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I came across an article in my organic chemistry textbook, which told about how benzyne was trapped within a molecular container and stabilized for short periods of time before undergoing a Diels-Alder reaction with the molecular container.

Obviously the container molecule must have had double bonds, which caused the powerful dienophile benzyne to react with it. While thinking about fully saturated molecular containers, dodecahedrane came into my mind:

enter image description here

The volume enclosed inside a dodecahedrane molecule was found to be approximately $2.8 \times 10^7 \text{pm}^3$. If I approximate the dodecahedrane to be a sphere, I get the diameter of the dodecahedrane to be approximately $5970 \text{pm}$.

Benzyne was conveniently prepared by diazotizing o-anthranilate, followed by heating simultaneously removes both $\ce{N2}$ and $\ce{CO2}$, and creating the benzyne molecule:

I am unsure on exactly how to find how much transverse space an anthranilate ion would occupy when stuffed into a dodecahedrane, but benzene itself is $496 \text{pm}$ wide, so I can suspect anthranilate maybe a bit wider, but nontheless can easily fit inside dodecahedrane.

And another advantage is, since both $\ce{N2}$ and $\ce{CO2}$ are linear molecules, they can easily escape through the pentagonal faces, leaving the benzyne trapped inside.

Edit: I made a mistake in calculating the diameter of the dodecahedrane. I now get it as $377 \text{pm}$, which is slightly smaller than a benzene ring width, so this compound won't work. 😭 Still if we used bigger molecule, it could fit in. Can this be done? Will this work? Are there any complications involved?

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    $\begingroup$ All this calculation, and you could find out it's too small just by counting bonds. $\endgroup$
    – Mithoron
    Commented May 27, 2017 at 16:20
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    $\begingroup$ Dodecahedrane is way too small to contain anything. $\endgroup$
    – Ivan Neretin
    Commented May 27, 2017 at 16:39
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    $\begingroup$ I'm also not sure how you did your calculation for the diameter. Did you square root instead of cube root? $V = 4\pi r^3/3$ $\endgroup$
    – orthocresol
    Commented May 27, 2017 at 16:42
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    $\begingroup$ Happens to the best of us. $\endgroup$
    – orthocresol
    Commented May 27, 2017 at 16:46
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    $\begingroup$ @IvanNeretin Actually it can contain He en.wikipedia.org/wiki/Dodecahedrane#Encapsulating_atoms $\endgroup$
    – Mithoron
    Commented May 31, 2017 at 19:58

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