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chem 282 final - SIMON FRASER UNIVERSITY E Organic...

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Unformatted text preview: SIMON FRASER UNIVERSITY E Organic Chemistry II, 2008-1 CHEM 282 Dr. Robert Britton FINAL EXAMINATION (50% of total course evaluation) 3:30 - 6:30 pm. Thursday, April 10, 2008 ANSWER ALL OF THE QUESTIONS. Your exam consists of 32 printed pages including this one — please check to ensure you have all pages. Fill in your name and student number below. The final pages of this exam include scrap paper, a periodic table and NMRIIR spectral data and you are permitted to use molecular models. Calculators will not be necessary in this exam. IF YOU NEED MORE SCRAP PAPER WE WILL PROVIDE IT. NAME: ........................................................................... STUDENT No. ' .......... . ................................................................. T.A. . C. Warford J. Heinonen A. Rojubally Marking Scheme Question No. Marks Score 1 36 2 8 3 5 4 3 5 10 6 10 7 6 8 15 9 10 10 8 11 22 12 18 TOTAL I151 1. a) Please draw TWO heteroaromatic molecules. (4 marks) (“ I H N / N / r’i _ 'b)'Please draw TWO-antiaromatic molecules. (4 marks) El/\ \/ ‘x "c) Please circle the more acidic molecule (the one that has the most acidic protons) and explain the difference in acidity between these two molecules. (4 marks) d) Please rank the following molecules from MOST to LEAST reactive in a hydrolysis reaction with NaOH in water. (2 marks) 0 O o o o o A CH F A k 0 io/ A0/ 2 o A B c 0 MOST LEAST REACTIVE D > A > C > g REACTIVE Question 1 continued..... e) Please rank the following molecules from MOST to LEAST acidic. (2 marks) 0 M /l‘\ O N(CH ) NH ice... H H ' 3 2 A '— 3 A B c D MOSTACIDIC 9 > A > D > C LEASTACIDIC f) What is the probability of finding an electron between the two orbitals in the LUMO of ethane? (2 mark) a 9) Why are F (EN. = 4.0) and Br (EN. = 3.0) both considered weak deactivating groups in electrophilic aromatic substitution reactions, even though their electronegativities are quite different? (2 marks) er: \/\0A ‘21“? ”91qu 0+0"? [or atrial/7A Ad/taka 5:3 mJOAQAC-e h) Please explain why pyridine is a better base (accepts a proton more readily) than pyrrole. (4 marks) ' ll“ H H a; N 0'? pyrrole pyridine 4M 77' Altm‘lq'c‘ W Question 1 continued..... i) For the foilowing pair of alkenes, circle the ONE that would react the fastest with HBr and draw the product of this reaction. (2 marks) j) Using a reaction coordinate diagram and considering the relative energies of both alkenes and the intermediate associated with these reactions,-please explain your choice of ‘fastest reacting alkene’ for part 1 i) (above). (5 marks) I s :t &A @44 A65 Question 1 continued..... k) What are the products of the following Diels-Alder reactions? Please circle the reaction that you expect would be faster. (5 marks) OCH3 2. Oxygen-18 (0“) is an isotope of oxygen. For each of the followin hydrolysis reactions, draw the products - pay attention to where the 0 ends up! For part b) draw a mechanism that explains your choice of products in the space below. (8 marks) 0 (3 . l ' 0‘3CH H 0+, H 0 V L? a) 3 a 2 d\b|—I 4 l . ‘ 3. Oxygen-18 (018) is an isotope of oxygen. In the following reaction, a 1:1 mixture of products is obtained (as shown) - please draw a mechanism that explains the formation of both products. (5 marks) 018 O 013 H‘" (catalytic) 0H CH30H (excess) OCHa + OCH3 ——r- 4. Please indicate (in the box provided) where the equilibrium would lie for the following reaction. In the space below, provide conditions that would favor the formation of the ketal. (3 marks) 0 CHsOH H3CO OCH3 sag-W Jj< ><i< Cm! @3014! H65 “sisal1 5. Acetaminophen (also known as paracetamol or Tylenol) is used to treat pain and fevers. The synthesis of acetaminophen is detailed in part below. Please answer the following questions relating to the synthesis. a) For step 1, two isomers are prodUced in approximately the same yield. In the boxes provided please draw the structures of the 2 isomers and indicate why one isomer will be more polar than the other. (4 marks) b) Please provide a name for each isomer. (2 marks) ' I more polar isomer: M— less polar isomer: M c) In the table provided, please provide (the number of singlets, doublets, triplets, doublet of doublets