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Homework Help: Most Exothermic Combustion of Hydrocarbon

  1. May 14, 2014 #1


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    1. The problem statement, all variables and given/known data

    Which would have the most negative heat of combustion? Consider only the best conformer.



    2. Relevant equations

    The more negative the heat of combustion, the more exothermic the combustion.

    We are given trans and cis configurations here. This is a sign to pay attention to steric hindrances.

    3. The attempt at a solution

    Okay, I attempted drawing out the cyclohexanes. I know that the chair conformation is the most stable cyclohexane conformation so I drew out two chairs.

    I know that cis means on the same size so I first drew the chair for cis-1,3-dimethylcyclohexane with two axial methyl substituents both pointing up. This however would create 1,3 diaxial interactions (i.e. crowding with the bulky methyl groups). However, depending on how the ring is numbered, one can have an up equatorial methyl group and an up axial equatorial methyl group. This would eliminate the strain caused by placing methyl groups too close too each other (steric strain).

    Similarly in trans-1,4-dimethylcyclohexane, we can have an up axial methyl group and a down axial methyl group. Or if we renumber the ring as to get the most stable/best conformer, we can have a down equatorial methyl group and an up equatorial methyl group. This conformation similarly lacks steric strain.

    So wouldn't the heats of combustions be about the same?
  2. jcsd
  3. May 24, 2014 #2
    "on the same size" ...
    Will the heats of combustion be "about" the same? Yes. Obviously. And they'll be "about" the same as any other C8H16 species.
    Re-numbering of a structure does NOTHING to the conformational energies. I hope you didn't really mean that.

    In analyzing conformational energy, you chose the most stable conformation of cyclohexane because you knew what it was. Hmmm. Do you know the story of the drunk and the streetlight? https://en.wikipedia.org/wiki/Streetlight_effect
    Your problem is to compare the lowest energy conformations of those two isomers. You seem to understand that. The issue you seem to be having is reducing that to practice.
    Here are some hints:
    Does adding one methyl to cyclohexane change the ground state conformation? In what way? (to answer this, requires you to understand that molecules lowest energy conformation is unlikely to be the ONLY contribution to the ground state (ie that molecules wiggle), but identifying the ground state is a great first step and in doing so, you will have evaluated some of the other possibilities.)
    What happens when you add a second methyl group? How significant are the Me-ring interactions (strain)? How about the Me-Me interactions?
    Once you have identified the ground state, you'll have to compare the differences you expect in terms of their contribution to the total energy.
  4. May 24, 2014 #3


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    Flipping the chair of course changes how the ring is numbered and the axial/equatorial configurations. I hope you know what you're talking about.

    Last edited: May 24, 2014
  5. May 24, 2014 #4
    Look at your picture again, flipping the ring changes the equatorial to axial and axial to equatorial but it does NOT change the numbering of the ring carbons nor does it change the "up" or "down" assignment. Notice how the hydroxyl is always on carbon #1 and is always "down" although the flip changes it from axial to equatorial. Same for all other substituents.

    I do agree with your analysis though, semantics aside, it doesn't seem very easy to compare the two structures. I might be forgetting some detail or other having to do with distances of substituents or something. But I'm getting the most stable conformer containing two equatorial methyl's for the trans-1,4 case and the most stable conformer having two equatorial methyl's for the cis-1,3 case. Are you sure you aren't supposed to compare cis-1,3 vs trans-1,3 or cis-1,4 vs trans-1,4?
    Last edited: May 24, 2014
  6. May 24, 2014 #5


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    You're right! I missed that the cis isomer had both methyls in equatorial positions; I don't know how I overlooked that. Thanks!
    Last edited: May 24, 2014
  7. May 25, 2014 #6
    So did you settle on a final answer? Has your instructor mentioned anything?

    I'd dig around for actual experimental measurements if I were you, otherwise I'd go with a simplified analysis of axial interactions and call it about even.

    EDIT: Found this trying to google for the measured enthalpies of combustion of the molecules you have listed. Check out the last question and posted answer. Not exactly peer reviewed, but it appears that atleast one O-chem lecturer agrees with your analysis. I sure wish lecturers were forced to site their sources for measurements given on an exam :smile:.
    Last edited: May 25, 2014
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