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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?

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    $\begingroup$ For an sp hybridized atom, the p orbitals that are used to form pi bonds will be perpendicular. That makes the pi bonds formed either side of an sp atom perpendicular to each other. $\endgroup$ – Jonathan Charmant Nov 27 '16 at 21:39
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    $\begingroup$ The answer to your question is here: Why are the two CH2 groups in allene perpendicular to each other? $\endgroup$ – orthocresol Dec 1 '16 at 14:13
  • $\begingroup$ Unfortunately, I think the answer to the general question is: learn chemistry. Because there are all sorts of factors that influence geometry... $\endgroup$ – Zhe Dec 1 '16 at 20:02
  • $\begingroup$ @orthocresol can you give reference to some other questions whee this problem has been discussed $\endgroup$ – Pole_Star Jan 30 '18 at 8:59
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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)

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    $\begingroup$ I think these rules would make $\ce{C60}$ planar... $\endgroup$ – Zhe Dec 1 '16 at 18:43
  • $\begingroup$ True. Interesting point. C60 might be described as having approximately planar geometry-it is basically a planar sheet cut and contorted into a spheroid. $\endgroup$ – Joseph Hirsch Dec 1 '16 at 19:15
  • $\begingroup$ 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. $\endgroup$ – Joseph Hirsch Dec 1 '16 at 19:22
  • $\begingroup$ Basically C60 is contorted graphite $\endgroup$ – Joseph Hirsch Dec 1 '16 at 19:27
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    $\begingroup$ I find that claim dubious, but instead of dwelling on that, here's another one. How about biphenyl? Planar or non-planar? $\endgroup$ – Zhe Dec 1 '16 at 19:38

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