Why does alkylamines have a higher boiling point than ammonia?

Question

Here are the boiling points of ammonia and some amines: $$ \begin{array}{|c|c|} \hline \textrm{Substance} & \textrm{Boiling Point} \\ \hline \textrm{Ammonia} & \pu{-33.34^\circ C} \\ \textrm{Methylamine} & \pu{-6.3^\circ C} \\ \textrm{Dimethylamine} & \pu{7^\circ C} \\ \textrm{Trimethylamine} & \pu{2.9^\circ C} \\ \textrm{Ethylamine} & \pu{16.6^\circ C} \\ \textrm{Diethylamine} & \pu{55.5^\circ C} \\ \textrm{Triethylamine} & \pu{89.28^\circ C} \\ \hline \end{array}$$

My question is, why every time you remove a hydrogen from ammonia and replace it with an ethyl group the resultant molecule has a higher boiling point? Since the hydrogens can make hydrogen bonds and the ethyl groups can only interact through London dispersion force, I would expect that the resultant molecule would had a lower boiling point.

Also, why is trimethylamine an exception to this trend?

Answer 1

When you compare ammonia to alkylamines you have to consider that the molar mass of your compound is also changing; methylamine has almost double the mass of ammonia and this will obscure the effect you're trying to look for. Other things (intermolecular forces, etc.) being equal, a compound with a higher mass will have a higher boiling point. Let's compare three alkylamines with the same molar mass.

$$\begin{array}{cr} \text{Compound} & \text{BP (}^{\circ}\text{C)}\\ \hline \text{Trimethylamine} & 2 \\ \text{Methylethylamine} & 36 \\ \text{n-Propylamine} & 49 \end{array}$$

From the table it's clear that changing a hydrogen-bondable hydrogen to an alkyl group reduces the boiling point, consistent with your suggestion that the stronger hydrogen bonding is being replaced with dispersion forces.