Why are amidines so much more basic than amides?

Both delocalize the added proton. Is it because the bond lengths in amidines are the same, or perhaps because oxygen is more electronegative than nitrogen and thus is more reluctant at keeping the proton? But how can that explane the huge difference in basicity between amidines and amides (resp $12$/$13$ versus $-1$/$0$ of $\mathrm{p}K_\mathrm{aH}$)?

• Since amides are protonated to oxygen, the question you're asking can be rephrased as 'why oxygen is easier protonated then nitrogen' . The easiest explanation is 'electronegativity'. When protonated, oxygen has to share an electron with proton, and it is much easier for nitrogen than for oxygen. – permeakra Jun 29 '14 at 22:32

1 Answer

Here is a diagram comparing what happens when we protonate an amide and an amidine.

Let's compare the protonated products in the two reactions. In the case of the amidine we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product. In the case of the protonated amide the resonance structure with the double bond to oxygen places a positive charge on oxygen. Therefor this resonance structure is a less significant contributor than the corresponding structure for the protonated amidine where the positive charge resides on the less electronegative nitrogen. Since we can draw more significant resonance structures for the amidine case, the protonated amidine is more stable than the protonated amide. This difference in protonated product stabilities explains why amidines are more basic than amides.

• Uhm, the proton adds to oxygen in the amide, not to the nitrogen? – Jori Jun 29 '14 at 15:27
• See edited version. – ron Jun 29 '14 at 16:16
• but can this electronegativity difference of 0.5 solely explain the guge pH difference of about 13? – Jori Jul 10 '14 at 14:41
• I didn't say anything about electronegativity, I argued based on the relative stabilities of the protonated forms. – ron Jul 10 '14 at 18:28
• Yes you did. Because oxygen is a more (0.5 Pauling) electronegative than nitrogen, the positive charge is less delocalized and thus is the resonance structure less stable. Right? – Jori Jul 10 '14 at 22:05