# Why are amidines more basic than amines?

Amidines can be thought of as carboxylic acid derivatives with the carbonyl oxygen replaced by a nitrogen and hydrogen atom; amidines exhibit conjugation.

Amidines are more basic (less acidic) than amines and amides by a factor of around $${10^{3}}$$ and $${10^{12}}$$, respectively (from pKas).

I would naturally think that conjugation would lower the energy level of MO containing the lone pair; thus, the lone pair would be more bound to the nucleus and less available to deprotonate.

This is an argument similar to explaining why pyridine is less basic than piperidine; $$\ce{N}$$ is $$sp^{3}$$ hybridized in piperidine but $$sp^{2}$$ hybridized in pyridine. Since $$sp^{2}$$ orbitals have less p character than $$sp^{3}$$ orbitals (2/3 vs 3/4), the lone pair on $$\ce{N}$$ in pyridine is at a lower energy level than the lone pair on $$\ce{N}$$ in piperidien, and pyridine is less basic than pyridine.

Although I realize that the positive charge on the amidine is stabilized by conjugation (as explored by a previous post comparing the basicity of amidines and amides), wouldn't my reasoning with MO theory outlined above suggest a lowering of the basicity of amidines?

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

• I feel that the stabilization caused by resonance in the conjugate acid of amidine will dominate here, as the canonical structures being generated are equivalent; so this will correspond to a lot of resonance energy being released – Yusuf Hasan Dec 27 '18 at 22:26
• Note that the lone pair that is delocalised in the conjugation is not the lp that is used to accept proton. – Tan Yong Boon Dec 28 '18 at 1:41
• Is that lone pair used to accept the proton because of delocalization in the protonated amidine is only exhibited when the lone pair on the nitrogen with only one H atom deprotonates an acid? – Ethiopius Dec 28 '18 at 1:47
• – Yusuf Hasan Dec 28 '18 at 5:30