This story starts with an organic chemistry tutorial, when a student asked for clarification of the Finkelstein reaction. This is a simple SN2 type displacement of an alkyl chloride or bromide, using sodium iodide in acetone solution, and resulting in an alkyl iodide. What was the driving force for this reaction he asked? It seemed as if the relatively strong carbon-chlorine bond was being replaced with a rather weaker carbon-iodine bond. But its difficult to compare bond strengths of discrete covalent molecules with energies of ionic lattices. Was a simple explanation even possible?
All is not as it seems however. The traditional explanation, found by the quick Google search linked above, is that the reaction illustrates Le Chatelier’s principle, whereby an equilibrium is driven over to completion by removal of one of the products (in this case sodium chloride or sodium bromide, which crystallize out of solution). Well, we have replaced one possible (and probably complicated) explanation based on bond strengths and ionic lattices by another based on the solubilities of an ionic material in a moderately polar solvent. But all we have done is ask a different question, which now becomes why is sodium iodide highly soluble in acetone, whereas sodium chloride and bromide are not? The answer to this is less easily found using Google!
A good start would be the crystal structure of any complex formed between acetone and sodium iodide. Fortunately, one such does exist, and it is shown below (sodium=yellow, iodine=purple).
(Acetone)3+NaI. NAIACE. Click for 3D.
The formula shows three acetone molecules for each sodium iodide. The carbonyl oxygen has two lone pairs of electrons, and each of these is used to coordinate a (different) sodium cation. This allows each sodium to be coordinated by a total of six lone pairs, giving it octahedral coordination. This sets up what in fact is quite a rigid scaffold, with the unusual feature of an approximately triangular shaped channel running down the lattice (two such are shown above). The size of this hole is determined by the methyl groups of the acetone, and it is into this cavity that the halide ion must fit.
As it happens, the iodide anion is exactly the right size to produce a perfectly snug fit up against those methyl groups (click on the image above to view this). If a chloride or bromide anion were to be fitted into the cavity, there would be empty space surrounding it. The cavity itself is too rigid to collapse around the halide anion to absorb this space. This means these halide anions are further away from the positively charged sodium than they would like to be such that they minimize their ionic lattice energies. Instead they avoid fraternizing with the acetone at all, and form a pure sodium chloride or sodium bromide lattice (where the two oppositely charged ions CAN approach at optimal distances). The result is that sodium chloride crystallizes out of solution, and the Finkelstein reaction proceeds to completion!
Acetone. NaI in spacefill mode. Click for 3D.
But that is not quite the end of the story. If you view the acetone.NaI lattice sideways (click on the diagram above to view this aspect), you will find that in fact there is still space in the scaffold after all! Each iodide anion has room above or below it, with space for exactly one more iodine atom to fit without having to change the shape of the scaffold. And indeed such a molecule has been reported[1],[2] but it is an odd one! The stoichiometry is now (acetone)3.NaI2, which implies that the iodide anion has been joined by an iodine atom. I2(-) is called a radical anion, and as such has an unpaired electron. Just like two iodine atoms can couple their unpaired electrons to form a covalent bond, so can two I2 radical anions, forming I42- [or I3–.I–] or on to infinity as a linear iodine polymer, of formula n[I42-], with all the I…I distances equal at 3.224Å (a system with no Peierls distortion). Straight rod-like polymeric chains of a single element might appear highly unusual, but curiously, another class of elements that exhibits this behaviour is Cu/Ag/Au and Ga,[3] the ultimate in thin wires!).
Acetone+NaI2. GADMOO. Click for 3D.
Finally, it is worth noting that the same phenomenon occurs with the dimethylformamide.NaI complex. In this example, only the NaI and not the NaI2complex has been reported.
DMF+NaI. Click for 3D.
This post has been cross-posted in PDF format at Authorea.
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hi there,
interesting reading :-)
Question:
if I'm trying to alkylate a hydrazine compound, and afraid of overalkylation, then using Finkelstein to make C-I in situ (from C-Br) will increase rate of fomation of overalkylated products, right?
so should I just use a large excess of my hydrazine? (the Finkelstein is essential in this case)
thank you
I am thinking it is better to add my hydrazine compound last.
