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Excluding compounds that have rings, there are three hydrocarbons that have the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{6}\). Write their structural formulas, and specify the hybridization of each carbon in these isomers.

Short Answer

Expert verified
1,3-Butadiene, 1-Butyne, and 2-Butyne are the hydrocarbons; their carbon atoms are \( sp^2 \), \( sp \), \( sp \) respectively.

Step by step solution

01

Analyze the Molecular Formula

The given molecular formula is \( C_4H_6 \). It indicates an acyclic hydrocarbon with four carbons and six hydrogens that must have multiple bonds because saturated hydrocarbons (alkanes) would have the formula \( C_nH_{2n+2} \). Here, \( C_4H_6 \) suggests the presence of a degree of unsaturation.
02

Determine the Degree of Unsaturation

The degree of unsaturation formula is given by \( (2n+2-m)/2 \), where \( n \) is the number of carbon atoms and \( m \) is the number of hydrogen atoms. For \( C_4H_6 \), it is \( (2(4)+2-6)/2 = 3 \). This means there are three degrees of unsaturation, indicating the presence of combinations of double bonds and/or triple bonds.
03

Identify Possible Structures

Consider what combinations of double and triple bonds can satisfy the three degrees of unsaturation without forming a ring structure. The possibilities include: 1. Two double bonds and no triple bonds (a diene). 2. One triple bond and one double bond (an alkyne with a double bond). 3. One triple bond and no double bonds (a simple alkyne).
04

Draw the Structural Formulas

1. 1,3-Butadiene: \( \text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2 \)2. 1-Butyne: \( \text{CH}\equiv\text{C}-\text{CH}_2-\text{CH}_3 \)3. 2-Butyne: \( \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 \)
05

Determine the Hybridization for Each Carbon

1. **1,3-Butadiene:** - Terminal \( \text{C} \): \( \text{CH}_2=\text{CH} \) are \( sp^2 \) - Internal \( \text{C}-\text{CH} \): \( sp^2 \) a second bond \( \text{CH}=\text{CH}_2 \) is also \( sp^2 \)2. **1-Butyne:** - Terminal \( \text{CH}_3 \), \( \text{CH}_2 \): \( sp^3 \) - \( \text{C} \equiv \text{C} \): Both carbons are \( sp \)3. **2-Butyne:** - Terminal \( \text{CH}_3 \): \( sp^3 \) - \( \text{C} \equiv \text{C} \): Both carbons are \( sp \)

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Key Concepts

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

Structural Formulas
A structural formula is a representation that shows the arrangement of atoms in a molecule. It depicts how atoms are bonded together in the molecule. For the hydrocarbons with the molecular formula \( C_4H_6 \), each isomer can have a different structure. Isomers are compounds with the same molecular formula but different structural arrangements. This variation in structure can significantly affect the properties and reactivity of each isomer.

1. **1,3-Butadiene**: This isomer has two double bonds. Each carbon atom in 1,3-butadiene is connected in a linear sequence, where alternating carbon atoms hold a double bond: \( \text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2 \).

2. **1-Butyne**: This compound has one triple bond and one single bond. The triple bond gives it a linear form at the center: \( \text{CH}\equiv\text{C}-\text{CH}_2-\text{CH}_3 \).

3. **2-Butyne**: Similar to 1-butyne, but the triple bond is between the second and third carbon atoms: \( \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 \).

Understanding the structural formulas helps in predicting how each molecule will behave during chemical reactions because each unique formula corresponds to a particular set of bonds, influencing molecular geometry and reactivity.
Molecular Formula
The molecular formula defines the number and type of atoms in a molecule. In this case, the formula \( C_4H_6 \) supports numerous structural forms known as isomers. The molecular formula provides basic information about the molecule's composition, but to understand how these atoms form a three-dimensional shape, one must look at the structural formula.

