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Muconin

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123.713A/B. Description of the Jacobsen synthesis of muconin. This is an example of total synthesis, retrosynthesis and asymmetric synthesis and shows the kind of information required in the assigment for this course.

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Muconin

  1. 1. synthesis of muconin ©Reinaldo Aguilar@flickr
  2. 2. O HO O OH C12H25 O O OH This molecule is called muconin. This document will act as an example of what is expected in the 123.713A/B assessment and it will discuss the synthesis of muconin. A glance at the mark scheme reveals that you should describe the retrosynthesis or the overall plan before outlining a possible synthesis. The emphasis is on the creation of the stereocentres within the molecule so this synthesis will not involve any chiral pool materials. I will go into more detail than you need to as I will use this as an opportunity to reinforce some of the principles we have covered.
  3. 3. O HO O OH C12H25 O O OH The Retrosynthesis Ideally we want a convergent synthesis. This requires the target to be split in half. The alcohol functionality is the obvious point to split the molecule. C–C bond formation at this point is simple, it involves nucleophilic addition to a carbonyl group.
  4. 4. O HO O OH C12H25 O O OH C–C addition to aldehyde O O O OH C12H25 O O OH fragment A fragment B M We could disconnect the molecule on either side of the alcohol. Two factors favour the disconnection shown: 1. Ease of generation of the nucleophile. An alkyl organometallic reagent should be simpler to prepare than an anomeric nucleophile (although there are a number of methods for achieving this). 2. Chelation control. A heteroatom 𝛂-to the carbonyl might allow stereocontrol in the nucleophilic addition step. This gives two fragments.
  5. 5. O O O OH C12H25 O O O OH C12H25 O O O OH C12H25 hetero-Diels-Alder challenging Fragment A is by far the more challenging half of the molecule. It contains 5 contiguous stereocentres with 4 of them build into tetrahydrofuran and tetrahydropyran (THP) rings. The formation of a 6-ring is often possible by a variant of the Diels-Alder reaction. In this case a hetero-Diels-Alder (HDA) reaction. It would still leave some stereochemistry to be installed but would be a good starting point …
  6. 6. O O O OH C12H25 O O O OH C12H25 O O O OH C12H25 hetero-Diels-Alder challenging … formation of the 5-ring is more challenging. One might argue that formation of C–O bonds should be easy as we have a host of nucleophilic addition reactions available to us but this would be a very linear process (making each by addition to a carbonyl). Alternatively, we could consider installing some by Sharpless Asymmetric Epoxidation or Dihydroxylation. This might be productive train of thought (consider the work of Tim Donohoe). But we’re going to stick to the HDA as we have already mentioned Diels- Alder reactions.
  7. 7. The idea will be to install at least one stereocentre of the THP ring by a catalytic asymmetric HDA reaction. This means we need to disconnect the tetrahydrofuran ring from the THP ring. It might be possible to create the requisite C–C bond with control of two adjacent stereocentres by using a [3,3]-sigmatropic rearrangement such as a Claisen or Cope rearrangement. Such rearrangements involve 6 atoms in a cyclic transition state, 5 of the atoms have been highlighted above. The 6th must be on the tetrahydrofuran side … O O O OH C12H25
  8. 8. O O O OH C12H25 FGI FGI protecting groups alkene O PG2O O OPG1 C12H25 To prepare for such a rearrangement we require more functionality in the molecule. Technically the two alkenes above would allow a Cope rearrangement but that doesn’t simplify the problem for us; we still have two rings that would need to be joined by a C– C bond. Instead, only one of these alkenes is required for the rearrangement. The other has been incorporated to aid construction of the tetrahydrofuran ring.
