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Final_3331_C - ‘-V I CAN m 7 afil Final Exam Name(PRINT...

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Unformatted text preview: ‘-V\- I . CAN). m 7, afil Final Exam Name (PRINT) Last, First Chemistry 3331 Signature December 5, 2001 88# Please circle the name of your professor and class time where appropriate. Dr. Bean (TIT h 10 AM) Dr. Cai Dr. Bean (T/Th 5:30 PM) TOTA L Note: Present your student in when you return the exam booklet I. NOMENCLATURE (12 pts; 3 pts each) Give an acceptable IUPAC name for each of the foliowing compounds. Be sure to indicate the stereochemical designations where appmpriate. HO CH3 3 HO::j::T4 IIIIIIIIII|||||||||||||||||||||||| H0 CH3 CH=CH2 CI II. FACTS: Iotal = 15 rooms 1. Place the following compounds in order of increasing frequency of the carbon - oxygen bond stretching vibration. (1 = lowest frequency. 3 2 highest frequency) (3 pts.) 0 O O O a Cl D 2. Place the following carbocations in order of increasing stability (lzleast stable, 3=most stable). (3 pts.) 9H3 + CH3 ‘ CH3 + t I + CH3--c——§3H-CH2 CH3—C—C-CH3 CHa—C—C-CHg l \ l l OH CH3 OH CH3 OH CH3 E D ' E 3. Place the following anions in order of increasing basicity. {1 z weakest, 3 = strongest base) (3 pts.) H / N S 0 II II II CHg—C\O_ OHS—0‘0“ CH3“C E] D D \— O 4. Label the pair of molecules as identical, enantiomers, diastereomers, or structural isomers. (2 pts.) FE 1 D :1. Piece the toltowing halides in order of their increasing reactivity in the SM process (i=least reactive, 3=most reactive). (3 pts.) Br Br l l Br—CHECHQCH=CH2 CHSQH20=CH2 CHSCHCH=CH2 6 Circle the alcohol(s) )that could be cleaved by periodic acid (Hi04).( (2 pts.) 51 or 630 H OH 0 3 (I: _c‘5:-“ CH3 OH “SC“I 1 OH H30 OH H 7. Draw the most stable conformation of the compound below in the box provided. (2 pts.) CH3 Had“ CH3 ill. 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 piace intermediate products beiow the reaction for partial cerdit. 1) Para 1} 2i Wig/ether " OH —---——h-— 3io=< 4) H2:<04, heat 2) 2) CH30H i? 3) Hose )Cl_fi©_ l\ o . N, 2) N%)®SH 0 )1 NBS light D 1) KMnOMHCl/warm wet—i 2) NaBH4/CH30H OH O 9H3 '—=’~ OH 1)PBT3 5) m 2 CH C ®K® CH3 ) I 3h 0 0 1> H :9 N359 6) -——--—-——‘-—‘—~—~—-—m——p— 2) H9504J'HQSO4/H20 7) _ 1)NafNH3 /_____ 2) Brango 1} BH3J'THF 2} HQOE/NaOH/H2Q 8) W 3) CrOS'pyridine‘HCl 4) )VMgar 5) H30+ Note: CrngyridineHCl = FCC 9} OH OH "OH OH 10) CE> CH3 CH3 OH Iv. mecnanism: to pomts For the toiiowing reaction, propose a detaiied, step by step mechanism to explain the formation of the product. Show aii intermediates and formal charges, and use curved arrows to indicate electron flow. OH 0 H2804 C=CH _...___.__._,., OH | 2 heat 3 CH3 CH3 v. aymnESIs: 1U Paints From alkanes, alkenes, or alcohols of two carbons or less, any oxidizing or reducing agents, and any inorganic reagents, synthesize the compound beiow. Eff OH I 0143an ——CH —-c:_cH3 CH3 ‘1 runsmumnc'e (73») VI. OPWII U§bUPy. IU PUI] Ila Carefuily examine the five infrared spectra and the seven compounds below. Place the letter of the compound in the box beSIde [ts spectrum. CH2 ?