A Grignard reagent formed would deprotonate H of the ethyl alcohol OH.

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1 216 S11-E2 Page 2 Name Key I. (9 points) Answer in the boxes below the following questions for the Grignard reagent C 3 -Mg. (1) (2 points) Is the carbon atom associated with magnesium electrophilic or nucleophilic? electrophilic nucleophilic [circle one] (2) (3 points) Explain why the Grignard reaction has to be carried out in an anhydrous solvent. The Grignard reagent produced would be destroyed by the reaction with water present in a non-anhydrous solvent. (3) ( points) Write the conjugate carbon acid (i.e., C-) of the C 3 Mg and explain why Grignard reagents are strong bases. structure: C explanation: The pka of C is ca. 0. Therefore, its conjugate base is highly basic. II. (6 points) Ethyl alcohol (C 2 ) quite often is present in technical-grade diethyl ether. If this grade of diethyl ether were used, what effect, if any, would the ethyl alcohol have on a Grignard reagent used? Explain using a chemical equation using methylmagnesium bromide as an example. explanation based upon the reaction mechansim: A Grignard reagent formed would deprotonate of the ethyl alcohol. C 2 C 3 R Mg C 2 C 3 R Mg III. (8 points) For each of the reactions below, draw in the boxes provided the structure of the expected major organic product. (1) 1. Mg (1.1 mol equiv) ether 2. C 3 C 3 (1.1 mol equiv) aq N Cl (acidic workup) enantiomer (2) Mg (1.1 mol equiv) ether aq N Cl (acidic workup) C 3 Mg enantiomer

2 216 S11-E2 Page 3 Name Key IV. (16 points) For the following reaction sequence, draw the correct structures in the boxes provided. Ph 3 P: (C 18 1 P) C 3 benzene, reflux Ph 3 P C 3 trans "salt" NaC 2 /C 2 C 3 major isomer: C Ph 3 P C 3 cis dichloromethane room temperature C P (ylide) Na C 2 C 3 minor isomer: C note: Ph = C 6 Ph 3 P= V (12 points) Show how many peaks you would expect to observe in the proton-decoupled 13 C NMR spectrum of each of the following compounds. Indicate your answers in the boxes provided. (1) C 3 -C 3 C chiral center! (2) (3) S N 3 S C cis () () C C C 2 C 2 6 C C 180 rotation (6) 3 C 3 C C 3 C 3

3 216 S11-E2 Page Name Key VI. ( points) The proton NMR spectrum of a mixture of acetonitrile (C 3 C N), methylene chloride, (C 2 Cl 2 ) and benzene (C 6 6 ) is integrated and results in step heights of 1, 6, and 18 mm for the signals at δ 2.10,.30, and 7.26 ppm, respectively. In what mole ratio are the three substances present? No explanation required. No partial credit will be given to this problem. acetonitrile : methylene chloride : benzene = : 3 : : : VII. (8 points) For each of the reactions given below, draw in the box the structure of the expected product. Make sure to indicate the stereochemistry of the product whenever applicable. (1) C 3 C 2 Cl 2 (solvent) room temperature C 3 endo product enantiomer (2) 3 C C 2 Cl 2 (solvent) room temperature C 3 endo product or C 3 enantiomer VIII. (10 points) Provide in the box below a step-by-step, curved-arrow mechanism for the formation of hydroxy acid 2 from diketone 1. Make sure to include the mechanism for the acidic workup as well. K ( mol equiv) 2 reflux 3 to p = mech arrows: 1 pt 1 This is fine, too. 1 mech arrows: 2 pts mech arrows: 1 pt mech arrows: 1 pt mixture of stereoisomers here interm str: 1 pt each mech arrows: 1 pt 2

4 216 S11-E2 Page Name Key

5 216 S11-E2 Page 6. Name IX. (continued) A B C D E F G X. (1 points) n the basis of its spectroscopic information provided below, propose the structure for the compound, C 12. Draw the proposed structure in the box on page 7 and give a brief explanation as to how the structure is deduced from the spectroscopic information provided.

6 Structure: 3 C C C C 3 3 C diastereotopic methyls chiral center! 0 unit of unsaturation Explanation (just assign the IR peak at 3368 cm -1 and all -1 peaks right next to the corresponding functional group/-1 nuclei): 0.92 ppm (doublet) 0.88 ppm (doublet) 3 C 3 C C 3. ppm (multiplet) C 1.6 ppm (octet) C cm ppm 1.3 ppm (doublet) 10

7 216 S11-E2 Page 8 Name Key XI. (1 points) n the basis of its spectroscopic information provided below, propose the structure of the compound, C Draw the proposed structure in the box on page 9 and give a brief explanation as to how the structure is deduced from the spectroscopic information provided. Make sure to assign all proton and C-13 NMR peaks.

8 216 S11-E2 Page 9 Name Key XI. (continued) Structure: 3 C C 3 3 C C 2 C 3 C 3 C 2 C 3 C 2 C 2 C 3 C 3 3 C C 2 C 3 3 C C 2 C 3 These should show only one sp 2 alkene C-13 peak. => -1 point Explanation (just assign the strongest IR peak and all -1 and C-13 peaks right next to the corresponding functional group/-1, C-13 nuclei): Units of unsaturation: 3; only 6 C-13 peaks! 22 ppm 2.0 ppm 3 C 12 ppm 16 ppm 172 cm -1.2 ppm C 2 C 3 60 ppm 1 ppm 1.2 ppm 2.0 ppm 3 C 172 cm ppm C 2 C 3.2 ppm 1 ppm 10

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