Question

Consider the following amines

From the above, identify the pair of amines with lowest \(pK_b\) and highest \(pK_b\) in aqueous solution.

• A. I, III
• B. IV, I
• C. II, IV
• D. I, II \bigskip

Hint:

For aromatic amines, resonance decreases basicity because the lone pair gets delocalized into the benzene ring. In aqueous solution: \[ \text{Secondary aliphatic amine} > \text{Tertiary aliphatic amine} > \text{Aniline derivatives} \] Lower \(pK_b\) means stronger base and higher \(pK_b\) means weaker base.

Answer 1

The Correct Option is B

Concept: The basic strength of amines is commonly compared using the \(pK_b\) value. \[ \text{Smaller } pK_b \Rightarrow \text{Stronger base} \] \[ \text{Larger } pK_b \Rightarrow \text{Weaker base} \] The basic character depends mainly on:
• Availability of lone pair on nitrogen.
• Electron-releasing (\(+I\)) effect of alkyl groups.
• Resonance delocalization of lone pair.
• Solvation of the protonated amine in aqueous solution.

Step 1: Identify the strongest base among the given amines.
Compound I is diethylamine: \[ (C_2H_5)_2NH \] Two ethyl groups exert a strong \(+I\) effect and increase the electron density on nitrogen. In aqueous solution, secondary amines are generally the strongest bases because they possess:
• Strong electron-releasing effect.
• Good solvation of the conjugate acid. Therefore, \[ (C_2H_5)_2NH \] is the strongest base among the given compounds. Hence it has the \[ \boxed{\text{lowest } pK_b} \]

Step 2: Identify the weakest base among the given amines.
Compound II is aniline: \[ C_6H_5NH_2 \] The lone pair on nitrogen participates in resonance with the benzene ring. \[ \ce{C6H5-NH2 C6H5=NH^{+}} \] Because the lone pair is delocalized, it becomes less available for protonation. Therefore aniline is considerably less basic than aliphatic amines. Compound IV is \(N,N\)-dimethylaniline: \[ C_6H_5N(CH_3)_2 \] Although the methyl groups show \(+I\) effect, the lone pair is still conjugated with the aromatic ring. Further, the protonated form is less effectively solvated in water due to the bulky methyl groups. As a result, \(N,N\)-dimethylaniline is weaker than aniline in aqueous solution. Hence it possesses the \[ \boxed{\text{highest } pK_b} \] among the given compounds.

Step 3: Arrange the bases approximately in decreasing order of basic strength.
\[ (C_2H_5)_2NH > (CH_3)_3N > C_6H_5NH_2 > C_6H_5N(CH_3)_2 \] Thus, \[ \boxed{\text{Strongest base} = \text{I}} \] and \[ \boxed{\text{Weakest base} = \text{IV}} \]

Step 4: Relate basic strength with \(pK_b\).
Since, \[ \text{lowest } pK_b \leftrightarrow \text{strongest base} \] and \[ \text{highest } pK_b \leftrightarrow \text{weakest base} \] we obtain: \[ \boxed{\text{Lowest } pK_b = \text{I}} \] \[ \boxed{\text{Highest } pK_b = \text{IV}} \] Therefore the required pair is \[ \boxed{\text{IV, I}} \] as listed in the options. \[ \boxed{\text{Answer = (B)}} \]