Q.Arrange the following in increasing order of their basic strength:
Concept understanding — Basicity Order Amines
Basicity of Amines: From Intuition to Precision
Imagine you have a nitrogen atom with a lone pair of electrons — that pair is like a key that can grab a proton (H+). The more willing the nitrogen is to share that pair, the stronger the base. That's the core idea.
The Intuition: What Makes a Nitrogen "Want" a Proton?
Amines are organic derivatives of ammonia (NH3), where one or more hydrogen atoms are replaced by alkyl groups (R). Alkyl groups are electron-donating — they push electron density toward the nitrogen. More alkyl groups = more electron density on nitrogen = stronger base.
So you'd expect: tertiary > secondary > primary > ammonia. That's the inductive effect argument.
But reality is more interesting. In water (the usual solvent for basicity measurements), the order is:
Secondary > Primary > Tertiary > Ammonia
Why the reversal for tertiary amines? Because basicity isn't just about the free amine — it's about the stability of the conjugate acid (the ammonium ion RNH3+) once the proton is grabbed.
The Precise Explanation: Three Factors at Play
1. Inductive Effect (pushes electron density)
Alkyl groups donate electrons through sigma bonds. More alkyl groups = more electron density on N = stronger base. This favours: tertiary > secondary > primary > ammonia.
2. Solvation Effect (stabilises the conjugate acid)
Once the amine grabs a proton, the resulting ammonium ion is positively charged. Water molecules surround it, stabilising the charge through hydrogen bonding. The more hydrogens on the nitrogen, the more H-bonds can form.
Primary ammonium (RNH3+) has 3 H's → best solvation.
Secondary (R2NH2+) has 2 H's → good solvation.
Tertiary (R3NH+) has 1 H → poor solvation.
This favours: primary > secondary > tertiary.
3. Steric Hindrance (blocks solvation)
Bulkier alkyl groups physically block water molecules from approaching the charged nitrogen. This further reduces solvation in tertiary amines.
The Net Result: The Actual Order in Water
| Amine Type | Inductive Effect | Solvation | Steric Hindrance | Net Basicity (pKb) |
|---|---|---|---|---|
| Ammonia (NH3) | Weakest | Best (3 H's) | None | 4.75 |
| Primary (RNH2) | Moderate | Good (2 H's) | Minimal | ~3.4 |
| Secondary (R2NH) | Strong | Moderate (1 H) | Some | ~3.2 |
| Tertiary (R3N) | Strongest | Poor (0 H's) | Significant | ~4.2 |
pKb is the negative log of the base dissociation constant. Lower pKb = stronger base. So secondary (pKb ≈ 3.2) is the strongest, then primary (≈3.4), then tertiary (≈4.2), then ammonia (4.75).
The Final Answer
The basicity order of amines in aqueous solution is:
Secondary > Primary > Tertiary > Ammonia
This is the standard order for simple alkyl amines (methyl, ethyl, etc.). The inductive effect dominates from ammonia to secondary, but the loss of solvation and steric hindrance in tertiary amines drops them below primary.
A Quick Check: Why Not Tertiary?
If you only considered electron donation, tertiary would win. But the ammonium ion R3NH+ has only one hydrogen to hydrogen-bond with water, and the three bulky alkyl groups physically block water molecules. The conjugate acid is poorly stabilised, so the equilibrium shifts back toward the free amine — making it a weaker base than expected.
This order applies to aliphatic amines in water. Aromatic amines (like aniline) are much weaker bases because the lone pair is delocalised into the benzene ring. Also, in the gas phase (no solvent), the order reverts to tertiary > secondary > primary > ammonia — confirming that solvation is the key reason for the reversal.
The Takeaway
Basicity is a tug-of-war between:
- Inductive effect (wants tertiary to win)
- Solvation + sterics (wants primary to win)
Secondary amines hit the sweet spot — strong inductive donation and decent solvation — making them the strongest bases in water.
The basicity order of amines in aqueous solution is one of the most frequently asked comparison-type questions in the NCERT Class 12 Chemistry chapter on amines, regularly appearing in CBSE boards, JEE Main and NEET. Anyone searching "basicity order of amines class 12 chemistry" or trying to understand why secondary amines outrank primary and tertiary amines will find the inductive-versus-solvation trade-off above is the standard NCERT-aligned explanation.
Why this formula?
Basicity Order of Amines: Why It Holds
Let's build this from first principles — understanding why amines have different basicities is essential for exams and for deeper chemistry.
1. What Does "Basicity" Mean Here?
Basicity of an amine is its ability to accept a proton (H+).
The stronger the base, the more readily it grabs H+.
The equilibrium is:
R3N+H2O⇌R3NH++OH−
The base dissociation constant Kb measures this:
Kb=[R3N][R3NH+][OH−]
A larger Kb means a stronger base.