“3 <3 HO-CHecHCHzCHa CH3(CH2)SC— OH CH3(CH2)4C"='N A B D C CH3 0 CHENHQ CH30H20H20EC— H E CH(CH3}2 G Micrometers 933' ‘ ' I ‘ I ‘ ‘ ’ I ' 400 _ 40m 3500 3200 2300 2400 2000 mo moo 1400 1200 :000 800 5‘00 Micrometers 1153' at? 553 Trunsmi tmnre (_%) m E: 4|» 133:: 4000 3600 3200 2 800 24130 2000 {800 1600 1400 t2EJO 1000 SUD 600 40 Wevenumbe: (cm-i} FE wavelength (lam) 2.5 3 3,5 4 4.5 S 5.5 6 6.5 percent Iransmilianue 4C an 3600 3200 2800 1400 300K) 1800 i600 Mm) 120‘} Wavennmber fem”) Micromecers Transnmlancc (9’s) 4000 3500 3200 2800 2400 IUD!) 180D i500 I 400 [2'90 Wavenumber (cm-1] FE m 500 3000 2500 2000 1800 1600 wavenumber (cm ' I) 7 MOO 8 10 lllZ i416 EODO 300 600 (1‘. . A. Alkyl C—I-I (stretching) 2853 41962 Isopropyl, —-—CH(CH3)1 1330 — 1385 ' .. and 1355_ 1370 terf—Butyl, ”C(Cfigjg 1385—1395 . and ~ 1365 B. Aikenyl c—H (stretching) 30104095 C=C (stretching) 1620—1680 R-—<:H=CH2 ' 985—1000 and 905 —920 REC: CH2 (out—of—plam . 880—900 cis-RCH=CHR CMH banding) 675—730 n-arzs—RCH=CHR 960—975 C. Alkynyl ‘ EC—H (stretching) ~ 3300 CEC (stretching) ' 2100—2250 D. Aromfifi'c' Ar—H {matching} 7 ~ 3030 Aromatic substitution type (CHI-I out—of—plane bendings) Monosubstituted 690—710 and 730—770 - 0 Disubsfituted 733376 In Disubsu'tuted = 680—725 and ‘750—8 10 p Disubsfituted 800—— 860 E. Alcohols, Phenols, and Cfirboxylié Acids 0—H (stretching) Alcohols, phenols (dilute solutions) 3590-4650 Alcohols, phc'nols (hydrogen bonded) EEOC-r3550 Carboxylic acids (hydrogen bonded] 2500-3000 F. Aldehydes, Ketones, Esters, and Qarbpxylic Acids ‘ C=O (stretching) 1630—1780 Aldehydes ‘ 1690—1740 Ketones ‘ 1680— I750 Esters 1735 -1750 Carboxyfic acids 1710—1780 Anfides 1630—1590 G. Amines ‘ NfiH ' 33004500 H. Niu'fles GEN ' - 2220-2260 (rm—s} (S) (5) (m) (s) (In) (V) (S) (s) (s) (s) {s} (S) (V) (V) (very 8) (very 5) (s) (3) (very 3) - (very 3) (sham. 2') (broad, 3) (bread. v) (s) (s) (S) {s} (a) (a) (m) (m) a Abbmvimiuus: s = strong, 00 =: medium w = weak. v = variable, ~ = approximatcly. @ Imzmm mo_mz.:_u_o \ umaogo 9.53 0." 41m. mrmZmzam z m 5 => Em Em 4m 5w 5E . s: 3 5 E» E» .5 E: a; mag—mm . ..>,. \/ - . .. .. .. .— n . z , .e _ a .... as... _ . .. as... as... . . .. __ .. .... m .o M / ,, .. .u 20 ,- M 93> LD> CD— .. 0‘!— 4h M “(0" CD m—Ih an a: F: 2-31: u' w “0'! "--l‘ 0‘"! w M m..- <23) an mm m 9'0 ch- . CH OJ (a CD 4% a) OCH nu . .9. ".3 ”mm w < 0.. gm: mm. 00 2. >m mm _. In .. com a..mm.mm au.m= ma.Mta. m..mmm ma.mmmo mm.maw.mmmmmm mm.... in- : . . ‘qa%.m.m um.mm. ammo» mw.ma .3. .3... an m—w M my man <... N.. 2.3 2.9 ...o n... ma: mm be On _: m: m... ...m v3. 3...... a... .33. a.» 3.8.... aw...._g...ms_§.3 .833 a... ......_..m..... :3... - .1- 3...... £82... 3...... .. mm... mm mu ....qu. .3 3—3... mm mm w . _..__#>:Iu .2 02 as. as. an..- 5 mm Om ..d an..- a... .....N......mu.l..m. 59...... ”8.3. .... ...... : 3.: ... .125: . . . .l «51!! nfl_m2::fi nU—LBIZq fl WWII . .. Imm : mo . ”.....HM...........H_.,..,.... ...... Om .3 Zn ..flm .5............n...._u....._ I... ” sew . ...... ma... .... .... .5..- :7:- 15015 3- nagicin 3.5!. _u u 5 .3 :11:- 93. 1.6:: .... 3.3.... a __ ...I... .... p. ... .. . ...... 31....2-2 ‘on an .9 3a 20 5. 32 mag “82 303% ‘néi. na:on_-a .q 335:... ... .3. .3: 4...... o. ....- nag-.610... 293% 1-53.. .9353. :2 c. 35:. {.523 9.39.3 _ 3m. 32 3.. am as c z... 3.53. mm gm bum 8.523 3.5 mag .. ... rllnc: ...
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