SolutionsA Complete Case: the Epothilones
1. Degrees of unsaturation
Epothilone C has the molecular formula C26H39NO5S. Calculate its degrees of unsaturation and distribute them among the groups you know from its structure. How does the result differ from that of epothilone A, C26H39NO6S?
(2 × 26 + 2 + 1 − 39) / 2 = 8. They are distributed as follows: three in the thiazole ring (the ring and its two double bonds), one in the side-chain double bond, one in the C12=C13 double bond, one in the lactone, one in the ketone and the eighth in the macrocycle.
Epothilone A has one more oxygen and the same number of hydrogens, so its index is also 8: oxygen does not count in the formula. What changes is the distribution, because in A the epoxide takes the place occupied in C by the double bond. A good reminder that the degree of unsaturation constrains but does not decide.
2. Choosing the technique
Say which technique you would use, and why, to settle each question: a) whether the two carbonyls are of the same kind; b) how many methyl groups there are; c) whether the ester is cyclic or open-chain; d) which of the two possible enantiomers is the natural one.
- IR: the ester and the ketone absorb at different frequencies and give two separate bands in the carbonyl region.
- DEPT: it classifies the carbons and counts the CH3 groups directly; the proton spectrum confirms this by the integration at high field.
- HMBC: if the proton on the carbon bearing the ester oxygen correlates with a carbonyl at the other end of the chain, the ester is cyclic. Neither IR nor the proton spectrum distinguishes the two possibilities.
- X-ray diffraction with a heavy atom, or a Mosher derivative; degrading the molecule and comparing a fragment with a standard of known configuration also works.
3. Why COSY is not enough
In epothilone B, COSY chains the protons into four loose pieces and no more. Explain why the chain breaks at each of the points where it does, and which experiment allows the pieces to be joined.
COSY correlates protons that couple to each other, so the chain breaks wherever there is no hydrogen to continue it: at the quaternary carbon bearing the two methyl groups, at the two carbonyl carbons (which have no hydrogen) and at the epoxide carbon bearing the methyl group, also quaternary.
The pieces are joined with HMBC, which correlates each proton with the carbons two and three bonds away: in this way the carbons without hydrogen are “jumped over”. It is also the experiment that proves the lactone is cyclic.
4. The resin in the fermenter
Why is an adsorbent resin added to the fermenter instead of extracting the broth at the end? Give three distinct advantages.
- The product is removed from the aqueous medium as soon as it is formed and stops inhibiting the bacterium that makes it, so the fermentation yields more.
- Adsorbed on the resin, the compound is protected from hydrolysis and from the enzymes in the broth.
- The resin is filtered off and eluted with little solvent: the product reaches the first purification step already concentrated, and evaporating hundreds of litres of water is avoided.