Among the following, which is the strongest base ? (A) 4-nitroaniline [$O_2N-C_6H_4-NH_2$] (B) Benzylamine [$C_6H_5CH_2NH_2$] (C) 4-methylaniline [$CH_3-C_6H_4-NH_2$] (D) Aniline [$C_6H_5NH_2$]

> [!TLDR] > Basicity of amines depends on electron density on nitrogen. Benzylamine has an alkyl group (electron-donating) attached to the amino group, making it the strongest base among the given aromatic amines. The correct option is **(B)**. ## Why Basicity Order Matters Here The question asks you to compare the basic strength of four amines. In organic chemistry, basicity is directly linked to how readily the nitrogen atom can donate its lone pair. The more electron-rich the nitrogen, the stronger the base. For aromatic amines, the key factor is **resonance** and **substituent effects** — electron-donating groups increase basicity, while electron-withdrawing groups decrease it. Let’s break down each compound. 1. **Aniline (D)** — $C_6H_5NH_2$ The lone pair on nitrogen is delocalised into the benzene ring through resonance. This makes the nitrogen less available to accept a proton, so aniline is a weaker base than aliphatic amines. 2. **4-nitroaniline (A)** — $O_2N-C_6H_4-NH_2$ The nitro group ($-NO_2$) is a strong electron-withdrawing group. It pulls electron density away from the nitrogen via both inductive and resonance effects. This drastically reduces the electron density on nitrogen, making it the **weakest base** among the four. 3. **4-methylaniline (C)** — $CH_3-C_6H_4-NH_2$ The methyl group is electron-donating (hyperconjugation + inductive effect). It pushes electron density toward the ring, which slightly increases electron density on nitrogen compared to aniline. So 4-methylaniline is a stronger base than aniline, but still weaker than benzylamine. 4. **Benzylamine (B)** — $C_6H_5CH_2NH_2$ …

Among the following, which is the strongest base ? (A) 4-nitroaniline [O₂N-C₆H₄-NH₂] (B) Benzylamine [C₆H₅CH₂N

Question

O_2N-C_6H_4-NH_2

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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? …

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): …

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Basicity of amines depends on electron density on nitrogen. Benzylamine has an alkyl group (electron-donating) attached to the amino group, making it the strongest base among the given aromatic amines. The correct option is (B).

Why Basicity Order Matters Here

The question asks you to compare the basic strength of four amines. In organic chemistry, basicity is directly linked to how readily the nitrogen atom can donate its lone pair. The more electron-rich the nitrogen, the stronger the base. For aromatic amines, the key factor is resonance and substituent effects — electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

Let’s break down each compound.

Aniline (D) — $C_6H_5NH_2$

The lone pair on nitrogen is delocalised into the benzene ring through resonance. This makes the nitrogen less available to accept a proton, so aniline is a weaker base than aliphatic amines.
4-nitroaniline (A) — $O_2N-C_6H_4-NH_2$

The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group. It pulls electron density away from the nitrogen via both inductive and resonance effects. This drastically reduces the electron density on nitrogen, making it the weakest base among the four.
4-methylaniline (C) — $CH_3-C_6H_4-NH_2$

The methyl group is electron-donating (hyperconjugation + inductive effect). It pushes electron density toward the ring, which slightly increases electron density on nitrogen compared to aniline. So 4-methylaniline is a stronger base than aniline, but still weaker than benzylamine.
Benzylamine (B) — $C_6H_5CH_2NH_2$ …

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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 27 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$O_2N-C_6H_4-NH_2$
›Reveal solutionSolution
Basicity of amines depends on electron density on nitrogen. Benzylamine has an alkyl group (electron-donating) attached to the amino group, making it the strongest base among the given aromatic amines. The correct option is (B).

Why Basicity Order Matters Here

The question asks you to compare the basic strength of four amines. In organic chemistry, basicity is directly linked to how readily the nitrogen atom can donate its lone pair. The more electron-rich the nitrogen, the stronger the base. For aromatic amines, the key factor is resonance and substituent effects — electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

Let’s break down each compound.

