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Spring 2007 Test 1 - Chem 228 Spring 2007 Test 1 Dr Wells...

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Unformatted text preview: / , Chem 228 Spring 2007 Test 1, Dr. Wells name: R Q ‘1 Score=opscan* 70;“20 + essay: 20 * 70220 + 3‘3 = [O O Nuke Essay A (10 points) You have a bottle of bromoethane. Devise a method for making ethoxyethane. Show each step. You may use any other reactants that you wish, but all of the carbons in the product must come from the starting alkyihalide. 1"] ]"| t ‘ .Ifi_(}_~¢#~ce" . F H" L H P “a" M) [I - M C \ | \ 1 ‘1 l’l M H ‘ 3 W33 bum: C L0 ‘4 j H M E v 6. Z HH-L.MC_H(E‘; ’- u ”I "' 1" 5/1 . H 111 x-r1 /\/ . v [ Jfl fleéhi‘afflrfl "OiLEL’H H) \ \K ' ‘ x a 5N2. 1/1 H “mu-1.x] a l // M 3"} H 1 L HEX“ Essay B (10 points) Complete the reaction below. Complete the mechanism for the reaction. - show each step with curved arrows - don’t forget charges - Show stereochemistry in each step O prom-- C/ \ n'H O H— __ ., )- Et/ Me H20 mechanism: / K3r , .. /C).r> . ‘ /OC\ . r: PfitIII/‘W” C C'IVIIIH 9—. .J‘l \ Et //\\Me PF EH , E + L It\“"/ Essay B (10 points) Complete the reaction below. Complete the mechanism for the reaction. - Show each step with curved arrows - don’t forget charges - Show stereoehemistry in each step /0\ H30+ aw (3— w”: '- P I H H20 E1 Me mechanism: H pImIVC 0‘wa Hag—H Et Me 19‘” ‘Jw/ ' $1“ 1’} 4 \7? Me Essay C (6 points) Mystery compound: Molecular formula: C7H12FIO. Spectral Data: IR: Set of intense peaks around 2900 cm'] sharp intense peak at 1680 cm]. lH NMR: 0.9? ppm 3H triplet; 1.61 ppm 2H sextet; 1.98 ppm 3H doublet; 2.45 ppm 2H triplet; 4.18 ppm 1H quintet; 4.33 ppm 1H doublet. ”’C NMR: peaks at 13.6, 16.9, 21.6, 37.1, 42.3, 43.9, 200.1 ppm l4 1" W O 1/" «Jr " 1 I I ll l l | l I _____ _ _'_ /I.|" l l ”H C —r—-* C“ L M L H L L L k I . r 3 l, l l ‘ L H l t H l: l 1 l 1 Essay D (4 points) The IH NMR of CHgLi consists of a singlet at — 1.4 ppm. Yes, negative 1.4 ppm. Explain why the chemical shift of the singlet is so low. l'l , .1 '\ .. r ,5. its)? 63 W ‘7’ it"iwC-‘EW Hf cub ' l J. 1'1... 1y ‘\ r - “l H I k h Li \B If; lfit \l/v 903”} \‘Nig r L mgr UK. e /\(H . I I $3.; ‘) plfiei'xb “2) SW Lt '}%L\1C\ci_'_ \2 Shwflhb I Essay C (6 points) Mystery compound: Molecular formula: CTHIZFIO. Spectral Data: IR: Set of intense peaks around 2900 cm'I sharp intense peak at 1680 cm". IH NMR: 0.96 ppm 3H triplet; 1.23 ppm 3H triplet; 1.7? ppm 2H quintet; 2.14 ppm 2H quintet; 4.18 ppm 1H triplet; 4.37 ppm 1H triplet. l3c NMR: peaks at6.8, 12.1,21.1, 25.1, 38.1, 45.3, 200.3 ppm 1/1 )A 1/1 Q 1” i4 ’ i 1 b 1 \\ 1 E ‘ 1/1 .,.x”~~—<;”L #QfiC-fkmkfl H % \ 1 1. l l 1“ 1: ”1' H H Essay D (4 points) The 1H NMR of CH3Li consists of a singlet at — 1.4 ppm. Yes, negative 1.4 ppm. Explain why the chemical shift of the singlet is so low. 1 . Oxymercuration is used primarily because '"_"“~. (5/ skeletal rearrangement is blocked “(BY B. the use of mercury generates toxic waste C, Hofmann products result D. Markovnikov products result E. secondary alcohols result 2. ER spectra can be used to distinguish dimethylbenzene from methylbenzene by examination of the cm"l region. 6 A. 4000 — 3500 B. 3500 — 3000 C. 2300 — 2000 D. 1800 , 1600 {0:900 w 600 3 . Rearrangement of the carbon skeleton may occur in the synthesis of alcohols from alkenes and dilute sulfuric acid because A. alcohols are unstable in acidic solution B. alcohols get deprotonated in acidic solutions B C. double bonds are stronger than C-O bonds @‘J carbocation intermediates may rearrange E. pseudo pi bonds are formed in the process. 