  9. 9. O PG2O O OPG1 C12H25 O PG2O O OPG1 C12H25C=C RCM FGI O PG2O O OPG1 C12H25 OMeO O PG2O O OPG1 C12H25 OO 1 2 3 1 2 3 [3,3]-sigmatropic rearrangement Ireland-Claisen rearrangement ring-closing metathesis C=C bond formation is relatively simple. There are many different possibilities, Witting reaction, Peterson olefination, Julia olefination, McMurry coupling and metathesis. The latter is both powerful but mild. Things are now looking much better. We have two alkenes with terminals 6- atoms apart. This is perfect for a rearrangement reaction. But if we did this we would still have a challenging C–C bond to make. The alternative …
  10. 10. O PG2O O OPG1 C12H25 O PG2O O OPG1 C12H25C=C RCM FGI O PG2O O OPG1 C12H25 OMeO O PG2O O OPG1 C12H25 OO 1 2 3 1 2 3 [3,3]-sigmatropic rearrangement Ireland-Claisen rearrangement ring-closing metathesis … is shown in the left hand corner; the Ireland- Claisen rearrangement. This converts a C–O bond into a C–C bond (during the rearrangement we break a bond between 1 & 1 and form one between 3 & 3). Therefore we need to carry out a series of functional group interconversions (FGI) to instal an ester or in the forward sense we must be able to convert an ester into an alkene. This is relatively simple (as I said, formation of C=C bonds is not hard).
  11. 11. And now all the hard work pays off with a massive simplification of our problem. The Ireland-Claisen rearrangement has the power to convert an ester (easily formed) into a C–C bond while creating two contiguous stereocentres. More importantly the new target (the right hand side) looks simple to disconnect. We have two C–O bonds that can easily be removed (ester and acyclic ether). This leaves … O PG2O O OPG1 C12H25 OMeO C–C rearrangement O OPG2 O O O C12H25 OPG1 12 3 1 2 3 Ireland-Claisen
  12. 12. O OPG2 O O O C12H25 OPG1 hetero-Diels-Alder O OPG2 O O O C12H25 OPG1 substrate control … a dihydropyran that can be prepared by asymmetric HDA, a functionalised acetate and a chiral diol. The latter could be prepared by a number of routes; reagent or catalyst controlled crotylation of a greasy aldehyde or substrate controlled alkenylation of a chiral aldehyde. This shows there are many correct answers. All would get you the marks (as long as you could rationalise the stereochemical outcome).
  13. 13. O OPG2 O O O C12H25 OPG1 esterification alkylation 2x C–O C12H25 OPG1 OH Br Br O O OPG2 OH So this slide shows what units we require to prepare fragment A. Doesn’t look to challenging at all anymore?
  14. 14. C12H25 OPG1 OH C–C Cram-chelation control C12H25 OPG1 O M C12H25 OH OH FGI C12H25 O C–O Jacobsen's hydrolytic kinetic resolution The alcohol stereocentre could be formed by Cram Chelation control as long as we can set up the initial stereocentre. Here it comes from a chiral diol that will be formed using Jacobsen’s powerful hydrolytic kinetic resolution but there are many other routes to the same compound. Other possibilities include asymmetric epoxidation of the alkene by either Jacobsen or Shi chemistry. Sharpless Asymmetric Dihydroxylation might also give the diol.
  15. 15. O OPG2 OH FGI O OPG2 O C–O C–C HDA O OPG2 OTMS H3CO hetero- Diels-Alder The dihydropyran comes from substrate controlled reduction of a enone, which in turn is derived from catalytic asymmetric HDA. There are a number of catalysts that can mediate this transformation. We have seen an example of a chromium ‘half-salen’ version already. There are a number of diol-based organocatalysts that can also control this reaction (Rawal kick started this area).
  16. 16. O O OH fragment B FGI O O OPG3M Fragment B This fragment is considerably easier to prepare; there are only two stereocentres and plenty of functionality to play with. First we have to find a method of creating the nucleophilic fragment. This could be achieved with a halide and conversion to the Grignard but such functionality is hard to carry through an entire synthesis so we have masked it here as an alkyne.
  17. 17. The key to this fragment is recognising that one (and in fact both) of the stereocentres could be formed from a chiral epoxide (and hence an alkene if we want to go back to cheap achiral starting materials). An alcohol four carbon atoms from an carbonyl as shown above is readily formed by ring opening an epoxide with an enolate. So all we have to do is find reliable chemistry to instal the double bond of the lactone as that then frees up the enolate … O OPG3 O ( )4 1 2 3 4 1,4-diO enolate and epoxide
  18. 18. O OPG3 O ( )4 FGI O OPG3 O ( )4 PhS C–C OPG3 ( )4 O O ≡ PhS O ( )4 enolate alkylation Sulfoxides readily undergo syn-elimination so can be used to install the double bond. Such functionality would also increase the acidity of the 𝛂-protons making enolate formation easier. I haven’t used the sulfoxide for the other steps as it adds unnecessary stereochemistry to the system (giving us a problem with diastereomers) but it can be easily added at the last minute (oxidation). Alternatively, we could use palladium chemistry to add the double bond.
