Acid-Base Properties of Nitrogen Heterocycles - Online Tutor, Practice Problems & Exam Prep

Aromatic heterocyclic amines exhibit weak basicity influenced by hybridization and aromaticity. Higher s character in hybridized orbitals correlates with lower basicity; for instance, sp³ hybridization (25% s character) results in a pK_b of 4, while sp² (33% s character) yields a pK_b of 8.8. Additionally, if a lone pair participates in the pi system, as in pyrrole, it cannot accept protons, further weakening basicity. Thus, understanding these factors is crucial for predicting the behavior of heterocyclic compounds in chemical reactions.

Acid-Base Properties of Nitrogen Heterocycles Concept 1

Hey everyone. So in this video, we're going to take a look at the acidic and basic properties of our heterocyclic compounds. Now, if we take a look here, we're going to say that aromatic heterocyclic amines are uniquely weak bases due to two factors, and those factors are 1, hybridization, and 2, aromaticity. If we take a look at hybridization, we're going to say here that the greater s character of the orbital containing the lone pair, then the lower the basicity. If we take a look here at our two heterocyclic compounds, we're going to say in the first one we have our lone pair here.

This molecule, these nitrogen this nitrogen here with this lone pair will not use them to donate into the ring at all to help establish any type of aromatic compound. So we're going to say it's going to be included in our hybridization. Here these lone pairs would have an sp3 hybridization. And remember, if you're sp3, that means you have 25% s character. If we look at pyridine here, or pyridine, we're going to say that these lone pairs, again, are not part of the aromatic system.

They belong to the nitrogen. Because of this nitrogen's hybridization will be sp2. Remember, if you're sp2 hybridized, you are 33% s character. What effect does this have? We can see here that we have an sp3 hybridization, and then we have an sp2.

If we look at our \(pK_b\) values here, the \(pK_b\) would be 4 for this one, and the \(pK_b\) for this one will be 8.8. Remember, our s character is greater here. Remember, the greater your \(pK_b\), that means the weaker you are as a base. So just remember, when it comes to hybridization, the more s character you're going to have for a given lone pair, then the weaker the base because the higher its \(pK_b\) value will be.

Acid-Base Properties of Nitrogen Heterocycles Concept 2

Now the second factor that we look at is aromaticity. Here we're going to say if a lone pair is part of our pi system and not available to accept a proton, it results in a weaker base. Here we have our two examples, pyridine versus pyrrole. We're going to say here that this nitrogen has a lone pair. Following our Huckel's rule, we know that we already have 6 pi electrons within this ring, which is one of our numbers that we want for aromaticity.

So the lone pair on this nitrogen is not part of that pi system. And if we take a look here at pyrrole, we have 4 pi electrons here that we can see, and the lone pair on the nitrogen would be donated to the ring so that we can get 6 as one of our numbers. Here we're going to say they are a part of the pi system. What effect does this have? If we look, we can see that since these lone pairs are part of the pi system, we have a higher pK_b value.

Remember, a higher pK_b value means that we're going to have a weaker base. These here are not part of our pi system, so they can be used to accept a proton, and, therefore, they'd have a lower pK_b and be a stronger base. So just remember, for our second factor here of aromaticity, if the lone pair is being used as part of our pi system, they're not available to act as a basic lone pair. Therefore, that will result in a weaker base overall.

Acid-Base Properties of Nitrogen Heterocycles Example 1

Hey, everyone. Here it says, draw a protonated form of imidazole, hint: draw both resonance structures. Alright. So remember, we need 6 pi electrons in order to make an aromatic structure. This nitrogen down here is single bonded, no double bonds.

Its lone pair will be used as part of the pi system. Therefore, they're not available to be used to accept a proton. The nitrogen over here, its lone pairs are on the outside. We don't need them because we already have our 6 pi electron system, so they are going to be free to accept an H⁺. So if I'm going to draw the protonated form, I'm going to show that nitrogen as being protonated.

This nitrogen here is not. Right now, this nitrogen is making four bonds, so therefore it's positively charged. This will be our first resonance structure. Now if we want to draw our second resonance structure, just remember that this lone pair here could resonate here to make a double bond, kicking this double bond to the nitrogen. This will result in our second resonance structure.

