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All carbon-to-carbon bond lengths are identical in benzene. Does this argue for or against the presence of \(\mathrm{C}=\mathrm{C}\) bonds in benzene? Explain.

Short Answer

Expert verified
The identical bond lengths argue against traditional \( \mathrm{C}=\mathrm{C} \) bonds; they indicate benzene has a delocalized structure.

Step by step solution

01

Understanding Benzene's Structure

Benzene is a chemical compound with the formula \( \mathrm{C}_6\mathrm{H}_6 \). It consists of a ring of six carbon atoms, each bonded to a hydrogen atom. The carbons can be thought of as being connected by alternating single and double bonds.
02

Theoretical Bond Lengths in Benzene

In typical organic molecules, a single carbon-carbon bond (\( \mathrm{C}-\mathrm{C} \)) has a bond length of approximately 1.54 Å, while a double bond (\( \mathrm{C}=\mathrm{C} \)) is shorter, around 1.34 Å. Therefore, one would expect alternating bond lengths in benzene if distinct single and double bonds exist.
03

Observing Carbon-Carbon Bond Lengths

In benzene, all the carbon-carbon bond lengths are found to be identical at about 1.39 Å, which is intermediate between typical \( \mathrm{C}-\mathrm{C} \) and \( \mathrm{C}=\mathrm{C} \) bond lengths. This observation is key to understanding benzene's structure.
04

Reality versus Expectations

The identical bond lengths imply that there are no discrete single or double bonds as in typical alkenes. Instead, the structure is best described by resonance or as a hybrid where the electrons are delocalized over all six carbon atoms uniformly.
05

Conclusion on Bonding in Benzene

The equal bond lengths argue against the presence of traditional \( \mathrm{C}=\mathrm{C} \) bonds. Instead, it supports a structure where the electron density is evenly distributed, leading to a consistent bond length throughout the ring.

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Key Concepts

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

Carbon-Carbon Bond Lengths
Carbon-carbon bond lengths in molecules provide insight into the type of bonds within that molecule. In general, single bonds (\( \mathrm{C}-\mathrm{C} \)) are longer, measuring about 1.54 Å, in comparison to double bonds (\( \mathrm{C}=\mathrm{C} \)), which measure around 1.34 Å. These typical values indicate that alternating bond lengths should exist in a molecule like benzene if there were indeed distinct single and double bonds.
However, in benzene, the bond lengths defy these typical expectations. Each carbon-carbon bond in the benzene ring measures approximately 1.39 Å. This value is intriguing because it's situated between the lengths of traditional single and double bonds.
The uniform bond length in benzene suggests a unique bonding arrangement that does not conform to the separate alternating pattern seen in other compounds. This uniformity hints at an equalization of bond characteristics across the entire molecule, which is key to understanding benzene's stable structure.
Resonance in Chemistry
Resonance in chemistry is a concept used to describe the way electrons are distributed across certain molecules. In benzene, the idea of resonance helps explain its structure and bond lengths. If benzene had alternating single and double bonds, each bond length would vary. But, because the bond lengths are uniform, benzene is said to exist as a resonance hybrid.
A resonance hybrid in benzene means that multiple structures could describe the placement of electrons. It's not that one structure with single and double bonds exists at any time but rather an average or composite of these structures functions continuously.
This results in the bonds having partial characteristics of both single and double bonds uniformly. The resonance concept helps us understand that electrons in benzene are not localized between specific atoms but are shared over the entire ring. This sharing results in increased stability and unique properties, such as equal bond lengths.
Delocalized Electrons
Delocalized electrons are a vital concept in explaining the structure of benzene. Unlike localized electrons, which are confined between two specific atoms forming a bond, delocalized electrons can move between multiple atoms. This movement occurs across the entire benzene ring.
In benzene, all the carbon-carbon bonds are identical due to this delocalization. The electrons are not fixed in place creating distinct single and double bonds but are spread evenly over all the carbon atoms. This even distribution leads to a bond length of 1.39 Å for each carbon-carbon bond.
  • This delocalization is responsible for:
  • Creating uniform bond lengths
  • Enhancing chemical stability
  • Contributing to benzene's unique planar and cyclic structure
The concept of delocalized electrons is critical to understanding why benzene behaves differently from other compounds with alternating bonds and triggers the realization that classical single and double bond labels do not apply in the same way within its structure.
Chemical Bonding
Chemical bonding is essential for understanding benzene's structure. It involves understanding how atoms share or transfer electrons to achieve stability. Benzene exhibits a special kind of bonding due to its resonance and delocalized electrons.
Traditional aromatic compounds like benzene don't adhere to the usual rules of alternating single and double bonds. Instead, the concept of resonance allows for a simultaneous sharing of electrons across all carbon atoms. This type of electron sharing creates bonds that are neither fully single or double in nature but are instead hybrids.
Hybrid bonding means that benzene has a stable, uniform bond structure throughout its ring. This stability affects benzene's chemical properties, making it less reactive than comparable alkenes that have clearly defined single and double bonds. Seeing benzene as a resonance hybrid leads to understanding its inertness and its vital role in various chemical applications.

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Most popular questions from this chapter

For each pair of bonds, indicate the more polar bond and use \(\delta+\) or \(\delta-\) to show the partial charge on each atom. (a) \(\mathrm{C}-\mathrm{O}\) and \(\mathrm{C}-\mathrm{N}\) (b) \(\mathrm{B}-\mathrm{O}\) and \(\mathrm{P}-\mathrm{S}\) (c) \(\mathrm{P}-\mathrm{H}\) and \(\mathrm{P}-\mathrm{N}\) (d) \(\mathrm{B}-\mathrm{H}\) and \(\mathrm{B}-\mathrm{Cl}\)

For each pair of bonds, identify the more polar one and use \(\delta+\) or \(\delta-\) to indicate the partial charge on each atom. (a) \(\mathrm{B}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{O}\) (b) \(\mathrm{O}-\mathrm{F}\) and \(\mathrm{O}-\mathrm{Se}\) (c) \(\mathrm{S}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{F}\) (d) \(\mathrm{N}-\mathrm{H}\) and \(\mathrm{N}-\mathrm{F}\)

Tetrasulfur tetranitride reacts with disulfur dichloride to form \(\mathrm{S}_{4} \mathrm{~N}_{3} \mathrm{Cl}\), a salt. $$ 3 \mathrm{~S}_{4} \mathrm{~N}_{4}+2 \mathrm{~S}_{2} \mathrm{Cl}_{2} \longrightarrow 4 \mathrm{~S}_{4} \mathrm{~N}_{3} \mathrm{Cl} $$ Write a plausible Lewis structure for the two reactants and the cation of the salt.

Consider the \(\mathrm{SCO}_{2}^{2-}\) ion in which each \(\mathrm{S}\) and \(\mathrm{O}\) atom is bonded to a central \(\mathrm{C}\) atom. Use Lewis structures and formal charges to write resonance structures and to determine which is the most plausible resonance form. Explain your choice.

Carbon-to-carbon double bonds, \(\mathrm{C}=\mathrm{C}\), react by addition. Cite experimental evidence that benzene does not have \(\mathrm{C}=\mathrm{C}\) bonds.

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