Question

I need the results and discussion of the lab please! I also need yield calculations and spectra interpretation.

MICHAEL ADDITION REACTION PURPOSE To use a Michael addition reaction followed by an Aldol Condensation reaction to produce a more complex molecule and demonstrate a tandem reaction, where a second subsequent reaction takes place after the first initial reaction

Answer 1

NaOH EtOH Reflux OET Trans-Chalcone Mol. Wt.: 208 26 Ethyl acetoacetate Mol. Wt: 130.14 0.166 g = 0.00127 mole 0.238 g = 0.00114 mole Final product Mol. Wt: 320.38 0.00114 mole= 0.365 g Results: When 0.238 g oftrans- Chalcone react with 0.166 g of ethyl acetoacetate, then 0.365 g of final product obtained at the end Discussion: Here, trans Chalocone is the limiting reagent which is consumed fully after the reaction and determines the amount of the final product Theoritical yield = 0.365 g Actual yield = 0.2155 g So, percent yield = (Actual yield/theoritical yield) x 100 = (0.2155 /0.365) x 100 = 59.04 % The structure of final product can be interpreted from it's 1 H NMR and IR IR shows two strong streching: one at 1732 cm-1 which is due to ester C=O streching and other at 1665 cm-1 which is due to alpha beta unsaturated ketone C=0 streching. I NMR SPECTRUM IH fo 6.39, 1H, $ All aromatic protons 4.15 2H, 3.8, 2H, M 3.0-3.1 2H, m CH: 2.2575 2.4346 1.12, 3H, t