Boiling point of: Halogenoalkanes vs alkanes vs alcohols

Question

We know that Boiling point (alcohol) > Boiling point (halogenoalkane) > Boiling point(alkane) with same number of $\ce C$ atoms, but is the following, good explanation of that?:

Since e.g. $\ce{CH_3CH_2-O^{\delta_1 \textbf{-}}-H^{\delta_1}+}$, similar with $\ce{ClCH2CH3}$ and with $\delta_2\textbf{-}(\text{at}\ \ce{Cl})\Longrightarrow \delta_1 \textbf{-}\stackrel{\text{bigger, electronegative}}{>}\delta_2 \textbf{-}$, thus electron pair is closer to oxygen atom, so more energy is needed to move that electron in alcohol than in halogenoalkane, so higher temperature is needed to reach boiling point.

Does this have any sence, if not then could somebody explain why does it not have sence?

Answer 1

The cause of this is the hydrogen bond that forms in the liquid state between the molecules, and debye bonds that form between polar molecules.

A hydrogen bond can only occur between an electronegative atom and a hydrogen atom, himself linked to an electronegative atom (F, O or N).

In order to evaporate the liquid, you have to provide enough energy to break the hydrogen bond. So the more bonds there are, the higher the boiling temperature. Soft debye bonds can also occur between two polar molecules and have to be broken too.

• The alkanes can not participate in any bonds, thus have the lowest boiling temperature.

• The halogen alkanes may not form hydrogen bonds, but they are polar, thanks to the very electronegative halogens. They form debye links, that are not very strong but sufficient to raise the boiling temperature.

• The alcohols have -OH extremities, very likely to associate themselves in hydrogen bonds, that are stronger than debye links, and thus alcohols have the highest boiling temperature.