Chem 100 lab 8 post lab, Lab Reports of Chemistry

Download Chem 100 lab 8 post lab and more Lab Reports Chemistry in PDF only on Docsity! Reduction of Vanillin Acetate CHEM 100L – Laboratory 8 Experiment 8 PostLab Assignment- Observations (4 pts) What we did Observations Adding a second round of ether Hard to see different layers by organic layer kept increasing each time Vanillin acetate weighing Was like white crystals and had a slight vanilla odor After drying the product and rotovap It was a clear liquid with a particular smell I can’t describe Results (18 pts) (1 pts) Product Identity: 4-(hydroxymethyl)-2-methoxyphenyl acetate Table 1- (3 pts) Product yield Mass of product (g) Theoretical yield of product (g) Percent yield of product (%) 0.768 g 0.51g 151% 0.768 g product/0.51 g = 1.51*100 = 151% Table 2- (7 pts) Comparison of key IR signals If the absorption peak(s) was present in the IR spectrum, fill in the approximate wavenumber(s). If the absorption peak(s) was absent in the IR spectrum, write “absent”. OH stretch sp2C-H stretches sp3C-H stretches aldehyd e C-H stretches C=C aromatic C=O stretch (ester) C=O stretch (aldehyde) Vanillin acetate absent 3024.8 cm-1 2964.64- 2917.15 cm-1 2847.49- 2749.34 cm-1 1600.2 cm-1 1745.65 cm-1 1675.99 cm-1 Product 3372.09 cm-1 2969.78 cm-1 2900-2800 cm-1 absent 1605.15 cm-1 1759.88 cm-1 absent CHEM 100L Lab 8 Vanillin Acetate Reduction Page 5 of 6 Authors: Dr. Matt Meyer & Dr. Michelle Leslie Last Revised 10/27/2021 by JML 1H NMR Results (7 pts total) As we are specifically focusing on what happens to the carbonyl containing functional groups in this experiment, the key signals from the starting material to pay attention to are the aldehyde hydrogen signal, and the signal for the CH3 hydrogens from the ester. (2 pts) From the provided 1H NMR spectrum of the starting material, identify each of these signals (by indicating the chemical shift, integration, and splitting). Aldehyde H: 9.942 ppm, 1H, Singlet Ester CH3: 2.338 ppm, 3H, Singlet (5 pts) Draw the structure of the product obtained in this reaction and assign all of the proton signals. You do not need to distinguish which aromatic proton is which.