Why is the $\ce{sp^2}$ hybridized nitrogen atom in guanidine the most basic?
I know that as a general guideline, $\ce{sp^3}$ hybridized atoms are more basic than $\ce{sp^2}$. Bases are electron pair donors; having the electrons further away from the nucleus makes donation easier.
On the other hand, protonating the double bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule. Protonating the other nitrogens does not allow for the same degree of resonance stabilization.
And finally, in one textbook, I found mention of a +M effect - what is the +M effect? And while we're on that, what's +/- I? I understand that I stands for induction, but what's +I? What's positive induction?
One more note, since the double bonded nitrogen is $\ce{sp^2}$ hybridized, does that mean it has a greater negative charge density than the other nitrogens by nature of having greater s-character? Could this + charge density be giving the double bonded nitrogen more nucleophilic character?
Answer
This may be another example of kinetics vs. thermodynamics. It is generally the case that $\ce{sp^3}$ hybridized lone pairs are more basic than $\ce{sp^2}$ hybridized lone pairs because the more s-character in an orbital, the more stable (unreactive) the lone pair. However in guanidine, ground state resonance structures (see picture below) tend to make the $\ce{sp^3}$ nitrogens less basic. Whether they are actually more or less basic than the $\ce{sp^2}$ nitrogen I don't know. But let me start by assuming that the $\ce{sp^3}$ nitrogens are more basic.
To whatever extent the $\ce{sp^3}$ hybridized nitrogens are more basic, they will protonate more rapidly than the $\ce{sp^2}$ nitrogen. However, due to resonance stabilization, the product formed by protonation of the $\ce{sp^2}$ nitrogen is thermodynamically more stable than the product formed upon protonation of the $\ce{sp^3}$ nitrogen. All of these protonations are simple equilibria between starting material and protonated product. In this case with guanidine there are two protonation equilibria occurring in parallel. Even though protonation of the $\ce{sp^3}$ nitrogen is kinetically favored due to its increased basicity, the product we observe is from protonation of the $\ce{sp^2}$ nitrogen due to its greater thermodynamic stability. Because both of these processes are in equilibria, thermodynamics controls the outcome. If the $\ce{sp^3}$ nitrogens are actually less basic than the $\ce{sp^2}$ nitrogen, then kinetics and thermodynamics favor the same outcome, kinetic protonation of the $\ce{sp^2}$ nitrogen and formation of its thermodynamically favored products.
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