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?

  • $\begingroup$ Trimethylamine's the only example here where your idea works. $\endgroup$
    – Mithoron
    Jun 13, 2022 at 15:14
  • 3
    $\begingroup$ "Because it's heavier" would probably be the over-arching answer here. Also, trimethylamine doesn't have any hydrogens that do hydrogen bonding. $\endgroup$ Jun 13, 2022 at 16:31
  • $\begingroup$ That's the point, hydrogen bonding is a stronger intermolecular force than London Dispersion forces. Yet, a molecule like ammonia (which has 3 hydrogen so it can make H-bonds) has a LOWER B.P. than methylamine which has only 2 hydrogens. I don't get why this happens, for me everytime you take an hydrogen out and replace it with a apolar carbon chain you should get a LOWER B.P. and not HIGHER. $\endgroup$
    – Anonymous
    Jun 14, 2022 at 14:53
  • $\begingroup$ And why is it that trimethylamine differs from the other examples? $\endgroup$
    – Anonymous
    Jun 14, 2022 at 14:53

1 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.


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