Q.Why is aniline less basic than ethylamine?

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.Which of the following groups when present at para position increases the basic strength of aniline?

   (a) $-\text{NO}_2$

   (b) $-\text{Br}$

   (c) $-\text{NH}_2$

   (d) $-\text{COOH}$

   ›Reveal solutionSolution

   An electron-donating para substituent raises aniline's basicity; among the options only –NH$_2$ is electron-donating.

   The basic strength of aniline depends on the availability of the lone pair on the nitrogen of the –NH$_2$ group. Groups that push electron density into the ring (electron-donating, +R/+I) increase the electron density on nitrogen and therefore increase basicity; electron-withdrawing groups do the opposite.

   Assessing the para substituents:

   • $-\text{NO}_2$: strongly electron-withdrawing ($-R,\,-I$) → decreases basicity.
   • $-\text{Br}$: electron-withdrawing by induction → slightly decreases basicity. …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Amines behaves as ______ due to presence of an unshared pair of electron on Nitrogen atom.

   ›Reveal solutionSolution

   Because of the nitrogen lone pair, amines behave as bases (and nucleophiles).

   In an amine, the nitrogen atom carries an unshared (lone) pair of electrons. It can donate this lone pair to a proton (H+) or to an electron-deficient centre. Donating the lone pair to H+ makes the amine act as a Bronsted/Lewis base; donating it t …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2025Set ANNUAL1 mark
   Q.Arrange the following in increasing order of their basic strength- C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2

   ›Reveal solutionSolution

   Aniline is the weakest (its lone pair is delocalised into the ring); among the ethylamines in water the accepted order is secondary > tertiary > primary, so diethylamine is the strongest and the increasing order is C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH.

   Basicity in aqueous solution is decided by the net effect of three factors: the electron-donating alkyl groups (+I effect, which raises the electron density on nitrogen and increases basicity), the steric hindrance to protonation together with the solvation (hydrogen bonding) of the resulting ammonium cation, and - for aniline only - resonance delocalisation of the nitrogen lone pair into the benzene ring, which sharply lowers its availability and hence its basicity.

   $C_6H_5NH_2$ (aniline) is the weakest base of the four, because its lone pair is delocalised into the aromatic ring and is much less available for protonation.

   Among the three ethylamines, the order actually observed in aqueous solution (the balance of +I effect, steric crowding and cation solvation) is:

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

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Identify the most basic among the given amines:

   (a) C6H3NH2

   (b) (C2H5)2NH

   (c) (C2H5)3N

   (d) C2H5NH2

   ›Reveal solutionSolution

   In water, basicity of these amines follows (C2H5)2NH > (C2H5)3N > C2H5NH2, so the most basic is diethylamine, (C2H5)2NH.

   Basicity of an amine in aqueous solution is governed by three competing effects: (i) the +I (electron-releasing) effect of alkyl groups, which raises electron density on N and favours protonation; (ii) solvation of the resulting ammonium cation by water through H-bonding, which stabilises the cation and favours basicity; and (iii) steric hindrance of bulky alkyl groups around N, which hampers solvation.

   • C2H5NH2 (1°): one ethyl group gives a modest +I effect; good solvation (two N-H bonds available). …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 mark
   Q.Arrange the following compounds in increasing order of their basic strength in aqueous solution: NH3, C2H5NH2, (C2H5)2NH, (C2H5)3N

   ›Reveal solutionSolution

   In AQUEOUS solution, basicity of amines depends not just on the electron-donating (+I) effect of alkyl groups but also on steric hindrance and how well the resulting ammonium ion is stabilised by hydrogen-bonding with water (solvation); this combination gives the order NH3 < 3 degree < 1 degree < 2 degree amine for the ethyl series.

   Factors affecting basicity in water:

   1. Inductive (+I) effect of alkyl groups: increases electron density on N, favouring basicity; increases with number of alkyl groups (3 degree > 2 degree > 1 degree > NH3 on this factor alone).
   2. Steric hindrance: bulky alkyl groups around N hinder the approach of a proton and hinder solvation of the resulting ammonium cation.
   3. Solvation (H-bonding) of the ammonium ion R-NH3+ (or its substituted forms) by water: more N-H bonds allow more hydrogen bonding with water, better stabilising the cation and boosting the observed basicity.

