For the same alkyl group, all the three primary, secondary and tertiary amines are more basic than ammonia in aqueous solutions. Their relative basicity, however, depends upon a combination of three factors: +I-effect of alkyl groups, the stability of the ammonium cation (formed after accepting a proton) due to H-bonding with water molecules and the steric effect of the alkyl groups which tend to reduce the extent of H-bonding. All these three factors are favourable for secondary amines thereby making them the strongest bases. Since methyl group has the smallest size, there is no steric hindrance to H-bonding. Consequently, stability of the ammonium cation due to H-bonding with water predominates over the +I-effect of the methyl group thereby making methylamine more basic than trimethylamine. If, however, the alkyl group is bigger than the methyl group, +I-effect of the alkyl group outweighs stability of the ammonium cation due to H-bonding there making tertiary amines more basic than the primary amines. Due to delocalization of lone pair of electrons of the nitrogen atom on the benzene ring, aniline is a weaker base than ammonia. The basic strength of the substituted anilines, however, depends upon the nature of the substituent. Whereas electron-donating groups tend to increase, electron-withdrawing groups tend to decrease the basic strength. The base strengthening effect of the electron-withdrawing groups and base weakening effect of the electron-withdrawing group is, however, more pronounced at p-than at m-position. However, due to ortho-effect, o-substituted anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing.   

What will be the increasing order of basic strength of the following compounds?

\(C_2H_5NH_2,\, \ (C_2H_5)_2NH,\, \ (C_2H_5)_3N,\, \ C_6H_5NH_2\)

\(C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH\)

The correct answer is option 2. \(C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH\).

To determine the increasing order of basic strength of the given compounds (\(C_2H_5NH_2\), \((C_2H_5)_2NH\), \((C_2H_5)_3N\), \(C_6H_5NH_2\)), we need to consider how readily each compound's nitrogen atom can donate its lone pair of electrons to accept a proton. This is influenced by the inductive and resonance effects of the substituents attached to the nitrogen.

\(C_6H_5NH_2\) (Aniline):

Resonance Effect: The phenyl group (benzene ring) has a resonance effect that delocalizes the lone pair of electrons on the nitrogen into the aromatic ring. This resonance effect decreases the electron density on the nitrogen, making it less available to accept a proton.

Basicity: Due to the resonance effect, aniline is the least basic among the given compounds.

\(C_2H_5NH_2\) (Ethylamine):

Inductive Effect: The ethyl group is an electron-donating group via the inductive effect, which increases the electron density on the nitrogen, making it more basic than aniline.

Basicity: Ethylamine is more basic than aniline because the lone pair on nitrogen is more available for protonation.

\((C_2H_5)_3N\) (Triethylamine):

Inductive Effect: Three ethyl groups provide a strong inductive effect, increasing the electron density on the nitrogen. However, the steric hindrance caused by the three bulky ethyl groups around the nitrogen can make it harder for the nitrogen to access protons.

Basicity: Despite the increased electron density, the steric hindrance slightly reduces the availability of the lone pair for protonation compared to diethylamine, making it less basic than diethylamine.

\((C_2H_5)_2NH\) (Diethylamine):

Inductive Effect: Two ethyl groups provide a significant inductive effect, increasing the electron density on the nitrogen without causing excessive steric hindrance.

Basicity: Diethylamine is more basic than both ethylamine and triethylamine because it has a high electron density on nitrogen with less steric hindrance compared to triethylamine.

Based on the inductive and resonance effects, the increasing order of basic strength is:
\(\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_3\text{N} < (\text{C}_2\text{H}_5)_2\text{NH}
\)

Therefore, the correct option is 2. \(C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH\)