322aF01final_key

322aF01final_key - CHEMISTRY 322aL/325aL ” Please K g!...

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Unformatted text preview: CHEMISTRY 322aL/325aL ” Please K g! FINAL EXAM Print Last Name DECEMBER 11, 2000 First Name SSN ‘ TA's Name Grader (1) (45) __ Lab Day 3: Time (2) (20) __ (3) (20) (4) (15) (5) (30) (6) (20) (7) (15) _ (8) (25) _.__ (9) (10) __.__ (200) first letter of last name I will observe all the rules of Academic Integrity while taking this exam. Chemistry 322aL/325aL -2- Name Final Exam (1) (45) Complete the chemical reactions below by providing the organic product or products (as requested) within each box. You do not need to show byproducts of the reactions or any mechanisms. Show or describe stereochemical details in the products wherever they are important. "Work-up' means adding water and bringing the solution to neutral pH] $H2CH3 H2 H "In. (A) (C\ Pd CH3 CH=CHCH3 single enantiomer CH3CH2CH2CH2BI' + K+ -CN CH3SOCI-I3 (DMSO) ?H conc. H2804 C CH HCHCH _"'"—’ ( ) 35H?) 3 heat major product (decolorizes cold KMnO4) KOC CH / (D) CH3 ( 3)3 (CH3)3CQH: \ Br H 1 % I ‘ product decolorizes Brz/CC14 (i) 3 eq NaNHz in liq NH3 CGHSCHBI'CHzBr (ii) workup C cu: C. E CH ir: sharp bands near 3300 and 2200 cm‘1 Chemistry 322aL / 325aL -3- 7 Name Final Exam V (1) Contd. Br H HBI‘ ‘ C HSQHZCHZCH 4 H3 / F=C\ no ero 'des H H ( p XI ) (F) Ox; ma < \vuc $0 :- chiral product of cold KMnO4, HO‘ CH (H) CH3CECCH3 H20 C4H602 single resonance in 1H NMR Chemistry 322aL / 325aL -4- Name Final Exam (1) Contd. 0H Br- (k) + PBf3 THF.BH3 (L) CH3CH2CH2CH2CH=CH2 (M) CH3 conc. HCl an12 CH CH /I-I 03H 3 2 <0) ;c=c\ + _ H CHZCH3 C02 (N) CH3CH2CH2CHZOH Chemistry 322aL /325aL -5- Name Final Exam (2) (20) (A) (10) In each reaction below, identify the Lewis acid (LA) and the Lewis base (LB) by writing these labels on the lines below the structures. Add nonbonding electron pairs as needed, and use the curved arrow formalism in each reaction to show movement of an electron pair. + _ CH3CH2'Q-ZDC 12 H + 9H2 CH3 (B) (10) For each chemical species below, add any missing formal charges immediately next to the atom where they belong. Note: there may be more than one charge in a structure. If there is no formal charge on any atom in the structure, write 1’none” on the line below the structure. Chemistry 322aL/325aL ~6- 7 Name Final Exam (3) (20) (A) (10) Circle the structures below that have one or more stereocenters, and identify lack each stereocenter with an asterick (*). CH3CFHg¥wIBrCH3 , CH3 Cl CH3- -CH2CHZOH CH3 (B) (10) Identify the following pairs of structures as constitutional isomers, enantiomers, diastereomers, or same stereoisomer. H ~$340H H ~$40H H LCAOH HO AC4 H CHZCH3 (2—HZCH3 D 5 qurc g: 0 M4 r S D 1 mafit‘e Mcv~s Chemistry 322aL/325aL Final Exam -7- . (4) (15) Answer the questions below. 3 M (A/ W (A) (6) Draw detailed structures of the following compounds in the boxes below. Your structures must show any stereochemical features indicated in the name. Name (E)-3—penten-2-ol 1,3-dimethylcyclohexene (B) (3) Provide a complete name for the compound below that includes the stereocenter. (C) (6) Carefully draw the conformational structures requested below. Poorly drawn structures will not receive full credit. Sb 3 H ,H M“ H H Y‘ a Newman projection of the C-C bond in the lower energy chair conformation ; 2 1L {Q 1,2-dibromoethane in an anti conformation 0f tr““S'Z'methylcydOhexan01 ! 