If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Conformational analysis of ethane

How to analyze the staggered and eclipsed conformations of ethane.

Want to join the conversation?

  • piceratops ultimate style avatar for user Costel Voica
    Is the potential energy difference between the eclipsed and staggered conformations bigger in other molecules (e.g. butane) ?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Ana
    What does degenerate mean?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • leaf orange style avatar for user Huayng  Fang
    For the potential energy graph, the energy difference between eclipse and staggered is 12 KJ/mol. However, the initial energy U = 0 J, and the energy of staggered and eclipse is 10 J and 22 J. I think they should be 10 KJ and 22 KJ. Right?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user RandomDad
    Why do we have to know about staggered and eclipsed conformation?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user Greacus
      I'm not sure, but this way you can predict how molecules are oriëntated in 3D space. This would mostly be staggered, because it has a lower potential energy, but when energy is added to the molecule eclipsed might be more common.
      Hope this helps a bit :)
      (1 vote)
  • old spice man blue style avatar for user Lucian Rex
    How do they calculate the kJ/m of a molecule's conformations?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine sapling style avatar for user Jakob Pankonin
    Currently learning this in college, so I want to make sure I have this right for redundancy's sake. On energy diagrams like this in general it is only the front carbon's constituents that are rotated, correct?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user horvatma
    Would a gauche interaction add additional energy?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user lennon4896
    Why the molecule with lower potential energy is more stable? Is it due to the electron repulsion or other theory?
    Or can I assume the potential energy as reserved energy for kinetic energy which allows the movement of atoms?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • female robot amelia style avatar for user him
      The stability of a molecule is indeed related to its potential energy, and this concept is deeply rooted in the principles of thermodynamics and quantum mechanics.

      In a molecular system, potential energy arises from the interactions between atoms and/or molecules. This potential energy is a function of their relative positions and the forces acting between them. When atoms are in a stable configuration, their potential energy is at a minimum. It is analogous to a ball sitting at the bottom of a valley; it has the lowest potential energy at that point.

      There are primarily two factors that contribute to the stability of a molecule:

      1. *Electron Repulsion*: In molecules, electron clouds around atoms repel each other due to their negative charge. When atoms come too close together, electron clouds start to overlap, leading to significant electron-electron repulsion. This repulsion increases the potential energy of the system. Stable molecules are those where the atoms are positioned in a way that minimizes this electron repulsion.

      2. *Chemical Bonding*: Stable molecules are formed when atoms share electrons to form chemical bonds. This process leads to a more energetically favorable state as compared to individual atoms. The energy released during bond formation contributes to the overall stability of the molecule.

      In quantum mechanics, the lowest energy state of a system is often referred to as the ground state. Molecules tend to adopt configurations that minimize their potential energy and reach this ground state, making them more stable.

      Now, about your mention of "reserved energy for kinetic energy": Potential energy can indeed be thought of as "stored energy" that can be converted into other forms of energy, including kinetic energy. When a molecule undergoes a chemical reaction, the potential energy of the initial molecules is converted into kinetic energy as atoms move and rearrange, and new bonds form or break. However, it's essential to understand that potential energy itself does not directly cause the movement of atoms. Instead, it plays a crucial role in determining whether a particular molecular configuration is energetically favorable and likely to undergo a chemical transformation.