4. Molecules absorb radio frequency radiation when A. bond vibration is energized B. ring warping is energized C. symmetric stretching is energized D. asymmetric stretching is energized @ the spin of the nucleus is flipped 5 . What causes the signals in 'H NMR spectra to split? A. The presence of deuterium (2H) and tritium (3H). B. The proton region is near the edge of the electromagnetic spectrum C. The small mass of the proton. Vi C13) The magnetic field of the neighboring protons. E. The polarity of 1H-C bonds. 6 . The hydroboration—oxidation process is best described as .’\ f\_ /' . . . . . <13): Racemic S. Scalemic (T./,Anti-markovnikov U. Markovnikov 01, Syn-addmon W. Anti-addltlon ARTW 1;. SUV / (QRTV D.SUW ESTV ’37 7. Ihe best way to perform the transformation below is (31 1)TfCl/base 2) NaOH H B. 1)THF:BH3 2) Par3 c. I)NaH 2)H20 D. 1) LDA 2) (3].‘]_3C()21-I3 heat \\\\“.‘C OH E. 1)TsCl/base 2)(CH3)3C0Na Me\“‘/ Et 3 8 . The best name for compound 8 would be (13) 2—butanol B. ethylhyclromethylmethanol C. 1—ethyl—1-hydro-1-methylmethanol D. 2-ethyl-2—hydro-2-methylmethanol E. l-methyl-l—propanol 9 . The proton-NMR spectrum for compound 9 contains A. two doublets; two triplet; one sextet B. three singlets; one triplet; one quintet CC} one singlet; one doublet; one triplet; one quintet; one sextet D. one singlet; two doublets; one triplet; one quartet E. two singlets; one doublet; one triplet; one quartet l0. Alcohols have higher boiling points than constitutional isomers that are ethers because _ . A. the alcohols have greater mass B. ethers have greater dipole-dipole interactions C. ethers are more flammable [2. alcohols are racemic CE) ofthe polarity of H-O bonds. 1 l . Reactions involving chromic acid that are carried out in aqueous solution convert primary alcohols to carboxylic acids because A. chromium reacts violently with water B. the route to esterification is blocked C. water is moderately acidic D. chromic acid is not soluble in water @t aldehyde hydrates are formed / K ‘t/ Consider the MAJOR product from reactions of compound G below for questions 12 - 15. U)THRBH3 ———h~ (2)1120; OH C€3 U)THRBH3 C,CH3 /—>~ CH: \ H (2)CH3C02H,heat / C-——-_'C / \ \ (1) Hg(OAc)2fTHF H H 1- ® (2)NaBH4, OH‘ dil HZSO4/H20 C83 CH3 CH3 CH3—C— CH3 CH3—Jj—CH3 \C/CH3 H H / | CH3 \ r l | l; / L/ CH3 (I'll—13 CH3—J:—CH3 CH3—C—OH l‘ l H | H— —C—H CH3- C— C—H 12. Compound 12would most likely be R . 13 . Compound 13 would most likely be D . AW 14. Compound 14 would most likely be Q“ . (/ \/ / 15. Compound 15 would most likely be Consider the reactions below for questions 16 - 20 16. In the first reaction the alkyl halide was A. hydrated B. dehydrated V“ g. oxidized S @- reduced _ E. catalyzed i 7. Compound 17 is a(n) ”x QIA A. Lewis reagent @carbanion C. carbocation D. carboxylic acid E. ester 18. Compound 18 is Br {j 19 Compound 19 is (3:3 §{J*</ \FKWO“ 20. The extent of the reaction to form compound 19 might be determined by infrared spectroscopy by _ . I peak at 1700 cm I II peak at 3300 cm __“___'I 1; / - ., ./ \III fi" if] Ill—flflfih f tAz watching IHdisappear and 11 appear i1 i {K it} B. watching I appear and II disappear ' Pi C. watching I disappear and II disappear W .1.” D. watching I appear and I] appear E. actually 1R would not be useful because neither I nor 11 would change. ...
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