322as04_final_key

322as04_final_key - CHEMISTRY 322aL Please g FINAL EXAM...

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Unformatted text preview: CHEMISTRY 322aL Please g FINAL EXAM Print Last Name MAY 5, 2004 First Name SSN TA's Name Grader (1) (45) _ . (2) (20) __ Lab Day & Time (3) (20) (4) (20) __ (5) (30) (6) (20) (7) (10) __ (8) (10) (9) (25) (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 3 a?” (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, including a racemic form of a chiral product. "Workup" means adding water and bringing the solution to neutral pH. (A) CH3CH2C EC: Na+ + CH3CH20H CH QHZQ ECH 4 {ABC H20 N a. @ 6? give all the products 5 1?" __ (B) CH3 CHzCI-I3 + KOH heat CHER CHM—ls ((11 = -C -/.,C/ C ethanol L 3 H3 HZC ECCH3 (C) excess H2 Pt C:H2CH2BI' D T ( ) CH4» + HC=C Na+ 3 CH2CH3 (100%) ee (E) CH3CH2q=CH2 CH3 H20 Chemistry 322aL/325aL -3- Final Exam (1) Contd. (F) (i) 0504, pyridine (ii) NaHSO3, H20 CH CH3 (1) hot KMnO4, KOH, H20 ((3) ii worku , H O+ CH3 () P 3 H3 + HBr ()H peroxides ( z: «A °‘ %\ a r #6 7590‘” product gi es 3 ppt with AgNO3 m ethanol VM9-WT9 (1H:S (as Q erUVK 0+ cc___§ .3, *TW (M‘(7 q [(7 6M m ,1;— 4°?M CH3 (1) 5 HCI 21102 _ + C6H5(1::I\CH2CH20H heat (100% ee) \ (RH C H 3CH1CH CH3 (J) CH3CH2CHCH3 + PBra racemic form V out»? c. ‘c—M-M product does not dissolve in cold conc. H2504 Chemistry 322aL/ 325aL -4- Name Final Exam (1) Contd. SeHs (K) (C (i) NaI-I CH3 é\OH (" CH CH CHI 1—1 11) 3 2 2 (100% ee) (1) THF.BH3/THF (L) M (11) H202, NaOH/H20 (i) Hg(OAc)2/THF, H20 (M) CH3CH2CH2CH=CH2 (ii) NaBH4, NaOI—I Chemistry 322aL / 325aL -5- Name Final Exam (2) (20) Answer the following questions. (A) (6) Provide complete Lewis structures for the following chemical species. Your structures must show all valence-level electrons, and use the dash notation for covalent bonds. H —c ~14 2 2 , _, A \4 formula is CH3 formula is CH3 formula is CH2 and is neutral and is negatively and is neutral charged (B) (6) Draw two resonance structures for ozone, O3, in the box below. Your resonance structures must show all valence-level electrons and formal charges, and obey the Octet Rule(Qv\ \g Q\Lc-\'\"O’V\ $\[S'l"2 M) 0(<( o/d What geometry is predicted by VSEPR theory for ozone accordi to the resonance structures? 3 Sgg Q;- o Kki QQK-¥f1\ 1“?!) M ‘— bmxx- " o r *r' \§o~\cc\ x:\ man" (C) (8) Indicate whether each compound is chira or achiral on the ine below each structure. Identify all stereocenters in each structure with an asterisk (*). A stereocenter is defined according to the usage in lecture and our text. ' @ ll CH3CH2€ECH=CH2 CH3CCH2CH2CH(CH3)2 CH3 Clxlr-4L\ 9 Oxtkgxs 0 @ CH3, Pm ”1*- HKC—Cfil—HH / HO air C6H5CH2CHC1COOH CL ‘3 :3, 3 Chemistry 322aL/325aL ~6- Final Exam Name (3) (20) Answer the questions below. (A) (10) Give either the complete name for the provided structure, or the detailed structure for the provided name, as requested below. trans—1-bromo-4-methylcyclohexane @’ (most stable conformation) (B) (10) Identify the pairs of structures below as constitutional isomers, enantiomers, diastereomers, or the same compound on the line below each pair. C=H20H QHZOH /\/\/ W céHs’g‘H CHXE‘H CH3 CeHs MW?- 5 {Mann-C omu— S EOOH TOOH (CH ) CHCH CHCH CH CH CH CHCH CHCH H | OH H I OH 3 2 2 2 3 3 2 2 3 .____ CH3 CH3 CH3 H OH HO | H CH3 CH3 ‘ *1? e. o m1 {-5 CH3CH2\ CH3CH2\ FH3 PR ;=% C1 CH3 C] A 3 3 :51 mos m‘ vs Chemistry 322aL / 325aL -7- Name Final Exam (4) (20) The photoinduced free radical chain bromination of propane at 125° C produces 1-bromopropane and 2-bromopropane in a mixture of 3% to 97%, as shown below. 