Question

Choose the correct order of decreasing basic strength of the following compounds in aqueous solution
(i) C\textsubscript{6H\textsubscript{5}NH\textsubscript{2} (ii) C\textsubscript{2}H\textsubscript{5}NH\textsubscript{2} (iii) NH\textsubscript{3} (iv) (CH\textsubscript{3})\textsubscript{2}NH}

• A. (i) $>$ (ii) $>$ (iii) $>$ (iv)
• B. (iv) $>$ (ii) $>$ (iii) $>$ (i)
• C. (ii) $>$ (i) $>$ (iii) $>$ (iv)
• D. (iv) $>$ (iii) $>$ (ii) $>$ (i)
• E. (ii) $>$ (iv) $>$ (iii) $>$ (i)

Hint:

In aqueous solution: - Ethyl amines: $2^\circ > 3^\circ > 1^\circ > NH_3$ - Methyl amines: $2^\circ > 1^\circ > 3^\circ > NH_3$ Secondary amines are always the champions in water!

Answer 1

The Correct Option is B

Concept: Basic strength in amines depends on the availability of the lone pair on the nitrogen atom and the stabilization of the resulting cation in aqueous solution.
• Aliphatic vs. Ammonia: Alkyl groups are electron-donating ($+I$), which increases lone pair availability. Thus, alkylamines are more basic than $NH_3$.
• Aromatic vs. Ammonia: In aniline ($C_6H_5NH_2$), the lone pair is delocalized into the benzene ring via resonance, making it significantly less available. Aniline is much weaker than $NH_3$.
• Aqueous Solution Factors: In water, basicity is a balance of $+I$ effects, steric hindrance, and salvation (hydrogen bonding).

Step 1: Compare aliphatic amines. In aqueous solution for methyl-substituted amines, the order is $2^\circ > 1^\circ > 3^\circ$. For (iv) $(CH_3)_2NH$ (secondary) and (ii) $C_2H_5NH_2$ (primary), the secondary amine is more basic due to the presence of two $+I$ groups and sufficient salvation. So, (iv) $>$ (ii).

Step 2: Place ammonia and aniline. Ammonia (iii) has no $+I$ groups but no resonance withdrawal, so it sits below aliphatic amines. Aniline (i) has resonance withdrawal, making it the weakest. Overall order: (iv) $(CH_3)_2NH$ $>$ (ii) $C_2H_5NH_2$ $>$ (iii) $NH_3$ $>$ (i) $C_6H_5NH_2$.