8
$\begingroup$

This question is a sort of follow-up to my earlier question.

I know about a method of solubilising potassium where the potassium atoms are "trapped" between layers of graphite, but I think that sodium naphthalenide exists only in solution (if a contradiction is found, please inform).

In the solution phase, say in THF, I would expect some THF molecules co-ordinated to the sodium, and an $\eta^4$ complex with naphthalene, involving one ring. An $\eta^{10}$ complex seems possible, involving both the rings, as the negative charge is distributed further, but I haven't read about any $\eta^{10}$ complex. An $\eta^1$ complex might be possible, but the distribution of negative charge will be lesser.

I expect sometehing similar to the tetrahedral form of butyllithium.

What is the structure of sodium naphthalenide?

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ Wikipedia says "It has not been isolated as a solid", which can be understood probably since the anion is a radical, so at the moment there is no single crystal structure (it actually should be doubted in such a case that something like a compound with name sodium naphthalenide really deserves to be called existing). Structures in solution are very difficult to determine, one of the many reasons is that complicated equilibria between different species can occur. Anyway the question is surely interesting, maybe one should look for computational studies (I could'nt find anything, yet). $\endgroup$
    – Raphael J.F. Berger
    Nov 6, 2016 at 22:03
  • 1
    $\begingroup$ That being said, there are quite a few known crystal structures containing naphtalene radical anion (or cation, for that matter). $\endgroup$
    – Ivan Neretin
    Nov 6, 2016 at 22:28
  • $\begingroup$ @Eashaan Godbole Not about sodium naphthalenide, but as partially related compound: the complex of THF, anthracene, and Mg -- useable, for example, to trigger Grignard reactions (though there are other methods as well); Acc. Chem. Res., 1988, 21, 261–267 (DOI: 10.1021/ar00151a002) by Bogdanovic himself. $\endgroup$
    – Buttonwood
    Jul 17, 2017 at 22:56

1 Answer 1

Reset to default
12
$\begingroup$

TL;DR: Naphthalene anion radical won't tolerate a presence of unshielded alkali metal cation nor in solution, neither in crystal. Reduction of naphthalene is possible strictly under air- and moisture free conditions by alkali-metals in the presence of coordinating solvents (cryptands). In this case solvent-separated cations don't form coordination bonds with naphthalenide radical.

On the other hand, further reduced (dianion) naphthalene readily forms various compounds not only with alkali metals, but also with lanthanides, typically $\eta^4$-complexes.

Various studies on stabilization of radical anions, especially in solid state, show that complexation (preferably with polar aprotic solvent or polydentate ligands) of counter-cations is essential:

$$\ce{C_nH_m + [M^0]_{\infty} + $x$ Solv + $n$ L <=> [M+(Solv_x|L_n)](C_nH_m^{.-})},$$

where $\ce{[M^0]_{\infty}}$ - any metal in oxidation state $0$ as a bulk reactant, $\ce{Solv}$ and $\ce{L}$ - solvent and ligand, correspondingly.

If cation is optimally solvated (complexated/sterically protected), radical counter anions $\ce{C_nH_m^{.-}}$ with extended $\pi$ systems are preferred because they lack extensively charged centers which favor contact ion formation (1). For example, crystals of bis(diglyme-O,O',O'')-sodium naphthalenide radical $\ce{[Na+(diglume)2](C10H8^{.-})}$ ([Na]) (1) has been isolated based on a principle of increasing enthalpy (calcd. based on crystal structures):

$$\Delta H (\ce{[Na+(THF)6]}) = \pu{-587 kJ mol^{-1}}$$ $$\Delta H (\ce{[Na+(DME)3]}) = \pu{-671 kJ mol^{-1}}$$ $$\Delta H (\ce{[Na+(diglume)2]}) = \pu{-677.3 kJ mol^{-1}}$$

Under certain conditions, the reduction of the naphthalene can proceed further to the naphthalene dianion, $\ce{C10H8^{2-}}$. For example, a complex with radical ion $\ce{[Li+(TMEDA)2](C10H8^{.-})}$ ([Li]) can reversibly be transformed into naphthalene dianion-complex $\ce{[Li+(TMEDA)]2(C10H8^{2-})}$ (2):

enter image description here

enter image description here

Of course, bigger cations require better shielding, commonly resulting in a usage of bulkier solvents or supramolecular assemblies, such as ([2,2,2]-cryptand)-potassium naphthalenide radical, $\ce{[K+(crypt-222)](C10H8^{.-})}$ ([K]) (3), in comparison to previously mentioned [Li] and [Na] naphthalenides (in the following crystal structures sphere-packing model is applied to metal atoms, and the rest is represented with sticks for simplicity):

enter image description here

It is also worth mentioning that naphthalene dianion forms numerous complexes with rare earth elements (4), e.g. with [Y] (see picture): $\ce{CpYC10H8(DME)}$, ($\eta^5$-Cyclopentadienyl)-(1,2-dimethoxyethane)-($2\eta^1:\eta^2(2\sigma,\pi)$-naphthalenide)-yttrium(III) with a $\ce{C6}$ ring being bent by $26.1^\circ$:

enter image description here

$\ce{La}$ and $\ce{Eu}$ complexes with bridged $\mu-\eta^4:\eta^4$-naphthalene are also obtained ($\ce{[LaI2(THF)3]2(C10H8)}$, $\ce{[EuI(DME)]2(C10H8)}$).

(1) Bock, H.; Arad, C.; Näther, C.; Havlas, Z. J. Chem. Soc., Chem. Commun. 1995, 23, 2393–2394. DOI: 10.1039/C39950002393
(2) Melero, C.; Guijarro, A.; Yus, M. Dalton Trans. 2009, 0 (8), 1286–1289. DOI: 10.1039/B821119C
(3) Rosokha, S. V.; Kochi, J. K. J. Org. Chem. 2006, 71 (25), 9357–9365. DOI: 10.1021/jo061695a
(4) Protchenko, A. V.; Zakharov, L. N.; Fukin, G. K.; Struchkov, V. T.; Bochkareva, M. N. Russ Chem Bull 1996, 45 (4), 950–953. DOI: 10.1007/BF01431330

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Please what does the notation $\ce{[M^0]_\infty}$ mean ? $\endgroup$
    – ParaH2
    Jul 17, 2017 at 21:03
  • 1
    $\begingroup$ @Hexacoordinate-C Any metal in oxidation state $0$ as a bulk reactant. $\endgroup$
    – andselisk ♦
    Jul 17, 2017 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.