I need to calculate the angles between CCC, CCH, HCH in the following molecules:

  • ethene ($\ce{H2CCH2}$),
  • propa-1,2-diene ($\ce{H2CCCH2}$), and
  • buta-1,2,3-triene ($\ce{H2CCCCH2}$).

If I'm not mistaken, the Lewis structures of all of these molecules have the carbons in the middle, connected by double bonds to each other, and the hydrogen atoms connected to the carbons in the edge. For example:

H           H
 \         /
  C = C = C
 /         \
H           H

The carbons at the edge are $\mathrm{sp^2}$ and the middle carbon is always $\mathrm{sp}$ .

This means that the angle CCC is always 180°. But how can I determine the angles CCH, HCH?

In addition, how can I determine if the hydrogen atoms in the three molecules lie on the same plane?


What are the bond angles in a normal trigonal planar arrangement? Treat the center atom as in H C H as a trigonal planar arrangement with the middle (carbon) atom in the center with a third atom attached on the left over position (another carbon). What angle maximizes their separation?

Your other question about the plane of the molecule is a bit more interesting. The pi bonds between the carbons ensure that the molecule is not rotating in space. To maximize the distance between the terminal hydrogen atoms, one set should be aligned up and down, while the other two on the opposite side should be rotated 90° in space, otherwise they would ellipse one another. In other words, one pair is horizontal while the opposite is vertical for example, depending how you want to imagine it.

  • $\begingroup$ Does the answer to your first question is 120 degrees? As for the second explanation, is there any picture or a diagram that can be explain this structure? I can't really understand it completely... Hope you'll be able to help Thanks ! $\endgroup$ – joshua Sep 17 '12 at 19:24
  • 1
    $\begingroup$ Yes, I get the same answer as you.. just out of curiosity though, why not google "trigonal planar" and the answer is literally the first thing your eye should notice. As for newman projections, I am sure you will enjoy some khan material. In your case, I recommend you keep watching every video until confusion is eventually obliterated. Good luck! $\endgroup$ – Leonardo Sep 17 '12 at 21:31
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    $\begingroup$ Be aware, that in $\ce{H2C=CH2}$ all hydrogens are in plane, and the same holds for $\ce{H2C=(C=C)_{$n$}=CH2}$ because of symmetry reasons. For any $\ce{H2C=(C)_{$n$}=CH2}$ the alignment of the hydrogens is perpendicular. You can explain this with the pi orbitals. $\endgroup$ – Martin - マーチン Feb 23 '15 at 4:16

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