Answer:
1-chloropropane exhibits dipole-dipole forces which are stronger than London dispersion forces, so it has a higher boiling point than methane and propane
Explanation:
[tex]CH_3CH_2CH_2Cl[/tex], 1-chloropropane, exhibits mainly dipole-dipole forces, chlorine is too big to be able to form hydrogen bonds. Dipole-dipole forces occur in this compound due to the fact that chlorine is strongly electronegative and polarizing, so the two molecules would form dipoles. Dipole-dipole forces are stronger than London dispersion forces, therefore, 1-chloropropane has a higher boiling point than methane or propane.
Both methane and propane are non-polar molecules in contrast to 1-chloropropane. This means the only predominant forces in the two molecules are London dispersion forces occurring due to bonded electron interactions. They are much weaker than the actual dipoles formed by a presence of an electronegative atom having lone pairs on it.
