Review the key concepts, formulae, and examples before starting your quiz.
πConcepts
Amines are derivatives of obtained by replacing one or more atoms with alkyl or aryl groups. Their physical state ranges from gases (lower aliphatic amines like ) to liquids and solids.
Boiling point order: Primary () > Secondary () > Tertiary (). This is because primary and secondary amines can form intermolecular hydrogen bonds due to the presence of atoms on , whereas tertiary amines lack such atoms.
Solubility: Lower aliphatic amines are soluble in because they can form hydrogen bonds with water molecules. Solubility decreases as the size of the hydrophobic alkyl part increases.
Basicity: Amines are Lewis bases because of the lone pair of electrons on the atom. In the gaseous phase, basicity follows the order: due to the effect of alkyl groups.
Basicity in Aqueous Solution: The order is influenced by the effect, solvation effect, and steric hindrance. For methyl-substituted amines, the order is . For ethyl-substituted amines, the order is .
Aniline is less basic than because the lone pair on is involved in resonance with the benzene ring, making it less available for protonation.
Carbylamine Reaction: Only aliphatic and aromatic primary amines react with and ethanolic to form foul-smelling isocyanides (). This is a test for amines.
Hinsberg's Reagent: Benzenesulphonyl chloride () is used to distinguish between , , and amines. amines form a precipitate soluble in alkali, amines form a precipitate insoluble in alkali, and amines do not react.
Electrophilic Substitution: In Aniline, the group is highly activating and ortho/para directing. Bromination with bromine water yields .
πFormulae
π‘Examples
Problem 1:
Arrange the following in increasing order of their basic strength in aqueous solution: , , , .
Solution:
Explanation:
In aqueous solution, basicity is determined by a combination of the inductive effect, solvation effect (hydration), and steric hindrance. For the methyl group, the steric hindrance is low, but the solvation of the cation is higher than the cation, making the amine () less basic than expected from the effect alone.
Problem 2:
Why does aniline not undergo Friedel-Crafts reaction?
Solution:
Aniline is a Lewis base and reacts with the Lewis acid catalyst used in Friedel-Crafts reactions.
Explanation:
The reaction forms a salt: . The positive charge on in the salt makes the benzene ring highly deactivated for electrophilic substitution.
Problem 3:
Convert Aniline to Fluorobenzene.
Solution:
Explanation:
This is known as the Balz-Schiemann reaction. Aniline is first converted to a diazonium salt, then reacted with fluoroboric acid () to form benzene diazonium fluoroborate, which decomposes upon heating to yield fluorobenzene.