Which of the following is a 3° amine? (i) 1-methylcyclohexylamine (ii) Triethylamine (iii) tert-butylamine (iv

NCERT Exemplar · Q1

Q. Which of the following is a 3° amine?

(i) 1-methylcyclohexylamine

(ii) Triethylamine

(iii) tert-butylamine

(iv) N-methylaniline

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✓ Free question

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 ( $NH_3$ ), 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 $RNH_3^+$ ) 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 ( $RNH_3^+$ ) has 3 H's → best solvation.

Secondary ( $R_2NH_2^+$ ) has 2 H's → good solvation.

Tertiary ( $R_3NH^+$ ) 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 (pK $_b$ )
Ammonia ( $NH_3$ )	Weakest	Best (3 H's)	None	4.75
Primary ( $RNH_2$ )	Moderate	Good (2 H's)	Minimal	~3.4
Secondary ( $R_2NH$ )	Strong	Moderate (1 H)	Some	~3.2
Tertiary ( $R_3N$ )	Strongest	Poor (0 H's)	Significant	~4.2

Note

pK $_b$ is the negative log of the base dissociation constant. Lower pK $_b$ = stronger base. So secondary (pK $_b$ ≈ 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 $R_3NH^+$ 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.

Watch out

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:

R_3N + H_2O \rightleftharpoons R_3NH^+ + OH^-

The base dissociation constant $K_b$ measures this:

K_b = \frac{[R_3NH^+][OH^-]}{[R_3N]}

A larger $K_b$ 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 ( $-CH_3$ , $-C_2H_5$ , 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:

$3^\circ > 2^\circ > 1^\circ > NH_3$

✗ Solvation Effect (Hydration of the Conjugate Acid)

When the amine accepts $H^+$ , it forms $R_3NH^+$ (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 $R_3NH^+$ → stronger base (because the equilibrium shifts toward protonation).

This suggests:

$1^\circ > 2^\circ > 3^\circ$ (since $1^\circ$ has 3 $H$ 's, $2^\circ$ has 2, $3^\circ$ has 1)

4. The Actual Order (Aqueous Solution)

In water, the observed order for aliphatic amines is:

\textbf{2° > 1° > 3° > NH}_3

Wait — that's not simply "more alkyl = stronger". Why?

The Reasoning (Step by Step)

From $NH_3$ to $1^\circ$ : Adding one alkyl group increases electron density (inductive effect) → stronger base.

So $1^\circ > NH_3$ .
From $1^\circ$ to $2^\circ$ : Adding a second alkyl group further increases electron density → even stronger base.

So $2^\circ > 1^\circ$ .
From $2^\circ$ to $3^\circ$ : Here, the solvation effect becomes dominant.

The $3^\circ$ 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 $3^\circ$ is weaker than $2^\circ$ .

Thus, the net order is:

\textbf{2° > 1° > 3° > NH}_3

5. The Key Formula(e) to Remember

For gas phase (no solvent):

Only inductive effect matters:

\textbf{3° > 2° > 1° > NH}_3

For aqueous solution (common in exams):

Both effects matter — solvation dominates for $3^\circ$ :

\textbf{2° > 1° > 3° > NH}_3

For aromatic amines (e.g., aniline):

The lone pair is delocalized into the benzene ring → much weaker base.

\textbf{Aliphatic amines > Aromatic amines}

6. Quick Exam Tip

If a question asks "basicity order of amines in water", always write:

2° > 1° > 3° > NH $_3$

And explain:

Inductive effect increases from $NH_3$ to $3^\circ$
But solvation of the conjugate acid decreases from $1^\circ$ to $3^\circ$
The balance gives the above order.

7. Summary Table

Amine Type	Inductive Effect	Solvation of $R_3NH^+$	Net Basicity (aq)
$NH_3$	Weakest	Best (3 $H$ 's)	Weakest
$1^\circ$	Moderate	Good (2 $H$ 's)	Moderate
$2^\circ$	Strong	Moderate (1 $H$ )	Strongest
$3^\circ$	Strongest	Poor (0 $H$ 's)	Weaker than $2^\circ$

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.

