1 Answer. 2 Answer A B C D

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1 216 W10-Exam #1 Page 1 of 9. I. (8 points) 1) Given below are infrared (IR) spectra of four compounds. The structures of compounds are given below. Assign each spectrum to its compound by putting the letter corresponding to the compound in the answer box above the spectrum. (4 points) A B C D 1 Answer C 2 Answer B

2 216 W10-Exam #1 Page 2 of 9. 3 Answer A 4 Answer D 2) Arrange the following four compounds (A, B, C, D) in order of their Rf values when analyzed by thin layer chromatography (TLC) on silica gel-coated plates using CH 2 Cl 2 as the developing solvent. No partial credit is given to this question. Answer: lowest Rf A < C < B < D highest Rf

3 216 W10-Exam #1 Page 3 of 9. II. (10 points) 1) Protection and deprotection steps are essential for some organic synthetic processes. For the following reaction, carboxylic acids are protected prior to the modification of the amino group and then deprotected under basic conditions. Provide in the boxes below the structure of the appropriate reagent or intermediate compounds. (6 points) 2) If we used 1g of the starting amino-dibenzoic acid (1) and obtain 0.8 g of the final N-acetylated dibenzoic acid (3), What are the theoretical yield and the percentage yield of the final product? Show you work. (4points) Total yield = mol equivalent of product / mol equivalent of limiting agent * mol of limiting agent * MW of product = (1/1)*(1 (g)/ 181 (g/mol)) * 223 (g/mol) = 1.23 g (2 points) % yield = experimental yield/total yield = 0.8 (g)/1.23 (g)*100 = 65 % (2 points)

4 216 W10-Exam #1 Page 4 of 9. III. (6 points) Cholic acid is one of major bile acids produced by the liver where it is synthesized from cholesterol. This cholic acid can be fully oxidized by oxidants such as PCC, NaOCl, Jones reagents (chromic acids) and Swern reagents (i) oxalyl chloride, DMSO ; ii) N(CH 2 CH 3 ) 3 ). The oxidized product can react with 2,4-dinitrophenylhydrazine (2,4-DNP) to make a DNP derivative. The following scheme shows the process of oxidation and the synthesis of the DNP derivative. Draw the structure of the product for each step.

5 216 W10-Exam #1 Page 5 of 9. IV. (12 points) Shown below is the procedure used for Experiment 3 (The esterification reaction) 1) Place g of p-aminobenzoic acid and 3.60 ml of ethanol in a 10-mL conical flask. With stirring, add 0.30 ml of concentrated sulfuric acid dropwise. In this step you can observe the formation of precipitate. Draw i) structures of two possible forms of p-aminobenzoic acid and ii) the structure of the precipitate. (4 points) 2) The precipitate will dissolve during the following reflux period, but if it will not dissolve in solution, what should we do? And explain why. (2 points) Add more sulfuric acid. Some amount of sulfuric acid is consumed for protonation of a starting material, and the rest of it is used as a catalyst for the esterification reaction. However, if sulfuric acid is not enough to protaonate the carbonyl oxygen atom of the carboxylic acid, esterification cannot take place. 3) Write the name of each part or chemicals of the reflux set-up below and show the flow the cooling water by indicating in and out, 6 points)

6 216 W10-Exam #1 Page 6 of 9. V. (12 points) In the experiment 3, you used an alcohol compound as a starting material and oxidized it to make a ketone. The following scheme illustrates the purification process of the product. 1) Fill answer boxes in the scheme, using the keywords in the box below the scheme (3 points) 2) Explain why we need the brine washing process in the purification above. (3points) Brine (saturated aqueous NaCl solution) is highly ionic, thus expelling non-ionic organic compounds or solvents out of the aqueous layer and absorbing water molecules in the organic layer. As the result, the organic layer becomes drier after this process. 3) After purification, you characterized your ketone product using TLC with hexane and ethyl acetate mixture as the eluent, and you got the following result. If you want to increase the resolution of the TLC spots, which solvent should be added to your eluent for running the TLC. And predict the result. (6 points) Hexane (2pts) TLC spots (4pts) 2pts for lower Rf value of a starting material than Rf value of a product 2pts for decreasing Rf values of both two spots

7 216 W10-Exam #1 Page 7 of 9. VI. (10 points) Shown below is the procedure used for Experiment 5 (The reduction reaction) 1) When you reduced benzoin compound and predict the products. (3points) 2) The following scheme illustrates the experimental procedure of the reduction of benzoin. Fill answer boxes in the scheme using keywords under the scheme (4 points). 3) Explain why we need the step 3 and 4 shown in the procedure above. (3 points) During the reaction, BH 3 (borane) can also reacts with ethanol to make B(OCH 2 CH 3 ) 3 (boronethoxide). This boronethoxide is not easily removed by purification processes. If we added water and heated the mixture, the boron ethoxide can be converted to B(OH) 3 (boric acid), which can be easily removed because it can be dissolved in water.

8 216 W10-Exam #1 Page 8 of 9. VII. (14 points) For each of the following synthetic reactions, draw the structure of an organic compound in each of the boxes provided. (8 points) And Match each of the following products to its expected IR frequency for the C=O bond stretching adsorption. 6 points) IR absorptions (vc=o): A cm -1, B cm -1, C cm -1

9 216 W10-Exam #1 Page 9 of 9. VIII. (8 points) For the aldol condensation, 1.5 g of benzaldehyde, 1.14 g of acetophenone, and 7.5 ml of 50 wt % NaOH (d=1.515 g/ml, MW of NaOH = 40 g/mol) were used. 1) What is the molar ratio between acetophenone and NaOH?.(4 points) Mol of acetophenone = 1 (g) / (g/mol) = 9.49( mmol) Weight of 1L of 50 wt % NaOH = 1,000 (ml) (g/ml) = 1,515 (g) Weight of NaOH = 1,515 (g) * 0.5 = (g) The molarity of 50 wt % NaOH = (g) / 40 (g/mol) = mmol/ml Thus, 0.75 ml of 50 wt %. NaOH = 0.75 ml mmol/ml = 14.2 mmol Therefore, acetophenone and NaOH are used in a 9.5 : 14.2 ratio, i.e., roughly 1 : l.5 molar ratio. (2 points for determing the mmol of NaOH, 1 points for mmol of acetophenone, and 1 pt for 1:1.5 ratio) 2) If you obtained 1.5 g of the product, what is the % yield? (4 points) Product = The limiting agent in this reaction is acetophenone. Total yield = (mol equivalent of product / mol equivalent of limiting agent)/ mol of limiting agent * MW of product = (1/1)*(1.14 (g)/ (g/mol))* (g/mol) = 1.98 g (2 points) % yield = experimental yield/total yield = 1.5 (g)/1.98 (g)*100 = 75.8 % (2 points)