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I am doing several questions involving judgment on the planarity of a compound.

Which of the following is not a planar molecule?

  1. $\ce{H_2C=C=CH_2}$
  2. $\ce{H_2C=C=C=CH_2}$
  3. $\ce{H_2C=C=O}$
  4. $\ce{NC-HC=CH-CN}$

I had the idea that the compound in with central atom has $\ce{sp}$ hybridisation is planar or the compound in which all the atoms has the same hybridization. But it is not working in this case. I am aware of finding out the hybridization of a atom in a compound but I feel trapped to decide the planarity of certain compounds. The question does not address this purpose. Anyone has any idea to solve this and many related questions?

Gaurang Tandon
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Pink
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1 Answers1

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I think these general rules work:

  1. If there is an $\ce{sp^3}$ hybridized carbon (or nitrogen), the molecular is NOT planar.

2) If there are no $\ce{sp^3}$ hybridized carbons (or nitrogens), and there is only one $\ce{sp^2}$ hybridized atom (carbon or nitrogen), it will be planar.

3) If there are no $\ce{sp^3}$ hybridized atoms, and there are 2 $\ce{sp^2}$ hybridized atoms (carbon or nitrogen) that are separated by an even number of double bonds and no single bonds, then the molecule will not be planar.

So a general simple rule is that:

the molecule will not be planar if there is an $\ce{sp^3}$ hybridized carbon (or nitrogen) atom or two $\ce{sp^2}$ hybridized atoms of carbon/nitrogen which are separated by an even number of double bonds and no single bonds. Otherwise, its structure allows it to be planar.

Even though the molecule will have a structure that allows for it to exist in a planar conformation, there may be some/many that do not persist in a planar conformation due to steric effects, or complex three dimensional geometries.

In the problems you listed above, using this rule:

  1. Not planar because there are no $\ce{sp^3}$ and the two $\ce{sp^2}$s are separated by an even number of double bonds.

  2. Planar because there are two $\ce{sp^2}$s but they are separated by an odd number of double bonds (3) (and no single bonds)

  3. Planar because there are no $\ce{sp^3}$s and only 1 $\ce{sp^2}$s that make 3 or more bonds (C or N). The orbital geometry is NOT planar because the $\ce{sp^2}$ oxygen is separated from the $\ce{sp^2}$ carbon by an even number of double bonds.

  4. Planar because 2 $\ce{sp^2}$s are separated by an odd number (1) of double bonds (and no single bonds)

Gaurang Tandon
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Joseph Hirsch
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    I think these rules would make $\ce{C60}$ planar... – Zhe Dec 01 '16 at 18:43
  • True. Interesting point. C60 might be described as having approximately planar geometry-it is basically a planar sheet cut and contorted into a spheroid. – Joseph Hirsch Dec 01 '16 at 19:15
  • Butatriene (which is listed above in the problem) is planar but that doesn't mean that the hydrogens will not be contorted out of plane. – Joseph Hirsch Dec 01 '16 at 19:22
  • Basically C60 is contorted graphite – Joseph Hirsch Dec 01 '16 at 19:27
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    I find that claim dubious, but instead of dwelling on that, here's another one. How about biphenyl? Planar or non-planar? – Zhe Dec 01 '16 at 19:38
  • Biphenyl is only conformationally aplanar because of steric interaction. – Joseph Hirsch Dec 01 '16 at 19:49
  • Also, in both C60 and other aromatics, there are not really any carbon-carbon single bonds and not really any carbon-carbon double bonds, so they are allowed to be a-planar by the rules above because no two SP2 hybridized carbons are separated by an odd number of double bonds. – Joseph Hirsch Dec 01 '16 at 19:58
  • Exactly, and steric interactions do a lot in terms of conformations that you haven't accounted for here. I'm just saying this is super-complicated. – Zhe Dec 01 '16 at 20:02
  • Since Benzene is planar we can see that the bonds between two carbon atoms can not be defined as double bonds, but your example of biphenyl gives me an idea for an exception that does violate the letter of the rules. – Joseph Hirsch Dec 01 '16 at 20:02
  • Let us continue this discussion in chat. – Joseph Hirsch Dec 01 '16 at 20:02
  • I am a little bit confused.According to this linka) is not planar but according to your rules,it is the other way.Also you mentioned in your answer that butatriene is non planar but in your comments,you mentioned it as planar.Please resolve the issue.Thanks. – Pink Dec 02 '16 at 08:08
  • Sorry, I got my evens and odds reversed in the prior edit. I have fixed them as well as the answers. So basically if two SP2s are separated by an odd number of double bonds (1, 3 etc) then their single bonds will be in the same plane, but if they are separated by an even number of double bonds (with no single bonds in between) they will be rotated $90^o$ – Joseph Hirsch Dec 02 '16 at 16:23
  • Note that b) butatriene is "textbook" planar, (it is a line of carbons with hydrogen fishtails), the hydrogens will have a tendancy to get twisted out of the same plane. – Joseph Hirsch Dec 02 '16 at 16:28
  • Likewise d) 1, 2 dicyanoethylene (both forms) is textbook planar but has a trans and a cis form and the cis form will probably twist mildly out of plane. – Joseph Hirsch Dec 02 '16 at 16:43
  • http://pubs.acs.org/doi/abs/10.1021/ed082p1329 – Joseph Hirsch Dec 02 '16 at 17:10
  • Shouldn't then [10]-Annulene be planar(all sp2, no sp3)? – crayon Apr 03 '18 at 15:00