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The most stable conformation of \(\mathrm{n}\)-butane is (a) eclipsed (b) gauche (c) anti-staggered (d) skew-boat

Short Answer

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(c) anti-staggered

Step by step solution

01

Understanding Conformations

Conformations refer to different spatial arrangements of atoms in a molecule that can be transformed into one another by rotation about single bonds. In the case of n-butane, a simple alkane, these conformations arise from rotations around the central carbon-carbon bond (C2-C3).
02

Evaluating Eclipsed Conformation

In the eclipsed conformation, atoms or groups on adjacent carbons are aligned with each other, maximizing steric hindrance and repulsive interactions. This results in the highest energy and makes it the least stable conformation.
03

Analyzing Gauche Conformation

The gauche conformation occurs when the methyl groups (CH3) are 60 degrees apart. While it is more stable than the eclipsed conformation, due to less repulsion, it is not the most stable due to some steric strain remaining.
04

Determining Anti-Staggered Conformation

In the anti-staggered conformation, the methyl groups are positioned 180 degrees apart. This alignment minimizes steric hindrance and allows for the most stable, lowest energy arrangement due to the favorable distribution of electron clouds.
05

Considering Skew-Boat Conformation

Skew-boat typically refers to a less common conformation and is not a standard term used for straight-chain alkanes like n-butane. It generally does not provide a stability advantage for simple alkanes.
06

Conclusion

After evaluating each conformation, the anti-staggered conformation is identified as the most stable due to the minimization of steric interactions and the optimal spatial arrangement of atoms.

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Key Concepts

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

n-Butane
n-Butane is a simple linear alkane containing four carbon atoms with the molecular formula \(C_4H_{10}\). This molecule consists of two end methyl groups (CH3) and two middle methylene groups (CH2), forming a chain structure.
As a member of the alkane family, n-butane has only single bonds, making it highly flexible for rotations around these bonds.
Understanding this structural flexibility is crucial for analyzing its conformations and stability.
The main focus when discussing n-butane is its ability to assume different spatial arrangements, known as conformations.
These arise from rotations around the central C2-C3 bond, allowing for multiple configurations with varying potential energies.
Molecular Stability
Molecular stability in alkanes like n-butane refers to the energy state of a specific conformation. A molecule will naturally prefer the lowest energy state because it represents greater stability.
Differences in molecular stability between conformations are often due to steric interactions and electronic repulsions.
These factors influence how atoms and groups within a molecule interact based on the spatial arrangement.
In n-butane:
  • The anti-staggered conformation offers the greatest molecular stability.
  • This is because it minimizes steric hindrance by keeping bulky methyl groups apart.
  • Eclipsed formations have increased energy due to greater repulsion between aligned atoms or groups.
  • Gauche conformations, while more stable than eclipsed, still present some steric strain.
The goal in identifying and evaluating conformations is to find the state that maintains the lowest energy, reflecting maximum stability.
Steric Hindrance
Steric hindrance occurs when atoms or groups within a molecule are positioned such that their electron clouds repel each other. This repulsion increases the energy of the molecule and impacts conformational stability.
In the analysis of n-butane:
  • The eclipsed conformation experiences maximum steric hindrance.
  • This is due to methyl groups being aligned, leading to overcrowding and increased repulsion.
  • The gauche conformation benefits from reduced steric hindrance compared to eclipsed, yet still faces some crowding.
  • The anti-staggered conformation has minimum steric hindrance.
  • Methyl groups are situated 180 degrees apart, avoiding overlap and thus reducing repulsion.
By minimizing steric hindrance, a molecule achieves a lower energy state, which corresponds to greater molecular stability.
Conformations
Conformations are different spatial arrangements of atoms that can be attained through rotations around single bonds within a molecule. In the study of n-butane, conformations result from rotating the C2-C3 bond.
Key conformations include:
  • Anti-Staggered Conformation: Methyl groups are 180 degrees apart, achieving the most stable arrangement due to reduced steric effects.
  • Gauche Conformation: Methyl groups are 60 degrees apart. Less stable than anti-staggered due to some residual steric hindrance.
  • Eclipsed Conformation: Methyl groups align directly behind one another. This results in the highest energy and thus is the least stable due to maximum steric hindrance.
Conformational analysis aims to understand which spatial arrangement reduces energy most effectively, indicating a prefered stable state for the molecule.

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Most popular questions from this chapter

The IUPAC name of compoud (a) \(1,2,3\)-tricyanopropane (b) Propane-1, 2,3 -tricarbonitrile (c) 3 -cyanopentane- 1,5 -dinitrile (d) Propane-1, 2,3 -trinitrile.

Which of the following does not show geometrical isomerism? (a) 1,1 -dichloro-1-pentene (b) 1,2 -dichloro- 1 -pentene (c) 1,3 -dichloro-1-pentene (d) 1,4 -dichlroro-2-pentene

Match the following Column-I (a) \(\mathrm{CH}_{3} \mathrm{CHDT}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{2} \mathrm{H}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) Column-II (p) No optical isomer (q) Two optical isomers (r) Four optical isomers (s) Three optical isomers (t) Two asymmetric carbon atoms.

Keto-enol tautomerism is observed in (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{3}\)

Match the following: List I List II 1\. \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{HCOOCH}_{3}\) (i) metamers 2\. \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}\) and (ii) position isomers \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) and (iii) tautomers \(\mathrm{CH}_{3}-\mathrm{CH}\left(\mathrm{NH}_{2}\right)-\mathrm{CH}_{3}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) (iv) functional isomer The correct matching is: \(\begin{array}{llll}1 & 2 & 3 & 4\end{array}\) (a) (iii) (iv) (i) (iv) (b) (i) (ii) (iii) (iv) (c) (iii) (ii) (i) (iv) (d) (iv) (ii) (ii) (iv)

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