2. The Key Factor: Electron Density on Nitrogen
The lone pair on nitrogen is what accepts H+.
More electron density on nitrogen → stronger base.
Why? Because a proton (H+) is attracted to negative charge. If the nitrogen's lone pair is more "available" (less tightly held), it binds H+ more easily.
3. The Two Competing Effects
✓ Inductive Effect (Electron-Donating)
Alkyl groups (−CH3, −C2H5, etc.) are electron-donating via the inductive effect.
They push electron density toward nitrogen.
- More alkyl groups → more electron density on N → stronger base.
This suggests:
3∘>2∘>1∘>NH3
✗ Solvation Effect (Hydration of the Conjugate Acid)
When the amine accepts H+, it forms R3NH+ (a positively charged ion).
This ion is stabilized by water molecules (hydration).
- More H atoms on the NH+ group → more hydrogen bonding with water → better stabilization of the conjugate acid.
- Better stabilization of R3NH+ → stronger base (because the equilibrium shifts toward protonation).
This suggests:
1∘>2∘>3∘ (since 1∘ has 3 H's, 2∘ has 2, 3∘ has 1)
4. The Actual Order (Aqueous Solution)
In water, the observed order for aliphatic amines is:
2° > 1° > 3° > NH3
Wait — that's not simply "more alkyl = stronger". Why?
The Reasoning (Step by Step)
-
From NH3 to 1∘: Adding one alkyl group increases electron density (inductive effect) → stronger base.
So 1∘>NH3.
-
From 1∘ to 2∘: Adding a second alkyl group further increases electron density → even stronger base.
So 2∘>1∘.
-
From 2∘ to 3∘: Here, the solvation effect becomes dominant.
The 3∘ amine has only one H on the NH+ group → poor hydration of the conjugate acid.
The inductive effect is still present, but the loss of solvation outweighs the gain in electron density.
So 3∘ is weaker than 2∘.
Thus, the net order is:
2° > 1° > 3° > NH3
5. The Key Formula(e) to Remember
For gas phase (no solvent):
Only inductive effect matters:
3° > 2° > 1° > NH3
For aqueous solution (common in exams):
Both effects matter — solvation dominates for 3∘:
2° > 1° > 3° > NH3
For aromatic amines (e.g., aniline):
The lone pair is delocalized into the benzene ring → much weaker base.
Aliphatic amines > Aromatic amines
6. Quick Exam Tip
If a question asks "basicity order of amines in water", always write:
2° > 1° > 3° > NH3
And explain:
- Inductive effect increases from NH3 to 3∘
- But solvation of the conjugate acid decreases from 1∘ to 3∘
- The balance gives the above order.
7. Summary Table
| Amine Type | Inductive Effect | Solvation of R3NH+ | Net Basicity (aq) |
|---|---|---|---|
| NH3 | Weakest | Best (3 H's) | Weakest |
| 1∘ | Moderate | Good (2 H's) | Moderate |
| 2∘ | Strong | Moderate (1 H) | Strongest |
| 3∘ | Strongest | Poor (0 H's) | Weaker than 2∘ |
Final takeaway:
Basicity is not just about "more alkyl = stronger". The solvation of the conjugate acid is the deciding factor in water. Always reason from both effects.
Concept: Basicity of amines depends on the combined effect of inductive effects (+I of alkyl groups), resonance delocalisation (in aromatic amines), and solvation of the conjugate acid in water. There is no single universal 2∘>1∘>3∘ rule — the observed aqueous order differs between the methyl and ethyl series.
(i) Aniline (C6H5NH2) is the weakest because the lone pair is delocalised into the benzene ring. Ammonia comes next (no +I alkyl groups). Benzylamine (C6H5CH2NH2) is stronger than ammonia — the CH2 spacer blocks resonance, leaving only a weak inductive pull from the ring — but weaker than a simple alkylamine. Among the ethylamines, secondary > primary.
Order: C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH
(ii) For the ethyl series in water, the observed order is 2∘>3∘>1∘: the strong +I effect of two/three ethyl groups outweighs triethylamine's poorer solvation, so triethylamine sits above ethylamine (Table 9.3: pKb (C2H5)2NH 3.00 < (C2H5)3N 3.25 < C2H5NH2 3.29). Aniline is weakest.
Order: C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH
(iii) For the methyl series in water, the order is 2∘>1∘>3∘ (the smaller +I of methyl cannot compensate trimethylamine's poor solvation). Benzylamine is stronger than aniline but weaker than all the methylamines.
