216%20F06-E2 - Chemistry 216 Second Examination December...

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Unformatted text preview: Chemistry 216 Second Examination December 12, 2006 Professor Masato Koreeda (2.0 hr, 120 points) Please CHECK OFF your lab section. Tuesday 130 131 132 133 230 231 232 Wednes— 150 day 151 152 153 251 252 Thursday 170 171 172 173 270 271 272 Therese Dorau Kapil Karki Joshua N eukom Robert Rari g Zachary Buchan Peter Mai Jonathan Fritz Kathryn MacKool Andrew Lewandowski Justin Lomont Walter Haberaecker Therese Dorau Jinhui Chen Joshua Neukom Jonathan Fritz Kapil Karki Robert Rari g Peter Mai J inhui Chen Zachary Buchan This exam has 14 pages including this cover page. The last five pages include a periodic table, tables of characteristic IR frequencies, representative H—1 and C-13 NMR chemical shifts, and H—l/H—l coupling constants and tables of electronegativity values for some elements, and bond dissociation energies for representative bonds, and include pKa values for representative acids Name Please print Signature Student ID # GSI Initial Name _—_’_ Page 2. I. (15 points) Draw the structures for the organic products containing an ethyl group (CH3CH2) which are expected upon treatment of each of the following chemicals with CH3CH2MgBr, followed by acrdic work—up. (1) methanol [CHgOHl (2) dry ice (C02) (3) benzophenone [PhC(=O)Ph] (4) propionic acid [C2H5C(=O)OH] II ( 4 points). The Grignard reaction has to be carried out under anhydrous conditions. Explain what would happen to the Gridgard reagent formed from bromobenzene (BrCéHs) and Mg if the solvent ether used for the reaction is contaminated with a small amount of water (i.e., more than one equivalent to the BrCGHS). Show in the box below the structure of the Grignard reagent to be formed from BrC6H5 and Mg and provide a chemical mechanism using the curved-arrow convention as to what effect the water can have to the Grignard reagent. III (11 points). In a number of experiments you performed in the lab, the product mixtures were extracted with diethyl ether or dichloromethane a few times. The combined organic layers were first washed with water and then with brine. The resulting organic solution then was dried over a drying agent such as anhydrous calcium chloride, sodium sulfate, and magnesium sulfate. Answer in the box provided the following questions. (1) (3 points) What is brine? (2) (4 points) What is the purpose of this washing with brine? (3) (4 points) What is the chemical basis for the use of an I anhydrous drying agent? Explain using magnesium sulfate as the example. Be specific if any chemical process is involved. Name Page 3. IV (20 points). For each of the reactions given below, draw in the box provided the structure of the expected major reaction product and propose a reasonable step—by—step mechanism for its formation using the curved—arrow convention. (I) O | O + PhsPQk ———> + Ph3P=O Mechanism: (2) 0 Ph Vkph NaOCH3 O HOCH3 Mechanism: Name Page 4. V. (4 points). The proton NMR spectrum of a mixture of acetone [(CH3)2C=OJ, p-dioxane (C4H802) , and dichloromethane (CHZClz), is integrated and results in step heights of 18, 12, and 2 mm for the signals at 6 2.17, 3.50, and 5.20 ppm, respectively. In what mole ratio are these three substances present? Provide your answer in the box shown below. No explanation required. No partial credit will be given to this problem. acetone :pdioxane :dichloromethane 2 VI. (12 points). For each of the reactions given below, draw in the box the structure of the expected organic major product. Make sure to indicate the stereochemistry of the product whenever applicable. (1) 0 room tem erature CHQCIZ O (solvent) 4 endo product (2) / \ 80 °C, 4 h \ ’0 toluene OCHZCHs endo product (3) O DUO (i; CK) (excess) + CHZCIZ, 45 °c 3 h l : 1 mixture endo product Note: Only