A 3° amine has three carbon groups directly attached to the nitrogen atom. Triethylamine, with three ethyl groups on N, is the only 3° amine among the options. The correct option is (ii).

The classification of amines — primary (1°), secondary (2°), and tertiary (3°) — depends entirely on how many carbon-containing groups are directly bonded to the nitrogen atom. It has nothing to do with the total number of carbons in the molecule, nor with the complexity of the carbon skeleton elsewhere. The nitrogen's immediate neighbours decide the degree.

Let's examine each option by counting the groups attached to N.

Option (i): 1-methylcyclohexylamine

The name tells you the amine is on a cyclohexane ring, with a methyl substituent at position 1. The nitrogen is attached to the ring carbon — that's one carbon group. The other two positions on N are occupied by hydrogens. So this is a primary (1°) amine.

Watch out

Don't be fooled by the "methyl" in the name — that methyl is on the ring, not on the nitrogen. Only groups directly on N count.
Option (ii): Triethylamine

"Triethyl" means three ethyl groups, and "amine" means they are all attached to nitrogen. The structure is $(CH_3CH_2)_3N$ . Nitrogen has three carbon groups and no hydrogens. That is the definition of a tertiary (3°) amine.

Tip

The prefix "tri-" in the name is a dead giveaway: trialkylamines are always 3°.
Option (iii): tert-butylamine

The name "tert-butylamine" means a tert-butyl group $(CH_3)_3C-$ attached to $NH_2$ . Nitrogen is bonded to exactly one carbon (the quaternary carbon of the tert-butyl group) and two hydrogens. This is a primary (1°) amine, despite the bulky, highly branched alkyl group.

Note

tert-Butylamine is a classic example of a primary amine that looks "tertiary" in its carbon skeleton — but the nitrogen itself is only singly substituted.
Option (iv): N-methylaniline

Aniline is $C_6H_5NH_2$ . "N-methyl" means one hydrogen on nitrogen is replaced by a methyl group. So the structure is $C_6H_5NH(CH_3)$ . Nitrogen is attached to two carbon groups (the phenyl ring and the methyl) and one hydrogen. That makes it a secondary (2°) amine.

✓Final answer

The correct option is (ii) Triethylamine .

Common Mistakes Students Make with 3° Amine Identification

The Concept: What is a 3° (Tertiary) Amine?

A tertiary amine has the nitrogen atom bonded to three carbon atoms (alkyl or aryl groups). The general structure is:

R_3N

where each $R$ is a carbon-containing group (not hydrogen).

Mistake #1: Confusing "tertiary" on carbon vs. nitrogen

The error: Students see "tert-butylamine" (option C) and think "tert" means tertiary amine — but tert-butyl refers to the carbon's substitution, not the nitrogen's.

tert-butylamine = $(CH_3)_3C-NH_2$ → nitrogen is bonded to only one carbon → 1° amine
A tertiary amine requires the nitrogen to be bonded to three carbons.

How to avoid: Always check the nitrogen's bonds, not the carbon's. Draw the structure:

Compound	Structure	Amine class
tert-butylamine	$(CH_3)_3C-NH_2$	1°
Triethylamine	$(CH_3CH_2)_3N$	3° ✓

Mistake #2: Forgetting that N-substituted rings count as carbon bonds

The error: In 1-methylcyclohexylamine (option A), students count only the cyclohexyl ring as one carbon bond and miss the methyl group.

Structure: cyclohexyl ring with $NH_2$ and a $CH_3$ on the same carbon
Nitrogen is bonded to one carbon (the ring carbon) → 1° amine

How to avoid: Count every atom directly bonded to nitrogen. If it's a carbon, it counts. If it's hydrogen, it doesn't.