Order: C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH
- C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH
- C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH
- C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH
Basicity of amines is set by the balance between the inductive effect (alkyl groups increase basicity), resonance (aromatic amines are far weaker), and solvation of the conjugate acid in water. That balance plays out differently for the methyl and ethyl series: in water, methylamines follow 2∘>1∘>3∘, but ethylamines follow 2∘>3∘>1∘. The final orders are: (i) C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH;
(ii) C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH;
(iii) C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH.
The Core Idea: What Makes an Amine Basic?
Basicity is about how readily the nitrogen atom donates its lone pair to a proton. In aqueous solution, the equilibrium is:
RNH2+H2O⇌RNH3++OH−
The stronger the base, the more it shifts right. Three factors compete:
- Inductive effect — Alkyl groups (−CH3, −C2H5) are electron-donating. They push electron density toward nitrogen, making the lone pair more available. More alkyl groups = stronger inductive push, and an ethyl group pushes harder than a methyl group.
- Resonance effect — In aniline (C6H5NH2), the lone pair on nitrogen is delocalised into the aromatic ring. This makes it much less available for protonation — aniline is a very weak base.
- Solvation and steric hindrance — In water, the protonated form is stabilised by hydrogen bonding with water. More hydrogen atoms on the nitrogen (i.e., fewer alkyl groups) means better solvation. Bulky alkyl groups also physically crowd the nitrogen.
The "one fixed order" trap
Because factors 1 and 3 pull in opposite directions, there is no single order that fits every alkyl series. For methylamines, the weak +I of methyl loses to solvation for the tertiary amine, giving 2∘>1∘>3∘ in water. For ethylamines, the stronger +I of ethyl compensates for the tertiary amine's poorer solvation, giving 2∘>3∘>1∘ — triethylamine is actually a stronger base than ethylamine in water. This is exactly what NCERT's Table 9.3 pKb data show. In the gas phase (no solvent) the inductive trend 3∘>2∘>1∘ holds for both series.
Basicity orders in water (NCERT)
Methyl series: (CH3)2NH>CH3NH2>(CH3)3N>NH3
Ethyl series: (C2H5)2NH>(C2H5)3N>C2H5NH2>NH3
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH
-
Identify the weakest — C6H5NH2 (aniline) has its lone pair delocalised into the benzene ring. This is a massive drop in basicity. It is by far the weakest here.
-
Next weakest — NH3 has no alkyl groups to donate electron density. It is a weaker base than any alkylamine.
-
Benzylamine — C6H5CH2NH2 has the amino group separated from the ring by a −CH2− spacer. The ring cannot delocalise the lone pair (too far away), but it does exert a weak electron-withdrawing inductive effect through the chain. So benzylamine is a weaker base than a simple alkylamine like ethylamine, but stronger than ammonia and much stronger than aniline.
-
Ethylamine vs diethylamine — (C2H5)2NH is secondary, C2H5NH2 is primary. In water, secondary > primary. So diethylamine is the strongest here.
Benzylamine shortcut
The −CH2− group insulates the nitrogen from the ring's resonance effect. So benzylamine behaves like an alkylamine, slightly weakened by the ring's inductive pull — above ammonia, below ethylamine.
Order: C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
-
Aniline is weakest — same reason as before. Lone pair delocalised into the ring.
-
Among the ethylamines — this is the ethyl series, so the aqueous order is 2∘>3∘>1∘. The two (or three) ethyl groups exert a strong enough +I push that triethylamine, despite its poorly solvated conjugate acid, stays above ethylamine. Diethylamine, which enjoys both a strong inductive push and reasonable solvation, tops the list.
Let the book's own data arbitrate
NCERT Table 9.3 (pKb, smaller = stronger base): (C2H5)2NH 3.00 < (C2H5)3N 3.25 < C2H5NH2 3.29 ≪ C6H5NH2 9.38. The numbers confirm: diethylamine > triethylamine > ethylamine > aniline.
Don't copy the methyl-series order here
Many students apply the memorised 2∘>1∘>3∘ rule and put ethylamine above triethylamine. That order is right for methylamines but wrong for ethylamines — the stronger +I effect of ethyl flips the 1∘/3∘ positions. For ethylamines in water: 2∘>3∘>1∘.
Order: C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2
-
Aniline is weakest — resonance delocalisation, as before.
-
Benzylamine — the −CH2− spacer prevents resonance but the ring still pulls electron density inductively. So it's weaker than any of the simple methylamines here, though far stronger than aniline.
-
Methylamines — the classic aqueous order for the methyl series: (CH3)2NH>CH3NH2>(CH3)3N. Dimethylamine (secondary) is strongest, then methylamine (primary), then trimethylamine (tertiary, demoted by poor solvation).
