Chemistry - AMINES Question with Solution | TestHub

(i) False: In aniline, the lone pair of electrons on the N-atom is delocalized over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in methylamine  $\text{C}{\text{H}}_3\text{N}{\text{H}}_2$  $+\text{I}$ effect of methyl group  $(\text{C}{\text{H}}_3)$  increases the electron density on the N–atom.Therefore, aniline is a weaker base than methylamine and hence its  $\text{p}{\text{K}}_{\text{b}}$  value is higher than that of methylamine.(ii) True: Methylamine being more basic than water, accepts a proton from water liberating  $\text{O}{\text{H}}^{-}$  ions.These  $\text{O}{\text{H}}^{-}$ ions combine with  $\text{F}{\text{e}}^{3+}$  ions present in  ${\text{H}}_2\text{O}$  to form brown precipitate, of hydrated ferric oxide. $\begin{array}{l}{\text{FeCl}}_3\to \text{F}{\text{e}}^{3+}+3\text{}\text{C}{\text{l}}^{-}\\ 2\text{F}{\text{e}}^{3+}+6\text{O}{\text{H}}^{-}\underset{}{\overset{}{\to }}2\text{F}\text{e}{\left(\text{O}\text{H}\right)}_3.\\ \text{H}\text{y}\text{d}\text{r}\text{a}\text{t}\text{e}\text{d}\text{f}\text{e}\text{r}\text{r}\text{i}\text{c}\text{o}\text{x}\text{i}\text{d}\text{e}\\ (\text{B}\text{r}\text{o}\text{w}\text{n}\text{p}\text{p}\text{t}.)\end{array}$ (iii) True: Aniline being a Lewis base reacts with Lewis acid  $\text{A}\text{l}\text{C}{\text{l}}_3$  to form a salt $\begin{array}{l}{\text{C}}_6{\text{H}}_5\text{N}{\text{H}}_2+\text{A}\text{l}\text{C}{\text{l}}_3\underset{}{\overset{}{\to }}{\text{C}}_6{\text{H}}_5{\text{N}}^{+}{\text{H}}_2\text{A}\text{l}\text{C}{\text{l}}_3^{-}\\ \text{L}\text{e}\text{w}\text{i}\text{s}\text{b}\text{a}\text{s}\text{e}\text{L}\text{e}\text{w}\text{i}\text{s}\text{a}\text{c}\text{i}\text{d}\end{array}$ As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel-Crafts reaction.