   NH3 has no alkyl groups at all (weakest +I effect) - it is the weakest base of the four.

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 mark
   Q.Why do amines react as nucleophiles?

   ›Reveal solutionSolution

   An available, reasonably accessible lone pair on N is exactly what defines a good nucleophile/Lewis base.

   In an amine, R–NH₂ (or R₂NH, R₃N), the nitrogen is sp³ hybridized (approximately) with one lone pair of electrons not involved in any bond. This lone pair is available to attack an electron-deficient (electrophilic) carbon or other centre, forming a new bond — the defining behaviour of a nucleophile (and, correspondingly, amines also behave as Lewis/Brønsted bases, accepting a proton with this same lone pair). This is why amines readi …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following is least basic ?

   (i) (CH₃)₂NH

   (ii) NH₃

   (iii) C₆H₅NH₂

   (iv) (CH₃)₃N

   ›Reveal solutionSolution

   Aniline is least basic because the lone pair on its N atom is delocalised into the benzene ring, making it far less available to accept a proton than in the aliphatic amines.

   Basicity of an amine depends on how freely available the nitrogen lone pair is to accept a proton (H⁺):

   • (CH₃)₂NH (dimethylamine, 2° aliphatic amine): two +I (electron-donating) methyl groups increase electron density on N → quite basic.
   • NH₃: no alkyl groups, moderate basicity, used as the reference.
   • (CH₃)₃N (trimethylamine, 3° aliphatic amine): three +I methyl groups would suggest highest basicity electronically, though steric hindrance and poorer solvation of the resulting cation slightly reduce it — but it remains far more basic than aniline. …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set 56/1/11 markMCQ
   Q.Assertion (A): Aniline is a stronger base than ammonia. Reason (R): The unshared electron pair on nitrogen atom in aniline becomes less available for protonation due to resonance. [Codes (A)-(D) as in the Assertion-Reason instruction.]

   ›Reveal solutionSolution

   Aniline is a weaker base than ammonia because the lone pair on nitrogen in aniline is delocalized into the benzene ring via resonance, making it less available for protonation. Therefore, Assertion (A) is false, and Reason (R) is true.

   To understand the basicity of amines, we need to consider the availability of the lone pair of electrons on the nitrogen atom. A base, according to the Brønsted-Lowry definition, is a proton acceptor, and according to the Lewis definition, it is an electron pair donor. Therefore, the more readily available the lone pair on nitrogen is to accept a proton (or donate to an acid), the stronger the base.

   Factors that increase the electron density on the nitrogen atom or localize the lone pair make it a stronger base. Conversely, factors that decrease the electron density on nitrogen or delocalize the lone pair make it a weaker base.

   Let's analyze ammonia and aniline based on these principles.

   1. Basicity of Ammonia ($\text{NH}_3$)

      • In ammonia, the nitrogen atom has one lone pair of electrons.
      • This lone pair is localized on the nitrogen atom and is readily available for donation to a proton ($\text{H}^+$).
      • There are no significant electron-withdrawing or electron-donating groups directly attached to the nitrogen that would extensively delocalize or concentrate this lone pair. Thus, ammonia serves as a reference point for basicity.

   2. Basicity of Aniline ($\text{C}_6\text{H}_5\text{NH}_2$)

      • Aniline consists of an amino ($\text{-NH}_2$) group attached to a benzene ring.
      • The nitrogen atom in the amino group also possesses a lone pair of electrons.
      • However, this lone pair is in conjugation with the $\pi$-electron system of the benzene ring. This leads to resonance.