0 1? CH3CH2CH3 + Brz M» CH3CH2CH2Br + CH3CHCI-I3 (3% ) (97%) (A) (3) Answer the following questions about the formation of 2-bromopropane. Provide the initiation step for this reaction in the box to the right. x— 3.; a CHB'CH CH3+ HE G r (ii) «3&4 QH3+ B 1"? CH3CH CH3 \— 8.: (i) CHSCHLCH Provide the two propagating steps 3 that lead to 2-bromopropane in the box to the right. (B) (7) Use the homolytic BDE values in the table in the Appendix to calculate the standard enthalpy change (AI-Io) for each propagating step in the formation of 2- bromopropane. Show your calculation. Which of the two propagating steps above is rate-determining? S ( \ What is the standard enthalpy change for the overall reaction: 0 _ CH3CH2CH3 + 131:2 ————-—> CH3CHBrCH3 + HBr | AH — -\5 kg“ gm: K—RS) (-H (v) (me) Q 82) DHZU2 rq\\ : (TL-,8) JPC.88\ ‘ (aqSB- (“R 6‘ : -—\§(,+‘"H 2 ~48 kid/Md “‘50 “ Sum DHFA )— BfionA : 451—3) 2 —\3 (“A/"‘0‘ Chemistry 322aL / 325aL -8- Name Final Exam (4) Contd. (C)(5) Calculate the relative reactivity of a secondary—H to a primary-H in propane (Rsec-H /Rprim-H) in the bromination reaction. You will need to use the information about the product mixture (see above) in your calculation. Place your answer in the box below. tz‘bf‘bmo Qa—OPKML] CV.) __ 1-H\ X RS£Q~U RPHM‘H Rag—H __ CH 8/ ,, CV1 CV1 RPP\M—\+ 5/} \ \ l (Rsec-H/Rprim-H) (D) (5) Explain this reactivity difference that leads to a predominance of 2-bromopropane in the product mixture. Include a discussion of the mechanism for the brornination reaction with reference to the rate-determining step. Your explanation must specifically describe how the site selectivity comes about, but an energy diagram is not needed. Tlms Fea<+wt4r clT-Q—levcmcg TLS;\1§_+S ‘l—lxk {mu-‘17 BQVV‘ “4‘ diQ‘an-«LMCK tn ‘\'\r\ CONPL-K—TM rafi~~l¢M 5-5195 \Dk‘v'\ Br“ “*5 S‘H‘a Cf‘g R\+lr\u- +kz PP\M-— H_ T\"'~ 'CGFWchD’K O‘G‘ +l,\9\ Pr—oclu g F ,4 M1 d . . _ 0‘ £33....) (HsCHLCHJ‘ HIE“ O\ C HSCHLCHZ-l- Br“ \E_D_S—_> QHSCQ (H‘s—t. PEPE Tye 31:14:;gfiiiiEbf°4 *hk Subtflwgltuhmi com?o¢—LLL¢\'*\’\ Jrkk ?w?3\ T‘OLQ-kc._\ ‘5 m%\2c"t~k< :E?EE%%::§Si:LjfiEEE QVWiLwQ;;I§ P“ P \ (AWL LL\,~ Pg+k KZ\ WOLLfid‘S kg—Flh CtSu\‘\—TIV\S \’l/\ C~ 426—5“? r'ccl‘k 0-9 QT‘MKAT'EOK off 2—— WonWFKxK~ Chemistry 322aL / 325aL -9- Name__________ Final Exam A] l w (5) (30) Circle the correct answer for each statement or question below. (A) The correct molecular formula and U- number (degrees of unsaturation) of the structure below are: (circle answer below) C5H802 3U c5114o2 3U I l CSI—ISOZ 4U H O (B) The numbers of valence electrons in each of the structures below are (circle answer below) C032" 502 NO; I 20 II 16 III 16 I . III 16 I II 111 I 24 II 18 III 16 I 24 II 18 III 17 (C) Among the following organic compounds of comparable size, intermolecular attractive forces 1n the condensed phases are weakest 1n 0 c(irc1e structure) ll CH3CH2CCH3 CH3C7H2COOH CH3CH2CH2CH20H MW: 72 (D) The order of acid strength (strongest to weakest) among the following compounds is (circle answer below) CHZZCHCH2CH20H HCE CCH2CH2CH2C1 1 11‘ I>III>II>IV C6H5CH2CH2COOH CH3CH2CH2CH3 III>II>I>IV III IV I>II>III>IV (E) The structure below that represents the lowest energy conformation of cis-l- isopropyl-4-methy1cyclohexane is New (CH3)2CH (circle structure) (CH3)2CHN (CH3)2CH Chemistry 322aL / 325aL -10- Name Final