  19. 19. O OPG3 O ( )4 FGI O OPG3 O ( )4 PhS C–C OPG3 ( )4 O O ≡ PhS O ( )4 enolate alkylation This allows the epoxide-enolate disconnection. How to we make this epoxide?
  20. 20. O ( )4 C–C ( )3 O ≡ ≡ ( )3 O MgBr Cl organometallic It is readily prepared from epichlorohydrin (the chemistry of which we covered in the third year). This can either be bought (if we are cheating) or can be prepared. In this synthesis it will be prepared by Jacobsen’s hydrolytic kinetic resolution again (for two reasons; firstly I think this one of the methods used commercially and the second will soon become clear …)
  21. 21. O O PhS OH PhS H3CO O C–C H3CO O SPh OH ≡ ≡O H3CO O SPh C–O lactone formation epoxide opening The lactone fragment is again prepared by the simple ring opening of an epoxide with an enolate; standard undergraduate chemistry. The only difficulty is sourcing the chiral epoxide. Once again, this is so simple it is undoubtedly commercially available. But in the spirit of the assignment we should make it so we could use Shi or Jacobsen epoxide or, of course … Jacobsen’s hydrolytic kinetic resolution. And now for the synthesis.
  22. 22. C12H25 i. mCPBA ii. hydrolytic kinetic resolution C12H25 OH OH C12H25 O i. CH(OCH3)3, H+ ii. DIBAL-H C12H25 OMOM OH i. TEMPO, NaOCl ii. CH2=CHMgBr, MgBr2C12H25 OMOM OH Synthesis Starting with Fragment A. The first important step in this reaction is the Jacobsen hydrolytic kinetic resolution. This reaction takes a racemic mixture of epoxide and hydrolyses one enantiomer far quicker than the other. This results in a mixture of diol and epoxide being formed. As these are different molecules they are readily separated. The only major draw back of this methodology is that the maximum yield is 50%.
  23. 23. H R H O t-Bu t-Bu N Co O N t-Bu t-Bu L O H R O t-Bu t-Bu Co O t-But-BuOH2 N N H H H H H O t-Bu t-Bu N Co O N t-Bu t-Bu L O O t-Bu t-Bu Co O t-But-BuOH2 N N H H disfavoured The hydrolytic kinetic resolution involves the cobalt-salen-mediated addition of water to an epoxide. Extensive kinetic studies have revealed that the reaction is second order with respect to the salen catalyst and the model above has been proposed to explain the selectivity. It is thought that both the nucleophile and the electrophile are both activated by the cobalt and it is the fact that both are now chiral that leads to such high selectivity in the reaction. Science 1997, 936 & J. Am. Chem. Soc. 2004, 126, 1360
  24. 24. The salen ligand basically acts as a sloped floor (or ceiling) that blocks approach of a substrate to three sides (The tert-butyl groups block approach from the front. The aromatic ring blocks approach from the side that is sloped downwards while the axial hydrogen of the diamine hinders approach from behind). Only if the two ‘open faces’ of the catalyst meet can the nucleophile attack. Further more the orientation of the epoxide is controlled by minimising its interactions with the various substituents. H R H O t-Bu t-Bu N Co O N t-Bu t-Bu L O H R O t-Bu t-Bu Co O t-But-BuOH2 N N H H H H H O t-Bu t-Bu N Co O N t-Bu t-Bu L O O t-Bu t-Bu Co O t-But-BuOH2 N N H H disfavoured H OH OH R ≡ R OH OH
  25. 25. C12H25 i. mCPBA ii. hydrolytic kinetic resolution C12H25 OH OH C12H25 O i. CH(OCH3)3, H+ ii. DIBAL-H C12H25 OMOM OH i. TEMPO, NaOCl ii. CH2=CHMgBr, MgBr2C12H25 OMOM OH The next step involves a selective protection of a secondary alcohol in the presence of a primary alcohol. This is a challenging reaction; primary alcohols are frequently more reactive than secondary alcohols due to reduced steric hindrance. In this case the selectivity is achieved by first forming an orthoformate and then selectively cleaving one bond (see next slide). In all honesty in your assignments I will not be too picky about protecting groups as long as there are no screaming errors. If you have selected them it normally means you are following a paper (someone else picked them for you).
