Question

Which of the two substance has a higher boiling point: butane (CH3CH2CH2CH3) or isobutane ((CH3)3CH)
Explain your choice using the theory of IMF’s.

Please, I am having trouble figuring this one out!

Answer 1

You've got no charges, so no ion-related forces. You can have dipole forces, or dispersion (London) forces. Any dipoles? Well...there *might* be a tiny one in isobutane, since it isn't perfectly symmetry, but C and H have such similar electronegativities that the C-H bond is generally considered nonpolar, and without polar bonds, you can't have polar molecules.

On first glance, the dispersion forces might also seem to be quite similar, since you have the same number of electrons in both molecules. But there's a wrinkle here (which is why you're being asked this question): the strength of dispersion forces actually depends on the polarizability of a molecule -- how easy it is for an electric field to cause shifts in the average locations of positive and negative charge. Roughly speaking, how "squishy" the molecule is to electric forces.

The most important factor in polarizability is the number of valence electrons in a molecule. The more electrons it has, the more polarizable it is. (That's because each electron contributes to the polarizability independently, so the more contributions you have, the more the total polarizability.)

But another factor is how the electrons are distributed. If they are distribute widely, over a larger area, then you can get stronger polarization. If, on the other hand, they are in a compact area, then the polarizability is lower.

In this case, since isobutane is a more compact molecule, the valence electrons are confined to a smaller area, and the polarizability will be lower. Hence the dispersion forces will be LOWER between isobutane than between butane. This is reflected in the lower boiling point of isobutane (-12 C versus -1 C for butane).