Animated Solution for Chemistry - Organic Chemistry: Arrange the following amines in the decreasing order of basicity :

The quest to understand the basicity of nitrogen-containing rings is one of the most fascinating journeys in organic chemistry. It’s not just about memorizing facts; it’s about visualizing the invisible dance of electrons. Today, we are going to dissect three iconic molecules: Pyridine, Pyrrole, and Piperidine. Our mission? To arrange them in decreasing order of their basicity.

Before we dive into the specific molecules, let's establish our ground rules. What exactly makes an amine basic? In the realm of Lewis acid-base theory, a base is an electron pair donor. Therefore, the basicity of an amine is directly proportional to the availability of the lone pair on the nitrogen atom.

If the lone pair is sitting comfortably on the nitrogen, ready to grab a passing proton ($H^{+}$), the molecule is a strong base. If the lone pair is busy doing something else, or if it's held too tightly by the nucleus, the molecule is a weak base.

Two major factors influence this availability: 1. Hybridization (s-character): The more s-character an orbital has, the closer the electrons are pulled toward the positively charged nucleus. This makes them less available for donation. 2. Delocalization (Resonance/Aromaticity): If the lone pair is involved in resonance, especially to satisfy aromaticity, it is practically unavailable to act as a base.

Let's apply these principles to our three contenders.

The Uncomplicated Base

Let's start with Piperidine (Molecule III). Take a close look at its structure. It's a saturated six-membered ring. There are no double bonds, no pi systems, no complications.

The nitrogen atom is bonded to two carbon atoms and one hydrogen atom, and it holds one lone pair. This means the nitrogen is $s{p}^3$ hybridized.

An $s{p}^3$ hybridized orbital has exactly $25\%$ s-character. Because this s-character is relatively low, the lone pair is not held excessively tight by the nucleus. Furthermore, there is no pi system for the lone pair to interact with, so it remains completely localized on the nitrogen atom.

This localized, highly available lone pair makes Piperidine a very strong base.

The Tighter Grip

Next up is Pyridine (Molecule I). This is an aromatic ring, similar to benzene, but with a nitrogen atom replacing one of the CH groups.

The nitrogen atom in Pyridine is part of a double bond, which means it is $s{p}^2$ hybridized. An $s{p}^2$ orbital has $33.3\%$ s-character.

Remember our rule? Higher s-character means the electrons are held closer to the nucleus. Because $33.3\%$ is greater than $25\%$, the lone pair in Pyridine is held more tightly than the lone pair in Piperidine.

But wait, is the lone pair involved in the aromatic pi system? No! The aromatic sextet is already complete with the three double bonds in the ring. The lone pair resides in an $s{p}^2$ orbital that is orthogonal (perpendicular) to the pi system. Therefore, the lone pair is still localized, but it is simply less available than in Piperidine due to the higher s-character.

Thus, Pyridine is a weaker base than Piperidine.

The Ultimate Sacrifice

Finally, we arrive at Pyrrole (Molecule II). This is a five-membered ring with two double bonds and a nitrogen atom.

At first glance, the nitrogen appears to be $s{p}^3$ hybridized because it has three single bonds and a lone pair. However, nature loves stability, and aromaticity is the ultimate form of stability. The ring has four pi electrons from the two double bonds. To reach the magical Huckel number of six pi electrons ($4n+2$ where $n=1$), the nitrogen atom rehybridizes to $s{p}^2$.

This allows its lone pair to reside in an unhybridized p-orbital, perfectly aligned to overlap with the other p-orbitals in the ring. The lone pair is completely delocalized into the ring to complete the aromatic sextet.

Because this lone pair is actively participating in maintaining the molecule's aromaticity, it is incredibly busy. If it were to donate its electrons to a proton, the molecule would lose its aromatic stability—a massive energetic penalty. Therefore, the lone pair is practically unavailable for donation.

This makes Pyrrole an exceptionally weak base.

Let's summarize our findings: Piperidine (III): $s{p}^3$ hybridized, localized lone pair, highly available. (Strongest Base) Pyridine (I): $s{p}^2$ hybridized, localized lone pair, moderately available. (Moderate Base) Pyrrole (II):* $s{p}^2$ hybridized, delocalized lone pair (aromatic), practically unavailable. (Weakest Base)

Therefore, the decreasing order of basicity is III > I > II.

This perfectly matches option (d). By understanding the profound effects of hybridization and aromaticity, you can conquer any basicity problem that comes your way!