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

- \emph{n-Butane} has the structural formula: $$ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 $$ - When viewed along the central C–C bond (C2–C3), different conformations arise due to rotation about this bond. - The key conformers are: - Anti: The two methyl groups are \(180^\circ\) apart (most stable). - Gauche: The methyl groups are \(60^\circ\) apart (less stable). - Eclipsed: The methyls and/or hydrogens are aligned (least stable — not counted here as stable conformers). - In \emph{n}-butane, due to symmetry: - There is 1 anti conformer. - There are 2 equivalent gauche conformers (mirror images of each other). - Hence, the molecule has 3 significant conformational isomers: - 1 anti + 2 gauche

Answer 2

The question asks how many anti and gauche conformers exist when we rotate around the central C2-C3 bond of n-butane. Looking at a Newman projection along this bond, the two methyl groups can end up at different angles to each other as the molecule rotates through a full 360 degrees, and only some of these angles count as the stable anti or gauche arrangements. Checking each option against the actual rotation profile settles the count.

1. One-anti & one-gauche: A full rotation of the back carbon relative to the front carbon passes through the anti position once, but it passes through a gauche type arrangement twice, once on each side of the anti position, because gauche conformations exist as a mirror image pair. Counting only one gauche form misses this second occurrence.
2. One-anti & two-gauche: As the back methyl group rotates through 360 degrees, it lines up directly opposite the front methyl group exactly once, giving one anti conformer with the methyls 180 degrees apart. It also passes through two separate 60 degree offset positions, one rotating clockwise from anti and one rotating counterclockwise, giving two distinct gauche conformers. This matches the actual rotational profile of n-butane.
3. Two-anti & one-gauche: There is only one arrangement where the two methyl groups sit exactly opposite each other in a full rotation, so counting two anti conformers overcounts this position, and undercounting gauche to just one misses its mirror partner.
4. Two-anti & two-gauche: This again overcounts the anti conformer, since the methyl groups can only be positioned 180 degrees apart once in a full 360 degree rotation.

Tracing the full rotation around the central bond shows the methyl groups line up directly opposite each other only once, giving a single anti conformer, while the 60 degree offset arrangement occurs twice, giving two gauche conformers.

So the correct answer is One-anti & two-gauche.