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U of M CHEM 211 - MWrite 3: Aldol Reaction

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5 April 2020MWrite 3: Aldol ReactionTo the biochemists at the University of Ghana, In an early step of the synthesis of Ivermectin, a drug used to treat onchocerciasis — alsoknown as river blindness, an intramolecular aldol reaction takes place. The traditional synthesisfor this reaction leads to undesired side reactions due to the need for strong acids and bases inthis step of the reaction. However, as you have identified, the catalyst triazabicyclodecene (TBD)can be used as a new strategy to perform this intramolecular aldol reaction while limiting theundesired side reactions.1 In my research, I have identified two potential mechanistic pathwaysfor an intramolecular aldol reaction using catalytic TBD, which I have labeled Mechanism A andB. Mechanism A is one where the catalytic TBD remains consistent throughout the entirereaction. Any time it is deprotonated — the removal of a hydrogen atom — at one nitrogen, it isprotonated — the bonding of a hydrogen atom — at the neighboring nitrogen. Because themolecule is symmetrical, this keeps TBD unchanged throughout the reaction. This mechanismbegins with the enol formation fromthe ketone of the 6-oxoheptanal bythe TBD deprotonating the alpha-carbon and protonating the oxygen ofthe ketone. This creates an enol,which means the carbonyl is nowalcohol, and the carbon of the carbonyl is now double bonded to the alpha-carbon. When lookingat the energy diagram for mechanism A, this step begins when starting material “a”, reacts in thetransition state “b”, and the enol is formed as intermediate “c”. A transition state is a point atwhich new bonds are formed while the old ones are broken, so the transition state at this point ofthe reaction is when the hydrogens are protonated and deprotonated.The next and final step of mechanism A is the formation of the aldol.All at once, the enol is deprotonated by the TBD, which pusheselectrons back down to re-form the carbonyl, which pushes theelectrons double bonded to the alpha-carbon over to the carbon of thealdehyde, which pushes electrons up to protonate the oxygen from theTBD. These bonds breaking and forming are transition state “d”, andthe final intramolecular aldol reaction product is product “e”.Mechanism B is a longer, more complicated reaction thanmechanism A, and TBD is used more than just to protonate anddeprotonate the 6-oxoheptanal and its intermediates. This reaction begins with the protonation ofthe oxygen of the aldehyde by TBD, followed by one of the nitrogens of the TBD reaching out to1 Shultz, Ginger. “Exploring Possible Reaction Pathways for a Catalyzed Intramolecular Aldol Reaction.” Chem 216 Canvas, University of with the carbon of the aldehyde. This corresponds to starting material “p” of the energydiagram reaching transition state “q” before becoming intermediate “r”. Intermediate “r” thenreacts with another molecule of TBD, this time at the ketone, where an enol is formed from thedeprotonation of the alpha-carbon and the protonation of the oxygen of the ketone in transitionstate “s” which results in intermediate“t”. In intermiediate “t”, the TBDmolecule bonded to the originalaldehyde then breaks the C-N bond andpulls away while also deprotonating thealcohol, reforming the aldehyde intransition state “u”, which results inintermediate “v”. In the final step of thismechanism, in transition state “w”, theenol formed from the ketone isdeprotonated by the TBD, which pusheselectrons back down to re-form the carbonyl, which pushes theelectrons double bonded to the alpha-carbon over to the carbonof the aldehyde, which pushes electrons up to protonate theoxygen from the TBD. This results in the final product “x”,with the TBD now unchanged from the start of the reaction.The reaction pathway I believe is most likely to occuris mechanism A. Not only is mechanism A less complicatedwith fewer steps, but it also requires less energy to complete inaddition to the catalyst remaining unchanged throughout thereaction. When looking at the energy diagrams, the highestamount of energy needed for mechanism A is 15 kcal/mol attransition states “b” and “d”, while in mechanism B, transition states “s” and “u” require 20kcal/mol. When looking at two different reactions, from the same starting materials, the morefavored reaction can be determined by the rate of reaction and the stability of the products.However, in these two reactions we don’t know the reaction times, and the products are thesame, which means that could not determine which is favored over the other. This means theonly way to look at which reaction is favored is to see which one requires the least amount ofenergy, or which reaction has the most stable transition states. It is by doing this that it can beseen that mechanism A requires less energy to progress, making it more favored, which makesme believe that it is the reaction that is most likely to occur. I hope this report helps in determining the feasibility of applying this reaction to the morecomplicated synthesis of ivermectin, and that it is useful in helping those with onchorerciasis. Thank you for reading,Lab

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