The pKb values (NCERT Table 9.3)
| Amine | pKb |
|---|---|
| (CH3)2NH | 3.27 |
| CH3NH2 | 3.38 |
| (CH3)3N | 4.22 |
| C6H5CH2NH2 | 4.70 |
| C6H5NH2 | 9.38 |
| Lower pKb = stronger base. The numbers confirm the order exactly. |
Order: C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH
- C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH
- C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH
- C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH
Method: Inductive Effect + Solvation Effect Analysis (for aliphatic amines) and Resonance Effect (for aromatic amines)
This is the standard approach for comparing basic strength of amines in aqueous medium. The one refinement that matters: the inductive and solvation effects pull in opposite directions, and their balance comes out differently for the methyl and ethyl series — so identify the series before applying an order.
The two aqueous orders to know (NCERT Table 9.3):
- Methyl series: (CH3)2NH>CH3NH2>(CH3)3N>NH3 — i.e. 2∘>1∘>3∘
- Ethyl series: (C2H5)2NH>(C2H5)3N>C2H5NH2>NH3 — i.e. 2∘>3∘>1∘ (the stronger +I of ethyl keeps the tertiary amine above the primary)
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH
Step 1 — Identify the type of each amine
- C2H5NH2 — 1° aliphatic (ethylamine)
- (C2H5)2NH — 2° aliphatic (diethylamine)
- NH3 — ammonia
- C6H5NH2 — aromatic (aniline)
- C6H5CH2NH2 — aralkyl (benzylamine)
Step 2 — Apply the rules
- Aromatic amines (C6H5NH2) are weakest due to resonance delocalisation of the lone pair into the benzene ring.
- NH3 has no +I alkyl group, so it is weaker than every alkyl/aralkyl amine here — but far stronger than aniline.
- Aralkyl amines (C6H5CH2NH2) sit between ammonia and the simple alkylamines: the CH2 spacer blocks resonance, but the ring's weak inductive pull keeps benzylamine below ethylamine.
- Ethylamines: secondary > primary, so (C2H5)2NH>C2H5NH2.
Step 3 — Arrange in increasing order
C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
Step 1 — Identify types
- C6H5NH2 — aromatic (weakest)
- C2H5NH2 — 1° aliphatic
- (C2H5)2NH — 2° aliphatic
- (C2H5)3N — 3° aliphatic
Step 2 — Apply the ETHYL-series aqueous order
This is the ethyl series, so in water: 2∘>3∘>1∘
So: (C2H5)2NH>(C2H5)3N>C2H5NH2
(Table 9.3 confirms it: pKb 3.00 < 3.25 < 3.29.)
Step 3 — Arrange
C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2
Step 1 — Identify types
- C6H5NH2 — aromatic (weakest)
- C6H5CH2NH2 — aralkyl (stronger than aromatic)
- CH3NH2 — 1° aliphatic
- (CH3)2NH — 2° aliphatic
- (CH3)3N — 3° aliphatic
Step 2 — Apply the METHYL-series aqueous order
This is the methyl series, so in water: 2∘>1∘>3∘
So: (CH3)2NH>CH3NH2>(CH3)3N — and benzylamine (pKb 4.70) slots in just below trimethylamine (pKb 4.22).
Step 3 — Arrange
C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH
Key Concept Summary
| Type | Order (increasing basic strength in water) |
|---|---|
| Aromatic | Weakest (lone pair delocalised) |
| Aralkyl | Intermediate (weak –I pull from the ring) |
| Methylamines (aq.) | 2° > 1° > 3° > NH3 |
| Ethylamines (aq.) | 2° > 3° > 1° > NH3 |
| Aliphatic (gas) | 3° > 2° > 1° > NH3 (only +I effect, no solvation) |
Exam tip: First check the medium (aqueous vs gas phase), then check which alkyl series you are ordering. In water the secondary amine tops both series, but the 1∘/3∘ positions swap between the methyl and ethyl series — quoting the book's pKb values (Table 9.3) is the safest justification.
🧠 Core Concept First
Basicity of amines depends on electron density on the nitrogen atom. More electron density → more available to donate → stronger base.
Key factors (in order of importance for these problems):
- Inductive effect — alkyl groups are electron-donating (+I), aryl groups are electron-withdrawing (–I and resonance). An ethyl group donates more strongly than a methyl group.
- Resonance effect — in aniline, the lone pair is delocalised into the ring, drastically reducing basicity.
- Solvation & steric hindrance — in aqueous solution, more H atoms on N allow better solvation of the conjugate acid, increasing basicity.
Because factors 1 and 3 oppose each other, the aqueous order is a compromise that differs by series: methylamines follow 2∘>1∘>3∘, ethylamines follow 2∘>3∘>1∘.
✗ Common Mistake #1: Forgetting that Aliphatic > Aromatic always
Example from (i):
C6H5NH2 (aniline) is much weaker than NH3, C2H5NH2, etc.
