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

Step 1: Identify the four conformations of n-butane

For n-butane ($\ce{CH3CH2CH2CH3}$), the Newman projection about the C2–C3 bond gives:

• Conformation I (Anti): Dihedral angle = 180°, two $\ce{CH3}$ groups are anti — most stable (lowest PE)
• Conformation II (Fully eclipsed): Dihedral angle = 0°, two $\ce{CH3}$ groups eclipsed — least stable (highest PE)
• Conformation III (Gauche): Dihedral angle = 60°, $\ce{CH3}$ groups gauche — second most stable
• Conformation IV (Partially eclipsed): Dihedral angle = 120°, $\ce{CH3}$ eclipses $\ce{H}$ — between gauche and fully eclipsed

Step 2: Order by potential energy (increasing)

Potential energy order (lowest to highest): $\text{Anti (I)} < \text{Gauche (III)} < \text{Partially eclipsed (IV)} < \text{Fully eclipsed (II)}$

Step 3: Match to answer choices

Increasing PE: I < III < IV < II → Option (a)

Key Points to Remember:

• Anti (180°) = most stable for n-butane; gauche (60°) has 3.8 kJ/mol more PE
• Fully eclipsed (0°) = least stable; 19 kJ/mol above anti
• Partially eclipsed (120°) is between gauche and fully eclipsed
• Stability order: anti > gauche > partially eclipsed > fully eclipsed