The trap: You memorise that tertiary amines have more alkyl groups donating electrons, so they must be the strongest bases. Then you pick trimethylamine as the most basic amine in aqueous solution — and lose a mark. The aqueous basicity order is not what you'd predict from inductive effect alone.
The concept: Basicity of amines is the tendency of the nitrogen lone pair to accept a proton. Two factors compete:
In the gas phase, only inductive effect operates: 3° > 2° > 1° > NH₃.
In aqueous solution, solvation of the conjugate acid matters. The bulky trimethylammonium ion is poorly solvated, destabilising it. Result: the aqueous order becomes 2° > 1° > 3° > NH₃ (NCERT Class 12 Chemistry Chapter 9, page 269).
Aromatic vs. aliphatic: Aniline (C₆H₅NH₂) is far weaker than methylamine because the nitrogen lone pair delocalises into the benzene ring (resonance effect). The basicity order across classes: aliphatic amines > NH₃ > aromatic amines. Aniline's pKb ≈ 9.4 versus methylamine's pKb ≈ 3.36 — a difference of six orders of magnitude in Kb.
Watch-out for NEET: When a question says "in aqueous solution," do NOT apply the gas-phase order. When it says "arrange in increasing basicity," check whether the set mixes aromatic and aliphatic — aromatic amines always sit below ammonia.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is the correct basicity order of amines in aqueous solution?
Answer: B. B is correct. In aqueous solution, solvation stabilisation of the conjugate acid favours secondary > primary > tertiary > ammonia (NCERT Class 12 Chemistry Chapter 9, page 269).
Why A is wrong: A reflects the gas-phase order (inductive effect only) and ignores solvation — a common trap where students apply 3° > 2° > 1° to aqueous conditions.
Why C is wrong: C places primary above secondary, which contradicts both inductive effect and solvation data.
Why D is wrong: D reverses the entire order; NH₃ is the weakest base among these due to absence of any +I group.
Aniline is a weaker base than methylamine primarily because:
Answer: B. B is correct. The nitrogen lone pair in aniline participates in resonance with the aromatic ring, making it less available for protonation (NCERT Class 12 Chemistry Chapter 9, page 269).
Why A is wrong: A is wrong — the –NH₂ group is not sterically hindered; aniline's reduced basicity is electronic (resonance), not steric.
Why C is wrong: C is wrong — methyl groups donate electrons (+I effect), not withdraw; this is why methylamine is a stronger base than NH₃.
Why D is wrong: D is wrong — it is aniline itself (free base) whose lone pair is stabilised by resonance, not the conjugate acid. Stabilisation of the free base makes protonation less favourable.
The pKb values of methylamine, dimethylamine, trimethylamine, and aniline are approximately 3.36, 3.27, 4.19, and 9.4 respectively. Which amine is the strongest base in water?
Answer: D. D is correct. Dimethylamine has the lowest pKb (3.27), meaning highest Kb, therefore strongest base in aqueous solution.
Why A is wrong: A is wrong — methylamine (pKb 3.36) is slightly weaker than dimethylamine (pKb 3.27) in water.
Why B is wrong: B is wrong — aniline (pKb 9.4) is the weakest base here; its lone pair is delocalised into the aromatic ring.
Why C is wrong: C is wrong — trimethylamine (pKb 4.19) is weaker than both primary and secondary methylamines in water due to poor solvation of its bulky conjugate acid (trap: applying gas-phase order).
Arrange the following in decreasing order of basicity in aqueous solution: (i) C₆H₅NH₂, (ii) C₂H₅NH₂, (iii) (C₂H₅)₂NH, (iv) NH₃
Answer: A. A is correct. Applying the aqueous basicity order: secondary aliphatic > primary aliphatic > NH₃ > aromatic amine. So (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂.
Why B is wrong: B is wrong — it places aniline (aromatic amine) above NH₃, but aromatic amines are weaker than ammonia due to lone-pair delocalisation into the ring (trap: aromatic vs aliphatic basicity confusion).
Why C is wrong: C is wrong — it places primary above secondary, contradicting the aqueous order where solvation + inductive effects make 2° > 1°.
Why D is wrong: D is wrong — same error as C (primary > secondary) and also incorrectly places aniline above NH₃.
In the gas phase, the basicity order of methylamines is 3° > 2° > 1° > NH₃. The reversal observed for tertiary amines in aqueous solution is primarily due to:
Answer: A. A is correct. The bulky trimethylammonium cation (conjugate acid) has fewer N–H bonds available for hydrogen bonding with water, so it is poorly solvated. This destabilises the protonated form, reducing apparent basicity in solution.
Why B is wrong: B is wrong — the +I effect of alkyl groups does not change between gas phase and solution; it is the solvation of the product that changes.
Why C is wrong: C is wrong — trimethylamine has no resonance stabilisation (no π system); the key factor is solvation of the conjugate acid, not the free base.
Why D is wrong: D is wrong — hydrogen bonding of the free base with water would actually stabilise it and is not the cause of reduced basicity; the issue is poor solvation of the protonated form.
Among the following, which is the most basic in aqueous solution?
(i) p-nitroaniline, (ii) aniline, (iii) p-methoxyaniline, (iv) p-methylaniline
Answer: C. C is correct. The –OCH₃ group at the para position donates electron density to the ring via +M effect (resonance donation), increasing electron density on nitrogen and enhancing basicity. Among the given aromatic amines, p-methoxyaniline is the strongest base.
Why A is wrong: A is wrong — the –NO₂ group is a strong electron-withdrawing group (–M and –I), which further delocalises the nitrogen lone pair and makes p-nitroaniline the weakest base in this set.
