Relative basicity of two molecules.

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SUMMARY

N-Acetylpiperidine is a weaker base than 1-aza-2-2oxobicyclo[2.2.2]octane due to the electron pair donation dynamics involving the carbonyl carbon. The nitrogen in N-acetylpiperidine loses its lone electron pairs to the acetyl carbon, reducing its basicity. In contrast, the bridgehead nitrogens in the bicyclic compound retain their basicity because the carbonyl's locked position prevents effective overlap between the nitrogen lone pair and the pi* orbital of the carbonyl. This results in the nitrogen behaving like a regular amine nitrogen, making the bicyclic compound more basic.

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  • Knowledge of molecular geometry and bond rotation
  • Basic concepts of nucleophilicity and reactivity in organic compounds
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Hi can someone explain to me why is N-Acetylpiperidine is a weaker base than 1-aza-2-2oxobicyclo[2.2.2]octane ?
Is it something about the ability to protonate the molecules ?

thank you
 
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N-acetylpiperidine loses its lone electron pairs to acetyl carbon (carbonyl carbon). However, the nitrogens on the bridgehead positions are not prone to reactions very much, so they retain their basicity to a great extent. This is I think why the latter compound is more basic than the former.
 
To add to what chem_tr wrote, you can think of it in terms of orbital symmetry. In order to draw the resonance structure with the nitrogen of the amide donating into the carbonyl (with a negative charge on oxygen) the orbital containing the nitrogen lone pair must be able to overlap with the pi* orbital of the carbonyl. In N-acetylpiperidine this isn't a problem because the N-C bond can rotate freely and attain the necessary planar geometry. In the case of the bicycle, the carbonyl is locked in a position that prevents overlap between the lone pair orbital and the pi* orbital. That is to say, the two orbitals are essentially perpendicular to one another (the N lone pair extends in the same plane as the carbonyl C-O bond, but the pi* orbital is in a perpendicular plane along the C-O bond). I think this will become clear if you build a model.

As chem_tr said, since this amide resonance structure cannot be drawn the lone pair on nitrogen behaves like a regular amine nitrogen and will be quite basic.

On a side note, the bicyclic compound (sometimes called 2-quinuclidone) is very difficult to make in the laboratory. Its synthesis has been reported once, but the report is widely disputed and the structural evidence provided is very lacking. The difficulty in isolating this compound likely arises from the reactivity of the N-C bond. The carbonyl group is essentially non-stabilized and will react much like an acid chloride would. Therefore, any nucleophile will cleave the C-N bond (e.g. if water is the nucleophile, the product would be an amino acid). There is an obvious problem with making such a reactive bicyclic amide: the amine nitrogen is a good nucleophile. Therefore polymerization of the substrate molecules is a big problem!
 

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