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Butan-1-ol has a boiling point of approx. 117.7 celsius, and Butan-2-ol has a boiling point of approx. 99.5 celsius. What causes this difference in boiling points?

My initial idea would be that Butan-2-ol essentially has a side-branch (the OH-group), whereas Butan-1-ol does not. This should disrupt the London dispersion forces and thereby reduce the strength. Another idea would be that the position of the OH-group being in the middle for Butan-2-ol creates fewer Hydrogen bonds than Butan-1-ol does. My third idea is that the position reduces the strength of the regular dipole-dipole bonds (when I mean regular I mean not hydrogen bonds), because the polarisation is weakened.

Am I correct in any of my assumptions? Is there something else to it?

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You have certainly understood the main idea.

In general the boiling point of a substance is determined by the way the molecules come together. The closer they are, the bigger are the intermolecular forces making it harder for one molecule to leave the entire group.

In this particular case the molecules are similar, since both substances have a hydroxyl group, thus being to able to create H-bonds. But in the case of 1-butanol, the molecule is linear while 2-butanol is not. This stereochemical difference affects the way the molecules attract each other. The linear molecules are able to come closer, thus creating stronger bonds or in other words raising the boiling point, because a higher amount of energy has to be applied in order for one molecule to obtain enough, to leave the liquid phase.

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