Q.Classify the following amines as primary, secondary or tertiary:
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.
Count how many carbon atoms are attached to the nitrogen: one C = primary, two C = secondary, three C = tertiary.
- naphthalen-1-amine: N bonded to one aryl carbon (plus two H) → primary.
- N,N-dimethylnaphthalen-1-amine: N bonded to one aryl C and two CH3 carbons → tertiary.
- (C2H5)2CHNH2: N bonded to the single CH carbon (plus two H) → primary.
- (C2H5)2NH: N bonded to two ethyl carbons (plus one H) → secondary.
✓Final answer
- primary,
- tertiary,
- primary,
- secondary.
An amine is classed by how many carbon atoms are joined directly to its nitrogen: one → primary (1°), two → secondary (2°), three → tertiary (3°). The number of hydrogens on nitrogen (2, 1, 0) mirrors this. Applying the rule gives (i) 1°,
(ii) 3°,
(iii) 1°,
(iv) 2°.
Concept
Amines are viewed as derivatives of ammonia, NH3, in which one, two or three of the N–H bonds are replaced by N–C bonds. The classification depends only on how many carbon atoms are bonded to nitrogen, not on whether those carbons are part of a ring or a chain, and not on the size of the group:
- Primary (1°): R–NH2 (one C on N)
- Secondary (2°): R2NH (two C on N)
- Tertiary (3°): R3N (three C on N)
Applying the rule to each compound
- naphthalen-1-amine, C10H7–NH2. The nitrogen is joined to a single carbon — the C-1 of the naphthalene ring — and still carries two hydrogens (–NH2). One carbon on nitrogen ⇒ primary amine (an aromatic/aryl amine).
- N,N-dimethylnaphthalen-1-amine, C10H7–N(CH3)2. Here the nitrogen is bonded to the aryl carbon of the naphthalene ring and to the two carbons of the two methyl groups. Three carbons on nitrogen, no N–H left ⇒ tertiary amine.
- (C2H5)2CHNH2 (pentan-3-amine). The –NH2 is attached to the central methine carbon (the CH), which itself bears two ethyl groups. Nitrogen sees only that one carbon and keeps two hydrogens; the ethyl groups are further out on the carbon skeleton, not on nitrogen ⇒ primary amine. (This part tests whether you count carbons on N, not carbons in the molecule.)
- (C2H5)2NH (diethylamine). Nitrogen is bonded to the two carbons of two ethyl groups and retains one hydrogen ⇒ secondary amine.
✓Final answer
- naphthalen-1-amine — primary (1°);
- N,N-dimethylnaphthalen-1-amine — tertiary (3°);
- (C2H5)2CHNH2 (pentan-3-amine) — primary (1°);
- (C2H5)2NH (diethylamine) — secondary (2°).
Method: Classifying Amines (Primary/Secondary/Tertiary) by Counting C-N Bonds
Core Concept
An amine's class depends ONLY on how many carbon atoms are bonded directly to the nitrogen atom - one carbon on N gives a primary amine, two gives secondary, three gives tertiary - regardless of whether those carbons belong to a ring, a chain, or how large the attached groups are.
Steps
- Locate the nitrogen atom in the given structure.
- Count only the bonds going from N directly to a carbon atom (ignore N-H bonds and ignore carbons further away in the molecule that are not bonded to N itself).
- Map the count to a class: 1 carbon on N gives primary (1 degree); 2 carbons give secondary (2 degree); 3 carbons give tertiary (3 degree).
- Repeat independently for every compound in the set - classification never depends on comparing compounds to each other.
Applying the Method to Each Sub-Part
- naphthalen-1-amine, C10H7-NH2: N is bonded to exactly one carbon (the aryl C-1 of naphthalene) and keeps two H -> primary.
- N,N-dimethylnaphthalen-1-amine, C10H7-N(CH3)2: N is bonded to the aryl C-1 AND to the two methyl carbons - three C-N bonds, no N-H left -> tertiary.
