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Homework Help: Rotation of double and triple bonds

  1. May 31, 2014 #1
    My textbook does not really go into the details of rotation about a bond.

    I know single bonds can rotate easily.
    What about double bonds and triple bonds?
    The pi bonds would be broken. Does that mean new compounds are formed?

    My textbook says rotation around triple bonds is easy because of circular symmetry around the electron density map and that if rotation happens, no new compounds are formed. I still do not visualize how breaking two pi bonds cannot refrain rotation and how breaking two bonds would not cause a new compound...
  2. jcsd
  3. May 31, 2014 #2
    Here is the extract.

    Attached Files:

  4. May 31, 2014 #3
    I think it is saying the group connected to the atom with the triple bond can rotate easily. The rotation of two atoms connected by a triple bond, a pi and sigma bond, is energetically unfavorable.

    In a pi bond, if you were to reduce the pi bond to a sigma bond temporarily and rotate the atom, you may have to deal with cis and trans configurations based on what groups are attached when the pi bond is reformed. This could result in a different compound.

    As far as I know, double and triple bonds do not rotate unless energy is added to the system.
  5. Jun 1, 2014 #4
    In the triple bond, since you have two pi bonds, I think it would be impossible to have cis trans problems because the pi bonds overlap correspondingly to the phase sign.
  6. Jun 1, 2014 #5
    Look here:
    In (a), the red has to match with the blue.

    Attached Files:

  7. Jun 1, 2014 #6
    The same thing goes for double bonds.
    I do not understand how rotation would occur even if enough energy were added!
  8. Jun 1, 2014 #7
    It is not that rotations around pi bond can't happen, it is just that there exists an energy barrier which is difficult to overcome under typical pressure/temperature conditions. You can take a look at the UC Davis ChemWiki for some discussion on the MO Theory take on this. My simplified explanation is that rotation around one or more pi bonds requires the pi electrons to be promoted into a higher MO orbital which requires energy, as ChiralWaltz stated above. For an example of this you can look for literature about NMR data collected on amide bonds as a function of temperature. I found this paper very quickly and quickly looked at the figures which should show the general idea.

    As far as the excerpt you posted from the book, here is my take interpretation. Triple bonds should have a large barrier of rotation if we extrapolate our knowledge about double bonds however because of the symmetry surrounding the triple bond, we have no way of distinguishing one rotamer from the other. If you imagine two identical cylinders joined by some mechanism along their long axes allowing for rotation, you should be able to notice that I can steal it from you for a minute and rotate along the axis, then give it back, and you would have no way of knowing if or by how much I have rotated it. This is because of the symmetry in such a geometry. This is analogous to the triple bond case of ethyne, for example.
  9. Jun 1, 2014 #8
    Sorry about the unclear post last night. I was going to bed because I couldn't even do basic algebra at that point.

    Yes, triple bonds consist of TWO pi bonds and a sigma bond. Please excuse what I said previously about Triple bonds being ONE pi and a sigma.

    Here is a simple three step mechanism to illustrate what I was talking about with energy levels.

    dimethyl fumarate to dimethyl maleate:

    This basic explanation is on page 2 at the top using the photochemical mechanism. Notice that by adding energy to the system, we are able to promote the pi bonded electrons into their anti-bonding molecular orbits (MO). This is allows us to break the pi bond temporarily, rotate into cis or trans configuration, and allow the electrons to fall back into its bonding pi MO. Thus we have created a new molecule through the photochemical rotation of a pi bond.
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