TBR ORGO Section 1, Passage 5, Q34 - Boiling points affected by cis and trans

[Liquidice07]

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So I am going through the TBR and I come to this question which asks me to rank 2 compounds based on their boiling points.

It gives me two molecules that are exactly the same, but one is TRANS and the other is CIS. I figure the molecule that is trans will have a HIGHER bp because it is lower in energy, more stable, thus, will take more heat for it to break apart.

However, the TBR states the trans molecule has a LOWER bp..."The boiling points of Compounds I and II are directly comparable, because they are
geometrical isomers. Compound I (the cis isomer) is polar, while Compound II (the trans isomer) is nonpolar.
This means that the boiling point of Compound I is greater than the boiling point of Compound II"
Their line of reasoning is the more polar a compound, the higher it's boiling point. That doesn't quite make sense to me. If it is highly polar, it will have more H-bonding but it should have more energy and be less stable.

Could anyone expound on this topic for me?
Thank you

 

[PingPongPro]

Boiling point is not a measure of a molecule stability. Remember that a molecules boiling point depends on the INTERmolecular forces, not INTRAmolecular stability.

More polar molecules DO have higher boiling points. I believe in TBR Orgo book, it talks about these characteristics that effect boiling point. It is probably confusing to you because TBR passages have a lot of questions that require info you haven't covered yet. Since you are only on section 1, you probably haven't gotten to it yet, since I think section 1 is mostly about INTRAmolecular stability.