Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Amines are regarded as derivatives of ammonia (), formed by the replacement of one, two, or all three hydrogen atoms by alkyl and/or aryl groups.
Structure of Amines: The nitrogen atom in amines is hybridized. Due to the presence of a lone pair of electrons, the geometry is trigonal pyramidal. The (where is or ) bond angle is slightly less than the tetrahedral angle of ; for example, it is in trimethylamine.
Classification: Amines are classified as primary (), secondary (), and tertiary () depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in the ammonia molecule.
Primary () amines: Only one hydrogen atom of is replaced by an alkyl/aryl group, resulting in the functional group (amino group).
Secondary () amines: Two hydrogen atoms of are replaced, resulting in the functional group (imino group).
Tertiary () amines: All three hydrogen atoms of are replaced, resulting in a tertiary nitrogen atom .
Aryl Amines: In these amines, the group is directly attached to the benzene ring. The simplest example is Aniline ().
📐Formulae
💡Examples
Problem 1:
Classify the following amine as primary, secondary, or tertiary: .
Solution:
It is a Primary () amine.
Explanation:
Even though the carbon atom attached to the nitrogen is a tertiary carbon (t-butyl group), the nitrogen atom itself is attached to only one carbon atom and has two hydrogen atoms ( group). Therefore, it is a primary amine.
Problem 2:
Give the IUPAC name and classification for .
Solution:
IUPAC Name: -Methylethanamine. Classification: Secondary () amine.
Explanation:
The nitrogen is attached to two alkyl groups (ethyl and methyl), making it a secondary amine. For naming, the larger alkyl group (ethane) is treated as the parent chain.
Problem 3:
Why is the bond angle in () less than the ideal tetrahedral angle ()?
Solution:
Due to lone pair-bond pair repulsion.
Explanation:
In hybridized nitrogen, the lone pair occupies one of the hybrid orbitals. According to VSEPR theory, lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions, which pushes the bonds closer together, decreasing the angle.