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Why is the $\ce{sp^2}$ hybridized nitrogen atom in guanidine the most basic?

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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?

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    $\begingroup$ There is no sp³ nitrogen in guanidine. $\endgroup$ – Jan Oct 2 '16 at 21:46
  • $\begingroup$ +I stands for inductive effect and compounds less electronegative than carbon(attached to carbon chain) can be identified as +I For eg. Alkyl groups,CH3... They can be described as donating groups Therefore it can be considered that these group generally carry negative charges.. $\endgroup$ – Bikram Singh Nov 8 '16 at 19:10
  • $\begingroup$ For the record, I think it's clearer if you represent lone pairs as dots instead of lines. It's a bit easier to read in the giant mix of lines that already exist. At least you circle your charges, so you can't confuse the line-lone-pair with a negative charge... $\endgroup$ – Zhe Nov 8 '16 at 21:16
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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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  • $\begingroup$ If I am interpreting your reaction coordinate correctly, you are saying that the protonation of the sp3 nitrogen has a lower activation energy but results in a thermodynamically unfavorable product; the protonation of the sp2 nitrogen has a higher activation energy but a more thermodynamically stable product? $\endgroup$ – Dissenter Jun 13 '14 at 22:35
  • $\begingroup$ Correct, and since they are all equilibria, thermodynamics, not kinetics, controls the outcome. $\endgroup$ – ron Jun 13 '14 at 22:38
  • $\begingroup$ What do you mean they are all equilibria? As in, we observe both the reactant and product to actually exist, so the activation energy barrier can be overcome, and thus we can generally ignore the activation energy and just say that the thermodynamically favored product is the likely product? $\endgroup$ – Dissenter Jun 13 '14 at 22:42
  • $\begingroup$ Yes, in protonation reactions the activation energy is usually low, so assume both pathways are in operation with guanidine. Then there are 2 parallel protonation reactions both starting from guanidine and leading to their respective protonated species. In such cases, no matter which of the 2 products is kinetically preferred (and therefor formed first), whichever product is thermodynamically preferred is what we will observe in our beaker of product. $\endgroup$ – ron Jun 13 '14 at 23:05
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    $\begingroup$ There are no sp³ nitrogens in guanidine. $\endgroup$ – Jan Oct 2 '16 at 21:47
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Your misconception is directly at the beginning of your question and turns up multiple times. There are no $\mathbf{sp^3}$ nitrogens in guanidine.

Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced.

This is also why the $\ce{-NH2}$ nitrogens are not basic. Their lone pair is not sitting in an orbital all alone waiting for a proton. It is actively resonating with the π system of the other two. The only available lone pair guanidine has is that of the $\ce{=NH}$ nitrogen sitting in an $\mathrm{sp^2}$ hybrid orbital. Thus, a second protonation requires somewhat superacidic media.

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The basicity of the double-bonded nitrogen is chiefly a result of the delocalization of charge permitted by the two single-bonded nitrogens with allylic lone pairs, as the resonance contributors you've drawn suggest. Notice that, upon protonation, the three nitrogens become chemically equivalent, with the positive charge shared equally among all three. If either of the single-bonded nitrogens is protonated, the resulting cation does not benefit from charge delocalization, which is a comparatively less stable arrangement.

Since the basicity is a function of the stability of the resulting guanidinium cation, it's a thermodynamic phenomenon. However, it may be worth mentioning that kinetic factors also likely favor protonation at the double-bonded nitrogen, since the lone pairs on the other two nitrogens are somewhat delocalized. This is the meaning of the $+M$ notation, which refers to the amine groups being electron-donating by the mesomeric effect (basically synonymous with resonance), while $-M$ would indicate groups which are electron-withdrawing. This is an effect active within the $\pi$-electron system of a molecule, which is probably better understood when considered from the standpoint of MO theory. The concept of resonance/mesomerism is a convenient abstraction that usually works well qualitatively to approximate the electron distribution in molecular orbitals (especially frontier orbitals). In reality, MO theory probably provides a more accurate rendering of the electron distribution, with $\pi$ orbitals that span multiple atoms. I say this to emphasize that resonance is not a physical process, but just a very expedient way of representing electron delocalization in $\pi$ orbitals.

Similary, $+I$ indicates a group being electron-donating by induction, while $-I$ denotes groups that are electron-withdrawing via induction. In essence, an electronegative group creates bond polarization, which can act to draw electron density away from an atom, while an electropositive group can do the opposite. This effect operates principally through the $\sigma$-bond skeleton of a molecule, and weakens as distance from the polarized group increases.

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