliminate) its basicity compared to an aliphatic amine. In acetanilide (C6H5-NH-CO-CH3), the nitrogen is attached to both the ring AND a carbonyl group; the nitrogen lone pair delocalises strongly into the C=O group (amide resonance, N-C(=O) <-> N+=C-O-), which is an even stronger, more effective delocalisation pathway than simple ring conjugation. This makes the nitrog …
- Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQQ.In benzene solution, what is the correct basicity order of the following amines?(a) CH3NH2 > (CH3)3N > (CH3)2NH(b) (CH3)3N > (CH3)2NH > CH3NH2(c) CH3NH2 > (CH3)2NH > (CH3)3N(d) (CH3)3N > CH3NH2 > (CH3)2NH
›Reveal solutionSolution
In benzene (an aprotic, non-hydrogen-bonding solvent), basicity of methylamines rises with the number of electron-donating methyl groups on nitrogen, giving (CH3)3N as the strongest base.
Step 1. In aqueous solution the observed basicity order of methylamines is (CH3)2NH > CH3NH2 > (CH3)3N because solvation of the protonated ammonium ion (by hydrogen bonding with water) matters enormously, and a bulky trimethylammonium ion is poorly solvated.
…
- Higher Secondary (+2 Stage) Examination 2024Set ANNUAL1 markQ.Between CH3CH2NH2 and CH3CONH2, which is more basic and why?
›Reveal solutionSolution
Basicity depends on how freely available the nitrogen lone pair is for accepting a proton. In an amide, that lone pair is tied up in resonance with the carbonyl group, making it far less basic than a simple amine.
In CH3CH2NH2, the N lone pair is fully localised on nitrogen and is further pushed towards N by the +I (electron-donating) effect of the ethyl group, making it very available for protonation — a good base.
In CH3CONH2, the nitrogen lone pair is delocalised into the adjacent C=O group by resonance:
CH3–C(=O)–NH2 ↔ CH3–C(–O–)=NH2+ …
- Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 markQ.Arrange the following amines in increasing order of basicity in aqueous solution: (CH3)3N, NH3, CH3NH2, (CH3)2NH
›Reveal solutionSolution
In aqueous solution, basicity of methylamines follows the order 2 degree > 1 degree > 3 degree > NH3, due to the combined effects of +I (electron donation), steric hindrance, and hydration/solvation of the resulting ammonium cation.
Basicity in water depends on THREE factors acting together: (i) the +I (electron-donating) effect of alkyl groups (increases basicity, more available lone pair on N), (ii) steric hindrance around N (bulky alkyl groups hinder solvation of the protonated ammonium ion, reducing effective basicity), and (iii) the extent of solvation/hydrogen-bonding stabilization of the ammonium cation formed after protonation (fewer alkyl groups on N means more N-H bonds available to hydrogen-bond with water, better stabilizing the cation).
For methyl-substituted amines in water, the interplay of these factors gives the experimentally observed order:
(CH3)2NH (2 degree) > CH3NH2 (1 degree) > (CH3)3N (3 degree) > NH3
…
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