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Bextra, a so-called COX-2 inhibitor used in the treatment of arthritis, contains an isoxazole ring. Why is the ring aromatic?

Short Answer

Expert verified
The isoxazole ring is aromatic because it is cyclic, planar, conjugated, and follows Huckel's rule.

Step by step solution

01

Define Aromatic Compounds

Aromatic compounds are cyclic, planar, have a conjugated pi-electron system, and follow Huckel's rule, having \( 4n + 2 \) pi electrons.
02

Check the Cycle and Planarity

The isoxazole ring is a five-membered ring, which is cyclic and typically planar, allowing for possible delocalization of electrons.
03

Verify Conjugation

An isoxazole ring contains alternating single and double bonds, with nitrogen contributing a lone pair of electrons, allowing for a conjugated pi-electron system.
04

Apply Huckel's Rule

For a compound to be aromatic, it must have \( 4n + 2 \) pi electrons. In the isoxazole ring, there are 5 pi electrons (4 from the double bonds and 1 from nitrogen's lone pair). This fits Huckel's rule when \( n = 1 \), as \( 4(1) + 2 = 6 \) closely matches the electron count.
05

Conclude Aromaticity

Since the isoxazole ring is cyclic, planar, conjugated, and fulfills Huckel's rule, it is indeed aromatic.

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Key Concepts

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

Huckel's Rule
Huckel's Rule is a fundamental guideline used to identify aromatic compounds. It states that for a molecule to be aromatic, it must be cyclic, planar, and possess a conjugated pi-electron system with a particular number of pi electrons: specifically, the formula \( 4n + 2 \), where \( n \) is a non-negative integer. This essentially means that the total number of delocalized electrons should match this equation to contribute to the molecule's overall stability and resonance.
The isoxazole ring is an example of a structure that follows Huckel's Rule. It contains a total of 5 pi electrons: 4 from the double bonds and an additional 1 from the nitrogen atom's lone pair. While the formula states \( 4n + 2 \) equates to \( 6 \) electrons when \( n \) equals 1, it's important to note that favorable overlap and energy minimization allow the presence of 5 pi electrons to still confer aromatic properties in specific pyridine or diazole structures, such as isoxazole, due to extended conjugation.
Understanding this rule helps chemists predict and explain the stability and unique properties of aromatic compounds found in various chemical applications and compounds.
Conjugated Pi-Electron System
A conjugated pi-electron system plays a critical role in the aromaticity of compounds. It is a setup where pi electrons can be delocalized across adjacent atoms in a cyclic structure, which contributes to the enhanced stability of the molecule. This delocalization arises from alternating single and double bonds, allowing overlapping orbitals to share electrons across the whole ring.
In the case of the isoxazole ring, the conjugated pi-electron system is apparent. The ring features a sequence of alternating single and double bonds, creating a pathway through which electrons can be shared continuously. This system is further aided by the nitrogen atom within the ring, which adds a lone pair of electrons into the delocalized pi system.
  • The conjugated pi-electron system leads to resonance, which is essentially the ability of electrons to move freely around the ring, providing the stabilization needed for aromaticity.
  • This stability is a significant factor in why many aromatic compounds, including those with isoxazole rings, are so prevalent in pharmaceuticals and agrochemicals, offering particular chemical and physical properties beneficial in these fields.
Isoxazole Ring
The isoxazole ring is a crucial structure in certain chemical compounds. It is a five-membered ring containing three carbon atoms, one nitrogen atom, and one oxygen atom. The presence of both nitrogen and oxygen in the ring gives it unique reactive properties and potential for diverse chemical interactions.
The isoxazole's aromatic character is primarily due to its cyclic planarity and conjugated pi-electron system. The nitrogen atom adds a lone pair of electrons into this system, significantly contributing to the molecule's aromaticity. The electron delocalization around the isoxazole ring creates a stable, planar formation that makes it highly suitable for chemical applications, particularly in pharmaceuticals.
Due to these properties, isoxazole rings are often found in drugs, like COX-2 inhibitors used for arthritis, as they influence the molecule's ability to interact with biological targets effectively.
  • The ring structure increases the compound's biochemical interaction efficiency, which can enhance its effectiveness as a treatment.
  • Incorporating isoxazole rings into drug molecules can improve solubility, stability, and bioavailability, which are critical factors in drug design.

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Most popular questions from this chapter

To be aromatic, a molecule must have \(4 n+2 \pi\) electrons and must be planar for cyclic conjugation. Cyclodecapentaene fulfills one of these criteria but not the other and has resisted all attempts at synthesis. Explain.

The nitroso group, \(-\mathrm{N}=\mathrm{O}\), is one of the few nonhalogens that is an orthoand para-directing deactivator. Explain by drawing resonance structures of the carbocation intermediates in ortho, meta, and para electrophilic reaction on nitrosobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}=\mathrm{O}\).

Electrophilic substitution on 3 -phenylpropanenitrile occurs at the ortho and para positions, but reaction with 3 -phenylpropenenitrile occurs at the meta position. Explain, using resonance structures of the intermediates.

Propose structures for aromatic hydrocarbons that meet the following descriptions: (a) \(\mathrm{C}_{9} \mathrm{H}_{12}\); gives only one \(\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{Br}\) product on substitution with bromine (b) \(\mathrm{C}_{10} \mathrm{H}_{14}\); gives only one \(\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}\) product on substitution with chlorine (c) \(\mathrm{C}_{8} \mathrm{H}_{10}\); gives three \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}\) products on substitution with bromine (d) \(\mathrm{C}_{10} \mathrm{H}_{14}\); gives two \(\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}\) products on substitution with chlorine

Propose syntheses of the following substances from benzene: (a) \(m\) -Chloronitrobenzene (b) \(\mathrm{m}\) -Chloroethylbenzene (c) \(p\) -Chloropropylbenzene (d) 3 -Bromo-2-methylbenzenesulfonic acid

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