# Question #50327

Feb 16, 2015

Boiling point is all about intermolecular forces, or, more precisely, about the strength of intermolecular forces.

The relationship between boiling point and intermolecular forces can be described like this: compounds that exhibit stronger intermolecular forces will have higher boiling points.

Start with ${\left(C {H}_{3}\right)}_{2} C H O H$, which is Isopropyl alcohol. Notice that you have an $\text{-OH}$ group attached to one end of the molecule.

This means that isopropyl alcohol molecules can form hydrogen bonds with each other. Hydrogen bonds are the strongest type of intermolecular bonds, which implies that this compound will have a relatively high boiling point.

$C {H}_{3} C {H}_{3}$, or ethane, has no capacity to form hydrogen bonds, i.e. it lacks a hydrogen atom bonded to $N$, $O$ or $F$. Moreover, the $\text{C-H}$ bonds are considered to be nonpolar. Nonpolar bonds make for a nonpolar molecule, which means that ethane molecules will only exhibit very weak London dispersion forces.

Finally, $C {H}_{3} C {H}_{2} C H O$, or propionaldehyde. Despite the fact that it has an oxygen atom attached to the molecule, it cannot form hydrogen bonds because the oxygen is attached directly to a carbon atom, not to a hydrogen atom.

This implies that the molecule will exhibit dipole-dipole interactions - the more electronegative oxygen will have a partial negative charge, will will induce a permanent dipole moment in the molecule.

As a conclusion, the three compounds are ranked in order of increasing boiling point like this

$C {H}_{3} C {H}_{3}$ $\to$ b.p = $- {88.5}^{\circ} C$;
$C {H}_{3} C {H}_{2} C H O$ $\to$ b.p = ${48}^{\circ} C$;
${\left(C {H}_{3}\right)}_{2} C H O H$ $\to$ b.p = ${82.6}^{\circ} C$