Q.Arrange the following:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
C2H5OH, (CH3)2NH, C2H5NH2
C6H5NH2, (C2H5)2NH, C2H5NH2.
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🔒 Start your 14-day free trial to unlock the full solution →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? …
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): …
Key idea: Basicity of amines depends on inductive effects, resonance, solvation, and steric hindrance. pKb is inversely related to basic strength (lower pKb = stronger base).
(i) pKb values: stronger base → lower pKb. Alkyl amines are more basic than arylamines due to resonance in arylamines. Among alkyl amines, secondary > primary. Among arylamines, N-methylaniline is more basic than aniline (alkyl group donates electron density).
Decreasing pKb: C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH
(ii) Increasing basic strength: weakest base first. Arylamines are weaker than alkylamines. Among arylamines, N,N-dimethylaniline is (slightly) more basic than aniline itself — the two N-methyl groups donate electron density inductively, and even though they add some steric hindrance to solvation of the conjugate acid, the net effect in water still favours N,N-dimethylaniline as the stronger of the two (its conjugate-acid pKa, ~5.1, is higher than aniline's, ~4.6). Among alkylamines, secondary > primary.
Order: C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH
(iii)(a) p-Nitroaniline has strong electron-withdrawing –NO₂ group (decreases basicity). p-Toluidine has electron-donating –CH₃ group (increases basicity).
Increasing basic strength: p-nitroaniline < aniline < p-toluidine
(iii)(b) Benzylamine (C6H5CH2NH2) is an alkylamine (no resonance with ring), so strongest. N-Methylaniline is more basic than aniline due to +I of –CH₃.
Increasing basic strength: C6H5NH2<C6H5NHCH3<C6H5CH2NH2
(iv) In gas phase, basicity depends only on inductive effect (no solvation). Alkyl groups donate electrons, so tertiary > secondary > primary > ammonia.
Decreasing basic strength: (C2H5)3N>(C2H5)2NH>C2H5NH2>NH3 …
Basicity of amines depends on the balance between inductive effects, resonance, solvation, and steric hindrance. pKb is inversely related to basic strength (lower pKb = stronger base). The answers are: (i) C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH; (ii) C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH; (iii)(a) p-nitroaniline < aniline < p-toluidine; (iii)(b) C6H5NH2<C6H5NHCH3<C6H5CH2NH2; (iv) (C2H5)3N>(C2H5)2NH>C2H5NH2>NH3; (v) (CH3)2NH<C2H5NH2<C2H5OH; (vi) C6H5NH2<(C2H5)2NH<C2H5NH2.
The Core Idea: What Makes an Amine Basic?
Basicity in amines comes from the lone pair on nitrogen being available to accept a proton. Anything that increases electron density on nitrogen makes it a stronger base (lower pKb). Anything that decreases it (by withdrawing electrons or delocalising the lone pair) makes it weaker (higher pKb).
Three main factors compete:
- Inductive effect – Alkyl groups push electrons toward nitrogen, making it more basic. More alkyl groups = stronger base, but only in the gas phase.
- Resonance / Delocalisation – If the lone pair is part of a conjugated system (like in aniline), it's less available for protonation, drastically lowering basicity.
- Solvation & Steric Hindrance – In water, the protonated ammonium ion is stabilised by hydrogen bonding. Bulky groups around nitrogen hinder solvation, reducing stability of the conjugate acid, and thus lowering basic strength in solution.
A common mistake is to assume that more alkyl groups always mean stronger base in water. In aqueous solution, the order for aliphatic amines is usually: 2∘>1∘>3∘>NH3 — because of the solvation effect. In the gas phase, the order follows purely inductive effects: 3∘>2∘>1∘>NH3.
(i) Decreasing order of pKb values
pKb is the negative logarithm of the base dissociation constant. Higher pKb = weaker base. So we need to arrange from weakest base (highest pKb) to strongest base (lowest pKb).
Step 1: Identify the compounds
- C2H5NH2 – ethylamine (1° aliphatic)
- C6H5NHCH3 – N-methylaniline (2° aromatic)
- (C2H5)2NH – diethylamine (2° aliphatic)
- C6H5NH2 – aniline (1° aromatic)
Step 2: Compare aromatic vs aliphatic
Aromatic amines are much weaker bases than aliphatic ones because the lone pair on nitrogen is delocalised into the benzene ring. So aniline and N-methylaniline will have higher pKb (weaker) than ethylamine and diethylamine.
Step 3: Within aromatic amines
N-methylaniline has an electron-donating methyl group on nitrogen, which slightly increases electron density compared to aniline. So aniline is weaker (higher pKb) than N-methylaniline.
Step 4: Within aliphatic amines
In water, diethylamine (2°) is a stronger base than ethylamine (1°) due to better inductive effect, but the solvation effect is less severe for 2° than for 3°. So diethylamine has lower pKb than ethylamine.
Step 5: Arrange from highest to lowest pKb …
Method: Electronic Effects + Solvation Analysis
This method uses inductive effect, resonance effect, solvation (hydration) effect, and steric hindrance to compare basicity. For pKb, remember: lower pKb = stronger base.
