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The molecule benzyne \(\left(\mathrm{C}_{6} \mathrm{H}_{4}\right)\) is a very reactive species. It resembles benzene in that it has a sixmembered ring of carbon atoms. Draw a Lewis structure of the molecule and account for the molecule's high reactivity.

Short Answer

Expert verified
The Lewis structure of benzyne shows four Carbon atoms with unpaired electrons. These unstable configurations contribute to its high reactivity as these free electrons often interact with other atoms or molecules to achieve stability.

Step by step solution

01

Drawing the Lewis structure.

Starting with the carbon atoms, as it is a six-membered ring similar to benzene, draw six Carbon (C) atoms in a hexagonal pattern. The Hydrogen (H) atoms will attach to two of these Carbon (C) atoms. The other Carbon atoms are connected directly to each other without any hydrogen. Each Carbon (C) atom has 4 valence electrons and takes 4 more to complete its octet. But since there are only 4 Hydrogen (H) atoms, only two of the Carbon atoms complete their octet and the other 4 Carbon (C) atoms will have two unpaired electrons each.
02

Accounting for the high reactivity.

The remaining four Carbon atoms left with unpaired electrons will make the molecule highly reactive. These unpaired electrons are often referred to as free radicals and are significant contributors to its reactivity. These free electrons often seek to bond with other atoms or molecules to make their electron configuration stable, hence leading to reactions.

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Key Concepts

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

Lewis structure
When constructing the Lewis structure of a molecule, you're visually representing the valence electrons around atoms. For benzyne, which is derived from benzene, this involves arranging the carbon atoms (C) in a hexagonal shape. Each carbon atom in the ring has four valence electrons, capable of creating bonds. In benzyne, two hydrogen atoms will attach to two carbon atoms, leaving the remaining four carbon atoms to bond only with each other. Since benzyne has a molecular formula of \( ext{C}_6 ext{H}_4\), not all carbon atoms will achieve a full octet. The incomplete octets contribute to the unique reactivity of benzyne.
Reactivity
The high reactivity of benzyne stems from its structural composition. With four carbon atoms not achieving a full octet, there are additional electrons which are not stable in their bonds. These electrons are generally energetic and seek to attain stability by forming additional bonds. This need for stability results in benzyne being highly reactive, often engaging in rapid chemical reactions with other molecules. The reactive sites on benzyne are prime for reactions, making it a molecule of interest in many organic chemistry explorations.
Reactivity in such molecules often leads to the study of reaction mechanisms where benzyne can be formed as an intermediate product.
Free radicals
Free radicals in chemistry are atoms, molecules, or ions with unpaired electrons. Benzyne's structure allows the formation of free radicals through its unpaired electrons resulting in high reactivity. These free radicals have the potential to interact with nearby molecules, being highly reactive compounds that often start chain reactions. Because of their reactivity, free radicals can have significant implications in organic chemistry, sometimes serving to start complex chemical reactions
due to their instability and desire to capture other electrons to create stable bonds.
Carbon ring
In benzyne, the carbon ring plays a fundamental role in determining the molecule's behavior and properties. Arranging carbon atoms in a six-membered ring is typical of aromatic compounds like benzene. However, unlike the stable benzene ring, benzyne's ring suffers from additional strain due to its lack of some hydrogen atoms and additional electrons seeking stability. This results in a different kind of reactivity in comparison to typical aromatic stability. In the carbon ring of benzyne, electron-sharing is uneven, which also affects the distribution of free radicals, leading to its characteristic high reactivity.
Valence electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in forming bonds with other atoms. In benzyne, valence electrons (from both carbon and hydrogen) are depicted in the Lewis structure to show possible bonds. Since carbon has four valence electrons, it seeks to acquire four more to fill its octet, usually by forming four covalent bonds. In benzyne, each carbon shares electrons unevenly due to the four carbons without complete octets. Thus, the distribution of valence electrons significantly influences the molecule's reactivity and bond formation.
Understanding these electrons not only helps predict reactivity but also how the molecule might engage with other compounds in a reaction.

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Most popular questions from this chapter

Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to \(\mathrm{Fe}^{2+}\) in the hemoglobin molecule. The molecular orbitals of CO have the same energy order as those of the \(\mathrm{N}_{2} \mathrm{~mol}-\) ecule. (a) Draw a Lewis structure of \(\mathrm{CO}\) and assign formal charges. Explain why CO has a rather small dipole moment of 0.12 D. (b) Compare the bond order of CO with that from molecular orbital theory. (c) Which of the atoms (C or O) is more likely to form bonds with the \(\mathrm{Fe}^{2+}\) ion in hemoglobin?

Both ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) contain the \(\mathrm{C}=\mathrm{C}\) bond. The reactivity of ethylene is greater than that of benzene. For example, ethylene readily reacts with molecular bromine, whereas benzene is normally quite inert toward molecular bromine and many other compounds. Explain this difference in reactivity.

The compounds carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) are similar in geometry and hybridization. However, \(\mathrm{CCl}_{4}\) does not react with water but \(\mathrm{SiCl}_{4}\) does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the \(\mathrm{Si}\) atom in \(\left.\mathrm{SiCl}_{4} .\right)\)

Which of the following species are tetrahedral? \(\mathrm{SiCl}_{4}, \mathrm{SeF}_{4}, \mathrm{XeF}_{4}, \mathrm{Cl}_{4}, \mathrm{CdCl}_{4}^{2-}\)

What is the angle between the following two hybrid orbitals on the same atom? (a) \(s p\) and \(s p\) hybrid orbitals, (b) \(s p^{2}\) and \(s p^{2}\) hybrid orbitals, (c) \(s p^{3}\) and \(s p^{3}\) hybrid orbitals.

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