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Stability of cycloalkanes

How to analyze the stability of cyclopropane, cyclobutane, cyclopentane, and cyclohexane.

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  • leaf yellow style avatar for user saransh60
    Table at shows cyclobutane has more heat of combustion than cyclopropane why at he said that cyclobutane has lower heat of combustion ?
    (6 votes)
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  • blobby green style avatar for user dkayastha
    ehat is the difference between torsional and steric conformation?
    (4 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      Both types of strain are caused by the repulsion between electrons.

      Torsional strain occurs when two atoms (or groups) on adjacent carbons get too close to each other. In torsional strain, the groups are separated by three bonds.
      The interaction corresponds to the eclipsed conformation of a molecule, e.g. 1.2-dibromoethane.

      Steric strain is the repulsive interaction between atoms (or groups) on carbon atoms that are not adjacent.
      An example would be the 1,3-diaxial interactions in cyclohexane rings . Here, the atoms are separated by four bonds.
      (14 votes)
  • leafers seed style avatar for user Gio
    For the heat of combustion on the cycloalkanes, why isn't the number of CH2 groups not included? I am understand what is in the cycloalkanes when measured the heat of combustion
    (2 votes)
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  • spunky sam red style avatar for user Sujay Rittikar
    How is strain related to stability? and, what are differences between steric strain and torsional strain? Pls someone explain it.
    (1 vote)
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    • piceratops seed style avatar for user RogerP
      Steric strain is the easier of the two to understand and it is due to bulky substituents repelling each other because they both want to occupy the same bit of space. Therefore, the conformation of the molecule changes or distorts, so as to move the bulky groups as far away from one another as possible.

      Torsional strain occurs when you have three connected bonds, and it is due to the electrons in one of the outer bonds repelling the electrons in the other outer bond. Take, for example, ethane which contains the bond sequence H-C-C-H. The ethane is free to rotate about the C-C sigma bond, but in the eclipsed conformation, the C-H bonds on either end of the molecule are aligned with each other. Torsional strain arises from the electrons in one C-H repelling those in the C-H bond on the other end of the molecule. This torsional strain is reduced when the molecule moves into the staggered conformation. (In fact, due to a concept known as hyperconjugation, there is actually added stability to the staggered form because of constructive orbital interactions involving the bonding and anti-bonding molecular orbitals on opposite ends of the molecule.)
      (4 votes)
  • orange juice squid orange style avatar for user RowanFarrell217
    Why do more stable molecules have lower heats of combustion?
    (2 votes)
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    • piceratops seedling style avatar for user Alexia Foy
      Remember that in organic chemistry, stable means lower energy; heat of combustion is the energy released as heat when the molecule is combusted with oxygen at standard conditions. Therefore, if the molecule is more stable it will have a lower heat of combustion.
      (1 vote)
  • female robot grace style avatar for user SULAGNA NANDI
    Why does a less stable compound have a higher heat of combustion?
    (1 vote)
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    • leaf red style avatar for user Richard
      In chemistry we associate more energetic matter as being less stable, and so having less energy makes matter more stable. The heat of combustion is the energy lost by the molecule upon being combusted. The energy lost by combustion was once entirely part of the molecule. So the molecules with the highest heats of combustion were before the ones with the most energy contained in their bonds essentially. And if they had more energy initially, that means they were more unstable than molecules with lower heats of combustion.

      Hope that helps.
      (2 votes)
  • duskpin ultimate style avatar for user Sabrina Kim
    It is said that cyclopentane is the most stable conformation. But in the chart, the cyclohexane's heat of combustion is lower than cyclopentane's. I thought the more stable conformation has a lower E. Then how does this work?

    Thank you in advance:)
    (0 votes)
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    • blobby green style avatar for user wr
      The video was saying that according to the old theory of planar cycloalkane structures, cyclopentane would be assumed to be the most stable due to having the bond angles closest to the ideal 109.5 degrees, and therefore the lowest bond strain.

      However, the heats of combustion show that cyclopentane is NOT as stable as cyclohexane - which shows that the planar theory is wrong - the molecules adopt lower-energy conformations in three dimensions. Cyclohexane can get closer to the ideal bond angles (think of the chair conformation of cyclohexane shows in previous videos) than cyclopentane can, hence the lower heat of combustion per CH2.
      (4 votes)
  • aqualine seed style avatar for user Martian
    Talking about stability, I've one question; Which isomer is more stable cis- or trans- alkanes?
    (1 vote)
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    • piceratops seedling style avatar for user Fola Okikiolu
      alkanes can't be cis or trans because they are able to rotate around their c-c sigma bonds. In the case for alkenes however, the trans conformation would be more stable since it has less steric hindrance when compare to the cis isomer. Although (i should have said this first) in the case of cyclohexane, the trans isomer would be more stable. This is because both the substituents can adopt an equitorial position which ultimately leads to less steric hindrance.
      (1 vote)
  • leafers sapling style avatar for user maxime.edon
    But you've said earlier that cyclopentane was a planar molecule... Didn't I follow the explanation correctly ?
    (1 vote)
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  • blobby green style avatar for user IKE N
    Isnt the stability of cyclohexane due to its bond angles being the same as the ideal tetrahedral bond angles of carbon bonded to four groups. (109.5)? your video says cyclohexane has 120degrees bond angles..
    (1 vote)
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Video transcript

