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, weakest base is 

CH₃NHCHO

The correct answer is option 4. \(CH_3NHCHO\).

Let us delve deeper into why \(CH_3NHCHO\) (Methylformamide) is the weakest base among the options provided:

Structure of Methylformamide \((CH_3NHCHO)\):

Methylformamide consists of a methyl group \((-CH_3)\) attached to the nitrogen atom and a formyl group \((-CHO)\) attached to the same carbon as the nitrogen. The formyl group \((-CHO)\) is highly electronegative due to the presence of oxygen and the partial double bond character between carbon and oxygen. This electronegative nature of the formyl group results in significant electron withdrawal from the nitrogen atom. The lone pair of electrons on the nitrogen atom is thus less available for protonation (accepting a proton, \(H^+)\), reducing the basicity of methylformamide.

Effect on Basicity:

Basicity of an amine compound is influenced by the availability of the lone pair of electrons on the nitrogen atom. Electron-withdrawing groups, like the formyl group in methylformamide, decrease the electron density on the nitrogen atom. This reduction in electron density decreases the ability of the nitrogen to donate electrons to form a bond with a proton \((H^+)\), hence decreasing its basicity. In contrast, compounds with electron-donating groups (such as alkyl groups) attached to the nitrogen typically increase the electron density on the nitrogen atom, enhancing its basicity.

Comparison with Other Compounds:

When compared to other compounds listed (such as benzylamine, N-methylbenzylamine, and nitroethylamine), methylformamide exhibits the lowest basicity due to the strong electron-withdrawing effect of the formyl group. Benzylamine and N-methylbenzylamine have alkyl groups attached to the amino group, which are electron-donating and thus increase the basicity of the nitrogen. Nitroethylamine, while having a nitro group that is electron-withdrawing, typically exhibits higher basicity compared to methylformamide because the nitro group's effect is generally less strong than that of a formyl group.

Therefore, \(CH_3NHCHO\) (Methylformamide) is the weakest base among the options provided because the formyl group \((-CHO)\) strongly withdraws electron density from the nitrogen atom, reducing its basicity significantly.