(i) Decreasing order of pKb:
C2H5NH2, C6H5NHCH3, (C2H5)2NH, C6H5NH2
Steps:
-
Identify base strength order first (stronger base → lower pKb).
- Aliphatic amines are stronger bases than aromatic amines (due to resonance delocalisation of lone pair in aniline).
- Among aliphatics: (C2H5)2NH (2° amine) > C2H5NH2 (1° amine) in aqueous medium (due to +I effect of two alkyl groups + better solvation of 2° ammonium ion).
- Among aromatics: C6H5NHCH3 > C6H5NH2 (methyl group donates electron density via +I and hyperconjugation).
-
Order of basic strength (aqueous):
(C2H5)2NH>C2H5NH2>C6H5NHCH3>C6H5NH2
-
Convert to pKb order (reverse of basic strength):
C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH
(ii) Increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH, CH3NH2
Steps:
-
Separate aliphatic vs aromatic.
- (C2H5)2NH and CH3NH2 are aliphatic → stronger bases.
- C6H5NH2 and C6H5N(CH3)2 are aromatic → weaker bases.
-
Compare within aliphatic:
- (C2H5)2NH (2°) > CH3NH2 (1°) in aqueous medium.
-
Compare within aromatic:
- C6H5N(CH3)2 has two methyl groups donating electrons → stronger than C6H5NH2.
-
Final increasing order:
C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH
(iii) Increasing order of basic strength:
(a) Aniline, p-nitroaniline, p-toluidine
Steps:
-
Identify substituent effect:
- −NO2 is strong electron-withdrawing (decreases basicity).
- −CH3 is electron-donating (increases basicity).
-
Order:
p-nitroaniline < aniline < p-toluidine
Answer: p-nitroaniline < aniline < p-toluidine
(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2
Steps:
-
Identify type:
- C6H5CH2NH2 (benzylamine) — aliphatic-like (no direct resonance with ring).
- C6H5NHCH3 (N-methylaniline) — aromatic with +I from methyl.
- C6H5NH2 (aniline) — aromatic.
-
Basicity order:
Benzylamine > N-methylaniline > Aniline
Answer: C6H5NH2<C6H5NHCH3<C6H5CH2NH2
(iv) Decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N, NH3
Steps:
- In gas phase, solvation is absent — only inductive effect matters. …
Here are the most common mistakes students make when solving basicity order problems for amines, along with how to avoid each.
1. Confusing pKb with Basic Strength
Mistake: Students often treat a higher pKb as meaning higher basic strength.
- Why it happens: pKb=−logKb. A smaller Kb means a weaker base, but a larger pKb.
- How to avoid: Remember the rule:
- Higher pKb → Weaker base
- Lower pKb → Stronger base
For part (i): You need decreasing pKb (weakest to strongest base). The correct order is:
C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH
2. Ignoring the Difference Between Aqueous and Gas Phase
Mistake: Applying aqueous-phase logic (alkyl groups increase basicity) to gas-phase questions.
- Why it happens: In water, solvation effects dominate. In gas phase, inductive effect (+I) is the only factor.
- How to avoid: For gas phase, more alkyl groups = more electron density on N = stronger base.
For part (iv): Gas phase decreasing basic strength:
(C2H5)3N>(C2H5)2NH>C2H5NH2>NH3
3. Forgetting Resonance in Aromatic Amines
Mistake: Treating aniline like an aliphatic amine.
- Why it happens: Students forget that the lone pair on N in aniline is delocalized into the benzene ring, making it less available for protonation.
- How to avoid: Always check if the N lone pair is part of a conjugated system. If yes, basicity drops sharply.
For part (iii)(b): C6H5NH2 is weaker than C6H5CH2NH2 (benzylamine) because the lone pair in aniline is resonance-stabilized.
4. Misapplying the +I Effect of Alkyl Groups in Aqueous Medium
Mistake: Assuming that more alkyl groups always mean stronger base in water.
- Why it happens: In water, steric hindrance to solvation reduces basicity for bulky amines like (C2H5)3N.
- How to avoid: In aqueous solution, the order is usually:
2∘>1∘>3∘>NH3
(due to balance of +I effect and solvation)
For part (ii): Increasing basic strength in water:
C6H5NH2<C6H5N(CH3)2<CH3NH2<(C2H5)2NH
5. Ignoring Substituent Effects on Aromatic Rings
Mistake: Not considering whether substituents are electron-donating or electron-withdrawing.
- Why it happens: Students focus only on the amine group and forget the ring substituents.
- How to avoid: Use the rule:
- Electron-donating groups (e.g., −CH3) → increase basicity
- Electron-withdrawing groups (e.g., −NO2) → decrease basicity
For part (iii)(a): Increasing basic strength:
p-nitroaniline<aniline<p-toluidine
6. Mixing Up Boiling Point Trends with Basicity
Mistake: Assuming stronger bases have higher boiling points.
- Why it happens: Both depend on intermolecular forces, but differently. …
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). …
- 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. …
- 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 electr …
- 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: …
- 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 o …
- 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. …
- 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 …
- 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 (cro …
- 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). …
- 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). …
- 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. …
- 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. …
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