[Kate]

Hi.

The question is: which of the following diols as the lowest boiling point:
1) cis 1,4-cyclohexanediol
2) cis 1,2-cyclohexanediol
3) trans 1,2-cyclohexanediol

I know the answer is 2 but I don't know why. Is it because the number of hydrogen bonds per molecule of cis 1,2-cyclohexanediol is less than in the other 2 diols?

[Enthalpy]

Hi,

I'd say as many hydrogen bonds at these three molecules. They are strong enough that the molecules assemble to satisfy them whenever possible. In addition, one versus two hydrogen bonds would make a huge difference in the boiling point, not just a few K.

Maybe possibly perhaps because the cis-1,2 makes less bad hydrogen bonds within one single molecule as a gas. Then, even if cis-1,2 and trans-1,2 made hydrogen bonds equally strong among several molecules in a liquid, evaporation would need less energy for the cis-1,2.

That's just a vague hypothesis from me. I tried to run Am1 on the single molecules, and nothing convincing comes out; the H and O arrange closer to an other for the lone cis-1,2 than the trans, while they repel an other for the cis-1,4 if I'm to believe software.

From a too quick search, the difference would be some 5K, and I doubt qualitative reasoning achieves this accuracy.

[Kate]

Well, apparently the lower boiling point of cis 1,2-cyclohexanediol is related with the greater intramolecular hydrogen bonding. But I agree with you here:

From a too quick search, the difference would be some 5K, and I doubt qualitative reasoning achieves this accuracy.

[Babcock_Hall]

It seems to me that 1) would have the highest boiling point, in that it can only make intermolecular hydrogen bonds.  However, I find it much more difficult to differentiate between 2) and 3) because both seem capable of forming intramolecular hydrogen bonds.

[Enthalpy]

Well, apparently the lower boiling point of cis 1,2-cyclohexanediol is related with the greater intramolecular hydrogen bonding.

Maybe - but I personally keep that only as an explanation attempt, because
- The intramolecular bonding must be weak since the positions and angles of the O-H-O are imperfect
- So the strength of the intermolecular bonding must dominate. Only if they match would the intra make a difference
- From the molecular formula, I'm astonished someone claims to make qualitative deductions. Even the molecules displayed in 3D are unconvincing.