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Conformational analysis of butane

How to analyze the staggered and eclipsed conformations of butane.

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  • duskpin tree style avatar for user khaagsma
    What is the difference between torsional strain and steric hindrance?
    (25 votes)
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    • orange juice squid orange style avatar for user Benita Dominic
      Torsional strain is the increase in potential energy in the eclipsed Newman projection due to the repulsion between the electron clouds of 2 neighboring carbon atoms.
      Steric hinderance is the strain that causes an increase in potential energy due to the repulsion between the electron clouds of 2 bulky substituents on carbons that are not neighboring .
      If u look at of this video....steric hindrance is caused due to the repulsion of electron clouds of hydrogens attached to carbons 1 and 4 (not neighbors).....whereas torsional strain is caused due to the repulsion of electron clouds of hydrogens 2 and 3(which are neighbors)
      (25 votes)
  • starky ultimate style avatar for user P.Vishesh.22
    Are there similar terms for anti and gauche when talking about the eclipsed confirmation?
    (4 votes)
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    • aqualine ultimate style avatar for user Grogan
      Yes, but only when you have a unique functional group to differentiate them.

      For example, ethane (C2H6), there would be no reason to differentiate between the eclipsed conformations (nor, the anti or gauche, which is why they are collectively referred to as staggered in that situation).

      However, in butane (C4H10, the one in this video), the front and back carbons in the Newman projections each have a methyl group, and we use those to identify anti or gauche. This means we an also identify one unique eclipsed situation. If the methyl groups eclipse each other, it's referred to as the totally eclipsed conformation (or sometimes the cis conformation, but this is not a the same as cis-trans isomers, so keep that in mind). If the methyl groups eclipse a hydrogen, it's an eclipsed conformation.

      The totally eclipsed conformation is even less stable/more potential energy state than the eclipsed conformations, which you can see in the potential energy graph for this video.
      (12 votes)
  • blobby green style avatar for user progress
    I am not in school right now, but i am studying at home. I was on chemguide.co.uk and the practie problem said to draw but-2-en-1-ol, and i drew the -oh on the left, but they claim it must invariably be drawn on the right. Can anyone please tell me why, i would greatly appreciate it.
    (3 votes)
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    • blobby green style avatar for user Cody Weiler
      It depends where the double bond is. In IUPAC naming, alcohol takes priority in naming over double bonds (i.e. you start counting carbons from the alcohol functional group). So, the alcohol needs to be on the end of the 4 carbon molecule and the double bond need to be between C2 and C3. Whether you start it on the right or the left doesn't matter as long as your double bond follows suit and switches sides as well.
      (3 votes)
  • leaf green style avatar for user Courtney Smith
    What does "degenerate" mean in this sense?
    (1 vote)
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  • blobby green style avatar for user Rebecca
    What videos is he referring to at ?
    (2 votes)
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    • piceratops ultimate style avatar for user Bailan
      He was talking about conformational analysis of ethane and propane
      Those are the last two videos prior to this one-conformational analysis of butane.
      In those videos, he found the energy cost between hydrogens, hydrogen-methane.
      (2 votes)
  • blobby green style avatar for user Daughtry St John
    Question about the eclipsed conformation that you showed around ... Can it still be eclipsed if the methyl group on the top carbon is eclipsing one of the bottom hydrogens? I thought they had to be "eclipsing" the same substituents.
    (1 vote)
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  • blobby green style avatar for user boytron3000
    Who does this voice over?
    (1 vote)
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  • leaf green style avatar for user alvinthegreatsh
    Wait, the diagram assumes that the hydrogens on the methyl groups don't rotate, right?
    (1 vote)
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  • purple pi purple style avatar for user douglaswarpengine
    your corbon atoms are not in a zigzag order, i don't understand why. ? shouldn't they be in a zig zag pattern.?
    (1 vote)
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  • aqualine ultimate style avatar for user SoraFBF
    "and remember, in some earlier videos we talked about the energy cost associated with a pair of eclipsed hydrogens as 4 KJ/mol" I'm watching all these videos in order - this is the last video under Stereochemistry for CHEMICAL PROCESSES under MCAT. In which video(s) was this mentioned? The energy costs here were not mentioned in any of the videos I have seen thus far.

