i am doing several questions including judgment on the planarity that a compound.

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Which that the complying with is not a planar molecule?

$\ceH_2C=C=CH_2$ $\ceH_2C=C=C=CH_2$ $\ceH_2C=C=O$ $\ceNC-HC=CH-CN$

I had actually the idea that the compound in with main atom has actually $\cesp$ hybridisation is planar or the compound in which every the atoms has actually the exact same hybridization. However it is not working in this case. I am mindful of finding out the hybridization that a atom in a compound however I feeling trapped to decision the planarity of particular compounds. The question does not deal with this purpose. Anyone has any kind of idea to solve this and also many associated questions?

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edited Feb 25 "18 in ~ 9:30

Gaurang Tandon
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I think these general rules work:

If over there is an $\cesp^3$ hybridized carbon (or nitrogen), the molecular is not planar.

2) If there room no $\cesp^3$ hybridized carbons (or nitrogens), and also there is just one $\cesp^2$ hybridized atom (carbon or nitrogen), it will certainly be planar.

3) If there are no $\cesp^3$ hybridized atoms, and also there space 2 $\cesp^2$ hybridized atoms (carbon or nitrogen) that are separated by an also number of double bonds and also no solitary bonds, climate the molecule will not it is in planar.

So a general an easy rule is that:

the molecule will not it is in planar if over there is one $\cesp^3$ hybridized carbon (or nitrogen) atom or 2 $\cesp^2$ hybridized atoms of carbon/nitrogen which space separated through an also number of double bonds and also no single bonds. Otherwise, the structure enables it to it is in planar.

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

In the problems you noted above, using this rule:

Not planar because there space no $\cesp^3$ and the two $\cesp^2$s room separated through an also number of dual bonds.

Planar since there room two $\cesp^2$s however they are separated by one odd number of double bonds (3) (and no single bonds)

Planar because there space no $\cesp^3$s and also only 1 $\cesp^2$s that make 3 or much more bonds (C or N). The orbit geometry is not planar due to the fact that the $\cesp^2$ oxygen is separated from the $\cesp^2$ carbon through an even number of twin bonds.

Planar since 2 $\cesp^2$s room separated by one odd number (1) of dual bonds (and no solitary bonds)