Why students go wrong:
They compare only inductive effects and forget that in aniline, the lone pair is delocalised into the benzene ring via resonance — making it far less available.
✓ How to avoid:
Always check: is the nitrogen directly attached to an aromatic ring? If yes → resonance delocalisation → very weak base. Place it last (or first in increasing order).
Correct order for (i):
C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2<(C2H5)2NH
✗ Common Mistake #2: Applying one fixed 2∘>1∘>3∘ rule to every alkyl series
Example from (ii):
Students memorise "in water: 2∘>1∘>3∘" from the methylamines and mechanically write C2H5NH2 above (C2H5)3N.
Why that's wrong here:
The aqueous order is a tug-of-war between the +I push (favours 3∘) and solvation of the conjugate acid (favours 1∘). For methyl groups the +I push is weak, so solvation wins and 3∘ drops below 1∘. For ethyl groups the +I push is stronger, so triethylamine stays above ethylamine. NCERT Table 9.3 confirms it: pKb (C2H5)2NH 3.00 < (C2H5)3N 3.25 < C2H5NH2 3.29.
✓ How to avoid:
Learn both series explicitly:
- Methyl (aq.): (CH3)2NH>CH3NH2>(CH3)3N>NH3
- Ethyl (aq.): (C2H5)2NH>(C2H5)3N>C2H5NH2>NH3 (In the gas phase, with no solvation, both series follow 3∘>2∘>1∘>NH3.)
Correct order for (ii) in aqueous medium:
C6H5NH2<C2H5NH2<(C2H5)3N<(C2H5)2NH
✗ Common Mistake #3: Forgetting that benzylamine is aliphatic in behaviour
Example from (iii):
C6H5CH2NH2 (benzylamine) is not like aniline — the nitrogen is not directly attached to the ring.
Why students go wrong:
They see a benzene ring and immediately assume "weak base".
✓ How to avoid:
Check the attachment:
- C6H5—NH2 → aniline (weak, resonance)
- C6H5—CH2—NH2 → benzylamine (no resonance; behaves like an alkylamine slightly weakened by the ring's inductive pull)
Benzylamine (pKb 4.70) is more basic than aniline by far, and slightly weaker than trimethylamine (pKb 4.22) — so in (iii) it sits just below the three methylamines.
Correct order for (iii):
C6H5NH2<C6H5CH2NH2<(CH3)3N<CH3NH2<(CH3)2NH
✗ Common Mistake #4: Placing benzylamine below ammonia in (i)
Why students go wrong:
They over-count the ring's electron-withdrawing pull and drop C6H5CH2NH2 below NH3.
✓ How to avoid:
The CH2 spacer insulates the nitrogen from the ring's resonance; only a weak inductive pull remains. Benzylamine (pKb 4.70) is slightly more basic than ammonia (pKb 4.75) — above NH3, below ethylamine, exactly as the printed NCERT answer for (i) has it.
📋 Quick Summary Table
| Amine type | Basicity (aqueous) | Key reason |
|---|---|---|
| Aniline (C6H5NH2) | Very weak | Resonance delocalisation of lone pair |
| Benzylamine (C6H5CH2NH2) | Just above NH3 | No resonance, weak –I pull through CH2 |
| NH3 | Weakest aliphatic | No alkyl groups |
| Methylamines | 2∘>1∘>3∘ | Weak +I; solvation demotes 3∘ |
| Ethylamines | 2∘>3∘>1∘ | Strong +I keeps 3∘ above 1∘ |
✓ Final Exam Tip
When asked "increasing order of basic strength":
- Separate aromatic from aliphatic — aromatic goes last (weakest).
- Identify the alkyl series: methyl → 2∘>1∘>3∘; ethyl → 2∘>3∘>1∘ (in water).
- Benzylamine = aliphatic in behaviour, just above NH3.
- Always check the medium — if not specified, assume aqueous; in the gas phase both series revert to 3∘>2∘>1∘.
Showing the 12 most recent of 13 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.Choose the correct decreasing order of basic strength of amines in aqueous solution: (A) NH3 > CH3NH2 > (CH3)2NH > (CH3)3N (B) (CH3)2NH > (CH3)3N > NH3 > CH3NH2 (C) (CH3)3N > NH3 > CH3NH2 > (CH3)2NH (D) (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (E) CH3NH2 > (CH3)2NH > (CH3)3N > NH3
›Reveal solutionSolution
Because solvation of the ammonium ion opposes the pure inductive trend, the aqueous basicity order of methylamines is (CH3)2NH>CH3NH2>(CH3)3N>NH3.
In water, base strength of amines depends on three competing effects: the electron-donating +I of methyl groups (raises basicity), steric hindrance to protonation, and stabilisation of the protonated cation by hydrogen-bonded solvation (fewer N–H bonds in more-substituted amines lowers solvation).
