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Secondary amines form hydrogen bonds, but having nitrogen in the middle of the chain rather than at the end makes the permanent dipole on the molecule slightly less. The lower boiling point is due to the lower dipole-dipole attractions in the dimethylamine compared with ethylamine.
Could someone please tell me why nitrogen, being in the middle of the molecule (as in secondary amines) causes the permanent dipole to be slightly less.
A primary amine has two $\ce{N-H}$ bonds and one $\ce{N-C}$ bond. A secondary amine has two $\ce{N-C}$ bonds and one $\ce{N-H}$ bond. Since $\ce{N-H}$ is more polarised than $\ce{N-C}$ we can expect ethyl amine to be more polar than dimethyl amine.
I would have gone for a less connected hydrogen bonding network as the primary engine. Diethyl amine boils at 55.6 °C, 1.11 debyes. n-Butylamine boils at 77.1 °C, 1.37 debyes (Organic Solvents, 4th Ed., Riddick, et al., 1986).
http://www.chemguide.co.uk/organicprops/amines/background.html
Both appear to be important.
Same as a branched alkanes and alkenes, there is less compactness (surface area exposed to the other compound); between the alkanes/alkenes, hence less van der Waals attraction between them; as compared to straight alkanes/alkenes. So, intermolecular forces of attraction is not too effective. This also applies here, having bulky groups (methyl, ethyl, propyl, …) attached to the nitrogen atom, make it more branched and less compactness (less surface area) , means less intermolecular forces, hence lower boiling point.
Guess that's the most obvious reason, because actually, the electron donating group such as methyl, ethyl and propyl, would have given more electrons to the nitrogen making it more negative, hence expecting to have a more electronegative nitrogen; which might result in stronger dipole-dipole bond, but this isn't the case. So, the only explanation that I could thought of is the one I explained above.