Aniline (D) — $C_6H_5NH_2$

The lone pair on nitrogen is delocalised into the benzene ring through resonance. This makes the nitrogen less available to accept a proton, so aniline is a weaker base than aliphatic amines.

4-nitroaniline (A) — $O_2N-C_6H_4-NH_2$

The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group. It pulls electron density away from the nitrogen via both inductive and resonance effects. This drastically reduces the electron density on nitrogen, making it the weakest base among the four.

4-methylaniline (C) — $CH_3-C_6H_4-NH_2$

The methyl group is electron-donating (hyperconjugation + inductive effect). It pushes electron density toward the ring, which slightly increases electron density on nitrogen compared to aniline. So 4-methylaniline is a stronger base than aniline, but still weaker than benzylamine.

Benzylamine (B) — $C_6H_5CH_2NH_2$ …
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CBSE 2026Set ANNUAL1 mark
$Q. Why is methanamine a stronger base than ammonia?$
›Reveal solutionSolution

Basicity of an amine depends on how available its nitrogen lone pair is to accept a proton; an alkyl group's electron-donating ( $+I$ ) inductive effect increases that availability compared with plain ammonia.

In ammonia, $NH_3$ , the nitrogen is bonded only to three hydrogens, which contribute no electron-donating effect of their own. In methanamine, $CH_3-NH_2$ , the methyl group is electron-releasing (+I effect): it pushes electron density through the $C-N$ sigma bond onto the nitrogen atom, increasing the electron density available in its lone pair.

A more electron-rich lone pair is a better proton acceptor (Lewis base), so protonation is favoured more strongly for methanamine than for ammonia:

$CH_3NH_2 + H^+ \rightleftharpoons CH_3NH_3^+ \quad(\text{favoured more than})\quad NH_3 + H^+ \rightleftharpoons NH_4^+$ …

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CBSE 2026Set ANNUAL1 mark
$C_2H_5NH_2$
›Reveal solutionSolution
Aqueous basicity of amines balances +I electron release (favours more alkyl groups), steric hindrance to solvation of the protonated ion (disfavours bulky/3° amines), and resonance delocalisation of the lone pair (drastically weakens aniline).

Factors at play

+I effect: each ethyl group pushes electron density onto N, making the lone pair more available and increasing basicity — this alone would predict $3^\circ > 2^\circ > 1^\circ$ .

Steric hindrance to solvation: basicity in water is effectively measured by how well the protonated ( $R_3NH^+$ ) ion is stabilised by H-bonding with water. A bulky, highly alkyl-substituted ammonium ion like $(C_2H_5)_3NH^+$ is harder to solvate, which lowers its effective basicity in water — this pulls $3^\circ$ amines down.

Aromatic ring delocalisation: in aniline, $C_6H_5NH_2$ , the lone pair on N is delocalised into the benzene ring by resonance, making it far less available for protonation — anilines are always much weaker bases than aliphatic amines. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is most basic?$
(a) C6H5NH2
(b) NH3
(c) C2H5NH2
(d) (C2H5)2NH
›Reveal solutionSolution

In aqueous solution, basicity of these amines follows secondary > primary > NH3 > aniline — aniline is weakest because its nitrogen lone pair is delocalised into the benzene ring, while a secondary alkylamine's two electron-donating ethyl groups make it the strongest base here.

Electron-donating alkyl (+I) groups on nitrogen increase electron density on N, making the lone pair more available to accept a proton, so alkyl-substituted amines are more basic than NH3. Between primary and secondary alkylamines in water, the extra +I contribution from the second ethyl group in (C2H5)2NH outweighs its slightly greater steric hindrance/solvation penalty, making it the strongest base of this set: (C2H5)2NH > …

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CBSE 2025Set ANNUAL1 mark
$C_6H_5NH_2$
›Reveal solutionSolution
Diethylamine is the strongest base (two electron-donating ethyl groups, still well solvated), followed by ethylamine, then unsubstituted ammonia, with aniline the weakest because the ring delocalises the nitrogen lone pair by resonance.

