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The compound 2-hydroxypyridine, a derivative of pyridine, is in equilibrium with 2-pyridone. 2-Hydroxypyridine is aromatic. Does 2-pyridone have comparable aromatic character? Explain.

Short Answer

Expert verified
Based on the analysis of the structures of 2-hydroxypyridine and 2-pyridone, both compounds possess aromatic character as they meet the criteria for being aromatic compounds (they are cyclic, have p-orbitals on all ring atoms, are planar, and follow Hückel's rule). Additionally, both molecules have 6 π electrons, and their aromatic character is thus considered comparable.

Step by step solution

01

Identify the structures of 2-hydroxypyridine and 2-pyridone

To analyze the aromatic character, we need to start by understanding the structure of both 2-hydroxypyridine and 2-pyridone. 2-hydroxypyridine has the following structure: ``` O \ N // / C-H \ \\ C====C / / C-H \ \\ N ``` 2-pyridone has the following structure: ``` O // N / / C-H \ \\ N / \ C====C ```
02

Determine the aromaticity

To determine if the molecules are aromatic, we need to check whether they meet the following criteria: 1. The molecule must be cyclic. 2. Every atom involved in the ring must have a p-orbital. 3. The cyclic compound must be planar. 4. The cyclic compound must follow Hückel's rule, having (4n+2) π electrons, where n is an integer (including 0). Both 2-hydroxypyridine and 2-pyridone are cyclic compounds. All atoms in the ring have a p-orbital, and both molecules are planar. For 2-hydroxypyridine, there are 6 π electrons (4π from the double bonds and 2π electrons from the nitrogen lone pair), which fits Hückel's rule (4n+2 = 6, with n=1). This indicates that 2-hydroxypyridine is aromatic. For 2-pyridone, there are also 6 π electrons as the nitrogen atom donates a lone pair of electrons to the ring, which also meets Hückel's rule.
03

Compare the aromatic character

Since both 2-hydroxypyridine and 2-pyridone meet the criteria for being aromatic compounds, they both possess aromatic character. Moreover, because both molecules have 6 π electrons and maintain the same ring structure, their aromatic character can be considered comparable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hückel's Rule
Understanding the foundations of organic chemistry can seem challenging, but breaking it down into simpler concepts helps in grasping the intricate ideas behind the molecules we study. One of the essential principles in determining the aromatic character of cyclic molecules is Hückel's rule. This rule is a straightforward yet powerful guideline for chemists. It states that a planar, cyclic molecule with conjugated π (pi) bonds is aromatic when the number of π electrons fits into the formula (4n + 2), where n can be any non-negative integer (0, 1, 2, 3, ...).

This seemingly simple formula is a gateway to predicting the stability and reactivity of aromatic systems. To apply Hückel's rule, we count the π electrons from double bonds and lone pairs (electrons that are not shared with another atom) on atoms within the ring that can contribute to the delocalized π system. When the count fits the 'magic number' from Hückel's formula, the molecule is aromatic. An aromatic molecule is notably more stable due to this special electron configuration, which is often referred to as aromatic stability or resonance stabilization.

If a molecule doesn't meet all the criteria including the (4n+2) π electron condition, it isn't considered aromatic. Such molecules might be non-aromatic, meaning they don't exhibit any special stability, or antiaromatic, which denotes instability usually because they have (4n) π electrons instead.
2-Hydroxypyridine
Dive into the world of heterocyclic compounds, and you'll find 2-hydroxypyridine, a molecule with a mix of characters. This compound, a derivative of pyridine, showcases the beauty of chemistry through its ability to exist in two forms that are in equilibrium with each other. The structure of 2-hydroxypyridine includes a six-membered ring with five carbons and one nitrogen atom. The additional elements are a hydroxyl (OH) group directly connected to the second carbon in the ring.

The reason behind the aromatic character of 2-hydroxypyridine lies in its electronic architecture. By applying Hückel's rule, we can see that 2-hydroxypyridine has a total of 6 π electrons – 4 coming from the double bonds within the ring and 2 from the lone pair on the nitrogen atom. Since this satisfies the (4n+2) π electron condition with n=1, it is considered aromatic. Not only does this give 2-hydroxypyridine its aromatic character, but also contributes to its stability and distinctive reactivity.
2-Pyridone
On the flip side of the equilibrium lies 2-pyridone, a compound almost identical to 2-hydroxypyridine yet different enough to have its unique characteristics. Like its counterpart, 2-pyridone contains a six-membered ring, but instead of a hydroxyl group, it boasts a carbonyl group (C=O) at the second position. This slight change has significant implications for the molecule's electronic properties and hence, its aromaticity.

