Animated Solution for Chemistry - Organic Chemistry: Arrange the following conformational isomers of n-butane in order of their increasing potential energy

When we talk about the conformations of an alkane like n-butane, we are essentially looking at the different spatial arrangements the molecule can adopt by simply rotating around its central carbon-carbon single bond.

Remember the golden rule of chemical stability: the more stable a molecule is, the lower its potential energy will be. Stability is governed by two main factors: steric strain (the physical crowding of bulky groups) and torsional strain (the repulsion between the electron clouds of adjacent bonds). Let's analyze the four given conformations of n-butane to determine their relative potential energies.

Let's look at conformation I first. Here, the two bulky methyl (${\text{-CH}}_3$) groups are positioned exactly opposite to each other, at a dihedral angle of $180^{\circ }$.

This is known as the Anti conformation. Because the bulky groups are as far apart as geometrically possible, the steric repulsion is at its absolute minimum. Furthermore, since it is a staggered conformation, the bonds are perfectly spaced out, resulting in zero torsional strain. This makes the Anti form the most stable conformation of n-butane, and thus, it possesses the lowest potential energy.

Now, observe conformation III. The methyl groups are positioned at a $60^{\circ }$ angle relative to each other.

This is known as the Gauche conformation. While it is still a staggered form (meaning there is no torsional strain from eclipsing bonds), the two methyl groups are close enough to interact. Their electron clouds repel each other, creating what is known as gauche steric strain. Because of this added repulsion, its potential energy is slightly higher than that of the Anti form.

Next up is conformation IV. If you look closely, you will see that a methyl group is directly aligned with a hydrogen atom.

This is the Partially Eclipsed form. The direct overlapping of bonds introduces significant torsional strain. Additionally, there is some steric strain because the methyl group is forced into the space of the hydrogen atom. This combination of strains pushes its potential energy significantly higher than the staggered Gauche form.

Finally, let's examine conformation II. Here, the two large methyl groups are completely eclipsing each other at a $0^{\circ }$ dihedral angle.

This creates a massive steric clash between the two bulkiest groups in the molecule, along with maximum torsional strain from all the eclipsing bonds. As a result, the Fully Eclipsed form is the least stable conformation possible for n-butane, giving it the highest potential energy of all.

If we plot these energy levels on a graph against the dihedral angle, we get a beautiful, undulating potential energy curve.

You can clearly see the hierarchy of stability: the Anti conformation sits comfortably at the global minimum. As we rotate the bond, we hit a local maximum at the Partially Eclipsed form, drop down to a local minimum at the Gauche form, and finally peak at the global maximum with the Fully Eclipsed form.

Based on our analysis, the increasing order of potential energy is directly opposite to the order of stability.

The order is: Anti (I) < Gauche (III) < Partially Eclipsed (IV) < Fully Eclipsed (II).

This perfectly matches option (d). Understanding these energy dynamics is a fundamental stepping stone in mastering organic chemistry!