Basicity of an amine depends on how available the nitrogen lone pair is to accept a proton, which in aqueous solution is governed by three competing effects: (i) the +I (electron-donating) effect of alkyl groups, which pushes electron density onto N and increases basicity; (ii) steric hindrance and the extent of solvation (H-bonding) of the resulting ammonium cation, which is reduced by bulky/more numerous alkyl groups and lowers basicity; and (iii) resonance delocalisation of the lone pair, which sharply lowers basicity when N is attached to an aromatic ring.

$(C_2H_5)_2NH$ (diethylamine, 2°): two ethyl groups give a strong +I effect, and being only disubstituted it is still reasonably well solvated in water — the strongest base of the set.

$C_2H_5NH_2$ (ethylamine, 1°): one +I-donating ethyl group, well solvated — a stronger base than plain ammonia but weaker than the disubstituted amine above. …
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CBSE 2024Set 56/3/11 markMCQ
$CH_3NH_2$
›Reveal solutionSolution
In aqueous solution, basicity of amines depends on a balance between the inductive effect (which increases electron density on nitrogen) and solvation of the conjugate acid (which stabilises it). For methyl-substituted amines, the order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > C_6H_5NH_2$ , so the correct option is (A).

The question asks for the increasing order of basicity in aqueous media — that’s the key. In water, basicity is not just about how much the nitrogen “wants” to donate its lone pair; it’s also about how stable the resulting ammonium ion is once it forms. Two effects compete here: the inductive effect of alkyl groups (which push electrons toward nitrogen, making it more basic) and the solvation effect (water molecules stabilise the charged ammonium ion by hydrogen bonding — more hydrogens on the nitrogen mean better solvation).

For aniline ( $C_6H_5NH_2$ ), the lone pair on nitrogen is delocalised into the aromatic ring, making it far less available for protonation. That’s why it’s always the weakest base among these four — no contest.

Now, among the methylamines, the trend in the gas phase (no solvent) is clear: more methyl groups → more electron donation → stronger base. So gas-phase order would be $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ . But in water, the story changes because the conjugate acid of trimethylamine, $(CH_3)_3NH^+$ , has only one N–H bond — it can form only one strong hydrogen bond with water. The conjugate acid of dimethylamine, $(CH_3)_2NH_2^+$ , has two N–H bonds, so it’s better solvated and more stabilised. This extra stabilisation outweighs the extra inductive effect of the third methyl group, making dimethylamine the strongest base in water.

Let’s walk through the reasoning step by step.

Identify the weakest base first.

Aniline (IV) has its lone pair conjugated with the benzene ring — resonance delocalisation reduces electron density on nitrogen drastically. It is by far the least basic. So IV must come first in the increasing order. That eliminates options (B) and (C), which place aniline later.

Compare the three methylamines in water.

The inductive effect of methyl groups increases electron density on nitrogen, favouring basicity: more methyl groups → stronger base, all else equal. But “all else” is not equal in water. The conjugate acid’s ability to be stabilised by solvation depends on the number of N–H bonds: each N–H can hydrogen-bond with water.

$(CH_3)_3NH^+$ has one N–H.

$(CH_3)_2NH_2^+$ has two N–Hs.

$CH_3NH_3^+$ has three N–Hs.

More N–H bonds mean better solvation, which lowers the energy of the conjugate acid and thus makes the base stronger. So solvation favours the opposite order: more hydrogens → stronger base.

The net effect in water is a compromise. …
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CBSE 2024Set D1 markMCQ
$Q. Which of the following is the most basic?$
(a) C6H5NH2
(b) (C6H5)2NH
(c) C2H5NH2
(d) (C2H5)2NH
›Reveal solutionSolution
Aliphatic amines > aromatic; secondary diethylamine is most basic here.

Basicity depends on availability of the nitrogen lone pair:

Aromatic amines C6H5NH2 (aniline) and (C6H5)2NH (diphenylamine) are weak bases because the lone pair is delocalised into the benzene ring(s); diphenylamine is the weakest.

Aliphatic amines are stronger bases due to the electron-releasing (+I) alkyl groups. …
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CBSE 2024Set B1 mark
$Q. Fill in the blank: Methyl amine is ______ acidic than ethyl amine.$
›Reveal solutionSolution

Ethyl amine is a stronger base than methyl amine because the larger ethyl group has a greater +I (electron-releasing) effect than methyl, so relative to ethylamine, methylamine is more acidic/less basic.

