If I were to compare the basic strength of 1-azabicyclo[2.2.1]heptane and triethylamine:

Structures of 1-azabicyclo[2.2.1]heptane and triethylamine

Can I say that 1-azabicyclo[2.2.1]heptane is more basic than triethylamine because the lone pair of electrons is less available in the latter due to rapid nitrogen inversion? Nitrogen inversion is not possible in the bicyclic amine.

There is not much difference in the inductive effect. In both cases, the nitrogen atoms are $\mathrm{sp^3}$ hybridized.



Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons. The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$) than the lone pair in pyridine (the lone pair is in an $\ce{sp^2}$ orbital), which is less acidic then the lone pair in a nitrile (the lone pair is in an $\ce{sp}$ orbital).

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(image source)

The rate of nitrogen inversion also correlates with, among other things, hybridzation. But in cases where these "other things", such as ring constraints, limit nitrogen inversion, then the rate of nitrogen inversion may not correlate with hybridization. So the rate of nitrogen inversion does not always correlate with hybridization and is therefore not a good indicator of lone pair "availability".


Looking at your two compounds, a reasonable first guess might be that the nitrogens in these compounds are both roughly $\ce{sp^3}$ hybridized. Therefore, even though one compound can undergo nitrogen inversion and the other can't, their hybridzations are similar and we would expect similar basicities.

We could refine this prediction by noting that in the bicyclic compound, due to geometric constraints, the carbon-nitrogen bonds around nitrogen are slightly pulled back a bit from the tetrahedral angle. This would increase the p-character in these bonds (as a bond angle moves from 109° to 90° the p-character increases with 90° being pure p). As a result, the p-character in the lone pair orbital decreases and the s-character increases. As mentioned above, more s-character in the lone pair orbital leads to lower basicity.

Therefore we might expect both compounds to have similar basicity with the bicyclic compound perhaps being a bit less basic.

This analysis is corroborated by the $\mathrm pK_\mathrm a$ values of the conjugate acids (a more positive number means that the amine is more basic). According to J. Org. Chem. 1987, 52 (10), 2091–2094, quinuclidine has a $\mathrm pK_\mathrm a$ of 10.90, and 1-azabicyclo[2.2.1]heptane a $\mathrm pK_\mathrm a$ of 10.53. The decrease in basicity can be attributed to the increased strain and greater s-character of the lone pair. Triethylamine itself has a $\mathrm pK_\mathrm a$ of 10.75 (Evans table), making it marginally more basic than 1-azabicyclo[2.2.1]heptane.

  • $\begingroup$ Ron, using your explanation, the former should be more basic. Strained systems would show more p character wouldn't they? Take cyclopropane which is actually sp5. If lone pairs are exposed with no N inversion, the compounds do tend to be slightly more basic. Compare piperidine with di or triethylamine $\endgroup$ – Beerhunter Oct 9 '15 at 20:47
  • $\begingroup$ "Strained systems would show more p character"...yes, the strained bonds. So the bonds to carbon around the nitrogen would have more p-character (right?) leaving more s-character for the lone pair. More s-character, lower basicity $\endgroup$ – ron Oct 9 '15 at 20:51
  • $\begingroup$ On the one hand I'm wrong; quinuclidine is not strained as the rings are 6 membered. On the other hand, quinuclidine is about 0.25 units more basic than triethylamine, using Evan's Pka's table. I think the explanation is almost there $\endgroup$ – Beerhunter Oct 9 '15 at 21:07
  • $\begingroup$ @Mithoron Does Bredt's rule is followed if we have any other atom(like, here N in this case) ? By Bredt's rule i mean "carbocation or double isn't stable at bridge head C atom" $\endgroup$ – user98685 Sep 15 '20 at 14:19
  • $\begingroup$ @safdar Does Bredt's rule is followed if we have any other atom(like, here N in this case) ? By Bredt's rule i mean "carbocation or double isn't stable at bridge head C atom" $\endgroup$ – user98685 Sep 15 '20 at 14:22

There is more steric hindrance to the approach of the N to some acid when you have a non bicyclic amine. With quinuclidine the nitrogen free pair of electrons are more "exposed" and available for bonding to a proton or other acid. With trimethylamine, there is bond rotation about all the single C-C and C-N bonds. The space around the N free electron pair is cluttered(spatially hindred) due to the ethyl groups. Tougher for a proton to get to the N electron pair.


if nitrogen atom on the first compound accepts H+.Since the lone pair is given up, there will be less lone pair bond pair repulsion only bond pair bond pair repulsion making the compound more stable in such strained systems.


The bridge head is the most unreactive, and thus if we see the conjugate acid of the first compound, it will be extremely unstable. Amine flipping does lower the basicity, but still in this context, the basicity of the second amine will be greater than that of the first.


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