"The structure of compound 1, the major compound, of the manuscript was mistakenly assigned. As a result the authors withdraw this manuscript."You heard it right, folks: An entire (published) manuscript, all down to one set of spectra.
So, what went wrong here? Here's the carbon-13 spectrum, from the SI:
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| Source: Jang group | JACS 2008 |
Another tweet (thanks, Neil!) clued me in to this Organometallics paper, in which they prepare the same compound. Compare the spectrum above to this one:
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| Source: Hor group | Organometallics 2011 |
So, what went wrong? One clue might be solvent; the first spectrum's taken in a highly polar solvent (d6-acetone), whereas #2 uses ol' NMR stand-by deuterated chloroform. Given the highly polar nature of the first compound, along with the extra signals (and perhaps a second benzyl group in the proton NMR), I'm guessing that spectrum #1 actually shows a quaternary ammonium salt, which might result from "over-benzylation" of the cinchonine starting material.
The real bummer here? I've looked through the rest of the SI, and most compounds appear spot on.
Certainly, the authors managed to perform a challenging radical addition with high selectivity. Even more curiously, the ammonium salt used to effect the transformation (1a) looks correct!
Tough pill to swallow. Kudos to the authors for making the right (tough) choice here, voluntary or not.
Update, 8/8/13: Over at Reddit, stop_chemistry_time has staged a fantastic, ongoing debate with me in the comments. Here's the link.
I've been following this since all of you were discussing it on twitter, and I just never got the argument for C2 and C6 having different resonances. Can you explain it in really basic terms?
ReplyDeleteIs there some hindered rotation of the ring that organic chemists know about that I don't? Because if not, C2 and C6 are completely equivalent (as are C3 and C5).
I get that the chiral centers in the molecule make the benzylic protons inequivalent, but I don't see how that translates to the planar ring with free rotation which will put the carbons in the *exact* same magnetic environment and should make them identical.
Andre,
ReplyDeleteIt all comes down to time (and kinetics, I guess). We think of free rotation as something that a molecule either has, or doesn't have, but it is not an either/or situation, it is more of a continuum. For NMR, it basically comes down to "can the two protons (or carbons) switch positions multiple times in the time between pulses in the NMR.
My favorite example of this phenomenon is the proton NMR of cyclohexane. At room temperature, you get 1 signal, because the ring can flip many times during the course of the pulse (1 secondish). However, if you cool the sample, the ring flip slows (kinetics and all), and the axial and equatorial protons resolve, giving two signals. This powerpoint deck has a good explanation and some nice spectra to explain the effect.
http://www.chem.queensu.ca/facilities/NMR/nmr/chem805/PPT09/5-NMR-conversion.ppt
Coming back to the original question, the benzyl group is probably pinned in by the groups surrounding it, so although the individual bonds have free rotation, the benzyl group gets "stuck" and can't rotate very fast at room temperature. Depending on the environment, this effect will probably go away if the sample were heated, and the two signals will coalesce into a single signal between the two observed signals.
Hope this helps!
So it is just a claim of restricted rotation in this case then. Okay. I was thinking there was something more subtle going on. I guess I just need to break out my modeling kit and make this guy to see the crowdedness of the central stereocenter.
ReplyDeleteA few of the results note "TE 999.9 K". What does this mean? And, is it applicable to the current discussion?
ReplyDeleteThis appears to be a measurement of temperature in degrees Kelvin, after looking at other results in the SI. But they are all about room temperature, and 999.9 K is well above room temperature.
See pages 36 and 37 of the SI at http://pubs.acs.org/doi/suppl/10.1021/ja807685r/suppl_file/ja807685r_si_001.pdf .
I am not familiar with the instrument, and couldn't locate a description of this at http://www.bruker.com .
The compound in JACS is definitely related to the one in OM. I am still not convinced that it is as simple as the contamination by the ammonium salt. The procedure says:
ReplyDelete"After filtration and evaporation of the solvent, the residue was purified by flash chromatography on silica gel (ethyl acetate / hexane = 3 / 7) affording the compound 1 in 86 % yield (3.3 g)"
I would expect this to remove the ammonium salt. Right? There must be more to it. I am still wondering what went wrong.
Could the student have cut the corners and skipped the chromatography? After all the conversion was spot-to-spot on TLC, hum, expect maybe for that little spot with an rf of 0.000...
Oops, the chromatography has actually been discussed on the reddit link above at the end of the article.
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