7 Chemistry 322aL /325aL 3%- Name Final Exam (5) (30) Circle the correct answer for each statement or question below. (A) The molecular formula for the structure below is (circle answer) ' \ ‘ CH14 C8H16 C9H18 (B) The species below that are predicted to have trigonal planar geometries around the bolded central atom by VSEPR theory are (note nonbonding electron pairs are not shown) (circle answer) NH3 3H3 CH3‘ CH; H2C=O 1, 11 and IV 11 and IV I II III IV V only IV (C) The chloroethenes below that have nonzero molecular dipole moments are H Cl H C1 C1 C1 H }—1 (circle answer) \ / \ / \ c=c \C=C/ C=C /C=C\ only 11 C1 0/ \H H/ \c1c1/ \c1 c1 I II III IV 1,11 andIV I and IV (D) The specific rotation of (R)-1¥chloro-2-methylbutane is [on]25 = -1.64° in chloroform using the sodium-D line. Under the same experimental conditions, a sample of 1- chloro—Z-methylbutane shows [0t]25 = +0.82°. The enantiomeric mixture of this sample is 25% R/75°/o S 40% R/ 60% S 50% R/50°/o S 75% R/25°/o S (E) The order of acid strength (strongest to weakest) of the following compounds is (circle answer) CH2=CH2 HC ECH CH30H CH3CH3 III>I>II>IV II>III>I>IV I H III IV III>II>IV>I (F) The order of base strength (strongest to weakest) of the following anions is (circle answer) NHZ' CHSCOZ' CH3CH2' CH3O' III>I>II>IV I>III>IV>II I H 111 IV I>III>II>IV o Chemistry 322aL/325aL -9- Name Final Exam (5) (Contd.) (G) The conformation of cyclohexane at the top of the 10.8 kcal/mol energy barrier for chair-chair interchange is (circle structure) O-NH (H) The rate expression of the reaction below is given by rate = k [RX][Nu‘]. The notation that describes the mechanism of this reaction is (circle answer) 5N1 R-X + Nu' R-Nu + X' E1 E2 (I) The major alkene product in the acid-catalyzed dehydration of 1,2- dimethylcyclohexanol is (circle answer below) H conc. H+ //CH2 H3 H3 H3 H3 """"" ' ' '> .(-H20) H3 H3 CH3 Ha (I) Which of the following statements are true about 13C NMR spectroscopy? (i) CMR spectra are run on samples containing naturally occurring 13C. (ii)The ratios of the signal intensities of the resonances provide useful information on the relative numbers of magnetically nonequivalent carbons in the structure. (iii) The hybridization state of the carbon (sp3, spz, sp,) greatly influences the chemical shifts in cmr spectroscopy. (iv) Spin-spin coupling between neighboring 13C nuclei is always a serious problem in cmr spectroscopy. (circle correct answer) all are true statements (i), (iii) and (iv) are true (i .. 'l. ar e (i) and (iii) are tru Chemistry 322aL / 325aL -10- Name Final Exam (5) Contd. (K) The proton decoupled 13C NMR spectrum of an alkene, C6H12, shows resonances at the following 8 values: 13.7, 17.7, 23.2, 35.3, 125.1 and 131,7. Which structure below is consistent with this cmr spectrum? (circle structure below) H H CH CH H CH3CH2:C_<:I CH3\ 7C=<C 3 3 2>=< CH3CH2 FH H H CH2CH3 CH3 ‘ (L) The stability order (most stable to least stable) of the carbon radicals below is H H H H (Cerle answer) I - I 1"1 I . I . I>II II>I>III CHs-c—CH-c-CHa, CH3_q_CH2_q_CH2 CH3—c-CH2-q-CH3 m H 1 CH3 CH3 CH3 CH3 CH3 CH3 > > I II III (M) If the relative reactivities of different types of H in the free radical chlorination reaction are 3°-H/2°-H/1°-H = 6.0/3.0/1.0, the expected relative yields of products I and III in