Final answer:
The boiling points of four compounds (propane, 2-methylpropane, n-butane, and n-pentane) are compared based on their intermolecular forces and molecular mass.Ethane (C₂H₆) has the lowest boiling point among the given nonpolar alkanes because it is the smallest molecule with the weakest London dispersion forces.
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).
Among the compounds listed, C₂H₆ would have the lowest boiling point. This is because all the compounds in consideration are nonpolar alkanes and exhibit only London dispersion forces. The intensity of these forces correlates directly with the size and mass of the molecule; hence, the larger the molecule, the higher the boiling point due to stronger dispersion forces. Therefore, the compounds can be organized by increasing boiling point as follows: C₂H₆ < C₃H₈ < C₄H₁₀. This order is a result of their molecular size, with ethane (C₂H₆) being the smallest and thus the one with the weakest intermolecular forces and the lowest boiling temperature.