CH251FINALAU2009KEY

CH251FINALAU2009KEY - Callam Chemistry 251 Autumn 2009...

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Unformatted text preview: Callam Chemistry 251 Autumn 2009 Final Exam 300 points H! 131 —— Tuesday, December 8, 2009 — 1:30 pm. — 3:18 pm. Key Name (PRINT) I have neither given or received aid on this exam [SIGN] Place an X in the column next to your recitation section. _ Failure to mark your correct recitation section will result in ~5 points from your exam total. .' _Teaching.Assistant. " _ SM2186 Christopher McDaniel _ _Mp1.oos m - __SM2186 _ Tuesday ~MP1008 m - Six-'12 186 Christopher McDaniel _ mm. - L 3 Chad Eichman _ — AVZM — KL330 _ AV ll ll Chrismpher McDaniel Christopher McDaniel N H 4:. Please read all of the directions for each section of the exam carefully. There are 13 pages including this cover page for the Final Exam. please check to make sure you have all of the pages. Make sure your answers are clear. Any structures or answers that are unclear or ambiguous are considered wrong. 1/13 Points Earned = ,Cailam Chemistry 251 . Autumn 2009 Part 1. Predict the Product. Predict the organic product(s] of the following reactions. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw all answers in skeletal form. You will be graded on the product you draw from the reaction no other information is needed for this question. (6 points each for a total of 120 points] 0 H 0‘ 3 NJLH 5.2” CI NaCN. . CH3 ""- (a) m I Br . / By 1. NaBH4, CH30H 2. HCI, H20 0 3. NaH b /\/\/u\ () Br m ' OH (C) ' H2804, heat ""-.. + W . 1. Mg“, ether 2. 6% - ,p- (d) W --. I \ 3. H20. HCI Br 0% 1. NaH (excess) l ‘ OH 2 B - ' A r (excess) l“- (e) __M—h*_ ' ' / HO 0 / 2/13 - Points Earned = Callam (f) Br D MD (9) U (M 3/13 Chemistry 251 CH3OH ‘——"—_——-—h heat NaOC(CH3)3 1. NaBH4, CH30H 2. H2O, HCI —'—'—-“_—""‘—F~—-—h- 2. H2O, HCI 1. Mg", ether Autumn 2009 [-10 g: /0 "‘- / GD Points Earned = Callam (k) (I) (m) 4/13 C15 SH GNOH Chemistry 251 HBr (excess) ——n—-——————_...—,. NaOCHZCH3 _—'—"—‘————->v 1. NaBH4, CH30H 2. H20 3. NaH 4. CH30H201 1. Bra, CCI4, hV 2. KOC(CH3)3 —""'""“—-‘—'~m-—> 1. PBr3 2. Mg°, ether 4. H20, HCI Autumn 2009 Se “‘16 o-—__/ Ho’\ Points Earned = Callam Chemistry 251 Autumn 2009 o c H . CH30H, H2804 5- 3 (m ___—__*_h____—* 0 . o H CF03, H2804, H2O CHSCHzMgBr Hmfifio MW CH3| Ha ac”; 91.439333? (q) 0‘0” D U 0 N/ I (r) Q/LOH s: /\ H “x H '4 A NH2 ' ‘ CH3 */ 8 OH -—. e m w —-—-———+ , (3’1er NH; O O )v 5/13 Points Earned = Callarn Chemistry 251 Autumn 2009 Part 2. Provide the Reagents. (5 points each for a total of 35 points) Provide the reagents or starting materials to perform the indicated transformations. More than one reaction may be required. You should number your steps if it is important for the synthetic transformation to take place l-Bv; ’ fly. C.ch O l-Uq0c(t_l§l5)§ O i. PCC,CH1CI1 OOH 2.. Wmfifir,/‘o‘r / 3. Hm. [41¢ OH l. N61 BHq ,CH3914 1¢HCHHLO -/\n/ 3. New, 6/ OW H- We» :- PM : (“1‘11 2 . HQ BDq )vcfl '3- Dzo'DCi u: D o O” -——2' a 01 ' , ' OH MOI-i O “0 1. H399 é/CHé 3 - Pa: 6/13 Points Earned = Callam Chemistry 251 Autumn 2009 Part 3. Mechanisms. Provide a mechanism for the following 2 reactions. Be sure to include all intermediates, formal charges and arrows depicting electron movement [20 points each for a total of 40 points] (a) H OH O _ _" Q a u H“? 55.0” 0 :rH- 0H OH 9 -—--—-'§ Pk\/‘LH E— Phdkn : “4% «3 914% n J .9 H '4 (-10 -—c.» won (426‘ 1413‘“ 14° ii “10;! 0 u 41‘ o’C‘I'I -oH P“ f) a _ 6H 0 Ha (1 I410 03) OH HSO,CH OH CL EH m ° 6 ‘ OH {4-0 15"" H “Hob—H CB #3 ’ ' - d —= rec --=~ w '5' c:- k“ u:- --., J “5 Col-l M HgCOI-i '2 ’H I Pencil ‘— We? \ 3 _7 ®\CH; H cot-l 5 p H 1 e 0% 7/13 Points Earned = Callam Chemistry 251 Autumn 2009 'Part 4. Synthesis. Provide a synthesis of the following ether. All of the carbons in the molecule below must come from alkanes containing six carbons or less. You may use any other reagents that you find necessary. [20 points] \/0 RV: ' B / V Nflol-l H Pcc O “V: “"4 O __'—’ "g" 0/ emu; \ I, 71% 3v 7—- H“. H 2.0 P I. No: l-l Kfi d e__..