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Chapter 1: Problem 39

Propose structures for molecules that meet the following descriptions: (a) Contains two \(s p^{2}\) -hybridized carbons and two \(s p^{3}\) -hybridized carbons (b) Contains only four carbons, all of which are \(s p^{2}\) -hybridized (c) Contains two \(s p\) -hybridized carbons and two \(s p^{2}\) -hybridized carbons

Short Answer

Expert verified
(a) Propene, (b) 1,3-butadiene, (c) But-2-yne.

Step by step solution

01

Understanding the terminology

Before proposing the structures, it's important to understand the hybridization states: \( sp^2 \) hybridization means the carbon is involved in double bonds, while \( sp^3 \) hybridization indicates single bonds. \( sp \) hybridization involves triple bonds or a combination of a double and a single bond.
02

Structure for condition (a)

For structure (a), start with two \( sp^2 \)-hybridized carbons, which means they are part of a double bond. For example, you can have an ethene unit \( (CH_2=CH-) \). Now, attach two \( sp^3 \)-hybridized carbons, which could be in the form of two single-bonded \( -CH_3 \) groups. A possible structure could be Propene \( CH_3-CH=CH_2 \).
03

Structure for condition (b)

For structure (b), all four carbons should be \( sp^2 \)-hybridized. This typically forms a cyclic compound or a linear chain with alternating double bonds. As an example, you can draw 1,3-butadiene \( CH_2=CH-CH=CH_2 \), a linear chain where all the carbon atoms have \( sp^2 \) hybridization.
04

Structure for condition (c)

For structure (c), two carbons should be \( sp \)-hybridized, meaning they are part of a triple bond. In addition, two \( sp^2 \)-hybridized carbons are needed. Drawing a molecule like but-2-yne (\( CH_3-C\equiv C-CH_3 \)), you have a triple bond between two carbons \( sp \) hybridized, and the adjacency with \( -CH_3 \) ensures the remaining two are \( sp^2 \) hybridized.

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Key Concepts

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

sp2 hybridization
In organic chemistry, \(sp^2\) hybridization is vital for understanding how carbon forms double bonds.When a carbon atom undergoes \(sp^2\) hybridization, it combines one s orbital with two p orbitals.This results in three \(sp^2\) hybrid orbitals and leaves one p orbital unhybridized.The three \(sp^2\) hybrid orbitals arrange themselves in a trigonal planar geometry, creating a flat structure with 120-degree bond angles.This type of hybridization is commonly seen in alkenes, such as ethene \((CH_2=CH_2)\).Here, the double bond forms as a result of the sideways overlap of the unhybridized p orbitals, which creates a pi bond, on top of the \(sp^2\) hybridized sigma bonds.
sp3 hybridization
When you hear \(sp^3\) hybridization, think of single bonds in molecules.In this process, a carbon atom's one s orbital merges with three p orbitals to form four equivalent \(sp^3\) hybrid orbitals.These orbitals arrange in a tetrahedral shape with 109.5-degree bond angles, allowing the carbon atom to form strong sigma bonds with four other atoms.This configuration is typical in alkanes, like methane \((CH_4)\), where each \(sp^3\) hybrid orbital forms a single bond with hydrogen atoms.Understanding this helps explain the flexibility and rotation possible around single bonds.
sp hybridization
\(sp\) hybridization is crucial for describing triple bonds or a mix of double and single bonds.Here, the carbon atom hybridizes one s orbital with one p orbital, leaving two p orbitals unhybridized.This forms two \(sp\) hybrid orbitals that align in a linear shape, 180 degrees apart.Typical in alkynes, such as acetylene \((C_2H_2)\), \(sp\) hybridization involves the carbon atoms connecting through a triple bond.This triple bond is made up of one sigma bond and two pi bonds, formed by the overlap of the unhybridized p orbitals, giving alkynes their distinctive characteristics.
molecular structures
Understanding molecular structures involves knowing how atoms arrange themselves based on their hybridization.Different hybridizations (like \(sp^2\), \(sp^3\), and \(sp\)) lead to various geometrical arrangements.
  • \(sp^2\) hybridization gives a planar shape, typical in flat molecules like benzene.
  • \(sp^3\) hybridized atoms result in tetrahedral structures, like the 3D shape of alkanes.
  • \(sp\) hybridization results in linear structures, seen in molecules like carbon dioxide \((CO_2)\).
By comprehending these fundamental structures, you can predict and understand the behavior of organic compounds in various reactions.
carbon bonding
Carbon bonding is fascinating because of carbon's ability to form strong covalent bonds with other atoms.This is largely due to hybridization, which allows carbon to adopt versatile bonding configurations such as \(sp^2\), \(sp^3\), and \(sp\).
  • The strong bond formed by \(sp^3\) hybridized carbon is pivotal in stable, saturated hydrocarbons.
  • The \(sp^2\) hybridization results in double bonds, contributing to molecules' reactivity and ability to engage in polymerization.
  • \(sp\) hybridization allows for triple bonds, giving rise to highly reactive molecules used in various chemical syntheses.
The diversity and strength of carbon bonding make it the backbone of organic chemistry, essential in forming life's molecular structures.

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Most popular questions from this chapter

What are likely formulas for the following molecules? (a) \(\mathrm{NH}_{?} \mathrm{OH}\) (b) AlCl? (c) \(\mathrm{CF}_{2} \mathrm{Cl}\) ? (d) \(\mathrm{CH}_{?} \mathrm{O}\)

There are three different substances that contain a carbon-carbon double bond and have the formula \(\mathrm{C}_{4} \mathrm{H}_{8}\). Draw them, and tell how they differ.

Draw structures for the following molecules, showing lone pairs: (a) Acrylonitrile, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N},\) which contains a carbon-carbon double bond and a carbon-nitrogen triple bond (b) Ethyl methyl ether, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O},\) which contains an oxygen atom bonded to two carbons (c) Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), which contains a chain of four carbon atoms (d) Cyclohexene, \(\mathrm{C}_{6} \mathrm{H}_{10}\), which contains a ring of six carbon atoms and one carbon-carbon double bond

There are two different substances with the formula \(\mathrm{C}_{3} \mathrm{H}_{6}\). Draw both, and tell how they differ.

What are likely formulas for the following substances? (a) CCl? (b) AlH? (c) \(\mathrm{CH}_{?} \mathrm{Cl}_{2}\) (d) \(\mathrm{SiF}_{?}\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{?}\)
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