Consider the following compounds:
\[
(C_2H_5)_2NH,\quad C_6H_5NH_2,\quad C_6H_5CH_2NH_2,\quad NH_3,\quad (C_2H_5)_3N
\]
The correct increasing order of the above bases on the basis of their basic strength is:
Show Hint
In aqueous solution:
\[
\text{Secondary amine}>\text{Tertiary amine}>\text{Primary amine}>NH_3
\]
Aniline is least basic due to resonance.
Concept:
Basic strength of amines depends upon the availability of the lone pair of electrons on nitrogen for protonation. The more easily nitrogen donates its lone pair, the stronger the base.
Factors affecting basic strength:
+I effect (electron donating effect) of alkyl groups increases basicity.
Resonance effect decreases basicity by delocalizing the lone pair.
Solvation effect stabilizes protonated species in aqueous solution.
Steric hindrance reduces effective protonation in tertiary amines.
Step 1: Analyze each compound. (i) C\(_6\)H\(_5\)NH\(_2\) (Aniline):
The lone pair on nitrogen is delocalized into the benzene ring due to resonance.
\[
\text{Lone pair participates in resonance} \Rightarrow \text{Less available for protonation}
\]
Hence, it is the weakest base among the given compounds.
(ii) NH\(_3\):
No alkyl group, no resonance. Moderate basic strength.
(iii) C\(_6\)H\(_5\)CH\(_2\)NH\(_2\) (Benzylamine):
The benzene ring is not directly attached to nitrogen; hence no resonance effect on lone pair.
The –CH\(_2\) group exerts +I effect.
Therefore, stronger base than ammonia.
(iv) (C\(_2\)H\(_5\))\(_3\)N (Tertiary amine):
Strong +I effect from three alkyl groups increases electron density on nitrogen.
However, steric hindrance and poor solvation slightly reduce its strength compared to secondary amine.
(v) (C\(_2\)H\(_5\))\(_2\)NH (Secondary amine):
Strong +I effect and better solvation compared to tertiary amine.
Generally strongest base in aqueous solution.
Step 2: Arrange in increasing order of basic strength.
\[
\text{Aniline}<NH_3<\text{Benzylamine}<\text{Tertiary amine}<\text{Secondary amine}
\]
\[
C_6H_5NH_2<NH_3<C_6H_5CH_2NH_2<(C_2H_5)_3N<(C_2H_5)_2NH
\]
\[
\therefore \text{Correct answer = (A)}
\]
