Answer :
To determine which compound is most basic in an aqueous solution, we need to understand the basicity of the given compounds. Basicity refers to the ability of a substance to accept protons (H+). For amines, the basicity is often influenced by the availability of the lone pair of electrons on the nitrogen atom and the influence of substituents attached to the nitrogen.
Let's evaluate each of the given compounds:
(A) [tex]$C_6H_5NH_2$[/tex] (aniline): Aniline has a phenyl group attached to the nitrogen. The lone pair of electrons on the nitrogen can delocalize into the aromatic ring, reducing the electron density on the nitrogen and making it less available to accept protons. Therefore, aniline is not very basic.
(B) [tex]$CH_3NH_2$[/tex] (methylamine): Methylamine is a primary amine with a methyl group attached to the nitrogen. The methyl group is an electron-donating group, which increases the electron density on the nitrogen, making it more available to accept protons. Thus, methylamine is relatively basic.
(C) [tex]$(CH_3)_3N$[/tex] (trimethylamine): Trimethylamine is a tertiary amine with three methyl groups attached to the nitrogen. The electron-donating effect of the methyl groups increases the electron density on the nitrogen. However, steric hindrance from the three bulky methyl groups can hinder the nitrogen's ability to accept protons. Although trimethylamine is basic, it is less basic than primary amines due to the steric hindrance.
(D) [tex]$(CH_3)_2NH$[/tex] (dimethylamine): Dimethylamine is a secondary amine with two methyl groups attached to the nitrogen. The methyl groups increase the electron density on the nitrogen, making it more basic. It has less steric hindrance compared to trimethylamine, which allows it to readily accept protons.
Comparing all the options, dimethylamine (D) has a favorable balance of electron-donating effects from the methyl groups and minimal steric hindrance, making it the most basic in an aqueous solution.
Therefore, the correct answer is:
(D) [tex]$(CH_3)_2NH$[/tex]
Let's evaluate each of the given compounds:
(A) [tex]$C_6H_5NH_2$[/tex] (aniline): Aniline has a phenyl group attached to the nitrogen. The lone pair of electrons on the nitrogen can delocalize into the aromatic ring, reducing the electron density on the nitrogen and making it less available to accept protons. Therefore, aniline is not very basic.
(B) [tex]$CH_3NH_2$[/tex] (methylamine): Methylamine is a primary amine with a methyl group attached to the nitrogen. The methyl group is an electron-donating group, which increases the electron density on the nitrogen, making it more available to accept protons. Thus, methylamine is relatively basic.
(C) [tex]$(CH_3)_3N$[/tex] (trimethylamine): Trimethylamine is a tertiary amine with three methyl groups attached to the nitrogen. The electron-donating effect of the methyl groups increases the electron density on the nitrogen. However, steric hindrance from the three bulky methyl groups can hinder the nitrogen's ability to accept protons. Although trimethylamine is basic, it is less basic than primary amines due to the steric hindrance.
(D) [tex]$(CH_3)_2NH$[/tex] (dimethylamine): Dimethylamine is a secondary amine with two methyl groups attached to the nitrogen. The methyl groups increase the electron density on the nitrogen, making it more basic. It has less steric hindrance compared to trimethylamine, which allows it to readily accept protons.
Comparing all the options, dimethylamine (D) has a favorable balance of electron-donating effects from the methyl groups and minimal steric hindrance, making it the most basic in an aqueous solution.
Therefore, the correct answer is:
(D) [tex]$(CH_3)_2NH$[/tex]
