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So I was reading about polyhedral hydrocarbons, because I dig the topic, and I read (admittedly on wikipedia entries... But they cited papers, I swear!) that prismane is explosive due to the massive ring strains of its faces, but cubane isn't, even though it has ring strain as well in its 90-degree edges.

The explanation that I read (this time on the appropriate paper) was that the symmetry break required for cubane to react leads to high activation energies for the transition states, which kinetically traps the molecule and prevents spontaneous reaction with the oxygen in the air. The result is that cubane, despite its ridiculous energy content, is remarkably stable.

Now... Prismane looks pretty symmetrical to me as well and I guess it has less symmetry than cubane, because prisms belong to lower-symmetry point groups than cubes. But is that the true explanation for why prismane is explosive? Higher energy content but lower symmetry?

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    $\begingroup$ Presumably you mean the C6H6 prismane not one of the larger prismanes. in which case the more obvious explanation is the much higher ring strain of carbons with 60° angles compared to the 90° angles in cubane. Symmetry doesn't seem like a good explanation for the barriers to reaction. $\endgroup$
    – matt_black
    Commented Jun 30, 2023 at 9:20

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According to Wikipedia, "Once formed, cubane is quite kinetically stable, due to a lack of readily available decomposition paths." In other words, the first step in decomposition requires a fair amount of energy in all available paths -- it's trapped in a "well" at the top of a hill, with no easy path out of that well.

Consider $\ce{H2 + Cl2 -> 2HCl}$. The $\ce{H2}$ and $\ce{Cl2}$ require initial energy to break their bonds to react.

Think, perhaps, not in more ways cubane is symmetrical, but rather that there is no easy path through the activation barrier. All it takes is one easy path out, and decomosition could take place. In the same way, diamond is metastable at STP... but there's no quick way out of that state.

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    $\begingroup$ Yes, I understand that. The reason why these paths are blocked is, according to the literature, a large loss of symmetry from the electron density of the cubane. Now, prismane is also very symmetrical but, unlike cubane, it is aggressively explosive. I am puzzled by that. Shouldn't prismane also benefit from high-symmetry decomposition blockages? $\endgroup$
    – urquiza
    Commented Jun 29, 2023 at 16:51
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    $\begingroup$ Cubane is symmetric in more number of ways than the number of ways in which prismane is symmetric. I suppose that is why. $\endgroup$
    – Proscionexium
    Commented Jun 29, 2023 at 17:33
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    $\begingroup$ @urquiza what Proscionexium suggests is interesting - "number of ways in which prismane is symmetric' could be the basis of a follow-up question. $\endgroup$
    – uhoh
    Commented Jun 30, 2023 at 0:12
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    $\begingroup$ @Mithoron When asked well, "amirite" questions can certainly be perfectly valid Stack Exchange questions. You do some reading, then some thinking, try to work it out, come to a tentative conclusion, then vet your work in Stack Exchange. Happens - and is well received - all the time. Sometimes they're so right that there is not much left to answer. Do you see anything actually missing here that needs to be said? $\endgroup$
    – uhoh
    Commented Jun 30, 2023 at 0:28
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    $\begingroup$ @uhoh That's not "amirite" question - you need to go further than OP to answer - how does symmetry or whatever give cubane high Ea of decomposition and it doesn't happen for prismane. $\endgroup$
    – Mithoron
    Commented Jun 30, 2023 at 12:54
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I suppose, the main reason prismane explosive, is bond angles, less natural for carbon.

As about symmetry, think about it that way: in cubane all C-C bonds and their energy levels are equal. So any disturbance distributes through the structure freely, as in model of a cube made of springs.

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    $\begingroup$ You mean to say that all C-C bonds are equivalent and degenerate in cubane which is not so in the concerned prismane? $\endgroup$
    – Proscionexium
    Commented Jun 30, 2023 at 11:21
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    $\begingroup$ I don't specifically meaned the word, but probably. I had in mind rather classical physics picture, consider two strings: |-----------------------| , |-------------=====| First one is symmetric, second not. Waves (of energy) could travel one string freely, other not. $\endgroup$
    – Alexey Birukov
    Commented Jun 30, 2023 at 13:13

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