Ch3ch2ch2cl exhibits hydrogen bonding\rm ch_3ch_2ch_2cl exhibits blank, which are stronger than blank, so it has a higher boiling point than \rm ch_4 and \rm ch_3ch_2ch_3., which are stronger than dipole-dipole forces\rm ch_3ch_2ch_2cl exhibits blank, which are stronger than blank, so it has a higher boiling point than \rm ch_4 and \rm ch_3ch_2ch_3., so it has a higher boiling point than ch4 and ch3ch2ch3.

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:

, 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.

the answer is learned

3

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