Mistake #3: Misidentifying aromatic amines

The error: In N-methylaniline (option D), students see "methyl" and "aniline" and assume it's tertiary.

Structure: $C_6H_5-NH-CH_3$
Nitrogen is bonded to two carbons (one phenyl, one methyl) and one hydrogen → 2° amine

How to avoid: Draw the full structure. Aniline ( $C_6H_5NH_2$ ) is 1°. Adding one methyl to nitrogen makes it 2°. Only adding two alkyl groups to nitrogen makes it 3°.

Mistake #4: Rushing and not checking all options

The error: Students see "triethylamine" and think "tri = three ethyl groups" but don't verify that all three are bonded to nitrogen.

How to avoid: Triethylamine is $(CH_3CH_2)_3N$ — three ethyl groups on nitrogen. That's correctly 3°. But always check the other options too — sometimes a trick option looks similar.

Quick Summary Table

Option	Name	Nitrogen bonds	Amine class
A	1-methylcyclohexylamine	1 C, 2 H	1°
B	Triethylamine	3 C, 0 H	3° ✓
C	tert-butylamine	1 C, 2 H	1°
D	N-methylaniline	2 C, 1 H	2°

Final Tip for Exams

Always draw the nitrogen's immediate neighbours. Count only the atoms directly bonded to N. If three are carbon → 3° amine. If two are carbon → 2°. If one is carbon → 1°. If zero → ammonia (not an amine).

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$NH_3$
›Reveal solutionSolution

Because solvation of the ammonium ion opposes the pure inductive trend, the aqueous basicity order of methylamines is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ .

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:

$(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$

The secondary amine is most basic; the tertiary amine drops because of poor cation solvation and steric crowding; ammonia is least basic.

✓Final answer
$The correct option is (D) .$
KEAM 2026Set eng-2026-04184 marksMCQ
$pK_b$
›Reveal solutionSolution

Weakest base = highest $pK_b$ ; 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 $pK_b \approx 9.4$ .

N-methylbenzenamine is slightly more basic than aniline ( $pK_b \approx 9.2$ ) because the electron-donating methyl offsets some delocalisation. The aliphatic amines (methanamine, ethanamine, dimethylamine) are much stronger bases ( $pK_b \approx 3.3$ ).

Thus benzenamine has the highest $pK_b$ .

✓Final answer
$The correct option is (E) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$_b$
›Reveal solutionSolution

Diethylamine (secondary amine) is the strongest base $\Rightarrow$ lowest $pK_b$ .

In aqueous solution basic strength reflects a balance of +I effect and solvation of the cation, giving the order secondary $>$ primary $\approx$ tertiary for aliphatic amines. Among the options, N-Ethylethanamine $(C_2H_5)_2NH$ is a secondary amine with two electron-releasing ethyl groups and good cation solvation, making it the strongest base (lowest $pK_b\approx3.0$ ).

✓Final answer
$The correct option is (D) .$
KEAM 2026Set eng-2026-04214 marksMCQ
$Q. The descending order of basic strength of the following amines is$
(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, $C_6H_5CH_2NH_2$ ): 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 answer
$The correct option is (C) .$
KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Which one of the following compounds is strongly basic in aqueous medium? (A) Benzenamine (B) N-ethylethanamine (C) Phenylmethanamine (D) N,N-Dimethylbenzenamine (E) Ammonia$
›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 $(\mathrm{C_2H_5)_2NH}$ 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 answer
$The correct option is (B) .$
KEAM 2025Set eng-2025-04234 marksMCQ
$Q. The order of basic strength of following amines is$
(i) $CH_3NH_2$
(ii) $(C_2H_5)_2NH$
(iii) $C_6H_5NH_2$
(iv) $C_6H_5NHCH_3$ (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 answer
$The correct option is (E) — (iii) < (iv) < (i) < (ii).$

ANSWER: E
KEAM 2025Set eng-2025-04274 marksMCQ
$_b$
›Reveal solutionSolution
Highest $pK_b$ 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 $pK_b$ .