___.__.. \H 0” 0L 2- A By 8"; H c -——-——9 A By 7’ tut, cu... If“?! 8/13 Points Earned = Callam Chemistry 251 Autumn 2009 Part 5. Explain. (3) Acid and Bases. Rank the following compounds in order of increasing pKa [1 = lowest pKa, 4 = highest pKa). Explain your rankings using pictures and words. (15 points] Br/YOH H—Cl H-Br C|/\/OH o 0 CD C9 G) 6) C- i9 Bra 09 y c‘ d j e tannin-ML L piss wouch (are; Bug‘qu cl Maze QLEc-rvzouu._ THAN C" STRUNCIEIL [Hyatt-nay. Y‘H'AU 5|» (b) Functional Groups. identify the functional groups in each ofthe molecules. (15 points} 0 O ,0 CI N \_’_~" 0 , Acts 251m SULF‘WE AMWE cumming Mona; , O ' O /\ N /L JL L 1 ° " "“ ” ° (5 ' H H spoxm 9 our: Agnew”, ETHErL «MM/Q . O O O O1 7< A/Nib 7 0M9 *0 o.\ SH I Pardok'me Ace-mu mm. cAKSoxYLm-e .50 CVANAf‘i 148mg 9/13 Points Earned : Callam Chemistry 251 Autumn 2009 (c) Conformations. For the following di—substituted cyclohexane draw the two possible chair conformations and Newman projections looking down the C1—C2 bond. Label the more stable chair conformation and estimate the energy difference using A values (CH3 = 1.70 kcal/mol, [CH3J2CH = 2.25 kcal/mol) and [CH3, CH[CH3]2 gauche = 1.5 kcal/mol]. [15 points) H3CVCH3 2? CH3 H E 11 2- \ . —-——-h --.-—- H ' 1. H H" [~5'ILcmlfmol “1'0 CH‘SQXQGJ panale ‘4 3.”! 5 Idealism! l c443 e) C“; (fir - . to H- C. H Gar—(4‘) 2..‘-[ 5' heal/{Mal 10/13 Points Earned = Callam Chemistry 251 Autumn 2009 Pa rt 6. Explain. Answer 2 of the following 3 explain questions. Please indicate with an X which ofthe 2 that you want graded. Failure to indicate your choice will result in the first two questions being graded. (a) Elimination Reactions. Using Newman projections down the appropriate bond predict the E2 elimination product[s] for each of the following compounds. (20 points] Br I Br NaOC(CH3)3 m 7 NaOC(CH3)3 / CH3 cl-Ile-l:J . H (3'; - ' “FLU-l; _ ,' pk ,- l " QH \ ~B.V;: :H . l 3 ~ “’ércu ' ' ' Pu L”; 3 H Br Br _ NaOC(CH3)3 , NaOC(CH3)3 : . / 3 W . CH3 - . (“amt m CH3 .. “"2045 m - Le: :x ’ 'Pk 13:”: r H “5‘ .i ‘H', "5 c" 2” '- 1 - (b) Grignard Reactions. The following synthesis does not proceed with any yield of the product when one equivalent of lithium acetylide [LiCCH) is used in the reaction. When two equivalents of lithium acetylide are added the reaction proceeds to give a very high yield of the product. Explain why two equivalents of lithium acetylide are needed. You answer should include a complete mechanism.[20 points] 0 1. Li 2 H OHy (2 equiv.) 2. H01, H20 7 HO HO \r j Frmsr ea Dos»; l}! a l, Rub easa ' H-o-H r4 mes—L; K—fi ‘3 11/13 - Points Earned = Callam Chemistry 251 (c) Epoxides. The following epoxide {X} reacts with the indicated nucleo the proposed intermediate is molecule [Y]. [20 points) C) §3H3 H20"S\@ CH3 (Excess) O & 12/13 Autu m n 2009 phile in excess to produce molecule [2), Points Earned = Callam Chemistry 251 Autumn 2009 Extra Credit. [10 points] For each of the following pairs of molecules identify them as one of the following: constitutional isomers (1}, same molecule [8], enantiomers (E), diastereomers [D], or none of the above [N] by placing-the appropriate letter in the space provided. (2.5 points each for a total of 10 points) CI RF (8) H30 .J3H3 F Cl S 0 CH3 NH2 _ ‘d’ H N”? “SCWCH3 H OH O OH OH CH3 - OH ‘ p (b) b 6.4 _._ CH3 _5_ "'3C Br Br . 7 (a H Br OH OH H I) . Br 0 HO H __ (C) H OH H OH H OH O OH OH _ Partial Periodic Table I II III IV V VI VII VIII 1 H He 2 Li Be B C N O F Ne 3 Na Mg ‘ AI Si P 5 CI Ar 4 K CS Ga Ge Ar Se Br Kr 13/13 ' Points Earned = ...
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CH251FINALAU2009KEY - Callam Chemistry 251 Autumn 2009...

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