      Resonance structures of Aniline:

      $\text{C}_6\text{H}_5\text{NH}_2 \longleftrightarrow \text{C}_6\text{H}_5\text{NH}_2^+ \text{ (with negative charge on ortho/para positions)}$

      The lone pair on nitrogen can be delocalized into the benzene ring, as shown by these resonance structures:

           NH2
          /   \
         C     C
        //     \\
       C       C
        \     /
         C---C
      
      (Initial structure with lone pair on N)
      
           NH2(+)
          /   \
         C     C(-)
        //     \\
       C       C
        \     /
         C---C
      
      (Lone pair moves into ring, negative charge at ortho position)
      
           NH2(+)
          /   \
         C     C
        //     \\
       C(-)    C
        \     /
         C---C
      
      (Negative charge moves to para position)
      
           NH2(+)
          /   \
         C     C(-)
        //     \\
       C       C
        \     /
         C---C
      
      (Negative charge moves to other ortho position)
      

      • Due to this resonance, the lone pair of electrons on the nitrogen atom is not entirely localized on nitrogen; it is delocalized over the nitrogen and the ortho and para positions of the benzene ring.
      • This delocalization makes the lone pair less available for donation to a proton.

   3. Comparison of Basicity …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set FZ1 markMCQ
   Q.The compound having most basic strength is:

   (a) $(CH_3)_2NH$

   (b) $CH_3NH_2$

   (c) $(CH_3)_3N$

   (d) $NH_3$

   ›Reveal solutionSolution

   In water, dimethylamine is the strongest base of the four → option (a).

   Concept. Basicity of an amine in water is decided by three factors: the electron-releasing (+I) effect of alkyl groups (raises basicity), steric hindrance around N (lowers it), and stabilisation of the protonated cation by H-bonding to water (more H on N = better solvation).

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 markMCQ
    Q.The compound showing highest basic strength in aqueous solution is -

    (a) (C2H5)3N

    (b) (C2H5)2NH

    (c) C2H5NH2

    (d) NH3

    ›Reveal solutionSolution

    In water, basicity of alkylamines is governed by a balance of +I (electron-donating) effect, steric hindrance, and how well the protonated ammonium ion is stabilised by hydrogen bonding with water; for ethylamines this gives the order secondary > primary > tertiary > ammonia.

    Electron-donating alkyl groups increase electron density on N (favouring basicity), which would suggest (C2H5)3N should be most basic. …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Which of the following compounds is the most basic?

    (a) benzylamine (C6H5-CH2NH2)

    (b) aniline (C6H5-NH2)

    (c) p-nitroaniline (4-NH2-C6H4-NO2)

    (d) m-nitroaniline (3-NH2-C6H4-NO2)

    ›Reveal solutionSolution

    Benzylamine is most basic because its nitrogen lone pair is not conjugated with the benzene ring, while all the anilines lose basicity because their -NH$_2$ lone pair delocalises into the ring (and nitro groups withdraw electron density further).

    In aniline, the nitrogen lone pair is conjugated (delocalised via resonance) into the aromatic ring. This delocalisation makes the lone pair much less available to accept a proton, so aniline is far less basic than an ordinary alkyl amine.

    In benzylamine, the -CH$_2$- group between the ring and -NH$_2$ breaks this conjugation — the nitrogen behaves essentially like an aliphatic amine, with its lone pair fully available for protonation. This makes benzylamine considerably more basic than aniline.

    The two nitroanilines are even less basic than aniline itself, because the electron-withdrawing -NO$_2$ group further reduces electron density on nitrogen: …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2022Set ANNUAL1 markMCQ
    Q.The amine having lowest pKb value is:

    (a) CH3NH2

    (b) (CH3)2NH

    (c) (CH3)3N

    (d) C6H5NH2

    ›Reveal solutionSolution

    In aqueous solution, basicity of methylamines follows (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2, because a balance of the electron-donating (+I) effect of methyl groups and steric hindrance to solvation (hydration) of the resulting ammonium ion determines the base strength; the secondary amine gives the best balance.

    Basicity of an amine in water depends on two opposing factors:

    1. +I (electron-donating) effect of alkyl groups - increases electron density on N, increasing basicity as more methyl groups are added (favours 3-degree > 2-degree > 1-degree).
    2. Steric hindrance and reduced solvation (hydrogen bonding with water) of the ammonium cation - a more crowded ammonium ion is less stabilised by water, decreasing basicity (favours 1-degree > 2-degree > 3-degree). …

    🔒 Start your 14-day free trial to see the full solution →