Final Answer key - Aut 2011 Chem 2373 Final Exam sec I ‘ ——————(14 Name the following compounds CHon a s 7 ~ b 0 Ho 1 H = en no

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Unformatted text preview: Aut. 2011 Chem 2373 Final Exam sec . I ‘ —————— _ (14) Name the following compounds: CHon / \ a) s 7 ~ b) 0) Ho 1 H = en no wider/20y ' *" “ ‘ ' ’ ~Ew5r (D \ ' 5 CHZCH3 z 3 5 3 ' 3 ‘I I a) ‘ 4’4'Ji°A/ard‘°¢*‘2'e”‘5:906 - ‘ 1&2! at’lterencé ’ 5 b) Ml’3-mefiil flier: -£—o/ 5 Iris {fligo/ofenfi/ 1+0) 5-! 2-—6u—£aneJlbL lint~ :1 ‘ _ ~ ' - v 0K 19 (x) 2. Multiple choice questions, mark (by an X) the correct answers, or all of the above (if applicable). Scoring: +4 points per correct answer, —2 for incorrect answers. A) HBr is a stronger acid than HF . The statement is wrong, HF is the stronger acid since fluorine has a much greater - 2/ .electronegativity. X . because the H—Br bond is weaker than the H—F bond. 4/ . X . because, in water, the small fluoride ion needs to be solvated; this increases the energetic 1/. cost of ionization. B) The following are correctly drawn, significant resonance contributors to the Lewis structure of nitrous acid (usually written as HNOZ in general chemistry). . - "0" " 2/ . + /o: ‘l' L)" . u — . H-146 —— - '.N\O_/H - :05: ~ “ ‘ ' no; t ’ "‘ 2/ ' . " / + Z - .H—N _ __- .N + H \c': \\O, / . v 2 [919; a y o fie. C) This quarter you have seen reactions in which'NH3 served as / MAX 3 Pt! . i . . 7%r {can of 3 . a Lew1$ base (nucleophile) . a Bransted acrd . a proton acceptor _ X . All of the above. 5f {'5 '(04) 3. Provide the structure of what is produced when cyclopentene is dissolved in concentrated sulfiiric . [7/ 1, acid (96% H2804). —> I 0 5 03 H /" : Oi“ 2”“ l Aut. 2011 Chem 2373 Final Exam p. 2' K E Y sec (14) Draw Lewis structures for: a) CH3CH2N02 and its conjugate base. .. . cH .O‘ CH3CHZ— bk . “5' ___._.._7 ‘ N\ a .9; not H no" _ 3 t a ‘ . - ? cafltc‘k‘ng Zf‘lx. Lawns W“5 1’) CH2N2 and its conjugate acid, H H + -- — + +H* + A} yzldzN H 1 “JEN?” H¥Na . H 2 \H'r 2 H 3 Inert Zp‘bs (Your BDE Table 'could be useful here.) Lift ‘ sh ow silt/J 2/13 //@ _ / (ms mm—se/lacifl/e) ¥or X=CL/ radical, shfilDI-li'l'j W‘Q‘E‘Ee—G’lrur ngr) whrda U endo‘fiflemlc (04) 6. C is— and trans-Z—butene fail to interconvert even on warming to quite high temperatures. Why is this true ? P {1/ fo‘hnrln m would brealt/ ' KIT—“()5th .— rLo overlA—P an': 90 . _ fir)- T—kuy tired} . PC a “tuna mowing“ Allmecl a'l‘ Clea/61» 5W5 Cglkm’“) (13) 7. Rank the following compounds and ions as to their relative acidity. Use 1 for the best proton donor, 6 for the weakest proton donor. ' ' L o _. + u / + + HONH3 CH2=CH-C-C_Hs H—F CH3NI_‘I3 H3O rank 3 ' I CHsNflz I 6 T ZF’cs 2‘75 Ant. 2011 Chem 237B Final Exam p. 3 [(EY ‘ sec 4 ‘ 8‘ Fill in the miSSing reagents: PTOductS or starting materials. Some arrows may re ‘ I multiple steps (and two or more reagents). @ . I. - Nah, H7, i ' z: ._..———9- —- CH3C..CH, NH3 .‘ 3 +2 12 1%” 9. Some reactions of achiral and racemic compounds produce mixtures of diastereomers or meso compounds even though stereocenters are generated. Indicate which of the following reactions will produce mixtures of diastereomers (by writing “diastereomers”) or meso compounds (write ‘ “meso”) after the reaction arrow. Scoring.- +4 for each Correct answer, —2 if incorrect. I You do not need to draw the product structures a — W rac... U ' )HZOL OH ‘ Izl- H10: stag I ‘1 4-?17 Br ca 3'; ‘ _ a We ,emmf, z ‘1 ‘ Hzo ' , - - a ma. ‘ (“HZC'H3 H7, N0 M as o Lfgts firms ann‘b'guer: ‘ M60 6/ Aut. 2011 Chem 237B Final Exam p. 4 K5 y sec (21) 10. Two possible reaction course diagrams (representing two distinct mechanisms) for the reaction shown immediately below appear further down this page. Which course will be followed? on: _ . v ' CH30H- solvent XV —-——'—._—+ heated C H3° ' Ms = CH3802 - (03) Which course will be followed? b 3 points (l 8) Now show the “structures” of each intermediate and transition state along a reaction path. You can choose either pathway; it doesn’t have to be the one you answered as the preferred route. , 8* 9M5 . z i . i S4, Hllofs/cavrtc'i' ‘ 5 (if/YMCAIZ of TS , 44rd ,nkrmea’. a mA-x.? 76 (09) 11. Indicate the relationship between the pairs of molecular structure depictions below. Your choices are: I (the two structures are Identical), C (the two forms are Qonformational isomers), D (the two compounds, or conformers, are Qiastereomers), and E (the two compounds, or conformers, are EnantiOmers). -If you indicate that the two forms are conformers, you should also indicate whetherthey are ‘enantiomeric’ or ‘diastereomeric’). - . 