Q.Among the following, which is the strongest base ? (A) $C_6H_5NH_2$ (aniline, drawn as a structure) (B) $H_3C-C_6H_4-NH_2$ (para-toluidine — benzene ring with $-CH_3$ and $-NH_2$ at para positions, drawn as a structure) (C) $C_6H_5-CH_2-NH_2$ (benzylamine — benzene ring with a $-CH_2-NH_2$ side chain, drawn as a structure) (D) $O_2N-C_6H_4-NH_2$ (para-nitroaniline — benzene ring with $-NO_2$ and $-NH_2$ at para positions, drawn as a structure)

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set 56/3/11 markMCQ
   Q.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 [$C_6H_5NH_2$]

   ›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.

   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) — $C_6H_5CH_2NH_2$

      Here, the amino group is attached to a $CH_2$ group, not directly to the benzene ring. The benzene ring is separated by a methylene spacer. This means the lone pair on nitrogen cannot participate in resonance with the ring. The nitrogen behaves like an aliphatic amine — its lone pair is fully available for protonation. The benzene ring still exerts a weak inductive electron-withdrawing effect through the $CH_2$ group, but this is much weaker than resonance. Hence, benzylamine is the strongest base.

   Watch out

   A common mistake is to think that because benzylamine has a benzene ring, it will be similar to aniline. But the $CH_2$ group breaks conjugation — the lone pair is not delocalised into the ring. That changes everything.

   Tip

   For quick comparison: any amine where the nitrogen is directly attached to an aromatic ring (aniline derivatives) will be weaker than an amine where the nitrogen is separated by at least one $sp^3$ carbon (benzylamine). The only exception is if the ring has very strong electron-donating groups.

   Basicity order for these compounds:

   $\text{Benzylamine} > \text{4-methylaniline} > \text{Aniline} > \text{4-nitroaniline}$

   ✓Final answer

   The strongest base is benzylamine, option (B).

2. 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:

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

   This is reflected in their base-dissociation behaviour: $CH_3NH_2$ is a noticeably stronger base than $NH_3$, consistent with the inductive argument (in aqueous solution, solvation of the resulting ammonium ion also plays a secondary role, but the $+I$ effect of the methyl group is the primary reason taught at this level).

   ✓Final answer

   Methanamine ($CH_3NH_2$) is a stronger base than ammonia because the electron-donating ($+I$) methyl group raises the electron density on nitrogen, making its lone pair more available to bind a proton.

3. CBSE 2026Set ANNUAL1 mark
   Q.Arrange the following in decreasing order of their basic strength: $C_2H_5NH_2$, $(C_2H_5)_2NH$, $(C_2H_5)_3N$, $C_6H_5NH_2$

   ›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

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

   Balancing (1) and (2) for the simple ethylamines in water gives the well-established order $2^\circ > 1^\circ > 3^\circ$ (the +I effect of a second ethyl group boosts basicity more than the added steric/solvation penalty, but a third group tips the balance the other way), with aniline last because of (3):

   $(C_2H_5)_2NH \;>\; C_2H_5NH_2 \;>\; (C_2H_5)_3N \;>\; C_6H_5NH_2$

   ✓Final answer

   $(C_2H_5)_2NH > C_2H_5NH_2 > (C_2H_5)_3N > C_6H_5NH_2$

4. 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 > C2H5NH2 > NH3 > C6H5NH2.

   Aniline, C6H5NH2, is the weakest base here because the nitrogen lone pair conjugates into the aromatic ring (delocalisation lowers its availability to a proton), and further the resulting anilinium ion is destabilised relative to alkylammonium ions.

   ✓Final answer

   (d) (C2H5)2NH.

5. CBSE 2025Set ANNUAL1 mark
   Q.Arrange the following in decreasing order of their basic strength: $C_6H_5NH_2$, $C_2H_5NH_2$, $(C_2H_5)_2NH$, $NH_3$

   ›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.

   • $NH_3$ (ammonia): no alkyl substituent at all, so no +I effect enhancement, but also no steric hindrance — intermediate.

   • $C_6H_5NH_2$ (aniline): the nitrogen lone pair is conjugated into the aromatic ring (resonance delocalisation over the ortho/para carbons), making it far less available to bind $H^+$; additionally the ring is electron-withdrawing by induction. This makes aniline much less basic than even ammonia.

