Exam2A - OZS’EZ CHEMISTRY 3331 Name Second EXAM(print...

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Unformatted text preview: OZS’EZ / CHEMISTRY 3331 Name Second EXAM (print) Last First Dr. R. Thumrnel S. S. Number Dr. C. Cai Please circle the name of your instructor on the left. March 22, 2002 Please read all directions carefully. Write all answers legibly in the appropriate spaces and THINK about what you are doing. Give only ONE answer for each question (100 pts total). 1. (20 pts) For each of the following reactions, draw the structure of the necessary starting material, or principle organic product in the empty sure to indicate STEREOCHEMISTRY where pertinent. 61) CH3 Na SH CH3 -—-———> H lIIu../ + Nal SH 9 CH3CH20 Na® _.__—————> E-2 c) 0“ st04 ——-———> heat H CH3 Major Product d . ) Br 6) GD H (CH 3)sC-O Na H3 |\ E-2 H CH3 2. (15 pts) The alkyl halide shown below undergoes competing SM and E1 reactions with methanol to provide the products A, B and C. Propose a reasonable and clear mechanism for the formation of these three products. Use curved arrow notation to indicate “electron flow”, and show all carbocation intermediates, their resonance forms, and all formal charges. CH3 C H3 CH3 H3 Br H H3 a 1 CHaoH 3 H3 / CH3 H3C \ CH3 _.._> — OCHa + H3 + __ H3C H3C C C racemic H300 H3 A B 3. (12 pts) Cyclopentane can be converted into 1,3-cyclopentadiene by the four steps outlined below. In the box, draw the structure of each reaction product in this synthetic sequence. 1 equiv. — + Brz (CH3)3C-O K light NBS l <—-——-—-——‘—'— 1,3—Cyclopentadiene 4. (8 pts) Benzyl ethyl ether can be prepared by the reaction of an alkoxide anion with an alkyl halide. Two different combinations of these reactants can be employed. In the boxes below show the two pairs of reactants which could be used to form benzyl ethyl ether. 5. (20 pts) For each of the names given below, draw the structure of the corresponding molecules, paying particular attention to stereochemistry where this is pertinent. (a) 4,4—Dibromo—1-methylcyclopentene 7 (b) Neopentyl chloride (c) cis, trans-2,4—heptadiene (d) Z—1,3-dibromo-2—pentene EC} (e) 4—(2—fluoroethyl)heptane 6. (12 pts) Complete the Fisher skeletons below, which already indicate the position of one of the bromine atoms, to show the correct strucure of S-2—bromo—3—methylbutane, the same compound, and its enantiomer. —-$— Br Br Br S—2-bromo-3-methylbutane same Compound Enantiomer 7. (10 pts) The NBS (N-bromosuccinimide) free radical substitution of 2,3-dimethyl-l—butene gives a mixture three products. Draw the structures of the three products. Are any of these molecules potentially chiral? (Yes or No) >_< NBS NBS = N-Br v2.2.6 28: .o. :6 Bosnia I|IIIIDIII|D I I l I llI'l|l!llllllill D i I t I I I I I I I l I I'DDIDIIIII||III a m o 2 O m 2w . . . g :b. 3.00 3.00 no.3 do :- 3 3 3 3 > m. .u m 0— >n n93 3.0m no.3 unbu uukm no.3 u: an nu n» um um ZEN: 9m 9m >m mm P. 3.. 3.09 3.: «who 2.9» 3.8 3.3 8.3 8 H 3.3 3.3 web» 3 .8 a. 20 . 3 3 mo 3 3 ma 0 ._.o m: 3: .un be On .3 m: mu Xm 3b.. 3.: 3.3 3.; Sub 3a.; 359 2: is.» 39.x 3:. Sub a; .8 .2 .3 .3 d 3 .3 8 S S 8. 2 a 8 Hm aw Om _—. any” >2 In A.— U—u W. Va >n 3: :3.» 5.0 :3.» so.» an.» 33 §b 39° 3... 2:.» 89° 38.3 38.2 38.3 mm ‘ we no 3 9» mm mm 3 mm mm .3 3 Om 3. Zn 3: 0.2.; U< Io m.. 4359—... Io; 29m 2;.» Iab 3.?» 399 3Nm 39o 3gb 3a.» 39o .390 mu .2 20 is n36 nun; * 533863 828 8 .8 :3 3a 2 “3 gm Pa “3.. 8.: 83 :3 3 no .2. um nunb S C 330 .390 no mu $9305 823 03 WW "3.. n3; ...
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