Basic strength directly depends on the availability of lone pair for H+.

  • Strong Acids have weak conjugate base.

  • For the same period
    Less electro negativity, more Nucleophilicity as more electronegative element has less tendency to give its electron pair.
  • Factors which affect the basicity of Amines
    1. Steric effects
    2. Inductive effect
    3. Solvation effect.
  • The base whose conjugate acid is more stable will be more acidic

Ex.1 Compare the basic strength of the following

  1. -CH3
  2. -NH2
  3. -OH
  4. F-


Ex.2 Which is more basic?


Ex.3  Which is more basic?

Forming conjugate acid

Comparison of Basicity of Ammonia and Alkyl Amines:

Ex.3 Compare the basic strength of the following:

Forming conjugate acid of the given base

Stability order of conjugate acid

Therefore basic strength

    (CH3)3 N > (CH3)2 NH > CH3NH2> NH3

(Vapor Phase or Gaseous Phase or in Non Polar Solvent)

In Aqueous solution or in polar solvent

    CH3NH2> (CH3)2 NH > (CH3)3 N > NH3

  • In aqueous solution, the conjugate acids form H-bonds (intermolecular) with water molecules and stabilize themselves conjugate acid of 1o amine which has largest no. of H-atoms form maximum H-bond with water and is most stable. Consequently 1o amine is most basic.
  • Due to steric effect 1o amine is considered more basic as compared to 3o amine as lone pair is hindered by three alkyl group and less available for H+.
    Considering the combined effect of the three (Inductive, solvation and steric effect) we can conclude that
    2o> 1o> 3o> NH3
  • Aromatic amines are least basic as their lone pair is in conjugation and less available for protonation.

Ex.4 Compare the basic strength of the following

In (a), If L.P. will be participate in Resonance, and then molecule becomes aromatic
Hence L.P. will have a greater tendency to take part in Resonance and will be less available for H+
This compound will be least basic.

Ex.5 Compare the basic strength of the following

ol. sp hybridized carbon being most electronegative will attract e- density from nitrogen and will make it less available for H+. Hence basicity decreases.
c> b > a

Ex.6 Compare the basic strength

a < b

Ex.7 Compare the basicity of the following compounds:

Sol. In part (a) the lone pair of nitrogen in Resonance therefore will be less available for H+ making it least basic among all followed by sp, sp2, sp3hybridised carbon atoms.
b> c > d > a

Ex.8 Compare the basicity of the numbered nitrogen atoms:

Sol. The planarity of ring will be destroyed if lone pair will part in Resonance.
Basicity order of Nitrogen follows the order

Ex.9 Compare the basic strength of the following

Sol. In part (a) NO2 is at p-position hence will attract e- density by both –M and –I
In part (b) NO2 is at m-position hence will attract e- density by –I only
Therefore is no such effect in part (c)
Availibity of L.P. no nitrogen in part (a) is minimum followed by b and then c.
c > b > a

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