  26. 26. C12H25 O O OCH3 C12H25 O O OCH3 AlHR2 C12H25 O O OCH3 AlHR2C12H25 OMOM OH This the probable mechanism for the selective protection. Orthoformate formation is followed by reduction with DIBAL. The Lewis acidic aluminium coordinates to the most accessible oxygen atom. This activates this oxygen atom as a leaving group. The lone pair of electrons on a second oxygen atom kicks it out giving an oxonium ion that is reduced by the DIBAL. Remember in your assignments to add the mechanisms to interesting reactions. This will show you are thinking about the chemistry and not copying it from a paper. You will get higher marks.
  27. 27. C12H25 i. mCPBA ii. hydrolytic kinetic resolution C12H25 OH OH C12H25 O i. CH(OCH3)3, H+ ii. DIBAL-H C12H25 OMOM OH i. TEMPO, NaOCl ii. CH2=CHMgBr, MgBr2C12H25 OMOM OH Mild oxidation of the primary alcohol to an aldehyde is followed by Cram Chelation Controlled addition of an alkenyl Grignard reagent. Remember what half this module was about? Asymmetric synthesis so you must go through the selectivity …
  28. 28. C12H25 OMOM O O H C12H25 O H H C12H25 OMOM OH H Mg Br Br OCH3 BrMg C12H25 OMOM OH ≡ This is an example of substrate control and Cram Chelation control. There are two ways of representing this. One uses the Newman projection (shown right). First translate the skeletal representation into a Newman projection. Then tie the two Lewis basic oxygen atoms together. The nucleophile approaches along the Bürgi-Dunitz angle. Finally, take the new Newman projection and convert it back to a skeletal drawing. The second method uses the skeletal representation and redraws the chelate as a 5-membered ring and attacks from the least hindered face.
  29. 29. Simple alkylation then gives one half of the Ireland-Clausen rearrangement precursor. C12H25 OMOM OH I ONa O C12H25 OMOM O CO2H
  30. 30. The dihydropyran moiety is going to be prepared by a hetero-Diels-Alder reaction of a reactive, electron rich, diene (known as Danishefsky’s diene) and an aldehyde (the protecting group PBB is para-bromobenzyl). The reaction is catalysed by the chromium-salen complex on the right. The rgiochemistry is most readily explained by considering the polarity of the bonds. For the aldehyde the carbonyl carbon is δ+. For the diene, the electrons are being pushed towards the top (as drawn above; consider the resonance forms). Matching the δ+ and δ– together gives the regiochemistry shown. TMSO OCH3 O OPBB catalyst O TMSO OPBB OCH3 H O O OPBB N Ot-Bu t-Bu N t-Bu t-Bu Cr O BF4
  31. 31. H Cr O L O N N O H t-Bu t-Bu t-Bu t-Bu H OPBB TMSO OCH3 H H H The stereochemistry can be explained by considering the sloped shape of salen catalysts. The aldehyde coordinates to the Lewis acidic chromium. It is orientated so that the metal is trans to the aldehyde substituent. It is suspected that aldehyde points away from the axial hydrogen atom to minimise non-bonding interactions as shown. The diene cannot approach from the front due to the tert-butyl groups. It can’t approach from the back due to the axial hydrogen and it can’t approach from the left (re) face of aldehyde due to the slope. So it must approach the si face as shown.
  32. 32. O H OPBBTMSO OCH3 H H H O OCH3 H TMSO H H H OPBB Approach from the si face is shown above and the resulting product has the correct stereochemistry at the CH2OPBB group. I have draw the endo transition state above but this is ultimately unimportant. The endo/exo selectivity only influences the methoxy stereocentre and as we shall see on the next slide this is removed …
  33. 33. O O OPBB O O O OPBB CH3H (H3C)3Si Treatment of the cycloaddition product with acid results in β-elimination of the methoxy group to give the enone as shown above. The precursor for the Ireland-Claisen rearrangement requires selective 1,2-reduction of the enone …
  34. 34. O O OPBB NaBH4, CeCl3 O HO OPBB O H OPBB O H vs. O H O H OPBB Classic Luche reaction conditions (NaBH4, CeCl3) result in exclusive reduction of the carbonyl with no 1,4-reduction. The origin of this chemoselectivity is described in the the notes for this course. Arguably the CeCl3 is a selective Lewis acid for the methanolysis of NaBH4. The resulting reagents are harder reducing agents. The diastereoselectivity is rationalised by axial attack (minimises torsional strain) on the more stable half-chair conformation. The hydride approaches from the least hindered face.