From previous reactions I've observed the KBr crashing out.
So I am thinking if I can generate most of the C-I initially, then I can just add a load of my hydrazine and the reaction should run more smoothly, instead of adding everything together at the beginning so that you generate a mixture of C-I and C-Br, and hence a more sluggish alkylation, and possible side products.
thoughts on this?
If I understand correctly, you want to use R-I to alkylate a hydrazine, because it is more reactive than R-Br? I suspect that rather than attempt a 1-pot reaction, you might be better off distilling your R-I in the first step so that it is pure, and has no R-Br, and then in a second step adding 1 mol of your hydrazine.
thank you, and yes, I plan to distill and then hit it with a large excess of the hydrazine
Hi all,
I'm writing my PhD thesis, and in one or two chapters the FS reaction is mentioned. As I am writing up all this, although I was aware of the precipitation explanation of the reaction, I set myself out in search of some background info. Normally I would do this via the "official routes", but now I chose simply to google. I am pleasantly surprised how this turned out! So thank you for all your comments, of which some of the mentioned refs will end up in my thesis!
Thank you all!!!
To confuse (or add food to thought), consider Poirier, et al in Synthesis 2006, 18, 3085-91 where Bu4NF hydrate reacting with primary iodides gives ratios of E/SN all over the place and in DMSO (RT) greatly favor elimination.
@Russ King:
Yes, the question of fluorination is, as always, very interesting. The matter with fluorides is complicated because both thermodynamics and kinetics of fluoride nucleophiles are heavily influenced by solvation AND hydration (in non-aqueous solvents). Only few truly dry AND soluble fluoride sources are known ("naked fluorides"). A standard reference on the influence of solvation on nucleophilicity/basicity of fluorides would be Landini et al, JOC 1989, 54, 328, where it is shown that the degree of hydration dramatically influences kinetics and selectivity (substitution vs elimination) of quaternary ammonium fluorides.
http://pubs.acs.org/doi/abs/10.1021/jo00263a013
In newer work, fluorides have been specifically solvated by tertiary alcohols, which gave ionic fluorides that show high nucleophilicity and relatively low basicity; see, e.g., JOC 2008, 73, 957.
http://pubs.acs.org/doi/abs/10.1021/jo7021229
There are again strong solvent effects, but it is not clearly discussed whether thermodynamics play a role in substitution of mesylates (the rate is CsI>CsBr>CsF in MeCN, but CsF>CsI>CsBr in t-BuOH; equilibration not investigated). Clearly, progress is still happening in this field (fueled by 18F PET applications). If anyone is interested in doing nucleophlic fluorination, he/she should definitely try the t-BuOH method. Maybe somebody should write a review about naked and nucleophilic fluorides again soon.
As I was looking for an explanation to why NaCl and NaBr *do* crystallize in acetone and not NaI, I came across this so clear explanation.
And as I was searching for a simple and elegant test tube experiment with organic compounds to illustrate solvent influence on kinetics of SN2, the diagrams suddenly gave me an idea : why not change the size of the cage in which iodide anions fit and even polymerize ?
So, I tried butanone instead of propanone.
And it *did* work ! It takes about 30% to 40% *less* time to form the same amount of NaCl in butanone than in acetone.
So, it seems, and repetitively, that the Finkelstein reaction is roughly 30 % faster in butanone than in propanone, using 0.4 mL of 1-bromobutane and 5 mL of a solution of 15 g of NaI in 100 mL ketone, at a temperature of 22 °C (did not try at a warmer or colder temperature).
Would that be consistent with the prediction of computation ?
Interesting result. As for consistency, there is no computation as such, only crystal structures. No doubt the addition two methyl groups do change the size of the cavity!
Hi,
Can this be explained based on HSAB theory?
Softer methyl group prefers to bond with softer Iodide and so the reaction moves forward? Precipitation of NaCl or NaBr adds to the product formation?
Fascinating explanation! but I cant find the original reference for the crystal structure of the first NaI-acetone solvate. Would you kindly give a link to it??? Thank you so much!