In hydrocarbons like \( C_4H_6 \), the number of carbon and hydrogen atoms tells us about the saturation level. Saturated hydrocarbons (alkanes) have only single bonds and follow the formula \( C_nH_{2n+2} \). However, \( C_4H_6 \) indicates unsaturation, meaning that some hydrogen atoms are removed and replaced by double or triple bonds, which affects the overall stability and reactivity.
Degree of Unsaturation
The degree of unsaturation tells us how many pairs of hydrogen atoms are "missing" compared to the fully saturated alkane. It signifies the number of rings or multiple bonds present. For \( C_4H_6 \), the formula for the degree of unsaturation calculates as \( (2n + 2 - m)/2 \), where \( n \) is the number of carbons and \( m \) is the number of hydrogens, resulting in 3 degrees of unsaturation.

The 3 degrees suggest that the molecule can have three structural features such as multiple bonds (double or triple) or rings (though rings are excluded here). These features indicate where the molecule might be more prone to reaction or instability, especially when exposed to catalysts or heat.
Hybridization
Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals. This mixing helps explain the shapes of molecules and the distribution of electrons in bonding. In hydrocarbons like \( C_4H_6 \), hybridization reveals the geometry.

- **1,3-Butadiene**: Each carbon in this diene possesses \( sp^2 \) hybridization due to the presence of double bonds, giving a planar shape to the molecule between the double bonds.

- **1-Butyne**: The terminal carbons of \( CH_3 \) and \( CH_2 \) are \( sp^3 \), showcasing a tetrahedral form. The carbons in the triple bond are \( sp \) hybridized, leading to a linear form.

- **2-Butyne**: Similar to 1-butyne, \( CH_3 \) groups are \( sp^3 \) hybridized, and the \( sp \) hybridization for the carbons involved in the triple bond also results in a linear configuration.

Hybridization not only influences molecular shape but also impacts chemical reactivity and properties.
Hydrocarbons
Hydrocarbons are organic compounds solely composed of carbon and hydrogen atoms. These molecules form the basis of organic chemistry and can manifest as alkanes, alkenes, alkynes, or aromatic compounds depending on the presence and type of bonding.

- **Saturated hydrocarbons**: Known as alkanes, contain only single bonds. Their general formula is \( C_nH_{2n+2} \). They are less reactive due to stable \( C-C \) and \( C-H \) bonds.

- **Unsaturated hydrocarbons**: Include alkenes and alkynes, featuring one or more double or triple bonds, respectively. This unsaturation increases reactivity, as seen in \( C_4H_6 \), a compound indicative of alkyne and alkene characteristics.

The presence of unsaturation, as in the given molecular formula, leads to possibilities of forming different isomers with distinct chemical and physical properties, useful in various chemical processes and industry applications.

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Most popular questions from this chapter

Write the structural formulas of the alcohols with the formula \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\) and indicate those that are chiral. Show only the \(\mathrm{C}\) atoms and the \(-\mathrm{OH}\) groups.

Draw all the possible structural isomers for the molecule having the formula \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{Cl}\). All isomers contain one benzene ring.

Name the classes to which the following compounds belong: (a) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Fill in the blanks in the given paragraph with the most appropriate term from the following: chiral, cis, constitutional isomers, enantiomers, resonance structures, stereoisomers, trans. Isomers are different compounds that have the same molecular formula. Isomers that have their atoms connected in a different order (branched versus unbranched chain, for example) or a different sequence of bond types \((\mathrm{C}=\mathrm{CCC}\) versus \(\mathrm{CC}=\mathrm{CC},\) for example) are termed Isomers with the same order of connections and sequence of bond types, but which differ in the spatial arrangement of the atoms are called \(.\) This is often seen in compounds where substituents may be on the same or opposite sides of a carbon-carbon double bond. Substituents on the same side are described as \(\quad ;\) those on opposite sides are . A different kind of isomerism characterizes a molecule, that is, a molecule with a structure that allows for two nonsuperimposable mirror-image forms. Two nonsuperimposable mirror images are of the other

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