1 (495)

Esta é uma pré-visualização de arquivo. Entre para ver o arquivo original

492 Organic Chemistry Solutions Manual Ho H N NH + Ho pyridoxal N Ho 0 H P Ho Ho H H P N N + N N Ho Ho + Ho H Ho Ho Problem 6 Concentrate now on the biosynthesis of scytalone in the first problem. You should have Pur identified it as a pentaketide. Now consider how many different ways the pentaketide chain might be folded to give scytalone. OH 0 Sup 0 0 0 ? scytalone have Purpose of the problem An exercise in the folding of the polyketide chain to make a simple pentaketide. the Suggested solution 0 0 This is not as obvious as it seems. At first, the simplest thing to do is to restore the carbonyl groups from the benzene ring and break the central bond so that we see the fifth carbonyl group clearly 0 (margin). We might then consider that the remaining OH group must have been formed by oH reduction of a carbonyl group and that the ring must have been closed by some sort of Claisen ester condensation. We can put the CO₂H group at any of the five sites so that gives us five possibilities. 0 0 0 0 0 CO₂H 0 0 0 0 OH OH 0 0 oH However, scytalone is not symmetrical - the right and left halves are very different so it matters which way round we wind the chain. This gives us another five possibilities. 0 0 0 CO₂H 0 0 Pur CO₂H 0 0 OH 0 OH OH 0 The folding of the polyketide Finally, and most easily overlooked, we don't have to wrap the chain round the outside of the two chain into scytalone was studied by rings - we can twist it through the middle so that the central bond is not made in the cyclization. double labelling by U. Sankawa et al., Tetrahedron Lett., 1977, 483; This gives us two more possibilities. There are 12 ways to fold the skeleton so that the results could only 487. be distinguished by the labelling pattern (particularly double labelling, p. 1426).