# Boiling point of: Halogenoalkanes vs alkanes vs alcohols

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?

• Oxygen is more electronegative, hence alcohols are more polar. Even more importantly, they also are capable of forming hydrogen bonds, which is not the case with haloalcanes. – Ivan Neretin Jun 15 '16 at 14:34