- [Voiceover] At one time it was thought that the cycloalkanes were all planar, so cyclopropaner was thought to be a flat triangle, cyclobutane was thought to be a flat square, cyclopentane was thought to be a flat pentagon and cyclohexan was thought to be a flat hexagon. And in terms of analyzing them, the idea of angle strain was introduced, and angle strain is the increase in energy that's associated with a bond angle that deviates from the ideal bond angle of 109.5 degrees, and this number should sound familiar to you, this was the bond angle for a carbon of tetrahedral geometry. So if you go through and you analyze these, the bond angle in here for this triangle must be 60 degrees, and 60 degrees is a long ways off from 109.5 degrees meaning cyclopropane has significant angle strain. For cyclobutane, this angle would be 90 degrees, and 90 degrees is still a ways off from 109.5 degrees so cyclobutane also has a large amount of angle strain, although not as much as cyclopropane. For cyclopentane, this bond angle is 108 degrees, and 108 degrees is pretty close to 109.5 degrees, closer than it would be for cyclohexane; This bond angle is 120 degrees. And so the theory was, cyclopentane is the most stable out of the cycloalkanes, because this bond angle is closest to 109.5 degrees. However that conclusion doesn't hold up if you look at the heat of combustion of the cycloalkanes. And first we'll start with cyclopropane. So if you define the heat of combustion as the negative change in the enthalpy, cyclopropane gives off 2,091 kilojoules for every one mole of cyclopropane that is combusted, and if you count the number of CH2 groups on cyclopropane, let's go back here and let's count them up. So here's one, two, and three on the drawings; That's why there's a three here. Now you can't analyze the cycloalkanes in terms of just the heats of combustion. So if we look at those we can see that they increase. 2,091 to 2,721, to 3,291 and then 3,920. But that's what we expect to happen because as we go from cyclopropane to cyclobutane, -pentane, and -hexane, we're increasing in number of carbons and we already know from the earlier video on heats of combustion, if you increase the amount of carbons that you have, you'd expect an increase in the heats of combustion. So you can't really compare the cycloalkanes directly in terms of just the heats of combustion, you have to compare them in terms of their heats of combustion divided by the number of CH2 groups, and that gives you a better idea of the stability. So if you take 2,091, which is the heat of combustion of cyclopropane, and divide that by the number of CH2 groups, which is three, you get approximately 697. So again, this is the heat of combustion divided by the number of CH2 groups in kilojoules per mole. And this is a much better way to compare the stability of the cycloalkanes. Notice cyclopropane has the highest value here, 697. Cyclobutane goes down to 680, cyclopentane is 658 and cyclohexane is approximately 653. If you remember back to the video on heats of combustion, this number here, 653 kilojoules per mole, is approximately the same value we got for a straight chain alkane when you add it on a CH2 group, so each additional CH2 group increased the heat of combustion by approximately 653, or 654 kilojoules per mole, and that tells us that cyclohexane is pretty much strain-free, cyclohexane is about as stable as an open chain alkane, and so we know that this idea of cyclohexane being flat must not be true, so cyclohexane isn't flat as we'll see in later videos. So cyclohexane is the most stable out of these cycloalkanes. Cyclopentane is a little bit higher in energy and therefore a little bit more unstable, and cyclobutane even higher than that, and finally cyclopropane at 697 for a heat of combustion per CH2 group, this is the most unstable, this is the highest heat of combustion, this is the highest in energy, and so let's analyze why cyclopropane has such a relatively high heat of combustion per CH2 group. Here we have a model of the cyclopropane molecule. If I turn it to the side you can see that all three carbon atoms are in the same plane. So cyclopropane is planar. You can also see that these bonds are bent. The significant angle strains means the orbitals don't overlap very well, which leads to these bent bonds. You can see the plastic is even bending in the model set. There's another source of strain associated with cyclopropane and we can see it if we look down one of the carbon-carbon bonds, so the front hydrogens are eclipsing the hydrogens in the back, and cyclopropane is locked into this eclipse confirmation. All this increased strain means that a three-membered ring is very reactive, and highly susceptible to ring-opening reactions. So I'm gonna take the ring here, I'm gonna break it and open up the ring so we can see that decreased the strain, those bonds even look straight now. Here we have the cyclobutane molecule, and you can see there is some angle strain here although not as much as in cyclopropane. If we turn it to the side you also see some torsional strain, the hydrogens in the front are eclipsing the hydrogens in the back. To relieve this torsional strain, cyclobutane can adopt a non-planar confirmation, this is called the puckered confirmation. And if you turn it to the side here so you're staring down one of the carbon-carbon bonds, you can see how that's relieved some of the torsional strain so the hydrogens in front are no longer eclipsing the hydrogens in the back. Finally we have the cyclopentane molecule, which has much less angle strain than cyclopropane or cyclobutane but if you turn it to the side you can see that the planar confirmation is destabilized by torsional strain, so we have some eclipsed hydrogens there. Some of that torsional strain can be relieved in a non-planar confirmation, so one of the non-planar confirmations would be to rotate the carbon up like that, and that's called the envelope confirmation. And you can see that four carbons are in the same plane. So this carbon right here, this one, the one in the back and this one in the back are the same plane is this fifth one here is up out of the plane. So this looks a little bit like an envelope, and so that's why it's called the envelope confirmation. Now that we understand the ability of cycloalkanes, let's do a quick problem. On the left we have ethylcyclopropane, on the right we have methylcyclobutane. They're isomers of each other, they both have the molecular formula C5H10. The first question is which isomer is more stable. Well, we're comparing a three-membered ring to a four-membered ring, and we know that cyclopropane is higher energy, there's more strain associated with it. So ethylcyclopropane must be the less-stable isomer. So this one is less stable, which makes methylcyclobutane the more stable isomer, so there's not as much strain in methylcyclobutane. So I've answered our first question. Our second question is which one has the higher heat of combustion. We know that the more stable compound has a lower heat of combustion. We know that from the heat of combustion video. That must mean methylcyclobutane has the lower heat of combustion because this one is more stable. Which means that ethylcyclopropane must have the higher heat of combustion. So the higher heat of combustion. The higher heat of combustion is the one that's less stable.