   ✓Final answer

   Decreasing basic strength: $(C_2H_5)_2NH > C_2H_5NH_2 > NH_3 > C_6H_5NH_2$

6. CBSE 2024Set 56/3/11 markMCQ
   Q.The order of increasing basicities of $CH_3NH_2$ (I), $(CH_3)_2NH$ (II), $(CH_3)_3N$ (III) and $C_6H_5NH_2$ (IV) in aqueous media is : (A) IV < III < I < II (B) II < I < IV < III (C) I < II < III < IV (D) II < III < I < IV

   ›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.

   1. 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.

   2. 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.

   3. The net effect in water is a compromise.

      The inductive effect and solvation effect pull in opposite directions as you add methyl groups. Experimentally, the order of basicity in water for methylamines is:

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

      Dimethylamine (II) wins because it has two methyl groups (good inductive effect) and two N–Hs (good solvation). Trimethylamine (III) loses to methylamine (I) because its solvation disadvantage (only one N–H) outweighs its extra inductive push.

   4. Arrange in increasing order.

      Weakest: aniline (IV).

      Next: trimethylamine (III) — weaker than methylamine due to poor solvation.

      Then: methylamine (I).

      Strongest: dimethylamine (II).

      So increasing order: IV < III < I < II.

   Watch out

   A common mistake is to assume that more alkyl groups always mean stronger base, even in water. That would give III > II > I, which is wrong. Always check solvation of the conjugate acid — the number of N–H bonds matters.

   Tip

   A quick memory aid for aqueous basicity of methylamines: “2 > 1 > 3 > 0” — dimethylamine > methylamine > trimethylamine > ammonia. Aniline is always weaker than ammonia, so it goes at the very end.

   ✓Final answer

   The correct option is (A) IV < III < I < II.

7. 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.

   • Among the two aliphatic amines, (C2H5)2NH (secondary) has more +I donation than C2H5NH2 (primary), so in aqueous solution diethylamine is the most basic of the four.

   ✓Final answer

   (d) (C2H5)2NH — diethylamine is the most basic.

8. 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.

   Ethyl (-C2H5) has a slightly stronger +I effect than methyl (-CH3), so ethylamine donates electron density to nitrogen more effectively than methylamine, making ethylamine the stronger base of the two. Correspondingly, methylamine, being the weaker base, is relatively more acidic (its conjugate acid is comparatively a stronger acid) than ethylamine.

   ✓Final answer

   More (methylamine is more acidic / less basic than ethylamine).

9. 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.

   Additionally, in aqueous solution the resulting ethylammonium ion is stabilised somewhat by the alkyl group and by solvation, further favouring protonation.

   ✓Final answer

   The ethyl group's electron-donating (+I) inductive effect increases the electron density on nitrogen in ethylamine, making its lone pair more available for protonation than ammonia's — so ethylamine is the stronger base.

10. CBSE 2023Set 56/1/11 markMCQ
    Q.Which of the following is least basic ? (A) $(CH_3)_2NH$ (B) $NH_3$ (C) Aniline, $C_6H_5NH_2$ (benzene ring bearing $-NH_2$, drawn as a structure in the paper) (D) $(CH_3)_3N$

    ›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.

    1. 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.

    2. 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 $NH_3$.

    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.

    4. 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.

       Resonance structures of aniline:

       $C_6H_5-NH_2 \leftrightarrow C_6H_5=NH_2^+$ (with negative charge on ortho/para positions)

       This delocalization stabilizes the neutral amine, but the conjugate acid (anilinium ion) does not benefit from such resonance. Hence, the equilibrium shifts toward the neutral form, making aniline a weak base.

       The $pK_b$ values confirm this:

       • $(CH_3)_2NH$: $pK_b \approx 3.27$
       • $(CH_3)_3N$: $pK_b \approx 4.20$
       • $NH_3$: $pK_b \approx 4.75$
       • Aniline: $pK_b \approx 9.38$

       A higher $pK_b$ means a weaker base. Aniline’s $pK_b$ is far higher than the others.

    Watch out

    A common mistake is to think that more alkyl groups always mean higher basicity in water. While true in the gas phase, solvation effects reverse the order for tertiary amines in solution. However, this nuance does not affect the ranking here — aniline is still the weakest by a large margin.

    5. Final ranking From most basic to least basic: $(CH_3)_2NH > (CH_3)_3N > NH_3 > C_6H_5NH_2$ So aniline is the least basic.

