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How would you arrange the following amines in order of decreasing base strength?

#NH_3#
#CH_3NH_2#
#(CH_3)_2NH#
#NH_2Br#

1 Answer
Mar 29, 2016

I also get:

How I would explain it is by using #"NH"_3# as a reference point, we could look at how the different substituents affect the electron density around the nitrogen.

The more electron density around the nitrogen, the more it has available to donate to something else, and thus the more Lewis basic it becomes because a Lewis base donates electrons.


METHYLAMINE

#"CH"_3"NH"_2# contains a methyl group, which is electron-donating. That is, it inductively donates electrons through the #"N"-"Br"# bond towards nitrogen, which "activates" the nitrogen's Lewis base behavior since there is more electron density around it.

The more electron density around the nitrogen, the more it has available to donate to something else, and thus the more Lewis basic it becomes because a Lewis base donates electrons.

Therefore, methylamine is more Lewis basic than ammonia.

DIMETHYLAMINE

Since #("CH"_3)_2"NH"# contains a second #"CH"_3#, it experiences a greater electron-donating effect from the methyl substituents and thus is more Lewis basic.

Therefore, dimethylamine is more Lewis basic than methylamine.

BROMOAMINE

For #"NH"_2"Br"#, note that bromine is NOT more electronegative than nitrogen (it's the other way around).

Instead, we should recognize that #-"Br"# is more inductively electron-withdrawing than #-"NH"_2#; the electronegativity difference between #"N"# and #"H"# (#~0.9#) is larger than the difference between that of #"Br"# and #"N"# (#~0.1#), so the #"N"-"H"# bonding-electron pairs are more polarizable.

The #"N"-"H"# bonding-electron pairs get polarized towards #"Br"#, so #"Br"# pulls electron density away from the nitrogen, leaving less electron density belonging to nitrogen to use.

Thus, it weakens the Lewis basicity of the nitrogen relative to that in #"NH"_3#.

RESULT

So, overall, we have: