Here's the explanation:
Delocalization does not completely remove the lone pair.
The lone pair on an aromatic amine is indeed involved in delocalization, contributing to the stability of the aromatic ring. However, it's not completely lost. Think of it as a shared resource:
* The lone pair spends a significant portion of its time within the ring's pi system. This contributes to the aromatic character and stability of the molecule.
* The lone pair retains enough electron density to participate in hydrogen bonding. Even though it's delocalized, it still has a partial negative charge and can interact with hydrogen atoms in other molecules.
The strength of the hydrogen bond is affected by delocalization.
The delocalization of the lone pair does affect the strength of the hydrogen bond formed by an aromatic amine. Since the electron density is distributed over a larger area, the hydrogen bond will be weaker compared to a non-aromatic amine. This is because the lone pair is less localized and available for interaction with the hydrogen atom.
Analogy: Imagine the lone pair as a person who works two jobs. They spend most of their time at one job (delocalization in the aromatic ring), but they still have some time left over to do some work at their second job (hydrogen bonding).
In summary:
* The lone pair in aromatic amines participates in both delocalization and hydrogen bonding.
* Delocalization does not completely remove the lone pair's ability to form hydrogen bonds, but it does affect their strength.
* The electron density of the lone pair is still available for interaction with hydrogen atoms, albeit to a lesser extent than in non-aromatic amines.
Let me know if you have any further questions!