the photochlorination of 2,4-dimethylpentane is H 1? c1 CH -¢-CH _ -CH ...... .3 ..................... -- 3 2 $ 3 hv 1 CH3 CH3 V H H H H H (Fl u I u CH3—QZ—CH2-ch-CH2CI + CH3-¢-CHCl-F-CH3 + CH3-c-CH2-c-CH3 CH3 CH3 CH3 CH3 CH3 CHa 1 III (circle answer) [11/ [III] = 6/ 1 4/ 1 1 / 2 (N) The most reactive ether below in reactions of the general type: R-O-R' + Nuz' ——> mm; + R0 is CH3 I (circle answer) . I E CH3CHZOCH2CH3 CH30g§H3 3 (O) The constitutional isomer of C4H80 with the highest boiling point is (Cirde answer) CH3CH2 CH3 1 S CH2=CH-O'CH2CH3 O Chemistry 322aL/325aL -11- Name Final Exam (6) (20) Photobromination of ethylbenzene yields only 1-brorno-1-phenylethane as shown below. No 1-bromo-2—phenylethane is formed. hv r C ——CH -CH -Br C 61—15 -CH2-CH3 + Br2 ——-.——-—> (361—15 —§H-CH3 + HBr 6H3 2 2 ethylbenzene 1-brom0-1-phenylethane 1-bromo-2-phenylethane not formed 131—10 Values in kcal /m01 (Defined as heat liberated when bond is formed from radicals) r C 6H5 _CH2'CH3 C 61"15 "-EH-CH3 C 6H5 "" CH2'CH2+BI Br-Br -46 T T T H—Br ~88 -85 -98 -51 '69 bond strengths for different C—H bond strengths for different C-Br (A) (4) Based on the DH0 values in the table, calculate the overall AHO in kcal/mol for the observed reaction that produces 1-bromo-1-pheny1ethane, and the reaction that is not observed that produces 1-bromo-2-phenylethane. Place your answers in the boxes. AHO for reaction producing 1—bromo-1-phenylethane:: Qua—HIGH; +— Br—Br ~——-> (A) sea n+3 +— \s- {sv- DH" ("8%) C~H~b3 Q— 5&5" (~88) O NP 2 Z b843,”; ~ 2 Dams: :. L—S\-sa)«L—es~%) NAG '1 " 8 Keel /M\ kcal/mOI AHO for reaction producing 1-bromo-2-phenylethane:: Cu Hyattst + Br—Br L34 LtASmLCHLi,‘ + H —g P WA" (was) ("Lt s) t— as) L~ 8?) (9 "‘ O A“ ‘ 1 Mia ‘ ébHrzr-d”. =C~m—229)—(£t8—%) as NAG : —- \Zs. \L‘q/wfi kcal/mol (B) (2) Do the calculated overall standard enthalpy changes for the two reaction paths explain the high degree of selectivity observed? Explain your answer in one or two sentences. Nuq‘hfibmi at.“ +\ALVMD&7—Kkm‘\qkkn\ Q¢00m\o\k Glam/k Mean—w; log—VAN (waive—t skim; ark \A‘L\OQ W—kc‘lm-n—l‘ Chemistry 322aL/325aL -12- Name Final Exam (6) Contd. (C) (4) Photobromination is a free radical chain reaction. Write out the two propagating steps for the chain reaction that produces the observed product 1-bromo-1- phenylethane. . (step 1) (Alchst +3.» ——9 chuggint +~ tt—Er . B (stepz) CouycucH$+ fivfir 7" cauydthits -\— Br’ (D) (2) Calculate the standard enthalpy change for the propagating step that is rate- determining. Place your answer in the box. EDS \S SAL E \ o ,_ o 0 \ - 3 An A D—H (was) — DH CQ~H 5H0: (“8 — L“ 9 2 ‘3 kcal/mol (E) (3) Write out the rate determining step for the photobromination reaction that would produce 1-bromo-2-phenylethane and calculate the standard enthalpy change for this step. Place your answer in the box. Rate-Determining Propagating Step: (aHgCH’LCH‘S +- Br —-—> (69%“sz 4‘ {4.4}? mi": DHOLH~BQ A NW Cc~(~(\ 4~ \0 mac _ 1 +—‘0 kcal/mol (F) (5)”ljlxplain in two or three clear sentences how the calculated enthalpy changes above account for the site selectivity in this reaction. How is the site selectivity in this reaction related to the stability of alkyl radical intermediates? Be specific. The sih aka-NH»! \s «seem-Md \n-J m n6“ cum to» ~ sci—03¢“ cuss-hm L:de skps +ku_+~ cum fl 0 . ruk~ cl L rm} x\ry\s Tl/x-Q CaKcLLK cc’bLCL QR \I C4.