Why B is wrong: B is wrong — aniline has no substituent to enhance basicity; both p-methoxyaniline (+M donor) and p-methylaniline (+I donor) are more basic.
Why D is wrong: D is wrong — p-methylaniline is more basic than aniline (methyl +I effect) but less basic than p-methoxyaniline, whose –OCH₃ provides stronger +M electron donation to nitrogen.
A student claims: "Since ethylamine has only one alkyl group and diethylamine has two, diethylamine must be exactly twice as basic." What is the fundamental flaw in this reasoning?
Answer: D. D is correct. Basicity (measured as Kb or pKb) is a thermodynamic equilibrium property, not a simple additive quantity. The difference in pKb between ethylamine and diethylamine is small (~0.1 units), reflecting the interplay of inductive donation and solvation — not a doubling.
Why A is wrong: A is wrong — alkyl groups clearly increase basicity of aliphatic amines via +I effect; the data shows CH₃NH₂ > NH₃.
Why B is wrong: B is wrong — diethylamine (secondary) is slightly more basic than ethylamine (primary) in aqueous solution, not less.
Why C is wrong: C is wrong — +I effects do not cancel; each alkyl group contributes electron density. The issue is that basicity is not simply proportional to the count of groups.
Consider the following amines: (i) (CH₃)₂NH, (ii) CH₃NH₂, (iii) (CH₃)₃N, (iv) C₆H₅NHCH₃. Arrange them in increasing order of pKb in aqueous solution.
Answer: C. C is correct. Lower pKb means stronger base. Basicity order (decreasing): (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > C₆H₅NHCH₃. So increasing pKb: (i) < (ii) < (iii) < (iv). N-methylaniline (iv) is weakest because the nitrogen lone pair still delocalises into the ring despite one methyl group.
Why A is wrong: A is wrong — it places trimethylamine as the strongest base (lowest pKb), applying the gas-phase order to aqueous solution (trap: gas-phase vs aqueous basicity).
Why B is wrong: B is wrong — this reverses the entire order, placing the aromatic amine as strongest base (lowest pKb), which contradicts the principle that lone-pair delocalisation into the ring reduces basicity.
Why D is wrong: D is wrong — it places trimethylamine between dimethylamine and methylamine in pKb, which does not match aqueous data where 2° > 1° > 3°.
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Pattern: Basicity ordering of amines in aqueous solution (P.CHE.U18.AMINE_BASICITY, observed NEET 2021, 2023, 2025).
Given
The following amines are to be arranged in decreasing order of basicity in aqueous solution:
(a) Aniline (C₆H₅NH₂), (b) Methylamine (CH₃NH₂), (c) Dimethylamine (CH₃)₂NH, (d) Ammonia (NH₃)
Required
Decreasing order of basic strength in aqueous solution.
Concept
Basicity = availability of nitrogen lone pair for protonation. Governed by: (1) inductive effect of substituents on N, (2) resonance (aromatic ring delocalisation), (3) solvation of conjugate acid in aqueous medium.
Formula / Data
pKb values: CH₃NH₂ ≈ 3.36, (CH₃)₂NH ≈ 3.27, (CH₃)₃N ≈ 4.19, C₆H₅NH₂ ≈ 9.4, NH₃ ≈ 4.75.
Substitution / Application
• Dimethylamine (2°): two methyl groups donate via +I, conjugate acid still has one N–H for solvation → pKb 3.27 (strongest).• Methylamine (1°): one +I group, two N–H bonds in conjugate acid for good solvation → pKb 3.36.• Ammonia: no +I group → pKb 4.75.• Aniline: lone pair delocalised into ring → pKb 9.4 (weakest).
Ordering
Decreasing basicity (= increasing pKb): (CH₃)₂NH > CH₃NH₂ > NH₃ > C₆H₅NH₂
Final answer
(c) > (b) > (d) > (a)
Common trap
Applying the gas-phase order (3° > 2° > 1°) to an aqueous-solution question. In water, the poorly solvated trimethylammonium ion makes tertiary amines weaker than secondary.
Similar NEET-style question
"Arrange the following in increasing order of basic strength: (i) C₆H₅NH₂, (ii) (C₂H₅)₂NH, (iii) C₂H₅NH₂, (iv) NH₃." [Answer: (i) < (iv) < (iii) < (ii)]
Aliphatic amines stronger bases than NH₃ (alkyl group +I donates e⁻ to N). Aromatic amines (aniline) weaker than NH₃ due to resonance delocalisation of lone pair into ring. Order in gas phase: 3° > 2° > 1° > NH₃; in aqueous: complex (steric + solvation).
-- NCERT Class 12 Chemistry, Ch. 9, p. 268Aqueous basicity order: 2° ≈ 1° > 3° > NH3. Aromatic amines much weaker due to resonance delocalisation of lone pair.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKb | -log Kb | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student treats aniline as stronger base than methylamine. Aniline weaker due to lone-pair delocalization into ring.
Basicity ordering across aromatic and aliphatic amines.
Aliphatic amines (CH3NH2) > NH3 > aromatic amines (C6H5NH2). Aniline lone pair delocalised into ring → less available for protonation.
Root cause: concept gap
Gas phase: 3° > 2° > 1° > NH₃. Aqueous: 2° > 1° > 3° > NH₃ (steric + solvation effects).
Root cause: concept gap
Aniline's lone pair delocalised into ring → less basic. Order: aliphatic (CH3NH2) > NH3 > aromatic (PhNH2).
More in Amines: 7 exam traps and mistakes · 2 question patterns from its other lessons.
2 questions from NEET 2025, 2026. Answers verified against NTA official keys.
The correct order of decreasing basic strength of the given amines is:
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
uses gas phase order in aqueous
Doesn't account for solvation reversal
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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