The lone pairs got kicked to the nitrogen as a lone pair. Here goes our hydrogen. This nitrogen is now making a double bond. It's making four bonds, so now it's positively charged. And this would represent our second resonance structure.

But again, it all starts out with drawing this first resonance structure. And remember, whether our lone pair acts as part of the pi system or not determines if it'll accept a proton. If it's not part of the pi system, it's free to accept a proton, which we saw the top right nitrogen do to give us our 1st resonance structure or 1st protonated form, and then we just draw its resonance structure to give us our second form. Alright. So these will be our two final answers.

Rank following heterocycles in order of increasing basicity.

• A. B < A < D < C
• B. C < D < B < A
• C. D < B < C < A
• D. D < C < B < A

Draw conjugate acid of 4-hydroxypyridine and choose the best reason why it’s more basic than pyridine.

• A. -OH group acts as EWG through resonance, which decreases its acidity.
• B. -OH is an EWG, which decreases electron density of N making it more electrophilic, therefore making it more basic.
• C. -OH is an EDG through inductive effect, which induces electrons towards the N.
• D. -OH is an EDG, which increases electron density of N making it more nucleophilic, therefore making it more basic.

Here’s what students ask on this topic:

What factors influence the basicity of aromatic heterocyclic amines?

The basicity of aromatic heterocyclic amines is influenced by two main factors: hybridization and aromaticity. Hybridization affects the s character of the orbital containing the lone pair. For example, sp³ hybridization (25% s character) results in a lower pK_b and higher basicity, while sp² hybridization (33% s character) results in a higher pK_b and lower basicity. Aromaticity affects whether the lone pair is part of the π system. If the lone pair is part of the π system, as in pyrrole, it cannot accept protons, resulting in weaker basicity. Conversely, if the lone pair is not part of the π system, as in pyridine, it can accept protons, resulting in stronger basicity.

How does hybridization affect the basicity of nitrogen heterocycles?

Hybridization affects the basicity of nitrogen heterocycles by altering the s character of the orbital containing the lone pair. In sp³ hybridization, the orbital has 25% s character, leading to a lower pK_b and higher basicity. In sp² hybridization, the orbital has 33% s character, resulting in a higher pK_b and lower basicity. The greater the s character, the more tightly the lone pair is held, making it less available to accept protons, thus reducing basicity.

What is the role of aromaticity in determining the basicity of nitrogen heterocycles?

Aromaticity plays a crucial role in determining the basicity of nitrogen heterocycles. If the lone pair on the nitrogen is part of the π system, it is not available to accept protons, resulting in weaker basicity. For example, in pyrrole, the lone pair is part of the π system, leading to a higher pK_b and weaker basicity. Conversely, if the lone pair is not part of the π system, as in pyridine, it can accept protons, resulting in a lower pK_b and stronger basicity.

Why is pyridine a stronger base than pyrrole?

Pyridine is a stronger base than pyrrole due to differences in aromaticity and hybridization. In pyridine, the lone pair on the nitrogen is not part of the π system, making it available to accept protons, resulting in a lower pK_b and stronger basicity. Additionally, the nitrogen in pyridine is sp² hybridized, with 33% s character. In contrast, in pyrrole, the lone pair is part of the π system, making it unavailable to accept protons, resulting in a higher pK_b and weaker basicity. The nitrogen in pyrrole is also sp² hybridized, but the involvement of the lone pair in the aromatic system reduces its basicity.

How does the pK_b value relate to the basicity of nitrogen heterocycles?

The pK_b value is inversely related to the basicity of nitrogen heterocycles. A lower pK_b value indicates a stronger base, as it means the compound is more willing to accept protons. Conversely, a higher pK_b value indicates a weaker base, as it means the compound is less willing to accept protons. For example, a nitrogen heterocycle with sp³ hybridization (25% s character) might have a pK_b of 4, indicating stronger basicity, while one with sp² hybridization (33% s character) might have a pK_b of 8.8, indicating weaker basicity.