3.6 : Konformationen von Ethan und Propan

A molecule of ethane, when viewed down the carbon-carbon bond, shows the C-H groups spaced at 60°dihedral angles. This is the staggered conformation of ethane.

The staggered form of ethane has the lowest energy. This is because the C-H bonds are furthest from each other, minimizing steric repulsion between the electrons in the bonds, and therefore, stabilizing the molecule.

Another factor that stabilizes the staggered conformation is the favorable interaction between the occupied, bonding molecular orbital and an unoccupied, antibonding molecular orbital.

Rotating the farther carbon by keeping the nearer carbon stationary generates an infinite number of conformations.

At 0° dihedral angles, the C-H bonds on the two carbon atoms are close and cover one another. This is the eclipsed conformation of ethane.

Due to an increased steric repulsion and absence of a stabilizing interaction, the energy of eclipsed ethane increases by 12 kJ/mol, and each eclipsing H-H interaction is assigned 4 kJ/mol.

The energy difference between eclipsed and staggered conformations is known as torsional strain or torsional barrier.

Rotating the molecule from 0º to 360º along the carbon-carbon bond generates several degenerate staggered and eclipsed states.

A sample of ethane gas, at room temperature, has approximately 99% of its molecules in the lowest energy staggered conformation.

The energy gained from molecular collisions is used to undergo internal rotation by overcoming the torsional barrier.  The molecule, thus, moves into a different staggered form after passing through the high-energy eclipsed state.

The next hydrocarbon — propane — also has two major conformers: eclipsed and staggered.

The eclipsed conformer has a torsional strain of 14 kJ/mol. Each pair of eclipsing hydrogens contributes 4 kJ/mol, while the eclipsing CH₃-H interaction contributes 6 kJ/mol.