[Il Divo]

Just a brief question regarding Intermolecular forces:

in general, would it be a safe assumption that the intermolecular forces between all non-polar compounds (Ex: Methane) are lower than the intermolecular forces between all polar compounds (Ex: Water, methanol) due to the former having only London Disperson forces, while the latter also possess permanent dipoles/hydrogen bonding?

And would the end result be that this lowers the boiling point of non-polar compounds, due to decreased intermolecular attractions?

Thanks in advance.

[Jorriss]

No, not at all. Large molecules can have enormous forces. It's the reason why large hydrocarbons are solids.

[gritch]

Just a brief question regarding Intermolecular forces:

in general, would it be a safe assumption that the intermolecular forces between all non-polar compounds (Ex: Methane) are lower than the intermolecular forces between all polar compounds (Ex: Water, methanol) due to the former having only London Disperson forces, while the latter also possess permanent dipoles/hydrogen bonding?

And would the end result be that this lowers the boiling point of non-polar compounds, due to decreased intermolecular attractions?

Thanks in advance.

In general yes, the order of strength of intermolecular forces goes: Loudon Dispersion (or Van Der Waal forces) <  induced dipole < dipole < H-bonding. There are other facts to take into account but for molecules of similar masses those with dipole moments or that can form H-bonds will tend to have a higher melting/boiling point.

No, not at all. Large molecules can have enormous forces. It's the reason why large hydrocarbons are solids.

And here we have some of these other effects coming into play. Large hydrocarbons have a large area by which Van Der Waal forces can interact, thus these large hydrocarbon tend to have higher melting points. The larger the surface area of a molecule the greater contribution one can expect from Van Der Waal forces. For very large alkanes with very large surface areas these contributions may be greater than those from polar functional groups.

Here's a nice example perhaps:
The boiling point of pentane (CH₃C₃H₆CH₃) is 36°C.
The boiling point of neopentane (2,2 - dimethylpropane, (CH₃)₄C) is 9-10°C.
The boiling point of 1-chloropentane (CH₃CH₆CH₂Cl) is 108°C.
The boiling point of pentanol (CH₃CH₆CH₂OH) is 137-139°C.
The unbranched alkane has a higher boiling point than its branched counterpart due to its greater surface area. The compound with a dipole moment has a greater boiling point than both these alkanes but the corresponding compound capable of hydrogen bonding has a greater boiling point still.