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Based on the Lewis structure of arginine, the terminal primary amine attached to the guanidine group seems to be $\mathrm{sp^3}$ hybridized and trigonal pyramidal. Also, in quantum mechanical computations, it is shown that all carbon-nitrogen bonds in the guanidine group are different lengths.

However, in the 3d structure below, the entire guanidine group is planar, indicating an sp2 hybridization. Why is $\ce{N}$ $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ hybridized? Why is the guanidine group planar? Why are all the bond lengths different if all the $\ce{N}$ atoms are $\mathrm{sp^2}$?

enter image description here

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    $\begingroup$ Have you considered resonance? $\endgroup$
    – electronpusher
    Commented Jun 18, 2019 at 20:04
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    $\begingroup$ The paper you linked with the reference to quantum mechanical calculations appears to be instead a description of a crystal structure doped with arginine. In that structure the C-N bonds are identical within error, and the guanidino group is described as planar, so all sp2 hybridized. $\endgroup$
    – Andrew
    Commented Jun 18, 2019 at 21:01
  • $\begingroup$ @Andrew, what do you mean "All C-N bonds are identical within error"? $\endgroup$
    – Cyclopropane
    Commented Jun 18, 2019 at 22:34
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    $\begingroup$ In the paper you linked, the bond lengths are given as 1.351, 1.340 and 1.322. That’s identical as far as I’m concerned. Compare to the C-N bond of the primary amine at 1.48. Clearly a difference in bond order. $\endgroup$
    – Andrew
    Commented Jun 18, 2019 at 22:55
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    $\begingroup$ Atoms cannot be hybridised, only orbitals can. Therefore it is incorrect to state N is sp² hybridised, even though it is used in that way (too) often. Consider it lab jargon. $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 18, 2019 at 22:55

1 Answer 1

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L-arginine is the most basic amino acid since it carries guanidyl group in addition to the $\alpha$-amino group. The arginine molecule is a zwitterion with the guanidyl group, rather than the primary $\alpha$-amino group, accepting extra proton from carboxylic acid group (Ref.1). Protonated or not, when consider guanidine as a molecule, there shouldn't be any $\mathrm{sp^3}$ nitrogen in the molecule because it is resonance-stabilized molecule. As a guanidyl group, this principle stays the same and it stays flat. There are few excellent answers to similar question you could find in here. In one answer it states that:

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 (Author: Jan).

And, accepted answer showed following diagram to depict the concept of resonance stabilization (Author: Ron):

Guanidine reseonance

Further, I'll attached recent crystal structure of L-arginine, which shows clear equivalent three nitrogen atoms with hydrohen bonding network (Ref.2):

Arginine Crystal

Two planes characterize the arginine molecule: one through the acid group and the other through the extended side chain, which contains the guanidyl group. The dihedral angle between these two planes is $74^\circ$ (Ref.1).

References:

  1. I. L. Karle, J. Karle, “An application of the symbolic addition method to the structure of L-arginine dehydrate,” Acta Cryst. 1964, 17, 835–841 (https://doi.org/10.1107/S0365110X64002250).
  2. Emilie Courvoisier, P. Andrew Williams, Gin K. Lim, Colan E. Hughes, Kenneth D. M. Harris, “The crystal structure of L-arginine,” Chem. Commun. 2012, 48, 2761–2763 (DOI: 10.1039/C2CC17203H).
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  • $\begingroup$ So the nitrogens in the guanidine group are all sp2 and planar, correct? $\endgroup$
    – Cyclopropane
    Commented Jun 18, 2019 at 22:17
  • $\begingroup$ Also, are all the nitrogen bond lengths the same in the guanidine group, or no? $\endgroup$
    – Cyclopropane
    Commented Jun 18, 2019 at 22:19
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    $\begingroup$ That answer by Ron, and especially the image, is terrible. Apart from some hand waving and a much too literal use of hybridisation, there is not much more to it. I doubt there is a large enough barrier (if at all) for the protonation step to be able to identify kinetics versus thermodynamics. $\endgroup$
    – Martin - マーチン ♦
    Commented Jun 18, 2019 at 23:05
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    $\begingroup$ @Mithoron: To your information, I didn't copy and paste and pretend it's my answer. Since you didn't know me, search google my name and find who I am. I'm a genuine researcher and do not disrespect me. You have no rights to order me. $\endgroup$
    – Mathew Mahindaratne
    Commented Jun 18, 2019 at 23:49
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    $\begingroup$ @Mithoron: If you know so much, why you don't give your own answer and mind your own business. There are moderators to do these kind of critical work. $\endgroup$
    – Mathew Mahindaratne
    Commented Jun 19, 2019 at 20:20

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