  35. 35. O HO OPBB C12H25 OMOM O CO2H O OPBB O O O C12H25 O MOM The precursor is formed by simple esterification.
  36. 36. O OPBB O O O C12H25 O MOM LDA, TMSCl, HMPA O OPBB O OTMS O C12H25 O MOM Substrate control (the existing stereocentre of the dihydropyran) only controls one of the two new stereocentres. The other is controlled by the geometry of the enolate as we shall see on the next slide). Normally, enolate (silyl ketene acetal to be accurate) formation with an ester favours formation of the E(OSi)-enolate (as shown by the Ireland model). This is due to the fact the ester (the alkoxy moiety) counts as a small substituent as the R group (the dihydropyran) can rotate out the way).
  37. 37. O OPBB O O O C12H25 O MOM LDA, TMSCl, HMPA O OPBB O OTMS O C12H25 O MOM In this example we need the Z(OSi)-enolate. Fortunately this is readily achieved by the addition of HMPA (hexamethylphosphoramide) to the reaction. This highly Lewis basic molecule coordinates the lithium and breaks up any aggregates or coordination. The reaction no longer can be modelled by a Zimmerman-Traxler-like 6-membered transition state and it gives the desired Z(OSi)-enolate. On warming (–78°C to rt) the rearrangement proceeds. It does not go through a chair- like transition state.
  38. 38. O O H O OTMS OPBB H R O O TMSO O H H OPBB R H H vs. O O TMSO O H H OPBB R H ≡O H O PBBO OTMSO R As you can see, the chair- like transition state suffers from considerable non-bonding interactions. Effectively the OTMS and ether groups suffer unfavourable 1,3-diaxial interactions with the dihydropyran ring. If the reaction proceeds through a boat-like transition state instead then we minimise these unfavoured interactions as shown on the right. The reaction then proceeds to communicate one stereocentre into two new stereocentres (with destruction of the original) as shown.
  39. 39. O H O PBBO OTMSO R O PBBO O OMOM C12H25 OMeO CH2N2 DIBAL O PBBO O OMOM C12H25 O i. CH2=PPh3 ii. TMSBr iii. TBSCl O PBBO O OTBS C12H25 Standard functional group transformations give the diene required for ring closing metathesis and formation of the tetrahydrofuran ring. Mild esterification with diazomethane gives the methyl ester. Low temperature reduction with DIBAL converts the ester into an aldehyde. Wittig reaction to give the alkene is followed by a change in protecting group.
  40. 40. Ring-closing metathesis forms the tetrahydrofuran ring. Hydrogenation removes the two alkenes and deprotects the primary alcohol. Selective oxidation with Dess- Martin’s periodinane gives Fragment A. O PBBO O OTBS C12H25 i. RCM ii. H2, Pd/C iii. Dess-Martin periodinane O O O OTBS C12H25 O I O OAc OAc AcO
  41. 41. Cl O N Ot-Bu t-Bu N t-Bu t-Bu Co O OAc H2O Cl O Cl OH HO MgBr TMS ( )4 Cl OH TMS ( )4 i. NaOH ii. TBAF ( )4 O ( )4 OTBS I i. LiI ii. TBSCl Fragment B The synthesis of the fragment B is more straight forward. Epichlorohydrin is commercially available as a racemic mixture of enantiomers. Another example of Jacobsen’s hydrolytic kinetic resolution permits isolation of the pure enantiomer we require. The selectivity was described earlier. After this it is standard functional group transformations. Opening of the more electrophilic epoxide (oxygen is more electronegative than chlorine and ring strain) gives the alcohol.