The net experimental order for methylamines in aqueous solution is:
(CH3)2NH>CH3NH2>(CH3)3N>NH3
The secondary amine is most basic; the tertiary amine drops because of poor cation solvation and steric crowding; ammonia is least basic.
✓Final answerThe correct option is (D).
- KEAM 2026Set eng-2026-04184 marksMCQQ.Which of the following amine has the highest pKb value in aqueous phase? (A) Methanamine (B) N-methylmethanamine (C) Ethanamine (D) N-Methylbenzenamine (E) Benzenamine
›Reveal solutionSolution
Weakest base = highest pKb; aniline, with lone-pair delocalisation into the ring, is the weakest here.
Basicity depends on availability of the N lone pair. In aniline (benzenamine) the lone pair is delocalised into the benzene ring, sharply lowering basicity, so pKb≈9.4.
N-methylbenzenamine is slightly more basic than aniline (pKb≈9.2) because the electron-donating methyl offsets some delocalisation. The aliphatic amines (methanamine, ethanamine, dimethylamine) are much stronger bases (pKb≈3.3).
Thus benzenamine has the highest pKb.
✓Final answerThe correct option is (E).
- KEAM 2026Set eng-2026-04204 marksMCQQ.Which of the following amine has lowest pKb value in aqueous phase? (A) Ethanamine (B) Methanamine (C) N-Methylmethanamine (D) N-Ethylethanamine (E) N, N-Diehtylmethanamine
›Reveal solutionSolution
Diethylamine (secondary amine) is the strongest base ⇒ lowest pKb.
In aqueous solution basic strength reflects a balance of +I effect and solvation of the cation, giving the order secondary > primary ≈ tertiary for aliphatic amines. Among the options, N-Ethylethanamine (C2H5)2NH is a secondary amine with two electron-releasing ethyl groups and good cation solvation, making it the strongest base (lowest pKb≈3.0).
✓Final answerThe correct option is (D).
- KEAM 2026Set eng-2026-04214 marksMCQQ.The descending order of basic strength of the following amines is(i) N-Methylbenzenamine(ii) N.N'-Dimethylbenzenamine(iii) Benzenamine(iv) Phenylmethanamine (A)(i) >(ii) >(iv) >(iii) (B)(iv) >(i) >(ii) >(iii) (C)(iv) >(ii) >(i) >(iii) (D)(iv) >(iii) >(ii) >(i) (E)(i) >(iv) >(ii) > (iii)
›Reveal solutionSolution
Benzylamine (nitrogen not on the ring) is the strongest base; for the ring-N anilines, basicity rises with the number of electron-donating alkyl groups: (iv) benzylamine > (ii) N,N-dimethylaniline > (i) N-methylaniline > (iii) aniline.
Reasoning
- (iv) Phenylmethanamine (benzylamine, C6H5CH2NH2): the –NH₂ is on an sp³ carbon, not conjugated with the ring, so the lone pair is fully available → most basic.
- Anilines have the N lone pair delocalised into the ring, lowering basicity. Adding alkyl groups (+I effect) partially restores basicity:
- (ii) N,N-dimethylaniline (two methyls) > (i) N-methylaniline (one methyl) > (iii) aniline (none).
Approximate pK_b values confirm this: benzylamine ≈ 4.7, N,N-dimethylaniline ≈ 8.9, N-methylaniline ≈ 9.2, aniline ≈ 9.4 (smaller pK_b = stronger base).
Descending basic strength: (iv) > (ii) > (i) > (iii).
✓Final answerThe correct option is (C).
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.Which one of the following compounds is strongly basic in aqueous medium? (A) Benzenamine (B) N-ethylethanamine (C) Phenylmethanamine (D) N,N-Dimethylbenzenamine (E) Ammonia
›Reveal solutionSolution
Diethylamine (secondary aliphatic amine) is most basic in water thanks to +I of two ethyl groups plus favourable solvation.
Basicity of amines in aqueous solution depends on the electron density on N (inductive effect) and on solvation of the ammonium ion. Aromatic amines (benzenamine/aniline, N,N-dimethylbenzenamine) are weak bases because the lone pair is delocalised into the ring. Among the aliphatic amines, N-ethylethanamine = diethylamine (C2H5)2NH is a secondary amine whose two electron-donating ethyl groups increase electron density on nitrogen, while the N-H's still allow good hydrogen-bonded solvation of the cation. This makes diethylamine more basic than ammonia and than the primary amine (benzylamine). Hence (B) is the strongest base.
