m05-part2 - PART B: Short Answer Questions. QUESTIONS MUST...

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Unformatted text preview: PART B: Short Answer Questions. QUESTIONS MUST BE ANSWERED ON THE ANSWER SHEET FOUND AT THE END OF THIS EXAM. Clear and concise answers are essential for. full marks. ‘ ' ' l (6 marks). The reaction of l-methyl-l-cyclohexene with bromine, Brz, in methanol, CH30H, gives a mixture of enantiomeric products of the formula C3H1 ZOBr. Using one of the templates on the answer sheet, draw the most stable conformation of one of the two enantiomers. - 2 (10 marks). Fill in the boxes on the answer sheet with the reagents or products for the following reactions; ' (a) ' . ' .\ ‘H .0 E H o (b) I - # Q # ,/ / i . o (c) 1. BH3 {pr 2.H202, NaOH __—> draw only one enantiomer (d) ‘ ' I Q (6) . .9 .0 —————»~ r9of11,‘ 3 (7 marks). On reaction with acid, 4-pyrone is protonated on the carbonyl-group oxygen to give a stable cationic product. Draw three more resonance forms on the answer sheet to assist in explaining why the protonated product is so stable. Be sure to place formal charges where appropriate. H 0 +0/ H+- l l -—-> I l O ' ‘0 4-pyrone 4 (15 marks). Using arrows to show the movement of electrons, write the mechanism for the reaction below. Be sure to draw the structures of all intermediates, including resonance forms where appropriate, but do NOT draw the structures of transition states. ———-—> ' 5 (10 marks). Consider each of the reactants below undergoing reaction with sodium ethoxide, NaOCH2CH3Z ' \ ' $ e 3 Br C>‘ Br 2’; I, I; l/ l (a) What term most accurately describes the" structural relationship between these two reactants above? 100f11 (b) On the answer sheet, circle the isomer that will undergo faster elimination. (c) Making sure to take stereochem’istry into account, draw the elimination products of this faster reaction. What is the structural relationship between the elimination products? 6 (12 marks). A solution of chiral (R)-C3H9Br in THF is optically active. Addition of Br; to one half of this solution produces no reaction. The introduction of H20 to the other half of the solution of (R)-C3H9Br leads to a diminishing of the solution's optical rotation as organic products of two different compositions are formed. Addition of more H20 to this solution does not increase the rate of optical rotation loss. Eventually, the solution obtains an optical rotation of 0°. ’ (a) Determine the degree of unsaturation of (R)-C3H9Br. (b) Draw the structure of (R)-C3H9Br that is consistent with all of the above information. (c) Draw the structures of the products that are formed in the reaction of (R)-C3H9Br' with H20. (d) In 20 words or less, explain why the solution displays no optical activity at the end of the reaction. ' 7 (10 marks). (a) On the templates of the answer sheet, draw the products expected from the free radical monochlorination of 2-methylbutane. Put each product, including stereoisomers, in a separate box. There may be more boxes than needed - do not draw any more molecules than necessary. (b) Identify any chirality centres in the products with an asterisk (*). (c) Circle the isomer(s) formed from the most stable intermediate of the reaction. — END OF THE EXAM — llof'll m' AuEm: m>=mEu=m B .3552 “9.55 gm: . . mag—5304‘ .2255 .caEufi .. .... . .. . . 5395 55.3 SE8 .55 22.9.2: :55: 55.82 .555 5:33 555 .oao_8_uo>_._.$m8_umm§ . mmGE . .0: Omecafl Egg 8. .243 $12.3 5:25. 8:83 8 SE a 5.3.. 3 Eu 8 .82.“. a .83“. 3 $0.3m 3 Saw a 2:2 a Sam. 8 , >wv— 523 5:5 s3: 53 55: 53.35 55 ago 5.93 g S395: .5582 $583: 5.8 $305. 3.— n> FE. gm 0—.— >D ch. 60 3m em En. 32 ..n. m0 =9=< . 5E : an: 2 8.8. 8 8m: 3 8§ s 8%. s 8.8. 3 33, a 8.3. 8 8.8. a 5.3: 3 as. 8 5.8— mm «3: 2.. 2..th . mm.me 2:02 Amctmm Em”: 82:935.. 255.? mcomofi: . 53.88%: g2: 53:2: . . _ . . sauna 5:38“. 0:: as: as: :3: 55 . «m E Efl a: .8“. 3. EN. Ema ,EEN. 333%. 3 sons. 3 53x 3.53 E3220; 5:55 fiat 2:32... 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O z o . m . om , .3 82m 2 age a 289 a 5.: N :2; w :2: m N858 e :8 5%: . N I! CHM 138HIS Final Examination Answer Sheet Student #: Demo Gp.: 1 (6 marks). Most stable conformation of one enantiomer 2 (10 marks). Reagents or roducts: (c) Only one enantiomer 3 (7 marks). Three more resonance forms: The protonated product is stable because CHM 138HIS Final Examination Answer Sheet Student #: 4 (15 marks). Reaction mechanism: 5 10 marks). (a) Structural relationship between these compounds: (13) Circle the isomer which (c) Structures of elimination products: undergoes faster elimination \ S \ l S <:>‘Br Z , , I, I 4 Relationship between products: Demo Gp.: W‘Jg or A.ns CHM 138HIS Final Examination Answer Sheet Student #: Demo Gp.: 6 (12 marks) a) Degree of unsaturation: (b) Structure of R-CngBI' (c) Structures of products (d) The solution displays no optical activity because 7 (10 marks). (a), (b) and (c) x Question Mark ...
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This note was uploaded on 04/25/2010 for the course CHM CHM138H1 taught by Professor Browning during the Fall '08 term at University of Toronto- Toronto.

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m05-part2 - PART B: Short Answer Questions. QUESTIONS MUST...

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