      In summary, the stability of a molecule is determined by its potential energy, which arises from electron repulsion and chemical bonding. Molecules tend to adopt configurations that minimize their potential energy, making them more stable. The potential energy can be converted into kinetic energy during chemical reactions, allowing atoms to move and rearrange.
      (1 vote)
  • blobby green style avatar for user Vic Lin
    At around , it is said the energy difference between Staggered/Eclipsed is caused by torsional strain but is it not due to the electrons repelling each other instead?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Yash
    Jay said that approx. 99% ethane molecules at room temperature have staggered conformation, while about 1% have eclipsed conformation. But what about the conformations in between? Rather than just occupying the extreme conformations, can't the molecules occupy the conformations midway?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] Here we have the eclipsed conformation of ethane, and if I turn it so we sight down the carbon carbon bond, we'll see the Newman projection for the eclipsed conformation. Notice I have some hydrogens in green here, and that's just to help us visualize rotation around the carbon carbon bond. So I'm going to rotate the front carbon and keep the back carbon stationary. So now we get the staggered conformation of ethane. I'm gonna keep rotating 60 degrees. So I'm gonna rotate again, and now we have an eclipsed conformation. I rotate again, and now we have a staggered conformation of ethane. I rotate another 60 degrees, and we get an eclipsed conformation. I rotate again, and we get a staggered, and you get the idea. One more rotation, and we get back to an eclipsed conformation. Here we have a graph of the potential energies of the conformations that we just saw in the video. So we started out right here with this eclipsed conformation of ethane, so we are at this particular potential energy. As we rotate to get to this staggered conformation, you can see there's a decrease in potential energy. So this staggered conformation has a lower potential energy than the eclipsed conformation. We rotate again, we get back up here to this eclipsed conformation. Notice that takes energy. So it takes energy to go from this staggered conformation to this eclipsed conformation. Going from the eclipsed to the staggered, that's a decrease in energy. Going from staggered up to this eclipsed, that's an increase in energy. Going from the eclipsed down to the staggered again is a decrease, and finally, back to the original eclipsed conformation would be an increase in energy. Notice that all of our eclipsed conformations here have the same potential energy. If I draw a line, this is all the same potential energy. Whoops, I didn't draw a very nice line there, but you get the idea. Therefore, we say that these are degenerate in terms of energy. Same thing for the staggered conformations. All of these staggered conformations, if I draw a line in here, have the same potential energy. So the staggered conformations are lower in energy than the eclipsed conformations. Actually, the difference is 12 kilojoules per mole. So if I write that in here, so 12 kilojoules per mole, that's talking about the difference in energy between the eclipsed conformations and the staggered conformations. Lower in energy means more stable. The easy way of thinking about that is imagine these things as hills, if I have a boulder, or a rock, or something down here at the bottom of the hill and I'm comparing that boulder or rock up here to a boulder at the top of the hill. In physics, you can set your potential energy equal to zero at the ground. So let's say that this is ground level. I say my potential energy, U, is equal to zero joules at this point. And so the boulder in this valley here, let's say it's 10 joules. Let's say this is 10 joules here at this point. And then it would take energy to push this boulder up this hill to this point. Let's say the final potential energy of the boulder at this point would be 22 joules. It takes energy in order to do that. This final position is less stable, and this is the higher potential energy. So higher in potential energy means less stable. Lower in potential energy means more stable. So why do we have a difference in energy between the staggered and the eclipsed conformation? Well this is called torsional strain. So this difference in energy, this 12 kilojoules per mole, is called torsional strain. The source of torsional strain has been a topic of debate. One of the current theories has to do with molecular orbital theory. I'm gonna go with one of the older ones, which talks about the electron pair repulsions. The electron pair repulsions are greatest when the bonds are eclipsed. So if you think about the electrons in this bond being close to the electrons in this bond, and you have that same thing over here and over here, so in space, these electron pairs, these bonding electron pairs are closer together in the eclipsed conformation than they are in the staggered. If I go down here to the staggered, you can see if you're thinking about these electron pairs they're relatively further away than they are in the eclipsed conformation. We know that electrons will repel. So electron pair repulsions are greatest when the bonds are eclipsed, and therefore, that's higher energy and the electron pairs are further away from each other when you're talking about staggered, therefore, more stable. The total energy cost between the two conformations is 12 kilojoules per mole. We have three pairs of eclipsed hydrogens. If I go back up to here, here's one pair of eclipsed hydrogens, here's another pair, and here's another pair. So if the total energy is 12 kilojoules per mole, and I have three pairs of eclipsed hydrogens, we could say that the energy cost for each pair of eclipsed hydrogens is four kilojoules per mole. So this would be four kilojoules per mole for this pair of eclipsed hydrogens, four kilojoules per mole for this one, and four kilojoules per mole for this one, adding up for a total of 12. So our total energy cost is 12, and now we can think about two hydrogens eclipsing each other as having an energy cost of four kilojoules per mole. We've just seen that the staggered conformation of ethane is more stable than the eclipsed conformation of ethane. If you want turn a staggered conformation into an eclipsed conformation, you would need energy. At room temperature, there's enough energy for the staggered conformation to turn into the eclipsed. Equilibrium is reached between the two conformations, and at room temperature, approximately 99% of ethane molecules have an approximately staggered conformation, whereas only about 1% have an eclipsed conformation. Again, that's due to stability. Staggered is more stable than eclipsed.