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The following quote is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha), from the chapter "Coordination Compounds", page 177:

...it must be noted that $\ce{NH2NH2}$ and $\ce{N(CH2CH2)3N}$ cannot act as chelating ligands due to the formation of a three membered ring and locked structure respectively.

But, as per the accepted answer to the question Are all complexes with a polydentate ligand examples of chelation?, when a ligand forms more than two bonds to the central atom, the result is a chelated complex. This is also inline with the IUPAC Gold Book's definition of Chelation:

The formation or presence of bonds (or other attractive interactions) between two or more separate binding sites within the same ligand and a single central atom.

As per my understanding, it doesn't seem to exclude any ligands as mentioned in my textbook.

I can understand that since $\ce{NH2NH2}$ forms a three membered ring, there would be ring strain in the structure and hence the process of chelation is not energetically feasible compared to four membered (or higher) rings. I think this would be the same for all such ligands with donors adjacent to each other, as I have not encountered any such species so far.

However, I'm unable to understand why $\ce{N(CH2CH2)3N}$ (1,4-Diazabicyclo[2.2.2]octane) cannot undergo chelation.

Molecular structure of 1,4-Diazabicyclo[2.2.2]octane

From its molecular structure, I can imagine that the three ethylene groups act as springs to accommodate any ring strain while bonding with the central atom and so there must be no hinderance to undergo chelation. The following molecular structure is what I think might be if $\ce{N(CH2CH2)3N}$ undergoes chelation with cobalt:

DABCO hypothetical chelation with cobalt

I felt given the size of the cage it would be difficult to fit the metal atom inside it and hence thought that it might be off-centred as shown in the image above.

So, it would be helpful if you could clarify whether my reasoning for the inability of $\ce{NH2NH2}$ to chelate is correct or not and why is it not possible for $\ce{N(CH2CH2)3N}$ to undergo chelation? Further, it would be helpful is you could tell why is there a contradiction between the book's statement and IUPAC definition? Is the statement from the book incorrect (because IUPAC can't be incorrect)?

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  • $\begingroup$ @Mithoron: I considered the metal atom to lie off-centred and not at the centre of the cage. If you can't imagine, I can try to draw a diagram. If you think that must be at the centre it would be helpful if you can explain why that's so? $\endgroup$ – Guru Vishnu May 21 at 15:48
  • $\begingroup$ And how you imagine turning lone pairs inside out? IIRC this is called piramidalisation and I doubt such structure was ever found for cases similar to that. BTW I guessed you'd think it somewhat off-center and it's still inside and doesn't make sense. $\endgroup$ – Mithoron May 21 at 16:11
  • $\begingroup$ @Mithoron: I've added the molecular structure of what I thought might be if DABCO undergoes chelation. Hopefully it clears the fact what (and why) I meant by "off-centred". After reading the answers, I can see somehow this is not possible. $\endgroup$ – Guru Vishnu May 22 at 5:24
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The simple answer for your query: the lone pair is practically immobile, and can't orient itself to chelate.

To understand this, we first have to understand the structure of DABCO. The rings force a conformational rigidity on both the nitrogens, and therefore the fluxional process of nitrogen inversion is inhibited in these structures. Now, as the accepted answer here states, in this process, the entire system swings from an $\ce{sp^3}$ hybridized state to a planar $\ce{sp^2}$ transitional state back to $\ce{sp^3}$. In this transitional process, the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.

However, due to the restrictions imposed by the structure, one can imagine that both the nitrogens of DABCO are perenially stuck in this planar transition state of the inversion process. Hence, the lone pair in each of them resides in an almost pure p orbital, while the bonding scheme for the central atom would be roughly $\ce{sp^2}$ .(Note: This just a suggestion for visualizing the geometry of the molecule. Please don't take it as an actual mechanism)

Hence, it would like this from a side view:

enter image description here

Blue:Nitrogen|Pink: p orbital containing lone pair (source)

So as you can see, there is not much possibility of having the metal atom lie off-center, as then the lone pairs will have to re-align either above or below the plane of the screen. But the lone pairs are stuck on the plane of the screen as shown.

Trying to fit the metal atom inside the cage will of course, be sterically hindered( although some other reagents like crown ethers are used for trapping cations in a such a fashion)

So although they can exhibit denticity of one like the other answer stated, chelation might be a far-fetched game for DABCO

As Martin pointed out in the comments,perhaps a little more introspection would be required to look at the orbital picture of this molecule. I will be citing a few findings from a research paper which has analysed the molecule based on frontier MO theory and thereby debunks a few of the classical valence bond hybridization predictions: Electron momentum spectroscopy of the frontier electrons of DABCO does not support an sp3 hybrid lone-pair description

Abstract The highest occupied molecular orbital (HOMO) and next-highest occupied molecular orbital (NHOMO) valence orbital electron density distributions of 1,4-diazabicyclo[2.2.2]octane (DABCO) have been investigated by electron momentum spectroscopy, a technique that probes the orbital-like nature of valence (frontier) electron transfer out of a molecule. In contrast, Pauling's widely used and taught valence bond (hybridization) model, which is equivalent to a localized molecular orbital description, does not correspond at all to the experimental measurements. It follows that, for considerations of electron transfer, the "lone pairs" of DABCO are not localized or hybridized, but rather exist as nondegenerate orbitals that are delocalized differently over the molecular framework

Introduction and background A first-order through-space treatment (2, 3) that the antisymmetric(AS) antibonding combination of 2p functions on the nitrogen atoms(in a simple LCAO-MO description) will comprise the HOMO, while the symmetric (S) bonding combination will comprise the NHOMO. However,when the through-bond interaction is turned on,the frontier orbital energy ordering is predicted to be reversed i.e. the HOMO is bonding(S) , while the NHOMO is antibonding(AS)

Results and Discussion The EMS binding energy spectrum confirms the earlier PES observations(4, 5) that there are two outer valence IPs(7.52 and 9.6 eV, respectively), and not one corresponding to the presence of degenerate nitrogen lone pairs in DABCO.