Exam (5) Contd. ; CA/ (F) The following hydrolysis reaction of (100% ee) 3-bromo-3-methylhexane H O CH3CHZCH2c(Br)CHZCH3 ——2-———> CH3CH2CH2¢(OH)CH2CH3 (100% ee) 3 proceeds by -the 5N2 mechanism and the stereochemical outcome is inversion. — the 5N2 mechanism and the stereochemical outcome is racemization.. - the 8N1 mechanism and the stereochemical outcome is inversion. - the 5N1 mechanism and the stereochemical outcome is racemization. (G) The Newman projection formula of the required conformation of 2-bromo-3- methylbutane in the E2 reaction below is (circle answer below) NaOEt Br (CH3)2CHCHBrCH3 ——————> (CH3)2C=CHCH3 EtOH H CH3 Br CH3 CH3 H Br (H) Ozonolysis of an hydrocarbon, C7H14, yields butanal and acetone (below). The alkene consistent with this result is (circle answer below) I / CH3CI-I2CI—12CHO CH33EPCH3 x butanal acetone \ (l) The stability order (most stable to least stable) of the alkyl radicals below is (circle answer below) H2 OLCH3 0% 3 > H 1>11>111 II>I>III I II II 1 Chemistry 322aL / 325aL -1 1— Name Final Exam (5) Contd. 2 65¢ A (I) An hydrocarbon A, C 4H6, undergoes catalytic hydrogenation with excess Hz to produce a product C4l-110. Compound A reacts with NaNHZ to produce a product C4I—I5Na. Which hydrocarbon below is consistent with these observations? (circle answer) CH2=CH-CH=CH2 E! CHSC =CCH3 (K) The slowest free radical halogenation reaction of alkanes is (circle answer) <iodination > bromination chlorination fluorination (L) The numbers and types of separable monochlorination products in the photochlorination of 2,4—dimethylpentane are (circle answer below) €11.13 Cl 4 products: 2 chiral and 2 achiral 2 4 uroducts: 1 chiral and 3 achiral CH l-lCI-l CHCH ————-> 3? 2 3 hv products: 1 chiral and 2 achiral CH3 products: all achiral (M) The best leaving group in a nucleophilic substitution reaction among the anions below is (circle answer below) /‘ _ _ Nu:‘/:\ R-L —————> Nu-R + L:' CH3C02 HO CHssog' (N) Which of the following synthetic methods will convert propene to isopmpyl alcohol? oxymercuration then demercuration direct l-l+ /HZO (circle answer below) I II II and III conc. H2804 then H20 hydroboration then oxidation III IV I and III II, III and IV (0) Which of the cyclic structures (1,11,111) below react with water by path B? CH; «3—1 9” \\\\}—1 CH3 ’6' ._ "SH 1:) A CH3". __ “‘AI—I CH3kf\—-— "AH +/C (circle answer) CH R C\ 3’4 Sr c} fig onlyI 1 , 11 CH ““H 3Lc\_f“l—l H OAC X I 11 111 ‘ 1/111 Chemistry 322aL/ 325aL -12- Name Final Exam (6) (20) There are four alkene isomers of C4H8. (A) (16) These isomers may be distinguished by the stereochemical outcome of their reaction with cold aqueous potassium permanganate. Identify these alkene isomers and their products by providing detailed structural formulas for A '9 H, using the prompts below the product boxes. 4/ QH All ’2“ cold KMnO4,KOH QHSCH CH CH- (4 CHSCHlCl-l : c H 3 ,_ H20 B C4I-11002 chiral contains one stereocenter cold KMnO4, KOH H20 D C4H1002 achiral contains two stereocenters .9:— ‘k :H cold KMnO4, KOH l-lhg — C(‘CH I + emu—shoppe \r H20 [4 0 O H F C4H1002 chiral contains two stereocenters H C4H1002 achiral contains no stereocenters (B) (4) The relative stability order of the four alkene isomers of C4H8 is (Use the letters A, C, E and F to identify the structure of the alkene) (most stable) CT > E > C. > A (least stable) Chemistry 322aL/325aL —13- Name Final Exam (7) (10) Use resOnance theory to explain the following observations. Be careful to show all valence-level electrons and all formal charges in your Lewis structures. (A) (5) The two oxygen atoms in the acetate ion, CH3C02- , are chemically equivalent. Explain how resonance theory accounts for this equivalency, and draw appropriate resonance structures in the box below to illustrate your answer T‘IVEK qu "VKJO Qt'~\‘\\lk\9’V\—‘V Vesommmcl SJTT‘ULth Q‘— x—wz an—haj‘k «3% Jews Cousme—Je Ame Mafia- la 44* Hkfii) cock C~o 390M \3 QR Us? \1 k\0.’