    ✓Final answer

    The least basic compound is aniline, option (C).

11. CBSE 2023Set 56/2/11 markMCQ
    Q.Among the following, which is the strongest base ? (A) $C_6H_5NH_2$ (aniline, drawn as a structure) (B) $H_3C-C_6H_4-NH_2$ (para-toluidine — benzene ring with $-CH_3$ and $-NH_2$ at para positions, drawn as a structure) (C) $C_6H_5-CH_2-NH_2$ (benzylamine — benzene ring with a $-CH_2-NH_2$ side chain, drawn as a structure) (D) $O_2N-C_6H_4-NH_2$ (para-nitroaniline — benzene ring with $-NO_2$ and $-NH_2$ at para positions, drawn as a structure)

    ›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

    The nitro group is a powerful electron-withdrawing group, both through resonance ($-R$ effect) and induction ($-I$ effect). It pulls electron density away from the ring and, by extension, from the nitrogen atom. This makes the lone pair on nitrogen even less available for protonation. Para-nitroaniline is the weakest base in this set, with $pK_b \approx 13$ (conjugate acid $pK_a \approx 1$).

    Watch out

    A common mistake is to think that because the nitro group is at the para position (not ortho), its effect is negligible. In fact, resonance operates through the entire conjugated system, and para-substitution allows direct resonance interaction with the amino group.

    Ranking the basicity

    Putting it all together, the order from strongest to weakest base is:

    $\text{Benzylamine} > \text{para-Toluidine} > \text{Aniline} > \text{para-Nitroaniline}$

    or in symbols:

    $C_6H_5CH_2NH_2 > CH_3-C_6H_4-NH_2 > C_6H_5NH_2 > O_2N-C_6H_4-NH_2$

    Compound	Key Feature	Relative Basicity
    Benzylamine	$-CH_2-$ insulates $N$ from ring	Strongest
    para-Toluidine	$-CH_3$ donates electrons	Moderate
    Aniline	Direct resonance withdrawal	Weak
    para-Nitroaniline	$-NO_2$ withdraws strongly	Weakest

    ✓Final answer

    The correct option is (C) $C_6H_5-CH_2-NH_2$ (benzylamine).

12. CBSE 2023Set 56/3/11 markMCQ
    Q.Among the following, which is the strongest base ? (A) $H_3C-C_6H_4-NH_2$ (para-toluidine — benzene ring with $-CH_3$ and $-NH_2$ at para positions, drawn as a structure) (B) $O_2N-C_6H_4-NH_2$ (para-nitroaniline — benzene ring with $-NO_2$ and $-NH_2$ at para positions, drawn as a structure) (C) $C_6H_5NH_2$ (aniline, drawn as a structure) (D) $C_6H_5-CH_2-NH_2$ (benzylamine — benzene ring with a $-CH_2-NH_2$ side chain, drawn as a structure)

    ›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.

    1. 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.

    2. 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.

    3. 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.

    4. Benzylamine (D) — $C_6H_5-CH_2-NH_2$

       Here, the amino group is attached to a carbon that is one bond away from the ring. The lone pair on nitrogen cannot conjugate with the aromatic π-system because the methylene ($-CH_2-$) group breaks the conjugation. The ring can still exert an inductive effect (slightly electron-withdrawing due to the sp² carbons), but this is weak and over a longer distance. The lone pair is essentially as available as in a simple aliphatic amine. Benzylamine is therefore the strongest base among the four.

    Watch out

    A common mistake is to think that because benzylamine has an aromatic ring, it behaves like aniline. But the key difference is the methylene spacer — it isolates the amino group from the ring’s π-system, so resonance delocalisation of the lone pair does not occur. Always check whether the nitrogen lone pair is directly attached to the aromatic ring.

    Tip

    For quick comparison in such problems:

    • Direct attachment to an aromatic ring → weaker base (resonance effect dominates).
    • Electron-donating substituents on the ring → slightly stronger base than aniline.
    • Electron-withdrawing substituents → much weaker base.
    • A methylene spacer between ring and amino group → aliphatic-like, strong base.

    So the order of basic strength is:

    para-nitroaniline (B) < aniline (C) < para-toluidine (A) < benzylamine (D).

    ✓Final answer

    The strongest base is benzylamine, option (D).