    Thanks!
    (1 vote)
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Video transcript

- [Voiceover] Here we have the butane molecule, and this is carbon one, carbon two, carbon three, and finally carbon four. And if we stare down the carbon two three bond, so here I'm rotating the molecule so we stare down the carbon two three bond. This is a staggered confirmation of butane. And if we rotate the front carbon, and keep the back carbon stationary, I rotate 60 degrees, and we're going to get an eclipse conformation. So I left it a little bit off so you can still see the bonds in the back. So from the eclipse conformation of butane I rotate again, and we get a staggered conformation. Rotate another 60 degrees here, and we get an eclipsed conformation, and I'm gonna turn it to its side, so we can see how close these methyl groups are in space. So if I rotate around, you can see with the model set, these hydrogens actually hit. So those hydrogens are close enough where they hit in the model set, and that's called steric hindrance, or steric strain. So back to the eclipsed conformation of butane. If we rotate again, then we get a staggered conformation, we're gonna rotate again to get another eclipsed again, slightly, a little bit off so you can still see the bonds in the back here, and rotate one more time to get back to our staggered conformation. Here's an energy diagram showing the different conformations we saw in the video. And these pictures are just stills from the actual video. We started with the staggered conformation of butane right here, which has a certain potential energy, and we went from this staggered conformation to this eclipsed conformation right here by rotating 60 degrees. It takes energy to go from the staggered conformation to this eclipsed conformation. The eclipsed conformation is higher in energy, the eclipse conformation is less stable. Remember, the higher the potential energy, the less stable the conformation. The lower the potential energy, the more stable, so the staggered is more stable than the eclipsed. The energy difference between these two conformations, let me go ahead and draw a dash line here. So this energy difference between these two conformations turns out to be approximately 16 kilojoules per mole. So it takes energy to go from the staggered conformation to this eclipse conformation. >From the eclipsed conformation we rotated 60 degrees and we got this staggered conformation. Notice this staggered conformation is a little higher in energy than our first staggered conformation. So if I draw a line right here, we can see there's an energy difference between our two staggered conformations. So the energy difference turns out to be approximately 3.8 kilojoules per mole. Going from this staggered conformation up here to this eclipse conformation takes energy. So if I draw a line here, so indicating the bottom, this energy difference is approximately 19 kilojoules per mole. So approximately 19 kilojoules per mole, and notice that this eclipse conformation is higher in energy. Let me change colors here. This eclipse conformation is higher in energy than this eclipse conformation. So if we draw a line right here, we can see there's an energy difference of approximately three kilojoules per mole. So this is the least stable conformation. This conformation, this eclipse conformation has the highest potential energy. >From this eclipse conformation we could go to this staggered, so that's a decrease in potential energy. Notice this staggered has the same energy as this staggered conformation, so they are degenerate. Going from this staggered conformation up here to this eclipsed, it would take energy, and notice this eclipse conformation is the same in energy as this one over here. So if I draw a line, you can see it's the same energy. So these two are degenerate. These two eclipse conformations are degenerate. And finally, going from this eclipse conformation back down to our staggered conformation this is lower in energy. Now let's look at the conformations in more detail, and we'll start with this staggered conformation of butane. And let's go ahead and number the carbons. If this carbon is number one, as we called it in the video. That carbon is attached to this carbon, which is carbon number two. We stare down the carbon two, three bond to get our Newman projection and in the video, you can't see carbon number three, because carbon number two is in front of it. But when you're drawing a Newman projection, you represent the carbon in back, in this case carbon number three, with a circle. So this circle here represents carbon number three. And finally, this would be carbon number four. Let's think about the dihedral angle between our two methyl groups, so between this