Question

State which of the following statements are true:

(i) $\text{p}{\text{K}}_{\text{b}}$ value for aniline is less than that for methylamine.

(ii) Methylamine in water reacts with ferric chloride to give a precipitate of ferric hydroxide.

(iii) Aniline does not undergo Friedel-Crafts reaction.

• A. (i), (ii), (iii)
• B. (ii), (iii)
• C. (i), (iii)
• D. (i), (ii)

Answer 1

Answer: (B)

(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 $C{H}_{3}N{H}_2+I$ effect of methyl group $\left(C{H}_3\right)$ 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.



(iii) True: Aniline being a Lewis base reacts with Lewis acid $\text{A}\text{l}\text{C}{\text{l}}_3$ to form a salt

$$\begin{gathered} {\text{C}}_6{\text{H}}_5\text{N}{\text{H}}_2+\text{A}\text{l}\text{C}{\text{l}}_3{\to }_{}^{}{\text{C}}_6{\text{H}}_5{\text{N}}^{+}{\text{H}}_2\text{A}\text{l}\text{C}{\text{l}}_3^{-} \\ \text{L}\text{e}\text{w}\text{i}\text{s}\text{b}\text{a}\text{s}\text{e}\text{L}\text{e}\text{w}\text{i}\text{s}\text{a}\text{c}\text{i}\text{d} \end{gathered}$$

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.