Question

Number of conformational isomers of \emph{n-butane:}

• A. One-anti \& one-gauche
• B. One-anti \& two-gauche
• C. Two-anti \& one-gauche
• D. Two-anti \& two-gauche

Hint:

Conformational isomers differ by rotation around single bonds. For alkanes like \emph{n}-butane, analyze using Newman projections for anti and gauche forms.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
n-Butane is CH3-CH2-CH2-CH3, and rotating around the central C2-C3 bond changes the relative position of the two terminal methyl groups. The main conformers of interest are anti, where the methyls are 180 degrees apart, and gauche, where they are 60 degrees apart.

Step 2: Key Formula or Approach:
Looking down the C2-C3 bond as a Newman projection, the back methyl group can be rotated through a full 360 degrees relative to the front one, and the task is to count how many times this rotation passes through the anti position and how many times it passes through a gauche position.

Step 3: Detailed Explanation:
Starting from the anti position at 180 degrees and rotating the back group by 60 degrees at a time traces out six positions in total across the full circle, at 0, 60, 120, 180, 240, and 300 degrees measured from the front methyl group.
The position at 180 degrees is the single anti conformer, with the methyls as far apart as possible and the least steric strain.
The positions at 60 degrees and 300 degrees are both gauche conformers, with the methyls closer together, and these two are non identical mirror images of each other, so they count as two separate gauche forms.
The remaining positions at 0, 120, and 240 degrees are eclipsed conformers, where substituents overlap directly, and these are higher energy arrangements not counted among the stable staggered conformers.

Step 4: Final Answer:
n-Butane has one anti conformer and two gauche conformers among its stable staggered forms.