NHOMO=Next Highest Molecular Orbital

What all this evidently shows is that both the lone pairs are in fact,not degenerate(as classical hybridization would have predicted both of them to be in $\ce{sp^3}$ hybrid orbitals.) In fact,both of them lie in inherently different orbitals with different ionization energies, that is,the HOMO and NHOMO, which are constituted by in-phase interaction of 2s orbitals and out-of-phase interaction of 2p orbitals respectively as a first order LCAO approximation

Here is an MO diagram for DABCO:

enter image description here secondary interactions with the in-phase combinations of the CÐC s and s orbitals which reverse the order of the HOMO and HOMO-1. The orbitals symmetries are specified in D3h(source)

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  • $\begingroup$ Thanks for the answer! I think we can also say that the p orbitals of nitrogen atoms are along the $\ce{N-N}$ axis of DABCO ring using symmetry arguments. This is evident when we consider the three dimensional molecular structure provided in the PubChem link in your post. So, if at all DABCO must undergo chelation it would be possible only when the metal atom is at the centre of the cage. I hope the rotation about the $\ce{N-C}$ bonds are restricted so that it can't align as shown in the question. $\endgroup$ – Guru Vishnu May 22 at 5:26
  • $\begingroup$ Yep,you summed it up pretty much correctly. The $\ce{N-C}$ bond rotation is restricted, so the alignment you have proposed is not possible as such. I doubt there are enough transition metal atoms that have the ability to fit into that cage and also coordinate with two donor atoms at the same time,but yeah, it's theoretically possible to chelate within the cage. But as Anirudha's answer cited from various sources,that DABCO's compounds are generally linear and monodentate. So it is $\endgroup$ – Yusuf Hasan May 22 at 7:31
  • $\begingroup$ The lone pairs are most likely in sp³ like type orbitals due to pyramidalisation of nitrogen at the bridge heads. $\endgroup$ – Martin - マーチン May 22 at 12:33
  • $\begingroup$ @Martin - マーチン♦ I can't find a source for it as of now,but I distinctly remember a question in our college exam which mentioned the lone pair of DABCO in a p type orbital,and the hybridization to be $\ce{sp^2}$. Also,I have learnt here on Chem.SE that geometry drives hybridization,not the other way around. The $\ce{sp^2/ p}$ orbital conjecture seems much more capable to be of explaining the bonding situation than the $\ce{sp^3}$ hybridization,as I feel that the latter can't account for a perpendicular lone pair. Am I wrong in postulating this? $\endgroup$ – Yusuf Hasan May 22 at 13:05
  • $\begingroup$ @Martin - マーチン♦ Ah,I found something! Electron momentum spectroscopy of the frontier electrons of DABCO does not support an sp3 hybrid lone-pair description $\endgroup$ – Yusuf Hasan May 22 at 13:17
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My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.

This then raises the question: is the author correct? Hydrazine derivatives such as Benzhydrazides and Salicylhydrazides 2 have been proven to show chelation. Pure hydrazine, however, is unlikely to show chelation due to excessive ring strain. Also, no literature has been found which shows hydrazine undergoing chelation. DABCO, on the other hand, forms complexes with Mercury halides3 with a denticity of 1. This ligand also does not show chelation, and fitting a large metal atom even off-center is highly unlikely due to large steric hindrance and the geometry of the lone pair, which is perpendicular to the structure and not favourable for chelation. Most complexes formed by DABCO tend to be linear in nature4 and no rings are formed.

In short, the author is correct in stating that hydrazine and DABCO do not form chelates.

References:

1: Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers & Blackwell Science, 1996.

2: Issa, R. M., et al. Coordination compounds of hydrazine derivatives with transition metals. I. Metal Chelates with Benzhydrazide and Salicylhydrazide. Zeitschrift fur anorganische und allgemeine Chemie, vol. 354, no. 1–2, Sept. 1967, pp. 90–97. doi:10.1002/zaac.19673540118.

3: Shan, Zeng-Mei, et al. Synthesis, Crystal Structures, and Characterization of Three Mercury(II) Halides Inorganic–Organic Hybrid Compounds with 1,4-Diazabicyclo[2.2.2]Octane Ligand. Inorganica Chimica Acta, vol. 366, no. 1, Jan. 2011, pp. 141–46. doi:10.1016/j.ica.2010.10.023.

4: Thorp-Greenwood, Flora L., et al. Three-Dimensional Silver-Dabco Coordination Polymers with Zeolitic or Three-Connected Topology. Crystal Growth & Design, vol. 14, no. 11, Nov. 2014, pp. 5361–65. doi:10.1021/cg501231v.

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  • $\begingroup$ Thanks for the answer! Regarding "It is most likely that the author of your adaptation has used his liberty to add these facts in your copy.": Only the text in first blockquote is from the textbook, others are my own interpretations :) $\endgroup$ – Guru Vishnu May 22 at 5:33
  • $\begingroup$ @GuruVishnu I understood that. I couldn't find the text in the first block quote in my copy of J D Lee, which is why I said that. $\endgroup$ – Aniruddha Deb May 22 at 8:22

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