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Why is [1.1.1] propellane more stable than other small members in the propellane family even though it seems to be under the greatest strain? What makes a small propellanes stable, or in general, any propellane stable?

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    $\begingroup$ Interesting question. Strangely, Wikipedia contradicts itself here: On the propellanes page it says: Surprisingly, the most strained member [1.1.1] is far more stable than the other small ring members ([2.1.1], [2.2.1], [2.2.2], [3.2.1], [3.1.1], and [4.1.1])., but on the page for [2.2.2] propellane it says This ([2.2.2] propellane) compound is unstable (although not as much as [1.1.1]propellane). The bond angles on the shared carbons are considerably strained: three of them are close to 90°, the other three to 120°. The strain energy is estimated to be 93 kcal/mol (390 kJ/mol).. $\endgroup$ Jul 1, 2017 at 15:07
  • $\begingroup$ Then what is the general consensus? $\endgroup$
    – AS_1000
    Jul 1, 2017 at 15:13
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    $\begingroup$ I think Carey A. F. Advanced Organic Chemistry: Part A: Structure and Mechanisms, 2000, p. 164 pretty much answers your question. $\endgroup$
    – andselisk
    Jul 1, 2017 at 15:13
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    $\begingroup$ @andselisk Can you write an answer for this. The link is not working on my phone. $\endgroup$ Jul 3, 2017 at 5:50
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    $\begingroup$ @Mockingbird Done, sorry for the inconvenience, that was a link to Google Books which can be wacky sometimes to get access to. $\endgroup$
    – andselisk
    Jul 3, 2017 at 6:25

1 Answer 1

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Fulfilling the request from @Mockingbird, there is a quote from (1, p. 90)

Surprisingly, [1.1.1]propellane is somewhat more stable to thermal decomposition than the next larger propellane, [2.1.1]propellane, indicating a reversal in the trend of increased reactivity with increased strain. To understand this observation, it is important to recognized that the energy of both the reactant and intermediate influence the rate of unimolecular reactions that lead to decomposition. In the case of propellanes, homolytic rupture of the central bond is expected to be the initial step in decomposition. This bond rupture is very endothermic for [1.1.1]propellane. Because relatively less strain is released in the case of [1.1.1]propellane than in the [2.1.1]- and [2.2.1]-homologs, [1.1.1]propellane is kinetically most stable.(2)

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Another manifestation of the relatively small release of strain associated with breaking the central bond comes from MP4/6-31G∗ calculations on the energy of the reverse ring closure.(3)

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The thermal decomposition of [1.1.1]propellane has been studied both experimentally and by computation(4)

The book chapter further discusses reactivity of [1.1.1]propellane, if OP is really interested in this area, I would suggest to grab the book in the library or focus on the papers (2-4) the book references to.

(1) Carey, F. A.; Sundberg, R. J. Advanced organic chemistry. Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
(2) Wiberg, K. B. Angew. Chem. Int. Ed. Engl. 1986, 25 (4), 312–322. DOI: 10.1002/anie.198603121
(3) Adcock, W.; Binmore, G. T.; Krstic, A. R.; Walton, J. C.; Wilkie, J. J. Am. Chem. Soc. 1995, 117 (10), 2758–2766. DOI: 10.1021/ja00115a011
(4) Jarosch, O.; Walsh, R.; Szeimies, G. J. Am. Chem. Soc. 2000, 122 (35), 8490–8494. DOI: 10.1021/ja994043v

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