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 $CH _3 NH _2+ I$ effect of methyl group $\left( CH _3\right)$ 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

$\text{C}_{6}\text{H}_{5}\text{N}\text{H}_{2}+\text{A}\text{l}\text{C}\text{l}_{3} \rightarrow _{}^{}\text{C}_{6}\text{H}_{5}\text{N}^{+}\text{H}_{2}\text{A}\text{l}\text{C}\text{l}_{3}^{-} \\ \text{L}\text{e}\text{w}\text{i}\text{s} \, \text{b}\text{a}\text{s}\text{e} \, \text{L}\text{e}\text{w}\text{i}\text{s} \, \text{a}\text{c}\text{i}\text{d}$

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.