\'\+. resonance structures for the acetate ion (B) (5) The cyclopropenyl cation (see below) is unusually stable for a secondary carbocation. Also, all three carbons are chemically equivalent. How does resonance theory explain this unusual stability and the chemical equivalency. Draw appropriate resonance structures in the box below to illustrate your answer. H H Thar—L aJ—q 4"ka equ‘twm\°~~\+ TLS'D'KC‘xQSc S‘l'T‘Uc—‘f‘xus r\ 4— (\(C\O'P|‘OE>Q’T\> Cc.:\—?‘D-\\ H Tluc \vttsrii \\&S 3 Qfiuctx). cYClopropenylcation Cur\o04\\l 9““ (“\K to?“ % Q\‘\ou~ & ' resonance structures for the cyclopropenyl cation Chemistry 322aL / 325aL -14— Name Final Exam - (8) (10) Mass spectra analysis indicates an unknown organic compound (A) has the molecular formula C8H100. Reaction of A in conc. sulfuric acid produces an hydrocarbon B, CSHS- The infrared and 1H NMR spectra of A are provided below. INFRARED SPECTRUM "" Relatlve Transmutta e 3000. 2000. 1000. Wavenumber (cm-1) (i) (2) Briefly indicate the key structural information that is available about A from the above IR spectrum. ”Vla— <—\I\&Hsc3“LA—Tg*1‘& \K o-\D$°rp-\>\av\5 sUfijts'Il' “Ax MA dflk Oct-0M5.“ L or UL’V‘Y’\C ”H ‘ (ii)(8) Use the information from the nmr spectrum below to assign a structure to A. Draw the structure in the box above the nmr spectrum. @eHi-CHl—OH structure of A Chemistry 322aL/325aL -15- ‘ Name Final Exam (9) (25) Provide reasonable syntheses, as requested below, for either (A) or (B) below and (C) (next page) using any needed chemical reagents. Below is a Table of Selected Reagents that may be helpful. You are not limited to these reagents in designing your syntheses. Common solvents and acids and bases may be included as needed. Table of Selected Reagents HBr HCl H1 800;; PBr3 Brz CH3SOzCl H2 / Pt H2 /Pd(poisoned) NaBH4 NaNH2 NaCN NaH NaOEt KOBu—t NaOH Hg(OAc)2 KMnO4 O3 BH3.THF H202 (peroxides) (CH3)3SiCl/ amine base MMPP (magnesium monoperoxyphthalate) You may use other reagents as well in your syntheses. Answer either (A) or (B) below. If you work both parts, only (A) will be graded. Part (C) on the next page must be done. (A) Provide a reasonable synthesis of trans-2-methylcyclopentanol from methylcyclopentane. (10) or (B) Provide a reasonable synthesis of cis~4-methylcyclohexanenitrile from trans-4-methylcyclohexanol. CH3 0d— 3 (AF) \ [Ta—9 NQOE-V BQIQJ“ \ _——§E\— OH- OH osoLCH—S gxl (H3 so zq KCE’ Mums gage. DH)- _ 6E3 Ch); CH3 #rm Chemistry 322aL/325aL -16— Name Final Exam (10) (Contd.) (C) (15) Provide a reasonable synthesis of 2,2,3,3-tetramethyloxacyclopropane _ )6 ¢H3 from 3,3-dimethy1butyne by way of CH3; — C =CH ' ' ' ' " CH3C|Z~CHCH3 a 2-substituted-2,3—dimethylbutane. CH3 Hint: There is a key rearrangement in this synthesis that must be shown. DEXA‘KMZ—R'“ 0 H \e 0e. Appendix Homolytic Bond Dissociation Energies (kcal/ mol) (given as energy released when bond is formed) CH3CI-I2CH2-H CH3QHCH3 H-Br CH3CH2CH2-B1- Cngl-ICH3 Br-Br Br -98 ...
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