Final answer:
The compounds are alkanes and nonpolar, so the only important intermolecular forces are London dispersion forces. Boiling points generally increase with increasing molecular mass, so propane has the lowest boiling point and n-pentane has the highest. The order of boiling points for the given compounds is propane < 2-methylpropane < n-butane < n-pentane.
Step-by-step explanation:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (-42.1°C) < 2-methylpropane (-11.7°C) < n-butane (-0.5°C) < n-pentane (36.1°C).