In scrutinizing 2-pyridone's aromatic nature, we count its π electrons: four from the double bonds and two from the nitrogen lone pair, which contribute to the ring's π system due to resonance, totaling again to six. Hückel's rule is thus fulfilled for 2-pyridone as well, qualifying it as an aromatic compound. In both 2-hydroxypyridine and 2-pyridone, the presence of six π electrons forms the basis of their comparable aromatic character, influencing how they interact with other molecules and participate in chemical reactions.
Pi (π) Electrons
Pi (π) electrons play a star role in the theatre of organic chemistry, particularly in aromatic compounds. They are the electrons found in p-orbitals that are able to participate in bonding above and below the plane of atoms in a molecule, forming what's called a π bond when they overlap. In the context of aromaticity, π electrons are the ones that create the delocalized electron cloud which spreads over the entire ring structure in a seamless dance of charge distribution.

These electrons are not locked in place; they're fluid and shared amongst atoms in a molecule. For a molecule to be aromatic, not only must it have π electrons, but it must also have the right number of them, according to Hückel's rule. The allure of aromatic molecules like 2-hydroxypyridine and 2-pyridone is largely due to their π electron systems which provide them with their enhanced stability and unique reactivity, setting them apart from non-aromatic compounds. Understanding π electrons is key to unlocking the secrets of why some molecules exhibit this special type of stability known as aromaticity, and how they will behave in chemical reactions.

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Most popular questions from this chapter

In certain clinical situations, there is need for an injectable \(\beta\)-blocker with a short biological half-life. The clue to development of such a drug was taken from the structure of atenolol, whose corresponding carboxylic acid (the product of hydrolysis of its amide) has no \(\beta\)-blocking activity. Substitution of an ester for the amide group and lengthening the carbon side chain by one methylene group resulted in esmolol. Its ester group is hydrolyzed quite rapidly to a carboxyl group by serum esterases under physiological conditions. This hydrolysis product has no \(\beta\)-blocking activity. Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (a) Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (b) Is esmolol chiral? If so, which of the possible stereoisomers are formed in this synthesis?

Given here are \({ }^{1}\) H-NMR and \({ }^{13}\) C-NMR spectral data for two compounds. Each shows strong, sharp absorption between 1700 and \(1720 \mathrm{~cm}^{-1}\), and strong, broad absorption over the region \(2500-3000 \mathrm{~cm}^{-1}\). Propose a structural formula for each compound. (a) \(\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{10} \mathrm{H}_{10} \mathrm{O}_{2}\) $$ \begin{array}{cc} \hline{ }^{1} \mathbf{H}-\mathrm{NMR} & { }^{13} \mathrm{C}-\mathrm{NMR} \\ \hline 2.49(\mathrm{t}, 2 \mathrm{H}) & 173.89 \\ 2.80(\mathrm{t}, 2 \mathrm{H}) & 157.57 \\ 3.72(\mathrm{~s}, 3 \mathrm{H}) & 132.62 \\ 6.78(\mathrm{~d}, 2 \mathrm{H}) & 128.99 \\ 7.11(\mathrm{~d}, 2 \mathrm{H}) & 113.55 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 54.84 \\ & 35.75 \\ & 29.20 \\ \hline \end{array} $$ $$ \begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 167.82 \\ 6.38(\mathrm{~d}, 1 \mathrm{H}) & 143.82 \\ 7.18(\mathrm{~d}, 1 \mathrm{H}) & 139.96 \\ 7.44(\mathrm{~d}, 2 \mathrm{H}) & 131.45 \\ 7.56(\mathrm{~d}, 2 \mathrm{H}) & 129.37 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 127.83 \\ & 111.89 \\ & 21.13 \\ \hline \end{array} $$

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Which compound gives a signal in the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) spectrum with a larger chemical shift, furan or cyclopentadiene? Explain.

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