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Does DIBAL-H reduce alcohol? I was reading an article by Rentsch and Kalesse [1] and I came across this reaction, but could not figure out how the final compound is formed:

Synthesis of 4

Reference

  1. Rentsch, A.; Kalesse, M. The Total Synthesis of Corallopyronin A and Myxopyronin B. Angew. Chem. Int. Ed. 2012, 51 (45), 11381–11384. DOI: 10.1002/anie.201206560.
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    $\begingroup$ Dibal-H is reducing the ester to the allylic alcohol which is oxidised to the aldehyde by Manganese Dioxide. $\endgroup$ – Waylander Dec 26 '20 at 19:52
  • $\begingroup$ The highlighting in your image seems to suggest you are stuck on a different step (c) than the DIBAl-H reduction (d). Could you clarify which step you are stuck on: (c) or (d)? If you understand (a) and (b), what do you think the structure is after (b)? $\endgroup$ – Ben Norris Dec 26 '20 at 21:58
  • $\begingroup$ i understand step (a) follows The Mitsunobu reaction, and steps (d) and (e) gives the aldehyde moiety in the final product, im stuck on how the C=C is formed in steps (b) and (c), the article says sulfer is oxidized to sulfone.Then, KHMDS deprotonate the carbon next to the sulfur and reacted with acetaldehyde $\endgroup$ – sarah.V Dec 27 '20 at 10:53
  • $\begingroup$ thank you, i guess im not understanding how the C=C is formed then and what would be the structure of step (b) and (c) $\endgroup$ – sarah.V Dec 27 '20 at 11:04
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    $\begingroup$ For the early steps, look at the Julia Olefination reaction: organic-chemistry.org/namedreactions/… $\endgroup$ – user55119 Dec 27 '20 at 15:35

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