final exam 4

final exam 4 - - 31:2 cham'533l Final Exam Name (PRINT)...

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Unformatted text preview: - 31:2 cham'533l Final Exam Name (PRINT) Last, First Chemistry 3331 Signature December 5, 2001 SS# Please circle the name of your professor and class time where appropriate. Dr. Bean (T/T h 10 AM) Dr. Cai Dr. Bean (T/T h 5:30 PM) TOTAL Note: Present your student ID when you return the exam booklet I. NOMENCLATURE (12 pts; 3 pts each) Give an acceptable IUPAC name for each of the following compounds. Be sure to indicate the stereochemical designations where appropriate. 2) HO é CH3 HO E CH3 CH=CH2 4) Cl ll. FACTS: Total = 18 points 1. Place the following compounds in order of increasing frequency of the carbon - oxygen bond stretching vibration. (1 = lowest frequency, 3 = highest frequency) (3 pts.) 0 O O O D E] E1 2. Place the following carbocations in order of increasing stability (1=Ieast stable, 3=most stable). (3 pts.) CH3 CH3 CH3 ' + + I I + CHa-Q—giH-CHz CHg—cl:—-c-CH3 OHS—o—c-CHa 0H CH3 OH CH3 OH CH3 D U D 3. Place the following anions in order of increasing basicity. (1 = weakest, 3 = strongest base) (3 pts.) H N’ " s 0 CH ('5 CH 3 c 5 _ \ —‘ _ \ — H — \ 3 0 3 o 3 o C] D D 4. Label the pair of molecules as identical, enantiomers, diastereomers, or structural isomers. (2 pts.) OH H 1 a 5. Place the following halides in order oft heir increasing reactivity in the 8N1 process (1=least reactive, 3=most reactive). (3 pts.) Br l|3r I CH30H20=CH2 D D D Br =CH2 6. Circle the alcohol(s) that could be cleaved by periodic acid (H IO4). (2 pts.) OH OH ‘\\\OH OH OH OH H ?H \\CHSCH \ C—C" 3 0H \ H C\‘ 3 / I OH H30 OH lll. REACTIONS (40 pts; 4 pts each) For each of the following multiple step reactions, draw the Final Major Organic product, or necessary reagents, or starting material in the box provided. Be sure to indicate the Stereochemistry where this is pertinent. You may place intermediate products below the reaction for partial cerdit. 1) PBr3 1) 2) Mg/ether OH ——-———> 4) H2804, heat 1) NBS, light 2) ~——————» 2) CH30H 4) 7) /? 1) KMnO4jHCI/warm ~————-———> 2) NaBH4/CH3OH 6 (9 2) (CH3)3CO K 2) HgSO4/H2804/H20 1) Na/NH3 2) Br2/H20 OH m OH 8)W 1) BH3/T HF 2) H202/NaOH/H20 3) Cr03'pyridine'HCI Note: Cr03'pyridine'HCI = PCC 1mg OH CH3 CH3 OH IV. Mechanism: 10 points For the following reaction, propose a detailed, step by step mechanism to explain the formation of the product. Show all intermediates and formal charges, and use curved arrows to indicate electron flow. OH 0 H2304 C=CH E) CH 1 2 heat 3 CH 3 CH3 ' V. Synthesis: 10 Points l§r OH I CH30H2 —CH~C—CH3 CH3 VI. Spectroscopy: 10 points Carefully examine the five infrared spectra and the seven compounds below. Place the letter of the compound in the box beSIde Its spectrum. CH2 (EH3 0 ll HO - CHZCHCHQCHS CH3(C H2)3C— OH CH3(CH2)4CE N A B D C CH3 0 CH2NH2 CH30HZCHQC_=_ C- H E CH(CH3)2 G F Micrometers 1 mnsmlrmnc’e (Vb) a s L a e ‘4 ' lD .. .. l . . , ‘ H I 4 0 4000 3600 3200 2800 2400 2000 1300 [600 1400 :200 1000 800 000 0 Micrometcrs 3 4 5 o 7 2 0 l0 ll 12 l3 M 15 20 100215——-—-1———————1—__~L__+ I IL.— 3 . + m we CD - Ch \I c; o Transxnittuncc l%) to c.