Lecture 14 Notes April 21, 2011 We are going to finish the synthesis of indolizomycin today, and also talk briefly about the material that will be covered on your third test. We stopped last time at this bicyclic intermediate: O
N
H
This then reacts with Me3OBF4, which is basically just a source of methyl cation, to generate the iminium intermediate that is then reduced with sodium borohydride (NaBH4). The resulting amine product can be protected with the acyl choride to generate a cationic nitrogen intermediate. This intermediate then undergoes a fragmentation reaction to break open the six‐five fused ring system. The protecting group on the nitrogen is abbreviated “TEOC.” It turns out to be very useful because when you deprotect it, it decomposes into mostly gaseous side products. Now we need to convert the alpha, beta‐unsaturated ketone into an epoxide/ protected alcohol: We will go through these steps in detail as well. First, the alpha,beta‐unsaturated ketone is treated with hydrogen peroxide in base. This generates the peroxide anion, OOH‐, which acts as a nucleophile and attacks the beta position. The resulting intermediate collapses to kick off OH‐ and generate the epoxide. This is good, although we ultimately want the epoxide in the other position. Basically, what you have done here is generate an epoxide where you had a double bond. However, you can’t directly add mCPBA or a standard epoxidizing agent to do this reaction, because those reagents are generally electrophilic and require a nucleophilic double bond. Our double bond is part of an alpha,beta‐
unsaturated ketone, and therefore acts very differently. Also in this case, you have an isolated double bond that would be more likely to react with a standard epoxidizing reagent. After that, you treat the ketone with hydrazine (NH2NH2), and that generates an imine‐type species. There is also acetic acid in the reaction mixture, so that will rearrange in a way that breaks open the epoxide ring. The acetate anion then removes a proton from the nitrogen, and you lose N2 (nitrogen gas) via a Wharton‐fragmentation reaction. Now you can treat the double bond with mCPBA. In this case, the OH group is in close proximity to the double bond, so it directs the reagent to that double bond as opposed to the isolated terminal bond. Also the OH group helps to direct the stereoselectivity of the epoxidation. It actually forms a complex with mCPBA directly, and that blocks the bottom face and causes the oxygen atom to be delivered from the top face. TBSOTf is used to protect the free OH group, and that gives you another key intermediate. The structure of mCPBA (meta‐chloroperbenzoic acid is shown below): In general, an 8 membered allylic alcohol will undergo epoxidation to form the epoxide from the opposite side of the double bond: Next in the synthesis of indolizomycin is to convert the double bond into a terminal aldehyde. This was actually done by a modification of the Wittig reaction that also added an additional carbon, so that the final product contained an alpha,beta‐unsaturated aldehyde. We are now ready for the Julia olefination reaction. This reaction gives an intermediate sulfone‐acetate product, that is then reduced with sodium metal (no mechanism!) to give the desired triene. In the final stages of the synthesis, the TBS group is deprotected with periodic acid and then oxidized to a ketone with TPAP (this particular oxidation is called a Rubottom oxidation). After that, the TEOC protecting group on the nitrogen is removed by treating it with fluoride. The fluoride attacks the silicon, and the rest of the protecting group fragments into ethylene gas (CH2=CH2) and carbon dioxide. This leaves a negative charge on the nitrogen atom. The nitrogen attacks the ketone, which re‐forms the six‐five ring system, and generates the OH group. This particular synthesis was a racemic synthesis of indolizomycin, and the OH group is 50% up and 50% pointed down. Now a few notes on your third test: You should NOT memorize the total syntheses that we have covered. I will ask you other questions that require you to understand the chemistry that we have learned in the context of these total syntheses. For example, you will need to know that the OH near a double bond directs mCPBA to oxidize that particular double bond with good stereocontrol. There will be no mechanisms on this test. The test will contain mostly retrosynthesis, and some forward synthesis problems. I will post a list of name reactions that I expect you to be familiar with. You may be asked about material that we have covered for the previous tests.