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.   

Among the following, the strongest base is 

C₆H₅CH₂NH₂

The correct answer is option 4. \(C_6H_5CH_2NH_2\).

Let us break down the basicity of each compound based on their structures and the effects of substituents:

1. C₆H₅NH₂ (Aniline):

Aniline is a benzene ring with an amino group (-NH₂) attached directly to it. The lone pair of electrons on the amino group can readily accept a proton (H⁺), making aniline a moderately basic compound. There are no substituents withdrawing electron density from the amino group, so it exhibits typical basicity for aniline derivatives.

2.p-NO₂-C₆H₄NH₂ (para-Nitroaniline):

In this compound, a nitro group (-NO₂) is attached to the benzene ring at the para position relative to the amino group. Nitro groups are strongly electron-withdrawing due to both inductive (-I) and resonance (-R) effects. The -I effect pulls electron density away from the amino group, making the lone pair on nitrogen less available for protonation. This reduces the basicity of para-nitroaniline compared to aniline.

3. m-NO₂-C₆H₄NH₂ (meta-Nitroaniline):

Here, the nitro group is attached to the benzene ring at the meta position relative to the amino group. Meta-substituted nitro groups have a weaker electron-withdrawing effect compared to para-substituted nitro groups. The -I effect is less pronounced at the meta position, so meta-nitroaniline is expected to be more basic than para-nitroaniline but less basic than aniline.

4. C₆H₅CH₂NH₂ (Benzylamine):

This compound has a benzene ring with an ethylamine group (-CH₂NH₂) attached to it. Ethylamine is an electron-donating group due to its alkyl chain. The alkyl group increases the electron density on the amino group, making the lone pair of electrons more available for protonation. Benzylamine is therefore more basic than aniline because of the electron-donating effect of the benzyl group.

Considering these effects, the basicity order from strongest to weakest is: \(\text{C₆H₅CH₂NH₂} > \text{C₆H₅NH₂ (Aniline)} > \text{m-NO₂-C₆H₄NH₂} > \text{p-NO₂-C₆H₄NH₂}\)

Therefore, the strongest base among the given compounds is \(C_6H_5CH_2NH_2\).