- i: U C) 0 \_/ C 6 o 4000 3600 3200 2300 2400 2000 I800 1600 [400 1200 1000 800 600 40 Wavenumber (cm—1) wavelength (#111) 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 8 9 10 1112 1416 100 1 ’ v 1 ‘ 3 ‘ 1 . 1 1 V ‘ 1 80 percent transmittance 1000 800 600 4000 .3000 2500 2000 1800 1600 1400 1200 wavenumber (cm ‘ ') i an _ . 1 V '1 30 1 _ n ' ' _ 1 :0 1 - 111 1 4000 3601’) 3200 1800 2400 2000 1800 1 600 1400 1200 1 000 300 600 400 Wavenumber (cm'l) Micrometers _ 2. 3 4 5 a 8 9 1o 11 12 13 14 15 20 100 5 > 7 I I ‘ l _V.,,...— 90 30 J \l O O t. O Transmittance (‘70) 1,; 0 O C 20 ‘ . 10 1 ‘ V 1) M 4000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000 800 600 40“ Wavenumber (cm-1) 10 (1‘. FREQUENCY GROUP RANGE (cm-1) INTENSITY" . A. Alkyl C—I—I (stretching) 2853 -—2962 (m—s) Isopropyl, —CH(CH3)2 1380— 1385 (s) ‘ .. and 1365—1370 (s) tert-Butyl, —-C(CH3), 1385—1395 (:11) . and ~ 1365 (s) B. Alkenyl C—H (stretching) 3010—3095 (In) c=c (stretching) 1620—- 1680 (v) R~CH=CH2 98541000 (s) and 905 —920 (s) R2C=CH2 (out-of-plane 880—900 (s) cis—RCH=CHR C_H headings) 675~73O (s) trans-RCH=CHR 960—975 (s) C. Alkynyl EC—H (stretching) ~ 3300 (s) cEc (stretching) ' 2100-2250 (v) D. Aromatic AI—H (stretching) ~ 3030 (v) Aromatic substitution type (C‘H out-of-plane bendings) Monosubstimted 690—710 (vary- 8) and 730—770 (very s) -o Disubstituted 735—770 (s) m Disubsfituted 680—725 (s) and '750—810 (very~ 8) -~ 17 Disubstituted 800— 860 (very s) E. Alcohols, Phenols, and carboxylié Acids 0—H (stretching) Alcohols, phenols (dilute solutions) 3590—3650 (Sharp, V) Alcohols, phenols (hydrogen bonded) 3200—3550 (broad, s) Carboxyfic acids (hydrogen bonded) 2500—3000 (broad, v) F. Aldehydes, Ketones, Esters, and Carboxylic Acids ' C=O (stretching) 1630.1780 (S) Aldehydes ‘ 1690—1740 (S) Ketones I 1680—1750 (S) Esters 1735—1750 (s) Carboxylic acids 1710—1780 (S) Amides 1630 - 1690 (S) G. Amines V N -H ‘ 3300—3500 (m) H. Nitrfles ' ' CEN ' ’ 2220—2260 (m) f . a Abmfiafions‘ 3 = Strong. m =. medium, w = weak. v = variable, ~ = approximately. Q flmxmm memzjlo \ umaoofi 9:54 0.“ 42m PE.sz 235 am 5. 4m 5w SE 4»: 6 Eu E> a» 5; 3mm» ‘ a» s» / . \ a; \ \ .82. $88 \ / \ K .. a m 4 m o 8 u . ,, x _. m n 2 g _u. 20 3.: we”; \ , \ 5.2 E... 2.82 5.82. 5.82.. 8.5. 3 3 , x a 3. a a A... 8 u 2m gm \ ,/ >_ w. v m n. E . WNMEN :8! . ,1? 11.2.35. 3.2;. 8.3“; $2.. 8.5 8.2;. ‘o no 3 mu 2 mm nm 3 um. 8 no a. an mu at um um “ x Om < 0.. .5: 1m 00 2. n: N: mm mm >m mm 9. E. .u rfifim‘. Sam _. . 89%. 2.2a 2.8: $3. Ea.” E: 3.um.mm..l.fl.ma 3.: :3. :33 3.3. . . 3 Q a 8 .3 A» 8 .2 8 .8 3 .8 {I an M ma < N_. 23 _so do mfz: Um be an ._.¢ “M ES. . m: @EN 383 2.3 3.82. 3.2. 825: 3.2. swag a: fig.“ :32; i .3 .3 W .3 fl .8 3 _ .8 .3 mo‘ mm: mm mm rm 7: "EV—Niom :42 >: In .2 2.8: 2.33.83 53.585 5.3. $3. a: £43.... 3 mm «mm 6.“ 6m . , \ 3. m >n ,. . .. Sn”: £982 a»: \ \ .Nw m ,. 0. 1:3... ,flm .3, 4’, 3,419 fig?“ . w 0m 3 Zn 3: mamcba .3 0< :0 m_. 43 <0 r: u.” éw E: E. . .3..,- an» :1? 25 was“. .32. as“: as. gas as. is u . 5 .30.... 2-: ! In. tam... . , , .- .3 3‘ c Zn 9.539: m: 2 Mm $252 8.88 825 8233 E: 33. >335 131:. 3523 5 3:33. .o :5 3: :E: o. .1. 09333.6: 9.. >353 32:1.- _ gun-«:07. 3: 3 3.70. $0.032: 9.323 vii... \.. ...
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final exam 4 - - 31:2 cham'533l Final Exam Name (PRINT)...

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