Conformational Analysis
Considering rotation about the central carbon-carbon single bond of n-butane, which conformation represents the global energy minimum on the potential energy diagram?
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Most stable conformation of n-butane is:
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About This Question
- Subject
- chemistry
- Chapter
- hydrocarbons
- Topic
- conformational analysis
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Anti(trans)conformation
Rotation about the C2−C3 bond of n-butane generates several conformations that differ in energy because of torsional strain and steric interactions between the two terminal methyl groups. The anti conformation places the two methyl groups at a dihedral angle of 180∘, which maximizes their separation and minimizes both steric and torsional strain, making it the most stable and therefore the global minimum. The gauche conformation, at a dihedral angle of about 60∘, is a local minimum but lies roughly 3.8 kJ mol−1 higher because the methyl groups are closer and experience steric repulsion, so it is not the lowest point. The fully eclipsed conformation has the methyl groups at 0∘ and suffers the maximum combined torsional and steric strain, representing the energy maximum, so it is incorrect. The partially eclipsed conformation is also a high-energy maximum where a methyl eclipses a hydrogen, so it cannot be the minimum. It is important to note that all these conformers interconvert rapidly at room temperature because the rotational barriers are small, yet the population is biased toward the anti form by the Boltzmann distribution. This analysis follows the standard NCERT discussion of ethane and butane conformers and explains why bulkier substituents widen the energy gap between conformers. A sanity check: the most stable arrangement should keep bulky groups farthest apart, which the anti form does.
This medium difficulty chemistry question is from the chapter hydrocarbons, covering the topic of conformational analysis. It appeared in the 2025 exam.
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