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04234 marksMCQQ.The order of basic strength of following amines is(i) CH3NH2(ii) (C2H5)2NH(iii) C6H5NH2(iv) C6H5NHCH3 (A)(ii) <(i) <(iv) <(iii) (B)(iii) <(iv) <(ii) <(i) (C)(ii) <(iii) <(iv) <(i) (D)(i) <(ii) <(iii) <(iv) (E)(iii) <(iv) <(i) < (ii)
›Reveal solutionSolution
The increasing order of basicity is aniline < N-methylaniline < methylamine < diethylamine: (iii) < (iv) < (i) < (ii).
Concept and Intuition
Aromatic amines are much weaker bases than aliphatic amines because the nitrogen lone pair is delocalised into the benzene ring. An added alkyl group increases basicity by electron donation. Among aliphatic amines, a secondary amine (diethylamine) is more basic than a primary one (methylamine).
Step-by-Step Solution
- Aniline (iii): lone pair delocalised → weakest base.
- N-Methylaniline (iv): +I of CH3 makes it slightly more basic than aniline, but still aromatic and weak.
- Methylamine (i): aliphatic primary amine, much more basic than the aromatic ones.
- Diethylamine (ii): aliphatic secondary amine, most basic here.
- Order: (iii) < (iv) < (i) < (ii).
Common Mistakes
- Ranking aniline above the aliphatic amines; resonance makes aromatic amines the weakest.
✓Final answerThe correct option is (E) — (iii) < (iv) < (i) < (ii).
ANSWER: E
- KEAM 2025Set eng-2025-04274 marksMCQQ.The amine with the highest pKb value is (A) Methanamine (B) N-methylmethanamine (C) Benzeneamine (D) N-Methylaniline (E) Ethanamine
›Reveal solutionSolution
Highest pKb means weakest base. Aromatic amines are far weaker than aliphatic ones; of the two aromatic amines, unsubstituted aniline is weaker than N-methylaniline, so aniline has the highest pKb.
Reasoning
Basicity depends on availability of the nitrogen lone pair. In aromatic amines the lone pair is delocalised into the benzene ring, sharply lowering basicity (raising pKb).
Approximate pKb values:
- (A) Methanamine (CH3NH2): 3.36
- (B) N-methylmethanamine ((CH3)2NH): 3.27
- (E) Ethanamine (C2H5NH2): 3.29
- (D) N-Methylaniline: ~9.2
- (C) Benzeneamine (aniline): ~9.4
Aliphatic amines (A, B, E) are strong bases (low pKb). Between the two aromatic amines, the electron-donating –CH3 in N-methylaniline slightly increases basicity, so it is a stronger base than aniline. Therefore unsubstituted aniline (benzeneamine) is the weakest base and has the highest pKb.
✓Final answerThe correct option is (C).
- KEAM 2024Set eng-2024-06074 marksMCQQ.The decreasing order of basic strength in aqueous solution of amines is (A) Dimethylamine > Methylamine > Trimethylamine > Ammonia (B) Methylamine > Dimethylamine > Trimethylamine > Ammonia (C) Trimethylamine > Dimethylamine > Methylamine > Ammonia (D) Ammonia > Trimethylamine > Dimethylamine > Methylamine (E) Ammonia > Dimethylamine > Trimethylamine > Methylamine
›Reveal solutionSolution
In aqueous solution the basicity order for methyl amines is dimethylamine > methylamine > trimethylamine > ammonia, reflecting the balance of the +I effect, solvation (H-bonding of the conjugate acid) and steric hindrance.
Basicity in water is governed by three factors: the electron-donating (+I) effect of alkyl groups (increases basicity), stabilisation of the protonated ammonium ion by hydrogen bonding/solvation (favours more N–H bonds), and steric hindrance (crowding in trimethylamine hinders both protonation and solvation). The combined effect gives the observed aqueous order (CH3)2NH>CH3NH2>(CH3)3N>NH3.
✓Final answerThe correct option is (A).
- KEAM 2024Set eng-2024-06084 marksMCQQ.The decreasing order of basic strength of amines in aqueous medium is (A) CH3NH2>(CH3)2NH>(CH3)3N>NH3 (B) (CH3)2NH>CH3NH2>(CH3)3N>NH3 (C) (CH3)2NH>(CH3)3N>CH3NH2>NH3 (D) (CH3)2NH>NH3>(CH3)3N>CH3NH2 (E) NH3>CH3NH2>(CH3)3N>(CH3)2NH
›Reveal solutionSolution
For methylamines in water the basic-strength order is (CH3)2NH>CH3NH2>(CH3)3N>NH3, because solvation of the ammonium ion and steric hindrance offset the +I effect.
Three factors govern basicity in water:
- +I (electron-donating) effect of methyl groups increases electron density on N (raises basicity).
- Solvation/stabilisation of the protonated cation by water — more N–H bonds allow more H-bonding (raises effective basicity).