methyl group and this methyl group. Well, that would be 180 degrees so hopefully you can see there's 180 degrees between our two groups. So the dihedral angle is 180 degrees. This conformation is called the anti conformation. And the anti conformation is lowest in potential energy, therefore, the anti-conformation is the most stable conformation for butane. And that's because we take these bulky methyl groups and we put them as far away from each other as we possibly can. And all of our bonds are staggered, so if you think about these bonds here, everything is staggered. So that makes this the most stable conformation. If we rotate 60 degrees from the anti conformation, in the video, I kept the back carbon stationary, and I rotated the front carbon. We would get this conformation. And this is an eclipsed conformation. So think about this bond eclipsing this bond and this hydrogen eclipsing this hydrogen. I didn't draw them as being completely eclipsed, just so we could actually see what's going on here. And remember, in some of the earlier videos we talked about the energy cost associated with a pair of eclipsed hydrogens, as four kilojoules per mole. So this energy cost is four kilojoules per mole. We also talked about the energy cost from a methyl group eclipsing a hydrogen. So in the video on propane. And this was approximately six kilojoules per mole, so six kilojoules per mole for a methyl group eclipsing a hydrogen. That's the same situation we have down here. A hydrogen and a methyl group eclipsing each other, so this should be an energy cost of six kilojoules per mole, too. If we add all of those up, so this would be four plus six is 10, plus another six is 16, we can see that's the energy difference between these two confirmations, so 16 kilojoules per mole higher. So this eclipsed conformation is higher in potential energy. Let's look at the other eclipsed conformation. So that's the one over here. This is the highest in potential energy, so this must be the least stable conformation for butane. If we look here, we have a pair of hydrogens eclipsing each other, so that should be four kilojoules per mole. We have another pair of hydrogens eclipsing each other so that's another four kilojoules per mole. And then we have two methyl groups eclipsing each other, so think about this bond, and this bond eclipsing each other, and these methyl groups being right on top of each other. So what's the energy cost associated with two methyl groups? A methyl group eclipsing a methyl group? We can figure that out, because we know the total energy cost is 19 kilojoules per mole. So this is what we don't know, so I'll call that X. So what is X? Well, if we add everything up, it should get 19, so 19 is the total. And we have four, and four, and X. So four plus four plus X is equal to 19. So obviously X is equal to 11. So 11 kilojoules per mole, is the energy cost associated with a methyl group eclipsing another methyl group. So there's torsional strain there, but there's also steric strain, or steric hindrance, which we saw in the video. These two methyl groups, the hydrogens can actually get close enough to touch in the video when you're using a model set. And that is destabilizing. So if you have increased steric hindrance, that destabilizes your conformation. And that's why this conformation, this eclipsed conformation is the highest in potential energy. It's the least stable. So if I draw a line over here, just remember that this eclipsed conformation is even higher in potential energy than this eclipsed conformation because of these two methyl groups being so close together. And finally, let's look at our staggered conformation. So our other staggered conformation is right here. Notice this staggered conformation is higher in energy than our anti. So if we look at our methyl groups, we think about the dihedral angle, so if I think about the angle between these methyl groups here, that is 60 degrees, and we say that this is the gauche conformation, so let me go ahead and write this as the gauche conformation. And for the gauche conformation, this is a little bit higher in energy than for the anti-conformation. And that's because these two methyl groups are closer together in space. We don't really have to worry about torsional strain here but we do have to worry about steric hindrance. So these hydrogens on these methyl groups can get pretty close to each other in the gauche conformation, and that has a destabilizing effect, therefore having a higher potential energy for this conformation. So the gauche conformation, while it's staggered, this is more stable than our eclipsed conformations. The gauche conformation is not as stable as the anti-conformation because these methyl groups are relatively close together in space.