Amine Basic Character

8 MCQs9-step worked example
Source: NCERT Polymers and Chemistry in Everyday LifePYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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:

  1. Electron-donating groups (alkyl groups via +I effect) increase electron density on nitrogen → stronger base.
  2. Steric hindrance and solvation — bulky groups around nitrogen make it harder for water molecules to stabilise the resulting ammonium ion through hydrogen bonding.

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 13, page 8).

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.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following is the correct basicity order of amines in aqueous solution?

MCQ 2Easy RecallPractice

Aniline is a weaker base than methylamine primarily because:

MCQ 3Easy RecallPractice

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?

MCQ 4Direct ApplicationPractice

Arrange the following in decreasing order of basicity in aqueous solution: (i) C₆H₅NH₂, (ii) C₂H₅NH₂, (iii) (C₂H₅)₂NH, (iv) NH₃

MCQ 5Direct ApplicationPractice

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:

MCQ 6Direct ApplicationPractice

Among the following, which is the most basic in aqueous solution?

MCQ 7Concept TrapPractice

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?

MCQ 8CalculationPractice

Consider the following amines: (i) (CH₃)₂NH, (ii) CH₃NH₂, (iii) (CH₃)₃N, (iv) C₆H₅NHCH₃. Arrange them in increasing order of pKb in aqueous solution.

Worked Example

Pattern: Basicity ordering of amines in aqueous solution (P.CHE.U18.AMINE_BASICITY, observed NEET 2021, 2023, 2025).

  1. 1

    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)

  2. 2

    Required

    Decreasing order of basic strength in aqueous solution.

  3. 3

    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.

  4. 4

    Formula / Data

    pKb values: CH₃NH₂ ≈ 3.36, (CH₃)₂NH ≈ 3.27, (CH₃)₃N ≈ 4.19, C₆H₅NH₂ ≈ 9.4, NH₃ ≈ 4.75.

  5. 5

    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).

  6. 6

    Ordering

    Decreasing basicity (= increasing pKb): (CH₃)₂NH > CH₃NH₂ > NH₃ > C₆H₅NH₂

  7. 7

    Final answer

    **(c) > (b) > (d) > (a)**

  8. 8

    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.

  9. 9

    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)]

Before solving, remember these

Key Fact

Basicity of amines

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, p. 8

Formulas

pKb of common amines (aqueous)

Aqueous basicity order: 2° ≈ 1° > 3° > NH3. Aromatic amines much weaker due to resonance delocalisation of lone pair.

SymbolQuantitySI Unit
pKb-log Kb-

Valid when

  • Aqueous solution; gas-phase order is 3°>2°>1°

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Aromatic amines weaker base than aliphatic

Category: Inorganic Exception

Student treats aniline as stronger base than methylamine. Aniline weaker due to lone-pair delocalization into ring.

When it triggers

Basicity ordering across aromatic and aliphatic amines.

How to avoid

Aliphatic amines (CH3NH2) > NH3 > aromatic amines (C6H5NH2). Aniline lone pair delocalised into ring → less available for protonation.

Trap

Hofmann vs Saytzeff product selectivity

Category: Organic Reaction Conditions

Quaternary ammonium hydroxide (Hofmann) elimination favours LESS substituted alkene (anti-Saytzeff). Halide elimination (E2) follows Saytzeff (more substituted).

When it triggers

E2 elimination question with quaternary ammonium hydroxide vs alkyl halide.

How to avoid

Hofmann: bulky leaving group + base → least substituted alkene (Hofmann product). Saytzeff: alkyl halide + base → most substituted alkene (Zaitsev product).

Past Year Questions

11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2023

The correct order of energies of molecular orbitals of N molecule, is 2

11s < *1s < 2s < *2s < 2p < (2p = 2p ) < (*2p = *2p ) < *2p z x y x y z
21s < *1s < 2s < *2s < 2p < *2p < (2p = 2p ) < (*2p = *2p ) z z x y x y
31s < *1s < 2s < *2s < (2p = 2p ) < (*2p = *2p ) < 2p < *2p x y x y z z
41s < *1s < 2s < *2s < (2p = 2p ) < 2p < (*2p = *2p ) < *2p x y z x y z
NTA Answer: Option 4(final)
NEET 2022

Match List-I with List-II. List – I List – II (Products formed) (Reaction of carbonyl compound with) (a) Cyanohydrin (i) NH OH 2 (b) Acetal (ii) RNH 2 (c) Schiff's base (iii) alcohol (d) Oxime (iv) HCN Choose the correct answer from the options given below

1(a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
2(a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
3(a) – (ii), (b) – (iii), (c) – (iv), (d) – (i)
4(a) – (i), (b) – (iii), (c) – (ii), (d) – (iv)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Chemistry Chapter 13, p.8

Test yourself on this topic with real past-paper questions:

Practice this topic →

Free NEET study resources

Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.