[Google555]

Predicting the relative boiling point of 2-methyl-1-propanol based on intermolecular forces, electron density?

Can someone help me out here?


How does the surface area, electron density cloud shape, and intermolecular forces affect the boiling point of
(CH3)2CHCH2OH 2-methyl-1-propanol

This is what I have so far:


It's polar molecule so IMFs: dipole-dipole and dispersion


surface area --no idea how that can be related to BP


And the e- cloud shape would be toward the molecular dipole ...kind of distributed more over the nitrogen
---dont know how that would affect boiling point though..my best guess is it would increase the dipole-dipole interactions the denser that cloud is over the N...but no idea really

[shelanachium]

Relative to what? Other isomers e.g. CH3CH2CH2CH2OH and (CH3)3COH? In this case surface area does matter. Though there is indeed dipolar attraction between the molecules because of the polar C-O and O-H bonds present, the non-polar hydrocarbon part is quite large and van-der-Waals forces as in nonpolar molecules are also significant. These forces are stronger the greater the surface area of the molecules, and the more branched the molecule the more compact it is, the less its surface area, and the lower the mutual attraction and hence the boiling-point. So in boiling point CH3CH2CH2CH2OH > CH3CH(CH3)CH2OH [2-methyl-1-propanol] > (CH3)3COH.

[invisiblegs]

We also seem to have missed an important intermolecular interaction.  I agree with dipole and van-der-Waals, but there is another interaction here to consider.  Think about the hydroxyl group.