As usual we are a bit late on this, but here is this month's molecule:

(4R,5S,6R)-4,6-Diacetylhygrophorone A12

which is known as (4​R,5​S,6​R)-4,6-diacetylhygrophorone A12, and is an antifungal natural product isolated from mushrooms of the genus Hygrophorus.

As far as I know this compound has not yet been synthesised in the lab, but has been isolated and its relative configuration assigned by NMR measurements in anisotropic media.[1]

The criteria are the same as for the last challenge I posted:

  • the synthesis must be enantioselective;[2] and
  • no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.)

Otherwise, any commercially available starting material (as usual, in the Sigma–Aldrich catalogue) is fair game.

(1) Schmidts, V.; Fredersdorf, M.; Lübken, T.; Porzel, A.; Arnold, N.; Wessjohann, L.; Thiele, C. M. RDC-Based Determination of the Relative Configuration of the Fungicidal Cyclopentenone 4,6-Diacetylhygrophorone A12. J. Nat. Prod. 2013, 76 (5), 839–844. DOI: 10.1021/np300728b.

(2) The absolute stereochemistry was not assigned in ref 1, so you may synthesise either enantiomer. Obviously, though, that doesn't make a difference to the design of the synthetic route, so I have arbitrarily decided to display this enantiomer (following ref 1).

  • 1
    $\begingroup$ I really want to find something that works with PLE... $\endgroup$
    – Zhe
    Dec 10, 2017 at 1:42
  • 1
    $\begingroup$ Is it fair game using known procedure for making an intermediate resembling the target and with most of the stereochemistry already setup? $\endgroup$
    – EJC
    Dec 18, 2017 at 0:04
  • 1
    $\begingroup$ @Marko Yes! Why not, especially if we can all learn something from it. $\endgroup$ Dec 18, 2017 at 0:17

2 Answers 2


Establishing the correct stereochemistry proved out to be an extremely difficult task, as I have found only one procedure for these systems. This first scheme is a published procedure: enter image description here

1) Diels-Alder reaction (Paul D. Bartlett, Fred A. Tate J. Am. Chem. Soc., 1953, 75 (1), pp 91–95);

2) Alpha-allylation and 3) ring closure by anionic addition to ester group to generate two regioisomers which are separated (Siwapinyoyos, Thebtaranonth, J. Org. Chem. 1982,47, 598-599);

4) Epoxidation of the non-conjugated enone and 5) opening of the epoxide by elimination (Chantarasiri, J. Chem. Soc., Chem. Commun., 1990,0, 286-288)

6) Kinetic resolution by enzymatic O-acetylation, followed by 7) hydrolysis of the acetate and vacuum flash pyrolysis at 450 0C and 0.1 mmHg. The acetate is needed in the target molecule, but I guess that it has to be hydrolyzed before the pyrolysis as to avoid side reactions. 8) is epoxidation with dimethyldioxirane and 9) opening of the epoxide with water provides a intermediate resembling the target molecule a lot (Klomkao, Tet. Asymm. 14, 24, 2003, 3885-3889).

From this molecule starts my approach: enter image description here

10) Acetonide protection of the diol should be selective (syn 1,2 hydroxyl groups react in the presence of anti hydroxyl groups and 1,3 hydroxyl groups). The free hydroxyl group is then acetylated.

11) The acetonide is hydrolyzed. The primary hydroxyl group is oxidized to the aldehyde and the tertiary one is protected as MOM ether.

The following 3 steps are used for protection of ketone in presence of aldehyde:

12) Aldehyde is protected in presence of enone first.

13) Enone is protected by Noyori conditions.

14) Aldehyde is deprotected.

15) Organocopper addition to the aldehyde in presence of the ester to form a mixture of diastereomers (surely it would show diastereoselectivity but this would need to be tried out).

16) If the diastereoselectivity is bad, the mixture of diastereomers is oxidized to the ketone which is then stereoselectively reduced by chelate-control with zinc-borohydride. This could be problematic as the acetonide oxygens could interfere with chelation, but I cannot predict this with confidence. If they don't interfere much the intermediate would look something like in the image attached at the bottom. The hydryde attacks opposite to the acetonide and the alkyl chain. If acetonide oxygens interfere, then a screening of reducing agents is required.

17) Acetylation of the hydroxyl group and MOM and acetal deprotection should provide the target molecule.

enter image description here

  • $\begingroup$ I wonder if I chose a molecule that was too tough... $\endgroup$ Dec 18, 2017 at 14:35
  • $\begingroup$ @orthocresol Seems so. The stereochemical relationship is difficult to establish. $\endgroup$
    – EJC
    Dec 18, 2017 at 15:56
  • $\begingroup$ I thought that the oxygens would provide a good handle to control stereochemistry. But I will admit, I haven't actually tried to come up with a route myself. $\endgroup$ Dec 18, 2017 at 16:24
  • $\begingroup$ @orthocresol when's the next challenge coming $\endgroup$
    – Laksh
    May 18, 2019 at 9:28
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    $\begingroup$ @Laksh I apologise, but I have been very busy in the previous months, too busy to do one of these. I will try to do one in June, possibly. $\endgroup$ May 18, 2019 at 11:16

A diastereoselective synthesis (total asymmetry obtained by a Sharpless kinetic resolution)

enter image description hereenter image description here

  • 2
    $\begingroup$ You have put in a tremendous effort into 3 Synthesis golfs (I noticed you had answered 2 similar questions recently). However, it will be much easier for the community to go through and vote for your answer were it slightly clearer. It would be great if you could use ChemDraw/ChemSketch (if you have them) or ChemDoodle to illustrate your reactions. I'm afraid I haven't got enough time right now, or I'd have done it myself. If you find it too difficult, you could wait for someone like @andselisk to come along and help. $\endgroup$ May 18, 2019 at 10:32
  • $\begingroup$ I would do it on ChemDraw and will at some point but don't have the time at the moment, sorry. $\endgroup$
    – Laksh
    May 18, 2019 at 10:37
  • $\begingroup$ Alright, I'll probably be free in 2 weeks, and if you couldn't find time to do it yourself yet, I'd be willing to do it. Thanks for replying! $\endgroup$ May 18, 2019 at 10:40

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