  42. 42. Cl O N Ot-Bu t-Bu N t-Bu t-Bu Co O OAc H2O Cl O Cl OH HO MgBr TMS ( )4 Cl OH TMS ( )4 i. NaOH ii. TBAF ( )4 O ( )4 OTBS I i. LiI ii. TBSCl Treatment with base forms a second epoxide. Ring-opening with iodide and protection of the alcohol to prevent reformation of the epoxide gives the iodide above.
  43. 43. O N Ot-Bu t-Bu N t-Bu t-Bu Co O OAc H2O O OH HO CO2HPhS LDA CO2HPhS OH H O PhS O ( )4 OTBS IO OTBS O ( )4 PhS The lactone is readily formed from an enantiomerically pure epoxide. Yet again, this epoxide can be accessed via non-selective epoxidation of an alkene follow by kinetic resolution. The enolate formation and epoxide ring opening gives the chiral secondary alcohol, which is readily cyclised by treatment with acid. Formation of a second enolate permits alkylation with the iodide formed on the previous slide.
  44. 44. O OTBS O ( )4 PhS O O O OTBS C12H25 i. Cy2BH ii. ZnEt2 O OTBS O ( )4 PhS EtZn O O OTBS O HO O OTBS C12H25 PhS Now we can join the two fragments together. Hydroboration of the alkyne gives an alkenylborane. Transmetallation converts the borane into an organozinc species. Addition of this reagent to the Fragment A aldehyde should give the desired alcohol. The diastereoselectivity can be rationalised by chelation control as shown on the next slide.
  45. 45. HO O H H O O H Zn HO nucH OH O HO O OTBS C12H25 R nuc As before there are two methods to predict the stereochemistry. I prefer the use of the Newman projection but it can be achieved with the skeletal representation. First draw the Newman projection then chelate the two Lewis basic oxygen atoms together. The nucleophile then attacks along the Bürgi-Dunitz angle. Then we have to unravel this representation. In theory we get the desired diastereomer (in practice the selectivity is quite poor).
  46. 46. The final steps are shown here. First, hydrogenation removes the unwanted alkene. Next we have to install the desired alkene in the lactone ring. This is achieved by oxidation of the sulfide with mCPBA to give the sulfoxide. Heating promotes syn-elimination and formation of the alkene. Potentially we could form an endo or exo alkene; both are in conjugation with the lactone. Both are trisubstituted. Arguably, reduced conformational freedom in the ring encourages elimination in the ring and possibly increases the degree of hyperconjugation (holds the C–H bond in place). So I think (and experiment agrees) that the desired product will be favoured. Acid then removes the protecting groups and finishes the synthesis of muconin. O O OTBS O HO O OTBS C12H25 PhS i. H2, Pd/C ii. mCPBA iii. heat O O OTBS O HO O OTBS C12H25 H O HO O OH C12H25 O O OH
  47. 47. Hopefully, this chemistry gives you more insight into planning total syntheses and asymmetric synthesis. Hopefully it also acts as an example of what is expected in your assignment. I confess I would not give myself full marks. This synthesis is more or less directly taken from a single paper (J. Org. Chem. 1998, 63, 4876 … by Jacobsen, hence the use of his methodology in many steps). Higher marks will be awarded for originality (designing your own) or from at least assimilating information from two or more syntheses. My excuse? I use this synthesis to cover other material and not just to act as an example of the assignment. O O OTBS O HO O OTBS C12H25 PhS i. H2, Pd/C ii. mCPBA iii. heat O O OTBS O HO O OTBS C12H25 H O HO O OH C12H25 O O OH
  48. 48. Notes based on the 1999 postgraduate course on Retrosynthesis delivered by gjr at the University of Sussex, UK. Overhauled in 2015 while listening to Neu! & Neu! 75 by Neu!, Maniac OST by Rob., The Leftovers OST & The Congress OST by Max Richter and Strategies Against Architecture IV by Einstürzende Neubauten. Hope it is of use to somebody. gareth
  49. 49. O O OH OH O H H AcO O O O AcO O O O OH O H H OCH3 HO O O O Ac O OHHN OO O O O HO HO OH O HO HO O O O O O OH OH H H
  50. 50. O O OH OH O H H 6 AcO O O O 2 AcO O O O OH 8 O H H OCH3 HO O O O Ac 5 O OHHN OO O 1 O O HO HO OH 7 O HO HO O O O 3 O O OH OH H H 4

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