Question

Arrange the following conformational isomers of n-butane in order of their increasing potential energy : 
 

• A. I < IV < III < II
• B. II < III < IV < I
• C. I < III < IV < II
• D. II < IV < III < I

Hint:

Remember the energy profile for butane: Anti (min) $\rightarrow$ Partially Eclipsed $\rightarrow$ Gauche $\rightarrow$ Fully Eclipsed (max). In competitive exams, "staggered" conformers (Anti, Gauche) are always lower in energy than "eclipsed" ones. \begin{center} Potential Energy: Staggered < Eclipsed \end{center}

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The potential energy of different conformations of n-butane is determined by torsional strain (repulsion between bonding electrons) and steric strain (repulsion between bulky groups when they are forced close together).
Step 2: Detailed Explanation:
We analyze the four Newman projections shown:
1. I (Anti Conformation): The two bulky methyl (\(CH_3\)) groups are at a dihedral angle of \(180^\circ\). This minimizes steric repulsion and torsional strain. It has the lowest potential energy.
2. IV (Gauche Conformation): The two methyl groups are at a dihedral angle of \(60^\circ\). There is some steric repulsion (van der Waals strain) between them, making it less stable than the anti conformation but more stable than eclipsed forms.
3. III (Partially Eclipsed Conformation): The \(CH_3\) groups eclipse hydrogen atoms (\(CH_3-H\) interactions). This involves significant torsional and steric strain.
4. II (Fully Eclipsed Conformation): The two methyl groups eclipse each other (\(CH_3-CH_3\) interaction) at \(0^\circ\). This results in the maximum possible steric and torsional strain. It has the highest potential energy.
Therefore, the increasing order of potential energy is: \(I<IV<III<II\).
Step 3: Final Answer:
The correct order is I < IV < III < II.