A common biologically active radical is the pentadicnyl radical, RCHCHCHCHCHR', in which the carbons form a long chain with \(R\) and \(R^{\prime}\) (which can be a number of different organic groups) at ench end. Draw three resonance structures for this compound that maintain carbon's valence of four.

Short Answer

Expert verified
Three resonance structures of the pentadienyl radical show the radical electron on the first, second, and third carbons from the left, respectively, with corresponding shifts in the positions of the double bonds along the carbon chain.

Step by step solution

01

Identify the Conjugated System

The given radical is a chain of carbons with alternating single and double bonds, which is a conjugated system. In a conjugated system, the electrons are delocalized, and this delocalization can be represented by drawing resonance structures. Start by drawing the backbone of the molecule with the radical at one end and leaving space to add different positions for the double bonds.
02

Draw the First Resonance Structure

For the first resonance structure, draw the radical electron at the first carbon position, denoted as R--C•. Then draw a double bond between the second and third carbon atoms, and alternate single and double bonds for the remaining chain. Make sure each carbon atom adheres to the tetravalence rule by having only four bonds.
03

Draw the Second Resonance Structure

In the second resonance structure, move the double bond between the first and second carbon atoms down the chain so that it is between the third and fourth carbon atoms. Move the radical electron to the second carbon atom from the left. Again, ensure that every carbon atom has four bonds.
04

Draw the Third Resonance Structure

For the third resonance structure, shift the double bonds one position further down the chain. The new double bond will be between the fourth and fifth carbon atoms, and the radical electron will reside on the third carbon atom from the left. Each carbon should continue to have only four bonds, maintaining its valence of four.

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Key Concepts

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

Conjugated Systems
Conjugated systems play a pivotal role in the stability and chemical reactivity of organic molecules. They consist of alternating single and double bonds between carbon atoms, which allows for a greater delocalization of π-electrons across the chain of atoms. This electron delocalization contributes to the molecule's stability and is a fundamental aspect of understanding resonance.

In the pentadienyl radical, the conjugated system is established by the chain of carbons with alternating bonds. When assessing the stability and possible resonance structures of these systems, one must consider how the electrons can 'move' through the molecule while maintaining the overall stability, and most importantly, the valence of the carbon atoms involved.
Radical Stability
Free radicals in organic chemistry are often unstable and highly reactive due to the presence of an unpaired electron. However, the stability of radicals is significantly influenced by their environment. In conjugated systems like the pentadienyl radical, the radical electron can be delocalized along the π system, which increases the stability of the radical.

In understanding steps to draw resonance structures, the radical electron's placement is key. The ability to distribute the single unpaired electron across a larger structure, sharing the 'burden' of reactivity, results in more stable intermediates. The different structures drawn in the exercise all aim to illustrate the radical stability by showing various electron distributions that adhere to carbon's valence.
Electron Delocalization
Electron delocalization is an essential concept that describes the spreading out of electrons across multiple atoms within a molecule. This is a characteristic of π-conjugated systems and contributes significantly to molecular stability, as seen in the context of resonance structures of organic compounds.

In the provided exercise, electron delocalization is depicted through the various resonance structures of the pentadienyl radical. The movement of the double bonds and the radical electron illustrates how the electron density is not localized to a single bond or atom but is rather spread across the molecular framework. This dispersion of electron density over the conjugated π system is what allows for the different resonance forms and ultimately enhances the stability of the molecule.
Valence of Carbon
The valence of an atom refers to its ability to form a specific number of chemical bonds. For carbon, this valence is four, meaning each carbon atom can form four covalent bonds. The valence rule is essential to drawing accurate resonance structures because each structure must maintain the tetravalence of carbon.

The step-by-step solution indicates the importance of adhering to this valence in each suggested resonance structure. By only allowing four bonds per carbon atom, the solution ensures that all resonance structures are chemically valid. Through these guidelines, the resonance structures successfully demonstrate the potential distribution of electron density while upholding the fundamental valence rule for carbon atoms.
Organic Chemistry Resonance
Resonance in organic chemistry is a concept used to describe the delocalization of electrons within a molecule over several atoms or bonds. It provides a way to represent molecules more accurately than a single static structure can depict. Rather than having fixed positions, certain electrons, particularly those in π bonds or unpaired electrons, are shared across adjacent atoms, giving rise to various contributing structures, known as resonance forms or structures.

The value of this concept shines in exercises like drawing multiple valid resonance structures for the pentadienyl radical. Each structure serves as a potential arrangement of electrons that collectively describe the true nature of the molecule. Students learn that the actual molecule is not any single drawn structure but a hybrid of all possible resonance forms, capturing the dynamic nature of electron distribution.

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Most popular questions from this chapter

Write the Lewis structures for the following reactive species found to contribute to the destruction of the ozone layer and indicate which are radicals: (a) chlorine monoxide, \(\mathrm{ClO}\); (b) dichloroperoxide, \(\mathrm{Cl}-\mathrm{O}-\mathrm{O}-\mathrm{Cl}\); (c) chlorine nitrate, \(\mathrm{ClONO}_{2}\) (the central \(\mathrm{O}\) atom is attached to the \(\mathrm{Cl}\) atom and to the \(\mathrm{N}\) atom of the \(\mathrm{NO}_{2}\) group); (d) chlorine peroxide, \(\mathrm{Cl}-\mathrm{O}-\mathrm{O}\).

Write the Lewis structures, including resonance structures where appropriate, for (a) the oxalate ion, \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) (there is a \(\mathrm{C}\) - \(\mathrm{C}\) bond with two oxygen atoms attached to each carbon atom); (b) \(\mathrm{BrO}^{+}\); (c) the acctylide ion, \(\mathrm{C}_{2}^{2-}\). Assign formal charges to each atom.

For each of the following ground-state ions, predict the type of orbital \((1 s, 2 p, 3 d, 4 f\), and so on) in which the electrons of highest energy will reside: (a) \(\mathrm{Fe}^{2+}\); (b) \(\mathrm{Bi}^{3+}\); (c) \(\mathrm{As}^{3+}\); (d) \(\mathrm{Os}^{+}\).

Write Lewis structures, including typical contributions to the resonance structure (where appropriate, allow for the possibility of octet expansion, including double bonds in different positions), for (a) sulfite ion; (b) hydrogen sulfite ion; (c) perchlorate ion; (d) nitrite ion.

Which of the following members of cach pair is the stronger Lewis base? Explain your reasoning. (a) \(\mathrm{CH}_{3}^{-}\)or \(\mathrm{CH}_{4}\); (b) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S}\); (c) \(\mathrm{NH}_{3}\) or \(\mathrm{NH}_{2}^{-}\).

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