Chloromethane \(\left(\mathrm{CH}_{3} \mathrm{Cl}\right)\) has a much higher boiling point than methane \(\left(\mathrm{CH}_{4}\right)\). Give two reasons for this.

In order to provide reasons for the difference in boiling points, we need to first understand the types of intermolecular forces present in both chloromethane and methane. Methane is a nonpolar compound, so the dominant forces between the molecules are weak London dispersion forces. Chloromethane, on the other hand, is a polar compound due to the electronegativity difference between carbon, hydrogen, and chlorine, resulting in stronger dipole-dipole interactions in addition to London dispersion forces.

When comparing the intermolecular forces present in the two compounds, we can see that methane only has the presence of weak London dispersion forces, while chloromethane has both dipole-dipole interactions and London dispersion forces. The presence of the stronger dipole-dipole interactions in chloromethane is the first reason for the higher boiling point. In general, stronger intermolecular forces result in higher boiling points because more energy is needed to overcome the forces holding molecules together in the liquid phase.

Another factor that we should consider when comparing boiling points is the molecular weight of the compounds. Chloromethane has a molecular weight of 50.49 g/mol, while methane has a molecular weight of 16.04 g/mol. In general, compounds with larger molecular weights have larger electron clouds and surface areas, which leads to stronger London dispersion forces. Although London dispersion forces are weaker than dipole-dipole interactions, the increased molecular weight of chloromethane contributes to the overall strength of its intermolecular forces, providing a second reason for its higher boiling point. In summary, chloromethane has a much higher boiling point than methane due to the presence of stronger dipole-dipole interactions and London dispersion forces that result from its polar nature and larger molecular weight.