\&.g( g \ Sufijcs-t— Gun Eck 0.4— (to Z—J—f \<<.a_\/\v\o\ 40v +\«¢ obscwc‘g; 'Pedrbx mmé. (,th Et‘g o-‘v— \7_-l\( kc<.(/rv\o\ 9,... 45kg o-‘(Lch Pct-HM 9% (beget—mes +H’BPECL: 3+ (“get—\ch +Y>v . 3 2/}: (GQYmCsLH-Wr E1: \1~W (rs _ The. AILfiN-Vu; tax EcL \dec‘s chVctg °~ dx‘L‘kN-NK V‘x 5+k\}\{$-\ 0+ ‘k—kk o.\\4k(\ Vuflkcg\g bzkfis .QhW-4' O QHYCHCflg (g Mm 3+°J0(<++\0~'V\ Ca'H-yQHéHL Chemistry 322aL/325aL -13- Name Final Exam ' (7) (15) Compound A, C9H10, reacts as shown in the scheme below to form compounds B and C. Assign structures to the three compounds based on the 1H NMR spectra shown below. Place your structures in the properly labeled boxes. Assign the resonances in each spectrum to the hydrogens in each structure by drawing arrows from the hydrogens inveach structure to the resonances in the corresponding spectrum. Chemistry 322aL/325aL -14— Name Final Exam ‘ (8) (25) The table of reagents below may be helpful in answering the following questions. You are not limited to these reagents in designing your syntheses, but they may be helpful. Also be sure to include common solvents and acids or bases, as needed, in your syntheses. Do not show mechanisms but each step in a synthesis (reagents and products) must be shown as a separate equation. Table of Selected Reagents H2/ Pt Hz/Pd(poisoned) NaBH4 NaNH2 NaH NaOEt KOBu—t NaOH Hg(OAc)2 KMnO4 O3 BH3.THF H202 (peroxides) MMPP (magnesium monoperoxyphthalate) You may use other reagents as well in your syntheses. (A) (15) Provide a reasonable synthesis for the following transformation using any needed reagents. H OH H" H CH CH H u" CH3CH2CH2BI' -----> '. 2 2 3 + 3 u” H HO OH HZCHZCI—I3 (zs 3R)-2 3-hexanediol (2R'3s)'2'3'hexanedi°1 1-bromopropane is the only as a racemicform source of carbon for this synthesis. g '1 Jo ' e @ (IX3QCL NKUH’zg/UKQHS 1 ‘ L~1\ H 09 3 Lu“ UH-S ‘ 5 eugenicng H H x ( C_.Q_HLC_Q.| cu M9 CZC 3 P9\$o'h¢d Cd. QH/ \ a QHLQkCHS Or (flog 0y. ,Ecxclm 4 COM KMflDy or Ho“. Hzo “cm—«$03, Hgo Chemistry 322aL/ 325aL -15- , Name Final Exam (8) Contd. (B) (10) Provide a reasonable synthesis for one of the following two transformations. Do not work both. If you work both and do not indicate which one is to be graded, only the first will be graded. Chemistry 322aL/325aL -16- Name Final Exam (9) (10) The oxymercuration reaction is highly regioselective and stereospecific as illustrated in the example below. tetrahydrofuran (A) (4) Sketch the formation of the key intermediate in this reaction as an equation or series of equations. Use curved arrows to show movement of electron pairs, and show formal charges. Draw the structure of the key intermediate in the box below. You do not need to draw resonance structures of this intermediate. éA key intermediate off» a (B)(4) Sketch the reaction of this key intermediate that leads to the oxymercuration addition product above as a series of equations. Be careful to show any proton transfer as a separate step. Your mechanism also should clearly show the stereochemical outcome above. CH km amatg Q0“ ' H @ 3+ '\ 2‘ Gum 'l‘ 3.0 Hébdn CH3 (C) (2) What factor seems to control the regiochemistry of this reaction? vogue drdeS mom dart“: OSH-x‘v( «rebut ?~\ Wuw-ctucht 'KKLKW (0 U ~ ...
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322aF01final_key - CHEMISTRY 322aL/325aL ” Please K g!...

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