Basicity of simple aliphatic amines increases as the alkyl group attached to nitrogen becomes a better electron donor (+I effect), because this raises electron density on nitrogen and better stabilises the positive charge on the protonated ammonium ion formed.

…

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CBSE 2024Set ANNUAL1 mark
$Q. Why is ethyl amine more basic than ammonia?$
›Reveal solutionSolution

An alkyl group's +I (electron-releasing inductive) effect pushes extra electron density onto nitrogen, strengthening its ability to accept a proton compared to plain ammonia.

Basicity of an amine depends on how readily the nitrogen's lone pair of electrons is available to accept a proton (H⁺) — the more available/electron-rich the lone pair, the stronger the base.

In ethylamine (CH3CH2-NH2), the ethyl group is an alkyl group with a +I (positive inductive) effect — it pushes electron density TOWARD the nitrogen atom through the sigma-bond framework. This makes the nitrogen's lone pair MORE electron-rich and more readily available to accept (bond to) an incoming proton, compared to ammonia (NH3), which has no alkyl group to donate electron density.

…

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CBSE 2023Set 56/1/11 markMCQ
$(CH_3)_2NH$
›Reveal solutionSolution
Basicity of amines depends on the availability of the lone pair on nitrogen for protonation. Alkyl groups are electron-donating (inductive effect), increasing basicity, while the phenyl ring in aniline is electron-withdrawing (resonance effect), drastically reducing basicity. Therefore, aniline is the least basic among the given options.

The core concept: what makes an amine basic?

An amine is basic because the nitrogen atom has a lone pair of electrons that can accept a proton ( $H^+$ ). The more available this lone pair is, the stronger the base. Two main factors affect this availability in the compounds listed:

Inductive effect: Alkyl groups ( $-CH_3$ ) push electron density toward nitrogen, making the lone pair more available.

Resonance effect: In aniline, the lone pair on nitrogen is delocalized into the benzene ring, making it much less available for protonation.

Comparing the alkyl amines: $(CH_3)_2NH$ and $(CH_3)_3N$

In the gas phase, basicity increases with the number of alkyl groups: $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ . However, in aqueous solution (the usual context for such questions), a different order emerges due to solvation effects.

$(CH_3)_2NH$ (dimethylamine) has two methyl groups donating electron density, and its conjugate acid is well-stabilized by hydrogen bonding with water.

$(CH_3)_3N$ (trimethylamine) has three methyl groups, but the bulky alkyl groups hinder solvation of the protonated form, slightly reducing its basicity in water. The typical order in water is: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ . So both $(CH_3)_2NH$ and $(CH_3)_3N$ are more basic than $NH_3$ .

Where does $NH_3$ stand?

Ammonia has no alkyl groups to donate electron density, so its lone pair is less available than in the alkyl amines. It is more basic than aniline but less basic than the alkyl amines listed.

Why aniline is the least basic

In aniline, the nitrogen’s lone pair is conjugated with the $\pi$ -electron system of the benzene ring. This resonance delocalization spreads the lone pair over the ring, making it much less available for protonation.
$C_6H_5-NH_2 \leftrightarrow C_6H_5=NH_2^+$
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CBSE 2023Set 56/2/11 markMCQ
$C_6H_5NH_2$
›Reveal solutionSolution

Basicity of amines depends on the availability of the lone pair on nitrogen. Electron-donating groups increase basicity; electron-withdrawing groups (especially through resonance) decrease it. Benzylamine is the strongest base because its lone pair is insulated from the ring by a $-CH_2-$ spacer, while aniline and its derivatives lose electron density through resonance. The answer is (C).

The basicity of an amine hinges on one simple question: how available is the lone pair on nitrogen to accept a proton? The more electron-rich the nitrogen, the more readily it bonds with $H^+$ , and the stronger the base.

