New version page

U of M CHEM 211 - Inquiry into the Possible Mechanistic Pathways for the Synthesis of Ivermectin by TBD Catalyzed Aldol Reaction

Documents in this Course
Load more

This preview shows page 1 out of 2 pages.

View Full Document
View Full Document

End of preview. Want to read all 2 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

Inquiry into the Possible Mechanistic Pathways for the Synthesis of Ivermectin by TBDCatalyzed Aldol ReactionBackgroundAn intramolecular aldol reaction catalyzed by TBD can convert 6-oxoheptanal (a ketoaldehyde) into 2-acetocyclopentanol (Shultz). Two mechanisms for this reaction, referred to as Mechanism A and Mechanism B, have been proposed. The goal of this project has been to determine the most likely pathway. Comparison of Mechanisms and Modeled Energy Diagrams of A and BProposed mechanism A occurs in 2 steps, and its modeled energy diagram reflects this - it has a starting material a, intermediate c, and product e. Between a and c is transition state (the state of a molecule currently breaking and forming bonds) b and between c and e is transition state e (Shultz). From a to c, the TBD protonates the ketone, then the non-protonated nitrogen of TBD molecule removes an alpha hydrogen. The deprotonated carbon double bonds to the carbon of the ketone, which “pushes up” the double bond of the protonated ketone, forming an enol. From c to e, the TBD deprotonates the enol, pushing the electrons back into a ketone; the electrons of the alpha carbon and ketone carbon double bond attack the carbon of the aldehyde. This pushes electrons from the aldehyde double bond onto the oxygen, which then deprotonates the TBD. Product e is formed. Although the 2 steps are different, very similar things happen at each step (protonation of oxygen and attack of carbon in a C=O bond), and the transition state energy is about the same (b and d). Proposed mechanism B occurs in 4 steps. The energy diagram shows starting material p, intermediates r, t, and v, and product x. Between p and r is transition state q, between r and t is transition state s, between t and v is transition state u, and between v and x is transition state w (Shultz). From p to r, TBD protonates the aldehyde while the TBD double bond attacks the aldehyde carbon; this forms r, where the starting material is attached to TBD. From r to t, a second TBD protonates the ketone, then the non-protonated nitrogen of TBD molecule removes an alpha hydrogen. The deprotonated carbon double bonds to the carbon of the ketone, which “pushes up” the double bond of the protonated ketone, forming an enol. From t to v, the attached TBD deprotonates the alcohol nearby it. The aldehyde reforms and pushes the TBD molecule off.From v to x, the TBD deprotonates the enol, pushing the electrons back into a ketone; the electrons of the carbon double bond attach to the carbon of the aldehyde. This pushes electrons from the aldehyde double bond onto the oxygen, which then deprotonates the TBD, forming product x. PredictionsMechanism A is more likely. The SM and products are the same for both mechanisms, so there will be the same difference of Gibbs free energy. Mechanism A has activation energy from a to c (value of b-a) of 15-0 = 15 energy units. Mechanism B has a much smaller initial activation, but step 2 (r to t) has a very high energy transition state s - and the difference from r to s is 20-3 = 17. The difference in energy of the transition state is due largely to sterics. Intermediate t of Mechanism B requires the large TBD molecule to attach to the keto aldehyde, which is difficult to do because of steric hindrance. The electronegative (slight negative charge) nitrogens of TBD are adjacent to the electronegative oxygen of the keto aldehyde, which makes this molecule even more unfavorable to form. Additionally, doing this makes the molecule as a whole less accessible to other molecules later in the reaction, which is why the activation energyfrom t to v is also very high. Transition state c of mechanism A, on the other hand, is an enol, which is highly reactive, so c itself is a high energy state, but because it does not have to negotiate steric hindrance, the activation energy to get from a to c is much lower as compared to from r to t.CitationsShultz, Ginger Victoria and Salvador, Tolani Kuam (Date of Publish Unknown). Exploring Possible Reaction Pathways for a Catalyzed Intramolecular Aldol Reaction. [Information provided by the U of M Chem 216 WN 2020 Instructors.] Retrieved on 4/1/2020 from


View Full Document
Loading Unlocking...
Login

Join to view Inquiry into the Possible Mechanistic Pathways for the Synthesis of Ivermectin by TBD Catalyzed Aldol Reaction and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Inquiry into the Possible Mechanistic Pathways for the Synthesis of Ivermectin by TBD Catalyzed Aldol Reaction and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?