- (C2H5)2CHNH2: the -NH2 is bonded only to the single central CH carbon; the two ethyl groups are attached to THAT carbon, not to nitrogen, so N still sees only one carbon -> primary (the trap here is counting the ethyl carbons as if they were on N).
- (C2H5)2NH: N is bonded to the two ethyl carbons and keeps one H -> secondary.
Key Exam Point
Sub-part (iii) is specifically designed to catch students who classify by "how many carbons are in the molecule" instead of "how many carbons are bonded to N" - always trace only the bonds directly touching the nitrogen atom.
Here are the common mistakes students make when classifying amines from drawn structures and condensed formulas, along with how to avoid each.
Mistake 1: Counting all the carbons in the molecule instead of the carbons bonded to nitrogen
Students see a big structure (like the naphthalene ring in part (i)) or a heavily branched formula and assume "many carbons = higher class."
- The Error: Calling naphthalen-1-amine "tertiary" because the ring has ten carbons.
- How to Avoid: The classification rule looks at one atom only — the nitrogen. Count the carbon atoms bonded directly to N:
- one C on N → primary (1°)
- two C on N → secondary (2°)
- three C on N → tertiary (3°)
- Example: In part (i), the nitrogen is bonded to a single ring carbon (C-1 of naphthalene) and two hydrogens → primary, no matter how large the ring system is.
Mistake 2: Classifying by the carbon skeleton next to the amine carbon (the "alcohol/haloalkane habit")
Students carry over the alkyl-halide/alcohol convention — where 1°/2°/3° describes the carbon bearing the functional group — and apply it to the amine.
- The Error: Calling (C2H5)2CHNH2 (part (iii)) a "secondary amine" because the CH carbon bearing the NH2 carries two ethyl groups (it is a secondary carbon).
- How to Avoid: For amines, the degree is a property of the nitrogen, not of the carbon it sits on. In (C2H5)2CHNH2, nitrogen sees only one carbon (the CH) and keeps two hydrogens → primary amine. The two ethyl groups are further out on the skeleton, not on nitrogen.
Mistake 3: Misreading −N(CH3)2 as "two substituents, so secondary"
- The Error: For part (ii), counting the two methyl groups of −N(CH3)2 and stopping there → "secondary."
- How to Avoid: Count every C–N bond, including the bond to the ring/parent chain. In N,N-dimethylnaphthalen-1-amine the nitrogen is bonded to the aryl carbon and to two methyl carbons — three C–N bonds, no N–H left → tertiary (3°). The "N,N-" prefix in a name is itself a signal that nitrogen carries two extra groups besides the parent.
Mistake 4: Thinking aryl amines classify differently from alkyl amines
- The Error: Treating a ring carbon on N as "not counting" (or counting it differently) because it is aromatic.
- How to Avoid: An aryl carbon bonded to nitrogen counts exactly like an alkyl carbon. Aniline (C6H5NH2) and naphthalen-1-amine are both primary amines — one C on N, two H on N — just aromatic ones.
Mistake 5: Not using the N–H count as a cross-check
- The Error: Deciding the class from the drawing alone and never verifying.
- How to Avoid: The hydrogens on nitrogen mirror the classification: 2 H → 1°, 1 H → 2°, 0 H → 3°. In part (iv), (C2H5)2NH has exactly one N–H → secondary, consistent with its two C–N bonds. If your carbon count and hydrogen count disagree, you have misread the structure — recount.
Quick Reference Table
| Part | Compound | C atoms on N | H atoms on N | Class |
|---|---|---|---|---|
| (i) | naphthalen-1-amine | 1 (aryl C) | 2 | Primary (1°) |
| (ii) | N,N-dimethylnaphthalen-1-amine | 3 (aryl C + 2 CH₃) | 0 | Tertiary (3°) |
| (iii) | (C2H5)2CHNH2 | 1 (the CH carbon) | 2 | Primary (1°) |
| (iv) | (C2H5)2NH | 2 (two ethyl C) | 1 | Secondary (2°) |
Final tip: Circle the nitrogen atom and draw only its four bonds before classifying. Everything outside those bonds — ring size, branching, chain length — is irrelevant to whether the amine is 1°, 2° or 3°.
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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