01. H. ' “6 Br ‘ “\CI’J.“5 : ' K/n. 3 ‘ II ' Aut. 2011 Chem 237B Final Exam p. 5 ' KE/ , sec (22) 12. Present schemes for completing two of the four syntheses shown below. (If you provide . answers for three or four, credit will be given only for the first two ‘answers ’,' so, be sure to cross out any answer you don ’t want to count.) You can use any inorganic reagents, acids or bases, typical solvents, sirane functionalized organic molecules of 3 of less carbons (including alkyl halides) and non—cyclic alcohols (methanol, isopropanol, t—butyl alcohol, etc.) that you need. ' , H A) /¢/H , H B) convert methylcyclohexane to Q ' -—> \ trans—Z-methylcyclohexanol' ~ ' H3 , O a s 0H . g .1 Cm —» \ \ You should use ’thinking backwards’ (retrosynthetic analysis) to derive the series of steps. You will get R partial credit for showing the final intermediate and reaction conditions, even if you can’t connect them V completer back to the starting material. ' 3, _ l 4 - ¢ NQNHz/Nyj 2.) CH3]: 3) [/2 Lind/arrai' A “health - l t; .— l "is!!ch ‘ 93174?“ I )4 Br . R/ H of SW” at L ’ OMS ,e‘Eo . . Autism Chem237B FinalExam p.6' 'KEZ _ ' sec (08) 14. Bromine and chlorine add aeross C =.' C bonds readily, but the corresponding reaction iodine (12) does not Work, Why ? (consult your BDE Table for ideas) 2' A . ' ' . ' X-‘X C" o . c—I'réa-nds wmk,.ffiur CI 515 x 81 LP. Wat-fro?" dwarf M4 “1:3: . Br 46 - as ' ' - ’b‘eééd' .‘Iv-I . L“! a: ~36 ‘3 However, the two reactions shown below are excellent high "yield syntheses. The addition of I—Cl actually goes faster than the addition of-Br'z. r ' _ fl 7 ,I-CL ‘ v ' 4 (I ~ . <————.— .—————> What do. these observatibns tell you about iodonium ions? C,éL Iado‘n Cum. for): do 797'“ ,‘ _ . ' C—OH ' {/14' ~—-—-—.- f" ’ ' N ,7,(10.) 15.. There is an overall 400—fold difierence in reactivity (reaction rate) with K1 in a 'K water/methanol mixture for three isomexic bromides (C5H13Br): l-bromo—4-n1ethylpentane, 2-bromo-3Lmethylpentane, 2—bromo-2-methylpentane. . _ ' ‘_ 5 ‘ a) whichismostreactive? ’irbmmo . Why? Paw-an flfier‘SA/Zv" 2;_D‘\v'v+—$ " . .F ‘ . . I » unmade/reek *— ‘ ' Br. I. _ > ‘ . r A . M L 2° but'wme HIV. 3" e '- 3 1 ' ‘ ‘ Mow 3N1 ~IS - zany!" 5342 ' '-_- eamfpobe.’ ' ' Ida-:3an encoder. . a Huerta flaweA A “3 PPS rolqr . a) ’lz-ffogd “iv— exyémm _ - «rot sms 1’) zifit‘ig’i’iiiagfi 3 WSNLOK n‘n «PG/(A mam Aut. 2011 Chem 237B Final Exam . p. 7 2 I sec (11) 16. Illustrate the use of the following reagents — PBr3 , concentrated (85%) phosphoric acid, and sodium acetate (N a02CCH3) by including them in a synthesis transformation. If there are stereochemical or other selectivities, be sure that your choiceof starting material serves to illustrate such features of the reaction. ' revs _c-oH -————» c-Bk 1W 2° . 397.33% Global/wit at 2° asowtaotot MC ’0~‘C[—CH3 in 0. diéflaam’b . I 0 or Mnaum (M, 0611A»wa w ap/KXOQA/VILT date/wag $64)er Ltft- 182$qu /q/3 :> A ~ . . . Ma Ma (09) 17. Concernmg the 1,3,5-tr1methylcyclohexane isomer shown , U Me, Draw two chair conformations for this isomer (below). Indicate which is more stable? fi/ ‘ i ' @ %@ more 54m 4: I a Estimate the energetic advantage (in kcal/mol) of the better of the twolcyclohexane flip conformations that you have drawn above. IA kcal/ mo l CW6! 3 “Q, ken/Q ’ 3/91‘3 24> tweet 2,94: 0 ejdfler‘ q—IS [chC 1196 ...
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This note was uploaded on 01/18/2012 for the course CHEM 237B taught by Professor Andersen during the Fall '11 term at University of Washington.

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Final Answer key - Aut 2011 Chem 2373 Final Exam sec I ‘ ——————(14 Name the following compounds CHon a s 7 ~ b 0 Ho 1 H = en no

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