In aromatic amines like aniline, the lone pair on nitrogen can delocalize into the benzene ring through resonance. This delocalization spreads the electron density away from nitrogen, making it less available for protonation. Any substituent on the ring will either amplify or dampen this effect depending on whether it donates or withdraws electrons.

Let's examine each compound systematically.

Step-by-step comparison

1. Aniline ( $C_6H_5NH_2$ ) — the reference point

The amino group is directly attached to the benzene ring. The lone pair on nitrogen participates in resonance with the $\pi$ -system of the ring, delocalizing into the aromatic cloud. This makes nitrogen less basic than aliphatic amines. The $pK_b$ of aniline is around $9.4$ (or $pK_a$ of its conjugate acid $\approx 4.6$ ), which is significantly weaker than methylamine ( $pK_b \approx 3.4$ ).

2. Para-toluidine ( $H_3C-C_6H_4-NH_2$ ) — electron donation helps

The methyl group at the para position is an electron-donating group (through hyperconjugation and weak inductive effect, often called the $+I$ effect). It pushes electron density into the ring, which in turn makes the nitrogen slightly more electron-rich. This partially counteracts the resonance withdrawal, so para-toluidine is a slightly stronger base than aniline. The $pK_b$ drops to around $8.9$ (conjugate acid $pK_a \approx 5.1$ ).

3. Benzylamine ( $C_6H_5-CH_2-NH_2$ ) — insulation is key

Here the amino group is separated from the benzene ring by a $-CH_2-$ spacer. This is crucial: the lone pair on nitrogen cannot participate in resonance with the aromatic ring because it's not directly conjugated. The nitrogen behaves almost like an aliphatic amine. The benzene ring exerts only a weak inductive effect (slightly electron-withdrawing through the $\sigma$ -bond), but this is far less significant than resonance delocalization. Benzylamine has a $pK_b \approx 4.7$ (conjugate acid $pK_a \approx 9.3$ ), making it much more basic than aniline.

Tip

Whenever you see a $-CH_2-$ group between nitrogen and an aromatic ring, treat the amine as essentially aliphatic. The insulating methylene group blocks resonance.

4. Para-nitroaniline ( $O_2N-C_6H_4-NH_2$ ) — strong withdrawal …
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CBSE 2023Set 56/3/11 markMCQ
$H_3C-C_6H_4-NH_2$
›Reveal solutionSolution
Basicity of amines depends on the availability of the lone pair on nitrogen for protonation. Electron-donating groups increase basicity; electron-withdrawing groups decrease it. Benzylamine is the strongest base here because its amino group is separated from the aromatic ring by a methylene group, preventing resonance delocalisation of the lone pair into the ring.

The question asks you to compare the basic strength of four aromatic amines. The key idea is simple: a base is stronger if its lone pair is more available to accept a proton. In aromatic systems, the lone pair on nitrogen can be delocalised into the benzene ring through resonance, which reduces its availability. Any substituent that either enhances or reduces this delocalisation will affect basicity.

Let’s examine each compound step by step.

Aniline (C) — $C_6H_5NH_2$

The lone pair on nitrogen is conjugated with the aromatic π-system. This resonance delocalisation makes the lone pair less available for protonation. Aniline is a weaker base than aliphatic amines (like benzylamine) because of this effect. Its $pK_b$ is about 9.4.

para-Toluidine (A) — $H_3C-C_6H_4-NH_2$

The methyl group at the para position is an electron-donating group (EDG) by hyperconjugation and inductive effect. It pushes electron density toward the ring, which slightly increases the electron density on nitrogen. This makes the lone pair more available than in aniline. So para-toluidine is a stronger base than aniline, but still weaker than an aliphatic amine because the resonance delocalisation is still present.

para-Nitroaniline (B) — $O_2N-C_6H_4-NH_2$

The nitro group is a strong electron-withdrawing group (EWG) by both inductive and resonance effects. It pulls electron density away from the ring, and through resonance, it further delocalises the lone pair on nitrogen into the ring and onto the nitro group. This drastically reduces the availability of the lone pair. para-Nitroaniline is the weakest base among the four.

Benzylamine (D) — $C_6H_5-CH_2-NH_2$ …
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Why Basicity Order Matters Here

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