Problem 55
Question
There is free rotation about the carbon-carbon \(\sigma\)-bond in ethane. This means that the energy difference between the staggered and eclipsed conformers is (a) \(3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(23 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(13 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(17 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-Step Solution
Verified Answer
The energy difference is approximately 13 kJ/mol, so the answer is (c) 13 kJ/mol.
1Step 1: Understand the Concept of Conformers
In ethane, the 3-bond between the two carbon atoms allows for free rotation about the bond, creating different spatial arrangements of the attached hydrogen atoms. These arrangements are called conformers. The two main conformers of interest are the staggered and eclipsed conformers.
2Step 2: Energy Comparison Between Conformers
The staggered conformer is more stable because it minimizes the repulsion between electron clouds of the hydrogen atoms, whereas the eclipsed conformer is less stable due to increased torsional strain from the alignment of hydrogen atoms.
3Step 3: Identify the Energy Difference
Empirical studies and quantum chemical calculations have determined that the energy difference between the staggered and eclipsed conformers of ethane is approximately 12-14 kJ/mol. When presented with multiple choice options, select the value that falls within or closest to this range.
4Step 4: Select the Correct Answer
Among the provided options, only option (c) 13 kJ/mol falls within the 12-14 kJ/mol range. Therefore, option (c) is the correct answer.
Key Concepts
Sigma BondTorsional StrainStaggered and Eclipsed Conformers
Sigma Bond
The concept of the sigma bond is fundamental to understanding molecular structures and interactions. In ethane, the carbon-carbon bond is an example of a sigma (\(\sigma\)) bond. A sigma bond is a covalent bond formed by the head-on overlap of atomic orbitals, specifically s-orbitals or hybrids like sp, sp2, or sp3. This type of bond allows for free rotation of the bonded atoms, which is why different conformers- or rotations - can exist.
- Ethane's \(\sigma\)-bond involves the overlap of sp3 hybrid orbitals from each carbon atom, creating a strong and stable bond.
- This single bond has a symmetrical distribution of electron density along the axis connecting the two nuclei.
- Because of this rotational symmetry, the orientation of the bonded hydrogens can change without breaking the bond.
Torsional Strain
Understanding torsional strain is essential when studying molecular conformation, especially in compounds like ethane. Torsional strain arises when atoms or groups closely interact in a molecule and repulsion occurs due to overlapping electron clouds. This form of strain is crucial in determining the stability of a molecule's conformation.
- In ethane, torsional strain is observed when the molecule is in the eclipsed conformation.
- Here, hydrogen atoms align with each other, causing electron repulsion due to their close proximity.
- Conversely, the staggered conformation reduces torsional strain as the hydrogen atoms are positioned further apart, lessening repulsive interactions.
Staggered and Eclipsed Conformers
Ethane can exist in different conformations due to rotation around its sigma bond. These variations are primarily the staggered and eclipsed conformers. Both have distinct energy levels due to their molecular arrangements.
- The staggered conformer of ethane is the most stable form. It occurs when the hydrogen atoms on one carbon atom are positioned between those on the adjacent carbon atom, maximizing distance and minimizing repulsion.
- The eclipsed conformer, on the other hand, occurs when hydrogen atoms on adjacent carbons line up, creating maximum torsional strain from their close proximity and alignment.
- The energy difference between these conformers results from the torsional strain in the eclipsed form and the decreased strain in the staggered conformer.
Other exercises in this chapter
Problem 52
Which among the following reagents converts cyclohexene into trans diol? (a) Baeyer reagent (b) \(\mathrm{OsO}_{4}\) (c) \(\mathrm{I}_{2} / \mathrm{CH}_{3} \mat
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The product formed on reaction of HBr with propene is (a) 1-bromoproapne (b) 2-bromopropane (c) 3-bromopropane (d) no reaction occurs
View solution Problem 56
Potassium salt of 2,3 -dimethyl maleic acid on electrolysis gives (a) ethyene (b) propyne (c) 2-butyne (d) 1 -butyne
View solution Problem 57
Which of the following shows acidic character? (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) (c) \(\mathrm{CH}_{
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