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I'm checking some curious syntheses and found cubane exists so decided to do something with it. The result is this thing:

enter image description here

According to the simulator I used these would be the properties:

Molecular weight 366.50 g/mol
Formula C26H26N2
Synthetic accessibility 4.69 (1 is very easy and 10 is very difficult)
Solubility in water between $\pu{1.21-17.0 mg/ml}$ ($\pu{3.30e-3 - 4.63e-2 mol/l}$) according to method.

Now I know that this is the “normal” way to make cubane and that at one point I get a carboxylic acid from cubane which I understand is the molecule used as a starting point to synthesize things containing cubane in their structure:

enter image description here

(One of the last steps of the cubane synthesis, at this point only a decarboxylation with cumene should be performed)

What would be the reactions necessary to synthesize the molecule I designed starting from either cubane or its carboxylic acid?

SMILES: N[C@H](C12C3C4C2C2C1C3C42)[C@H](C12C3C4C2C2C1C3C42)NC12C3C4C2C2C1C3C42

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  • $\begingroup$ "According to the simulator I used..." which one, aizynthfinder? $\endgroup$
    – Buttonwood
    Commented Nov 19 at 20:42

1 Answer 1

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I have rotated the lower half of your structure by 180° about the central C-C bond to provide a different perspective as shown in the red box (structure 1). Cubane carboxylic acid 2 is converted to cubylamine 3 via Hofmann rearrangement while the acid 2 is also transformed by reduction and anhydrous oxidation to cubane carboxaldehyde 4. McMurry[1] self-coupling of 4 leads to alkene 5 presumably of the trans-configuration given the bulk of the cubane moieties. Peracid oxidation of achiral alkene 5 forms racemic epoxide 6, which undergoes ring opening with sodium azide affording hydroxy azide 7.
Treatment of the azide with triphenylphosphine forms an N-cubyl phosphanimine with loss of nitrogen. This intermediate reacts further to form aziridine 8 and triphenylphosphine oxide.[Nice mechanism problem!]. Note that both carbon atoms of epoxide 6 have undergone inversion of stereochemistry in forming aziridine 8.[2] Finally, opening of the aziridine ring in 8 with cubylamine 3 provides the 1,2-diamine 1.[3]

  1. J. E. McMurry and M. P. Fleming, J. Am. Chem. Soc., 1974, 96, 4708.
  2. D. Tanner, C. Birgersson And H. K. Dhaliwai, Tetrahedron Letters, 1990, 31, 1903.
  3. R. Mueller, R. Gust, H. Schoenenberger and U. Klement, Chemische Berichte, 1991, 124, 2381.
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  • $\begingroup$ Man, it's a very complete answer, thank you, I guess now I have to read the bibliography you left, find out what the products of each reaction are soluble in and how to purify them and start trying to synthesize this thing. $\endgroup$ Commented Sep 10 at 13:53

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