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I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

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  • $\begingroup$ I guess that NH3 will form bond perpendicular to the ring. So the carbanion is non-planar. $\endgroup$ – JM97 Sep 19 '16 at 16:38
  • $\begingroup$ But I need to know whether the ring will be planar or not $\endgroup$ – EdmDroid Sep 19 '16 at 16:46
  • $\begingroup$ The ring will be planar but the NH3 will come out of the plane due to hindrance because of CH3 group $\endgroup$ – Prakhar Sep 19 '16 at 17:20
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    $\begingroup$ Assuming that your premise is correct, I would suppose it is just because it wants to allocate the lone pair as much s-character as possible. So, the bond angles change, and the ring loses its planarity. I think that you might get better answers if you try searching for the phenyl anion, because that is what your compound basically is (sans the methyl group). I did a quick search, but couldn't find literature on it. $\endgroup$ – orthocresol Sep 19 '16 at 17:24
  • $\begingroup$ And it would want to allocate more s-character to the lone pair to keep the electrons as close as possible to the nucleus? $\endgroup$ – EdmDroid Sep 19 '16 at 19:13
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I have given one of these molecules a little spin on the computer. With whatever I tried, I was unable to find a non-planar solution. I really tried my best. Below is an animation (sorry for the poor quality) of an attempt which uses a chair like starting geometry. If there was any non-planar geometry lower in energy, this would have found it.

animation of the optimisation

The level of theory is DF-BP86/def2-TZVPP and the program used is Gaussian 09 rev. D.01.
There is really not much more I can add to the discussion.

I was told by my chemistry teacher that the carbanion formed in the picture above is non-planar. I don't understand why. It looks like benzene with an extra electron pair at one of the carbons so shouldn't it be planar?

To be absolutely, brutally honest here: I have no clue why your teacher had this idea. If this wasn't a "naked" carbon, we could talk about it, but it is essentially a deprotonated phenyl ring, and as such, you are absolutely right in assuming the lone pair being in phase with the ring. The π system is hardly different from benzene. To fill this post a little bit more, have those fancy π-orbitals.

total pihalf pimore half pia quarter pi?another quarter pi
(I could not find the completely anti-bonding pi-orbital.)

And for good measure, here is the HOMO, which is the lone pair at the carbon:

HOMO

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    $\begingroup$ I don't know that it would make a difference, but did you include dispersion corrections? $\endgroup$ – hBy2Py Feb 24 '17 at 15:09
  • $\begingroup$ @hBy2Py No. I don't think it'll make a difference. How would they affect the geometry? I think the sigma type orbital would still be that and therefore we have a nice Cs symmetry... $\endgroup$ – Martin - マーチン Feb 24 '17 at 15:13
  • $\begingroup$ Yeah, I would think it really shouldn't matter, since the only departures from mirror symmetry in the electron density should occur on the ammonium and methyl hydrogens. $\endgroup$ – hBy2Py Feb 24 '17 at 15:21
  • $\begingroup$ I was told the same thing by my chemistry teacher. Thats what brought me here. $\endgroup$ – prog_SAHIL Aug 5 '18 at 10:12
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I cannot find any published structures — neither computational nor crystal structures — on the actual phenyl anion that is the centre of your question. However, quite a few similar structures are published for the benzyne radical anion and for phenyl lithium and related compounds — the bond between carbon and lithium can generally be considered rather ionic considering the electronegativity difference so it is a reasonable argument to assume phenyl lithium’s structure to be akin to the phenyl anion’s.

Rather unimpressively, Nash and Squires report the calculated structure of the ortho-benzyne radical anion to be $C_\mathrm{2v}$ symmetric.[1] This automatically also means a planar benzyne ring. This may have been an artifact from the calculation, but considering the system’s potentially high symmetry it also makes most sense. The benzyne ring in itself is distorted but planarity is retained.

Likewise, the crystal structure of $\ce{[LiPh(m5dien)]}$ was published by Schümann, Kopf and Weiss (others have also been published but this was my first ‘hit’).[2] Their research also indicates a planar phenyl ring. To quote the paper:

The benzene ring is approximately planar; but the hexagon is distinctly distorted (angle at $\ce{C{1}}$ $113.1$ instead of $120^\circ$, [sic!] while those at $\ce{C{2}}$ and $\ce{C{6}}$ on the other hand, with values of $123.9$ and $125.1^\circ$, [sic!] are significantly enlarged).

Thus, this too is a primarily planar system with most distortions in plane around the anionic carbon to increase the lone pair’s (or $\ce{C-Li}$ bond’s) $\mathrm{s}$ orbital contribution.

In the absence of any further evidence, I would also assume your two structures to be primarily planar; however, featuring significant distortions at the anionic carbon centre. Theoretical calculations (I shall bounty this question in the hope of receiving some) may provide a more conclusive answer.


References:

[1]: J. J. Nash, R. R. Squires, J. Am. Chem. Soc. 1996, 118, 11872–11883. DOI: 10.1021/ja9606642.

[2]: U. Schümann, J. Kopf, E. Weiss, Angew. Chem. Int. Ed. 1985, 24, 215–216. DOI: 10.1002/anie.198502151.

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