Strength of LDFs in a large molecule

[Claire Li 2J]

Hello,
I remember in lecture, Dr. Lavelle mentioned that a very very large non-polar molecule can have extremely strong London dispersion forces, making it have a higher boiling point than a smaller polar molecule that can have dipole-dipole interactions. Is there a cut off for the necessary size of a non-polar molecule to have a higher boiling point than a small polar molecule? For example, would C10H22 have a higher boiling point compared to CH3OH?

Thanks!

[AnthonyEdwardsLec1]

Hi, I believe there's no specific cut-off per se of the size per se that would make the non polar molecule have a higher boiling point than the small polar molecule. However I believe you could calculate the potential energy of the dipole-dipole molecule and than calculate the polarizability of the non-polar molecule to find the strength of the LDF force and then compare t e two values. If the London force is strong enough in C10H22 and has more potential energy than the polar molecule like CH3OH, you would be able to determine if it has a higher boiling point. However I believe we don't have enough knowledge in this course to carry out those calculations and wouldn't be expected to know that.