one of the two starting dienes undergoes the reaction. Name Page 5. VII. (12 points) Following are the 1H and proton—decoupled 13C NMR spectra (all run in CDCl3) for each of the four compounds with the formula C7H14O. Draw the structure of each compound in the box provided. No explanation is needed. No partial credit will be given to this problem. (1) [SC r T“! *1 r‘j‘ 'T 1—‘r—r I 220 200 180 180 140 120 100 80 60 40 20 O (2) - 13; l ' ' l ' l ‘ 200 180 160 140 120 100 80 60 4O 20 0 VII. (continued) Name (3) *‘1—'_I‘T—""l"'—f_"1 180 160 140 120 100 80 60 40 20 (4) 200 180 160 140 120 100 80 60 40 20 ppm 200 Name Page 7. VIII. (15 points) On the basis of its spectroscopic information provided below, propose the structure of the compound (C9H16O4). Draw the proposed structure in the box on page 8 and give in the box on page 8 a brief explanation as to how the structure is deduced from the spectroscopic information provided. Make sure to assign the IR peak mentioned and all 1H NMR peaks. IR: 3 strong absorption at 1745 cm'l; no absorptions between 4,000-3,000 cm“. 100MHz proton NMR spectrum CDCl solution 3 T—Tfifi—l—Tfi—T—fi—Tflfi—T—Tf'r’ Proton-coupled C-l3 NMR spectrum in CDCl3 __l i told Proton-decoupled C-13 NMR spectrum in CDCl3 404; 200 180 160 140 120 100 80 60 40 2O 0 '3 76pm Name Page 8. VIII. (continued) Proposed structure: Explanation: IX. (15 points) On the basis of its spectroscopic information provided below, propose the structure of the compound (CQHHNOZ). Draw the proposed structure in the box given on page 9 and give in the box on page 10 a brief explanation as to how the structure is deduced from the spectroscopic information provided. Make sure to assign all 1H NMR peaks. IR spectrum (liquid film): [00 5D TIINSHITYINCEI '4I l D00 3000 noon 1 5 M 1000 5 IO IRVENUIII Ell 'II Proton—decoupled C-13 N MR spectrum in CDCl3: Name Page 9. IX. (continued). 1H NMR spectrum in CDC13: ,10 9 8 7 6 . 5 4 Proposed structure: Explanation: X. (12 points). Show how many peaks you would expect to observe in the proton—decoupled 13C NMR spectra of the following compounds. Indicate your answers in the box provided. Do not count the 13C NMR peaks of the solvent and TMS. (1) H30 CH3 (2) NH2 CH3 (3) CH3 H3C+<:>{, VCHS H30 CH3 OH E E “O E (4) OCH3 (5) OCH3 (6) OH o o so 0 0 "CH3 Ofiiw 0 "CH3 OCH3 OCH3 E E CHE Page 10 8 A 1 3A 4A 5A 6A 7A 4003 6 B O 10 HnmnFm 10.91 12.01 16.00 19.00 20.13 13 14 15 17 18 Egg ~aagm~ 26.93 23.09 30.97 35.45 39.95 19 20 21 3' 2 d '- B 29 30 31 32 34 3 36 Elam fiiflfiHfiE939Efifififl 39.10 40.03 44.96 47.39 50.94 52.00 54.94 55.35 53.93 53.69 93.55 65.39 69.72 72.59 74.92 73.96 79.90 33.90 37 38 39 41 42 43 £5 46 4 ‘8 ‘9 SO 54 Rb Sr Y E Nb HEE Xe 95.47 97.62 33.91 92.91 95.94 (93) 102.9 106.4 107.9 112.4 114.3 119.7 127.6 131.3 56 56 57 5841 72 3 7‘ 5 8 73 79 80 B1 82 83 84 85 86 “mmflflflflfifimflmwflfiflfimr 132.9 137.3 133.9 ANIDES 173.5 190.9 133.9 136.2 190.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) B7 98 89 90403 104 106 7 Fr Fla Ac mm. (223) 226.0 227.0 IDES (261) (263) flfififlfifiafififlfifififlflfl 140.1 140.9 144.2 (145) 152.0 157.3 153.9 162.5 164.9 167.3 169.9 173.0 175.0 fififlaflfiafifififlfifiwm 232.0 233.0 237.0 (243) (247) (247) (251) (252) (257) (253) .7 H 1008 2A 3 4 Li Be | "'1 "T—i———_I I , i I , 1500 cm' 1000 cm" 400 cm' 4000 “.4 3000 cm" a 2000 cm" 2200 i u 00 I: sips 0H \‘bc. 03:15: ;' 5p2 C-H ‘1 c=c ' : O=C-H C=N I," 5p 0-11 V , , 0-H, N-H \ fingerprint region carbonyl absorption frequencies roughly follow observed reactivity trends [or carbonyl compounds: 0H absorption broadens and moves to lower 1) when H-bonded: acid hnlide > anhydride > ester > aldehyde > ketone, acid > amide non-H-bond: sharp. 3600 cm“ H-bond alcohol: brood, 3500-3200 cm" H-bond acid: very broad. 3300-2800 cm" conjugation lowers v by ca. 20 cm‘ TABLE 10.2 Characteristic Infrared Absorption Frequencies J . C—H alkanes 2960—2850 (3). . 