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 $pK_b$ ).

Approximate $pK_b$ values:

(A) Methanamine ( $CH_3NH_2$ ): 3.36

(B) N-methylmethanamine ( $(CH_3)_2NH$ ): 3.27

(E) Ethanamine ( $C_2H_5NH_2$ ): 3.29

(D) N-Methylaniline: ~9.2

(C) Benzeneamine (aniline): ~9.4

Aliphatic amines (A, B, E) are strong bases (low $pK_b$ ). Between the two aromatic amines, the electron-donating $–CH_3$ 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 $pK_b$ .

✓Final answer
$The correct option is (C) .$
KEAM 2024Set eng-2024-06074 marksMCQ
$Q. The decreasing order of basic strength in aqueous solution of amines is (A) Dimethylamine > Methylamine > Trimethylamine > Ammonia (B) Methylamine > Dimethylamine > Trimethylamine > Ammonia (C) Trimethylamine > Dimethylamine > Methylamine > Ammonia (D) Ammonia > Trimethylamine > Dimethylamine > Methylamine (E) Ammonia > Dimethylamine > Trimethylamine > Methylamine$
›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 $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ .

✓Final answer
$The correct option is (A) .$
KEAM 2024Set eng-2024-06084 marksMCQ
$CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N > NH_3$
›Reveal solutionSolution
For methylamines in water the basic-strength order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ , 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:
$(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$
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 answer
$The correct option is (B) .$
KEAM 2024Set pha-2024-06104 marksMCQ
$Q. 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 ( $C_6H_5NH_2$ ): lone pair delocalised into the ring $\Rightarrow$ weakest (pKb $\approx$ 9.4).

Ammonia ( $NH_3$ ): pKb $\approx$ 4.75.

Benzylamine ( $C_6H_5CH_2NH_2$ ): the $CH_2$ insulates N from the ring; slightly more basic than ammonia (pKb $\approx$ 4.66).

Ethylamine ( $C_2H_5NH_2$ ): alkyl $+I$ effect makes it the strongest (pKb $\approx$ 3.25).

So increasing basicity: $C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2$ .

✓Final answer
$The correct option is (B) . Aniline weakest (resonance), ethylamine strongest (+I), benzylamine just above ammonia.$
KEAM 2022Set eng-2022-P1-A14 marksMCQ
$Q. Among methanamine, ethanamine, benzenamine, N-methylaniline and N, N-dimethylaniline, the weakest and the strongest base in aqueous phase, respectively are (A) benzenamine and methanamine (B) N-methylaniline and ethanamine (C) N, N-dimethylaniline and ethanamine (D) benzenamine and ethanamine (E) N-methylaniline and methanamine$
›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 answer
$The correct option is (D) — benzenamine and ethanamine.$

ANSWER: D
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$pK_b$
›Reveal solutionSolution
Statement (C) is wrong: Hofmann degradation of ethanamide gives methylamine, not ethylamine.

Concept and Intuition

The Hofmann bromamide reaction converts an amide $RCONH_2$ to an amine $RNH_2$ 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

$CH_3CONH_2 \xrightarrow{Br_2,\ NaOH} CH_3NH_2$ (methylamine), one carbon fewer.

Statement (C) claims ethylamine — incorrect.

Check others: aniline is a weaker base than methylamine (higher $pK_b$ ) — (A) true; ethylamine is water-soluble, aniline sparingly so — (B) true; ethylamine + $HNO_2 \rightarrow$ ethanol — (D) true; aniline forms a Lewis salt with $AlCl_3$ so no Friedel–Crafts — (E) true.

Common Mistakes

Forgetting the carbon loss in Hofmann degradation and expecting $RCONH_2 \rightarrow RCH_2NH_2$ -type retention.

✓Final answer
$Br_2$

ANSWER: C

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Q.Which of the following is a 3° amine?

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