and quartets you would expect for the LEAST POLAR ISOMER. (2 marks) (I) In the box provided, please provide a structure for acetaminophen. (2 marks) more polar less polar OH T H >10“); HN03, H2804 ééu ' step1 é A A; 1' H2. Pd singlets doublets triplets quartets doubletof 2. O O -. doublets AOJK 1 equivalent Acetaminophen 6. ibuprofen is the active ingredient in a number of painkillers. A synthesis of ibuprofen is detailed in part below. OveriUnder each arrow, please provide conditions that would effect the desired transformations. Note: each box indicates one ‘step' or one set of conditions - and you do not need to include work-up steps. (10 marks) coza ibuprofen 7. What conditions would you use to promote the reaction shown below? Please provide a mechanism for this reaction in the space below. (6 marks) O _ 683‘» L134 0 + Jk/ ———-—————> H ‘ . A B — r oxolX’“ Job/N 06 O u 6/ +U 0‘01? oarr 10 8. Consider the spectral data for compounds A and B below. Please draw the structure for compounds A (3 marks), B (3 marks), C (2 marks), D (3 marks) and E (3 marks) in the boxes below. Also, for compound E indicate if this material is chiral or not. (1 mark) COMPOUHCI A I C7H7Bl’ compound B I CaHaoz 1H NMR data 1H NMR data 7.5 ppm (doublet, 2 H) 9.1 ppm (singlet, 1H) 7.3 ppm (doublet of doublet, 2H) 7.5 ppm (doublet, 2 H) 7.1 ppm (triplet, 1H) 7.3 ppm (doublet, 2H) 3.0 ppm (singlet, 2H) 3.0 ppm (singlet, 3H) IR data 1700 cm'| (strong) 1250 cm'1 medium /H / fiP/‘KH Ock3 compound B 1. Pth 2. butyllithium (n-BuLi) compound C compound E 11 9. Please provide a name for each of the following molecules. (2 marks each, 10 marks) m (DIS) chklarwé— Maw—'2 w / . Cl 0 0 gm —§‘ PM elk/J '— b (a _ proxcflchlxlp‘iaom O O ULOJQ l)((\Zo Cc. Omtb b1—é/t . MO/\ 1% t ' §~ M4451 LU’L““°Qk '- OH Q” 2,4me s—Wlujl Flam! H C 3 12 10. Please provide products for the following reactions. As a hint, the chemical formula for each product is provided. (8 marks) Br 1. Mg, Cul 3. H20 (work-up) ' 1. NH3 (1 equiv.) - NaBH(OAc)3 (excess) HOAc 2. NaOH. H20 (work-up) 11. Please fill in the product or reagents for each of the reactions detailed below. (2 marks each, 22 marks) Br conditions F) ‘59 +0 on.“- o H 6) (geek) 1- . NaOH | NH conditions 2. HCI, H20, heat 4—.—-—-—-———-— 3. catalytic HCl distill off H20 l3 Question 11 continued ..... O excess NaOH b) EXCESS '2. H20 E j/ \ ———-———> C) : .Br ‘5 ”,2. L) (0mm, MIG)” [17/0 conditions HzCrO4 H20 OH +—-———. product 63L: Al H~\‘ z) [4363’ [the OH conditions 1 CF3003H HO excess pyridinium ohlorochromate (P00) 0 i HO-Cr-Cl .CHzclz N ll NaOH, H20. heat product 14 12. Using the starting material indicated and any other reagents you need, please design a synthesis for 3 of the following molecules. (18 marks) -‘OH ‘\ O ARM—>- Br N 0 H 0 D O 0 “lg/“‘1 “‘96; 3 m G /l3~ “ ,S 5 I: r /g" AME! 3 QfJK-OQH] dig)“ at 13 (MI _ \ 0L (3 \TJLU H; X 9" [4M Question 12 continued ...... l6 SCRAP PAPER 2,) (MI lHHH (”a \3LAIIA m ¢ “‘ 20"”: A approximate frequency of lR Signal functional group 3650-3200 cm“ 0—H (alcohol) 330042500 cm"1 O—H (carboxyiic acid) 3500-3300 cm‘1 N—H 3350-3250 cm" C—H (alkyne) 3200-2800 cm" _ - C——H (alkane) 2260-2220 cm" GEN 22602100 cm'1 CEO 1760 cm'1 :0 (carboxylic acid) 1740 cm“ o=c (ester) V 1715 cm“1 o=c (ketone) 1700 cm'1 - o=c (aldehyde) 1680-1600 cm"1 c=c 1600 W1 © 1250 -1 050 cm" c--—o Periodic Table of the Elements June 4::an mu» g}! - M- 11 12 “I s: a “may a El 055% it: ‘5: gm l-M 211:» new ztO-IM ‘7. am- New 3511 «9:1: emu «um Jam: Inn 41M 3mm rm: :35 8r Y “Zr Nb y; :10: :80 “an “Pd Ag [:0ka Te" aw it” “sum “aw um ”IMO-Y was. +201“ 4|!!! "but sganu emu" -I-a-1H “155;,“ "IN“ , an» «Inn tluau ml “uh rum» new , .‘3 “WM gum” {gentle-0:521 m.“ 99mm mlvlnrgamA'fu-nnuz 352.1233: 6" m «mm M *5 EMIEIIWWW I mam-2 Emu-a ,7 U ‘ . \ mmmmwmmdmwlmim flhMMNWoPMWMW WWWWMW‘CEMM“ mmhnmdwmiwrsm 18 chemical shift type of proton (Ppm) (underlined) O 9 - 10 ppm /u\ carboxylic acid 01:1 0 A aldehyde fl : 6 5 - 8 ppm 5 benzene H. 4 5 ' 6 5 ppm =< alkene O 4 0 - 4 5 ppm io—c ester flC—OH alcohol 3'0 ‘4-0 ppm flC—OCH3 ether HC_X X = F, 0', Br, I alky! hallde 2 O -4 0 ppm /\0H allyllc CH3 benzylic O A adjacent to carbonyl Ctis = fl alkyne 1.2 - 2-5 Ppm /\< 3° (tertiary g methine) H H- H 2° (secondary g A methylene) 0.3 - 1.2 ppm /\/CH3 10 (primary g methyl) ...
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