1470—1350 (5) Page 11- C—H alkenes 3080—3020 (m) 1000—675 (3) C—H aromatic 3100—3000 (v) 870—675 (v) C—H aldehyde 2900, 2700 (m, 2 bands) C—H alkyne 3300 (s) CEC alkyne - 2260—2100 (v) CEN nitrile ' 2260—2220. (v) C=C alkene 1680—1620 (v) C=C aromatic 1600—1450 (v) C=O ketone 1725—1705 (.r) C=0 aldehyde 1740—1720 (5) C=C a.B-unsaturated ketone 1685—1665 (.r) C=O aryl ketone 1700—1680 (3) C=O ester 1750—1735 (3) C=O acid 1725-1700 (s) C=0 amide . 1690-1650 (5) , O—H alcohols (not hydrogen bonded 3650-3590 (v) O—H alcohols (hydrogen bonded) 3600—3200 (s, broad) ' 1620—1590 (v) 0—1-1 acids 3000-2500 (5, broad) 1655-1510 (.r) N—H amines 3500-3300 (m) N—H amides 3400—3200 (11:) C—0 alcohols, ethers, esters 1300—1000(3) C—N amines, alkyl 1220-1020 (w) C—N amines, aromatic 1360—1250 (5) N0, nitm 1560-1515 (3) 1385—1345 (r) s = strong absorption w = weak absorption m = medium absorption v = variable absorption TABLE 11.] Typical Chemical Shifts for Types of Hydrogen Atoms, Seen in Proton Magnetic Resonance Spectra Type of Hydrogen Atom V Type of Hydrogen Atom RCH, 0.9 RCHIR acyclic 1.3 RCH=CR2 5.3 cyclic l.5 ArH 7.3 RJCH 1.5—2.0' 0 R3C=$CH3 |.8 RICIH 9.7 RI RNHz 1—3 (H) ArNHz 3—5 RCCHJ 2.0—2.3 O ArCH, 2.3 RgNHR 5_9 RCECH 2.5 R0“ H RNHCHJ 2—3 Aron 4_7 RCH2X(X=CI, Br, 1) 3.5 . O 0 H H RCOH 10—13 ROCH3, RCOCHJ 3.8 ‘The chemical shift values are given in ppm relative to tetramethylsilane at 5 0.00 and are for the hydrogen atoms shown in boldface in the fonnulasi The values for hydrogen atoms on oxygen and nitrogen are highly dependent on solve"l I‘nwen'm'ion, an!l temperature. TABLE 11.3 Coupling Constants Seen in Proton Magnetic Resonance Spectra Page 12 TABLE 11.4 Chemical Shifts for Carbon Atoms in Carbon-13 Nuclear Magnetic Resonance Spectra 9“, Type at VCS‘rlxini 53:}? ‘ ' _ If Carbon Atom RCHZQH3 13—16 RCH=£H2 115—120 REHZCH3 16—25 R£H=CH2 125—140 R39}! 25-38 RCEN 117—125 0 ArH 125—150 gHggR ~30 0 0 R(”:0R' 170—175 EHSCHJOR ~20 o ' RCHZCl 40—45 R(“IOH 177-1l35 RCHlBr 28-35 0 RCHZNH2 37—45 R(”IH 190—200 RCHZOH 50-64 0 RC 5 SH 67-70 R(“2R’ 205—220 RE 2 CH 74—85 *The chemical shift values are given in ppm relative to tetnmethylsilane at 5 0.00 and are for the carbon atoms shown in boldface in the formulas. Tab|e 1.2 Electronegativity Values for Some Elements 1 2 13 14 15 15 Table 2.4 Average Bond Energies [kJ/mol (above) and kcal/mol (below)! Single Bonds Muniple Bonds C=O (aldehydes) I76 17 Page 13. Page 14. ACID pKa CONJUGATE BASE ACID pKa CONJUGATE BASE I” w ( S trongest Acid) ( Weaken Base) -'o‘- 30-. n. no a. a. .' C u . "G HQ—g—QH _9 H.O_-—:S:—Q:e 9.9: \9H 10.2 9;.0;C\p.; .53.. -.0.- .-O-. .. .. G) “e "O" H—I : :1: 6.. N H I! e .. —9 .. .O/ \C/ 10 2 G ./N\ -. .. .. e ' I \ ‘ q ,c e H—g : _7 :91: H H H \H H@ .. G .- o / \ 1 . CH CHa— : CH3CH2/O\H —2.4 CHJCHZ/ \H CH3CH2 H 05 3 S’ :26) o- g H 15 0 fl / \ / \ / \ /. . C u H H —1.7 H H CH3 CH3/ \Nle . . H | (9 G) H e 0§N\§H -13 @ O’N‘b' : @ H/O\H 15.7 H—9. "0" O .. ., g g @ CH3CH1 0\H 17 CH3CH2—:6 16 C13C’ \OH 0.64 CHC/ \g)‘; ' o- .. 0 .’ '. H-.F.= 3.2 we a. H i 9 .° '. . . / \C’ 19 / \C'. (3 .fi. CH3 H! \H CH3 H/ \H C - c .. \9.” 4.2 ‘qze $1 20 $1 c1——c—H Cl_$:e H ,H H G G I c / C\\ Pkg / \\C—-N/' H_CEC"H 26 H-CEC :@ _ -q‘-H 4.6 _ (H . . H _. ._ H2 35 cha 'c9 'c? e 7‘ P'i @ CH3/ \:0:H 4.8 CH3/ ‘9: H/NéH 36 H/‘?’ z 30.. .‘O'. H 9 II “ H H ‘ —c- c ..G - - H6<C\'o'u 6.5 Ho’ ‘9: }C=C 36 H/ \H . .. H H .. ., . .. O O , H / C,H E ii a 4 Gr 41 Q a / \ / \ / \ l \ H \ . CH3 H/C\ CH3 CH3 H/Cb CH3 \ C\/C/\H H/C\H _ H H—CEN 9.1 exam 6 H H 43 ' H—IlJ—H H—N/ H A 9.4 H H H e / “(3—0” 10.0 </ 56-30:: 9 — -—— (Weakest Acid) (Strongest Base) ...
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216%20F06-E2 - Chemistry 216 Second Examination December...

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