Carrying out a nucleophilic aromatic substitution reaction with a carbon nucleophile, the iodo and fluoro nitrobenzenes were both active. Proceeding through the characteristic Meisenheimer intermediate the fluoro is expected as you get inductive stabilisation, why is the iodo system also active and not the other two halides?
The mechanism is complicated by the possibility of additional proton transfers. I am currently speculating that high polarisability function of the iodo group offers additional stabilisation or that the iodo activity may be due to trace transition metals, although addition of Pd(PPh3)4 does not seem to singificantly alter the observed yields.
I did not go in details on the ongoing discussion but you should consider an electron transfer to XArNO2 as an alternative mechanism, if X is F,Cl or Br the resulting radical anion will be stable and nothing will happen but for X = I fragmentation to give a aryl radical could be possible. Coupling of aryl radical with Carbanions Derived from beta-Dicarbonyl compounds are reported. Rossi, R. a; Pierini, A. B.; Peñéñory, A. B. Chem. Rev. 2003, 103, 71–167. page 137
If the meisenheimer complex forms in all cases and the energetically limiting step is the elimination then the observed trend would be I>Br>Cl>F and the fluoro would not be active.
If the meisenheimer complex formation is energetically limiting then the observed trend would be F>Cl>Br>I and the iodide would be inactive.
The problem is that empirically you can't have it both ways for a single set of reaction co-ordinates without an alternative mechanism.
I think the limiting step is the formation of the Meisenheimer complex, as F reacts better than Cl and Br. However, the Meisenheimer explanation of the aromatic nucleophilic substitution is only based on electronic factors, and doesn't consider steric factors. Iodine is a HUGE atom, and steric decompression compensates the unfavorable electronic factor.
I now realize that Pavel has given the same explanation, and that I've read the reference he gave already.
I-O interactions were observed for hypervalent iodine, which is not of importance in this particular case. Also rate limiting formation of the Meisenheimer intemediate cannot take into account leaving group ability.
However the steric decompression argument makes the most sense, in the developing transition state, the movement of iodine out of plane of the sterically demanding nitro group could easily provide the required energy.
A crystal structure of a related compound also suggests that the oxygens of the nitro functionality are out of plane.
I've been out of the game a long time but in my experience the simplest is often worth considering in which my thought is steric hinderance prevents the Br & Cl reaction but Iodine is the most weakly bonded and possibly the existing bonding is stressed due to its size and "negativity". When you look at it there's not a lot of room there.
I think this agrees with Pavel.
It would be interesting to know if the mechanism for both was the same and if there are any byproducts, e.g. Sn2 for Fluorine vs Sn1 for Iodine.
Some O-arylation was also observed by crude NMR in both the F and I cases but not isolated. Also some self condensation of the diketone. Nothing offering any particularly helpful clues.
Steric hindrance has a relatively minor role in nucleophilic aromatic substitution. The problem with nucleophilic attack at iodine to form a hypervalent intermediate is that it doesn't lie on a reaction co-ordinate that leads to carbon-carbon bond formation.
I am satisfied that the size of the iodine atom introduces additional steric strain in the starting material that is relieved by formation of the Meisenheimer complex. In effect the higher energy of the starting material reduces the relative activation energy for the formation of the adduct in the rate limiting step.
The reactivity of the fluoride system is well estabished as to be due to electronic induction.
I did not go in details on the ongoing discussion but you should consider an electron transfer to XArNO2 as an alternative mechanism, if X is F,Cl or Br the resulting radical anion will be stable and nothing will happen but for X = I fragmentation to give a aryl radical could be possible. Coupling of aryl radical with Carbanions Derived from beta-Dicarbonyl compounds are reported. Rossi, R. a; Pierini, A. B.; Peñéñory, A. B. Chem. Rev. 2003, 103, 71–167. page 137
Thank you Javier, I think that the SRN1 is a far better explanation, as an alternate reaction pathway is far more consistent with the discontinuous observations. Electron transfer likely to be spontaneous for the o-iodonitrobenzene owing to the relative stability of the diketo radical and the pi acceptor ability of o-iodonitrobenzene also the iodo reaction has less decomposition in the dark.
Carbon nucleophiles do react with 2-iodonitrobenzene via the SRN1 mechanism:
Galli, C. "Evidence for a Non-Chain SRN1 Reaction Occuring on a Nitroarylhalide." Tetrahedron 44:5205-5208 (1988)
All the data I have would suggest that the 2-iodonitrobenzene does indeed react via a non-chain SRN1 reaction with 1,3-cyclohexanedione. Particularly given that better yields were observed in the dark with a strong preference for DMSO over the other trialled solvent systems.
Mystery solved, then.
I wonder if iodide would catalyze the fluoride displacement via ArF -> ArI -> ArNu, letting you use milder conditions.
The synthetic pathway in which this reaction was situated has since been abandoned. Although the mechanistic considerations are interesting and perhaps worth revisiting with an appropriate radical stable, 2-substituted 1,3-dione; alpha aryl quaternary centres being quite synthetically useful.
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