Question

Which from following compounds has lowest \(\text{pK}_\text{b}\) value?

• A. N -Ethylethanamine
• B. Propan-2-amine
• C. \(\text{NH}_3\)
• D. Benzenamine

Hint:

In aqueous solution: \(2^\circ>1^\circ>NH_3>\text{aromatic amine}\) due to +I effect and solvation.

Answer 1

The Correct Option is A

Concept:
\[ \text{pK}_b = -\log K_b \] Lower \(\text{pK}_b\) $\Rightarrow$ higher \(K_b\) $\Rightarrow$ stronger base.
Basic strength of amines depends mainly on:

• +I (electron donating) effect of alkyl groups
• Availability of lone pair on nitrogen
• Solvation effects

Step 1: Identify types of amines.

• (A) N-ethylethanamine → Secondary amine \((2^\circ)\)
• (B) Propan-2-amine → Primary amine \((1^\circ)\)
• (C) \(\text{NH}_3\) → No alkyl group
• (D) Benzenamine → Aromatic amine

Step 2: Compare electron donating effects. Alkyl groups increase electron density on nitrogen via +I effect: \[ 2^\circ>1^\circ>NH_3 \]
Step 3: Consider aromatic amine. In benzenamine (aniline), lone pair is delocalized into benzene ring due to resonance: \[ \Rightarrow \text{less available for protonation} \Rightarrow \text{weak base} \]
Step 4: Final order of basic strength. \[ \text{Secondary amine}>\text{Primary amine}>NH_3>\text{Aromatic amine} \]
Step 5: Conclusion. Strongest base has lowest \(\text{pK}_b\): \[ \text{N-ethylethanamine has lowest } \text{pK}_b \]