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. |
Arrange the following amines in the decreasing order of their basic strength.
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I > II > III > IV I > IV > II > III II > I > IV > III I > IV > III > II |
I > IV > III > II |
The correct answer is option 4. I > IV > III > II. The given compounds are:
Let us break down the reasoning for arranging the amines in decreasing order of basic strength based on their structures and the effects of substituents: Aniline (I): Aniline is a benzene ring with an amino group (-NH₂) attached to it. It is a moderately basic compound due to the lone pair of electrons on the nitrogen atom, which can readily accept a proton (H⁺). Aniline itself has no substituents that withdraw electron density from the amino group, so it is the most basic among the given compounds. 4-Chloroaniline (IV): In 4-chloroaniline, the chlorine atom is attached to the benzene ring in the para position relative to the amino group. Chlorine is an electron-withdrawing group due to its strong inductive effect, which pulls electron density away from the amino group through the benzene ring. This reduces the availability of the lone pair of electrons on the nitrogen for protonation, making 4-chloroaniline less basic than aniline. 3-Chloroaniline (III): In 3-chloroaniline, the chlorine atom is attached to the benzene ring in the meta position relative to the amino group. The meta position is less electron-withdrawing compared to the para position. Therefore, 3-chloroaniline is more basic than 4-chloroaniline but less basic than aniline and 2-chloroaniline. 2-Chloroaniline (II): In 2-chloroaniline, the chlorine atom is attached to the benzene ring in the ortho position relative to the amino group. Ortho-substituted chloro groups can exert a significant steric hindrance effect due to close proximity to the amino group, which can reduce the basicity compared to meta-substituted compounds like 3-chloroaniline. Additionally, the ortho position's electron-withdrawing effect is stronger than the meta position but less than the para position. Thus, 2-chloroaniline is the least basic among the given compounds due to both steric and electronic effects. Therefore, the correct decreasing order of basic strength among the compounds is: This order reflects the combined effects of electronic withdrawal through chlorine substitution and potential steric hindrance on the basicity of the amino group in these substituted anilines. |