Review the key concepts, formulae, and examples before starting your quiz.
πConcepts
Amines act as Lewis bases due to the presence of a lone pair of electrons on the nitrogen atom, which can be donated to an electron-deficient species.
Basicity is quantitatively expressed by the basic dissociation constant . A larger or a smaller value indicates a stronger base.
In the gaseous phase, the basicity of amines follows the order of the (inductive) effect: Tertiary amine () > Secondary amine () > Primary amine () > .
In aqueous solution, the basicity is determined by a combination of the effect, solvation effect (hydrogen bonding with water), and steric hindrance of alkyl groups.
For methyl-substituted amines in aqueous solution, the order of basicity is: .
For ethyl-substituted amines in aqueous solution, the order of basicity is: .
Aromatic amines like aniline () are significantly less basic than ammonia because the lone pair on nitrogen is delocalized into the benzene ring through resonance, making it less available for protonation.
Substituents on the benzene ring affect basicity: Electron-donating groups (e.g., , ) increase basicity, while electron-withdrawing groups (e.g., , , ) decrease basicity.
πFormulae
π‘Examples
Problem 1:
Arrange the following in increasing order of their basic strength in aqueous solution: , , , and .
Solution:
Explanation:
In aqueous solution, the basicity is a result of effect, solvation effect, and steric hindrance. For methyl groups, the steric hindrance in and the reduced solvation of its cation make it less basic than the primary and secondary amines, despite having more alkyl groups.
Problem 2:
Why is aniline () a weaker base than methylamine ()?
Solution:
In aniline, the lone pair of electrons on the atom is in conjugation with the -electron system of the benzene ring. This delocalization (resonance) makes the lone pair less available for donation to a proton. In , the effect of the methyl group increases the electron density on .
Explanation:
Resonance structures of aniline show that the lone pair is distributed over the ortho and para positions of the ring, decreasing its availability. has no such delocalization.
Problem 3:
Between -nitroaniline and -toluidine, which is more basic and why?
Solution:
-toluidine () is more basic than -nitroaniline ().
Explanation:
The methyl group () in -toluidine is an electron-donating group ( and hyperconjugation), which increases electron density on the group. The nitro group () in -nitroaniline is a strong electron-withdrawing group ( and effects), which significantly reduces the availability of the lone pair on nitrogen.