- Steric hindrance — bulky groups around N hinder protonation and solvation (lowers basicity).
The net result for methylamines in aqueous medium is:
(CH3)2NH>CH3NH2>(CH3)3N>NH3
The secondary amine wins the balance; the tertiary amine is depressed by poor solvation and steric crowding, but all methylamines are still more basic than ammonia.
✓Final answerThe correct option is (B).
- KEAM 2024Set pha-2024-06104 marksMCQQ.The correct increasing order of basic strength is (A) $NH_3 < C_2H_5NH_2 < C_6H_5NH_2 < C_6H_5CH_2NH_2$ (B) $C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2$ (C) $C_6H_5NH_2 < C_6H_5CH_2NH_2 < NH_3 < C_2H_5NH_2$ (D) $C_6H_5CH_2NH_2 < NH_3 < C_2H_5NH_2 < C_6H_5NH_2$ (E) $C_6H_5NH_2 < NH_3 < C_2H_5NH_2 < C_6H_5CH_2NH_2$
›Reveal solutionSolution
Basic strength order (pKb): aniline (9.4) < NH3 (4.75) < benzylamine (4.66) < ethylamine (3.25).
Basic strength depends on availability of the N lone pair.
- Aniline (C6H5NH2): lone pair delocalised into the ring ⇒ weakest (pKb ≈ 9.4).
- Ammonia (NH3): pKb ≈ 4.75.
- Benzylamine (C6H5CH2NH2): the CH2 insulates N from the ring; slightly more basic than ammonia (pKb ≈ 4.66).
- Ethylamine (C2H5NH2): alkyl +I effect makes it the strongest (pKb ≈ 3.25).
So increasing basicity: C6H5NH2<NH3<C6H5CH2NH2<C2H5NH2.
✓Final answerThe correct option is (B). Aniline weakest (resonance), ethylamine strongest (+I), benzylamine just above ammonia.
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.Among methanamine, ethanamine, benzenamine, N-methylaniline and N, N-dimethylaniline, the weakest and the strongest base in aqueous phase, respectively are (A) benzenamine and methanamine (B) N-methylaniline and ethanamine (C) N, N-dimethylaniline and ethanamine (D) benzenamine and ethanamine (E) N-methylaniline and methanamine
›Reveal solutionSolution
The weakest base is benzenamine (aniline) and the strongest is ethanamine.
Concept and Intuition
Aromatic amines are weaker bases than aliphatic amines because the lone pair on nitrogen is delocalised into the ring. Among aromatic amines, N-methyl and N,N-dimethyl substitution increases electron density on nitrogen, so plain aniline is the weakest. Among aliphatic amines in water, ethanamine is more basic than methanamine due to a better balance of inductive and solvation effects.
Step-by-Step Solution
- Aromatic set: aniline < N-methylaniline < N,N-dimethylaniline in basicity; aniline is weakest.
- Aliphatic set: in aqueous phase ethanamine > methanamine.
- Aliphatic amines exceed aromatic amines overall, so ethanamine is the strongest.
- Hence weakest = benzenamine, strongest = ethanamine.
Common Mistakes
- Assuming methanamine is the strongest base; in water ethanamine is more basic than methanamine.
✓Final answerThe correct option is (D) — benzenamine and ethanamine.
ANSWER: D
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.Which one of the following is not correct with respect to properties of amines? (A) pKb of aniline is more than that of methylamine. (B) Ethylamine is soluble in water whereas aniline is not. (C) Ethanamide on reaction with Br2 and NaOH gives ethylamine. (D) Ethylamine reacts with nitrous acid to give ethanol. (E) Aniline does not undergo Friedel-Crafts reaction.
›Reveal solutionSolution
Statement (C) is wrong: Hofmann degradation of ethanamide gives methylamine, not ethylamine.
Concept and Intuition
The Hofmann bromamide reaction converts an amide RCONH2 to an amine RNH2 with the loss of one carbon (the carbonyl C leaves as carbonate). So a two-carbon amide yields a one-carbon amine.
Step-by-Step Solution
- CH3CONH2Br2, NaOHCH3NH2 (methylamine), one carbon fewer.
- Statement (C) claims ethylamine — incorrect.
- Check others: aniline is a weaker base than methylamine (higher pKb) — (A) true; ethylamine is water-soluble, aniline sparingly so — (B) true; ethylamine + HNO2→ ethanol — (D) true; aniline forms a Lewis salt with AlCl3 so no Friedel–Crafts — (E) true.
Common Mistakes
- Forgetting the carbon loss in Hofmann degradation and expecting RCONH2→RCH2NH2-type retention.
✓Final answerThe correct option is (C) — Ethanamide with Br2/NaOH gives methylamine, not ethylamine.
ANSWER: C
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