Exam3Fall2004 - CHEMISTRY 322aL/325aL Please EXAM NO 3...

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Unformatted text preview: CHEMISTRY 322aL/325aL Please EXAM NO. 3 Print W OCTOBER 27, 2004 A l: x K First Name USC ID No. TA's Name Grader (1) (15) __ x Lab Day & Time (2) (15) “ ‘ (3) (20) ‘ x (4) (10) “ ~ (5) (10) a E (6) (15) a K (7) (15) % “ (100) first letter of last name I will observe all the rules of Academic Integrity while taking this exam. M. Chemistry 322a / 325a -2- Name Exam No. 3 3 lack (1) (15) Complete the chemical reactions below by providing the missing organic product or products within each box. You do not need to show the byproducts of the reactions, or any mechanisms. Be careful to show stereochemical details in the products wherever they are important. Identify products produced as a racemic form. (A) c =CH CH3 H geHs (B) H(f\ CH2CHZBr + CH3O'Na+ methanol CHZCH3 , (100% ee) ’ 7 ; ' ( ‘C C_6H5 (C) Br’lC-‘H + CH3S‘Na+ CHZCH3 3L (100% ee) 4/: 0W H Br (D) (CH3)2CH H NaOH H20 / acetone acetom‘trile if (EH3 ? C H O (E) C1CH2§CHZ( |\Br 2 CH3 CH3 acetone, heat (100% ee) C7H150X note a halogen atom remains (0% ee) {Mm 0K / Chemistry 322a / 325a E N y3 Name xam o. (2) (15) aveack —-L M; aclu ”/6 90% Cl rc/c/ (A)(10) Circle the compounds 1n the box below that are chiral. (ll/Mfr <0 ~14) C6H5CH2CH=CCICH2COOH NH2 C6H5CH2ClI-ICOOH (B) (5) Identify each pair of structures 1n the box below as enantiomers, diastereomers or the same stereoisomer. Place your answer on the line below each pair. 9H3 @1ch3 CH3 L thmilbmd 18 a6 Chemistry 322a / 325a -4 Exam No. 3 g a (3) (20) Circle the answer that correctly completes each question or statement below. - Name (A) Which isomer (cis or trans) of 1,2—dimethy1cyclohexane is shown below, and what is the magnitude of the steric strain in the structure in gauche interactions (number of G)? (circle answer below) CH3 trans isomer and 2 G trans isomer and 3 G H cis isomer and 2 G _ cis isomer and 3 G . CH3 (B) The lowest energy chair conformation of trans-l-tert-butyl-3-methylcyclohexane is (circle answer below) H H CH3 H3 wma WCHSMH H H H (CH3)3C (CH3)3C (C) How many stereoisomers are possible for 2,3—dimethyl—4-phenylhexane? ¢H3 (circle answer below) CH3CH2CHCI-ICHCH3 C6H5 CH3 2 @ 6 8 (D) Optically pure (100% ee) (S)-2~ioclobutane has a specific rotation of + 15.900. A sample of 2-iodobutane that has a specific rotation of 4.590 is a mixture of 90% S and 10% R 55% S and 45% R 45% S and 55% R 10% S and 90% R (E) The name of the stereoisomer below is (circle answer below) 9H3 . t -, — .-- I n nentane Hr $‘ Br 25,3R)-2,3-dibromopentane HVCVBr ( , - , -oibromopentane (2R,3R)-2,3-dibromopentane CHZCH3 Chemistry 322a / 325a Exam No. 3 -5- (3) Contd. 2 MC 4/ (F) For an unsymmetrical compound with two stereocenters (carbon A and carbon B), there are four stereoisomers with the possible configurations at the stereocenters shown in the table. The stereoisomers that are related as enantiomers are Name configurations at (circle answer below) steremsomer carbon A carbon B only I and IV 111 S S onl II and III I and IV and also II and III I and II and also III and IV S R III R S IV R R (G) In each pair below, the more reactive chemical species in a nucleophilic substitution reaction with CH3I is (circle answer below) 1 NH3 , 11 CH3C02' “3' NH’Z' “£02" I” 1 11 _ 1 NH3, 11 HO I NHZ‘, 11 CH3C02' pairs (H) The order of reactivity (fastest to slowest) of the following oxygen leaving groups in a nucleophilic substitution reaction will be [-5 K: r: (circle answer below) R—OH R_OH2 R O 1 > 11 >111 alcohol alkyl oxonium alkoxide ion 111 > 1 > 11 ion I >111>11 I II In (I) The order of reactivity (fastest to slowest) of the alkyl bromides below in an 5N2 reaction will be (circle answer below) I>II>II I CH3CH2CH<CH3)CH2BI CH3CH2CH2CH2CH2BT (CH3)3CCH28I I H 1" 111>1>11 III>II>I (J) A mixed solvent of acetone-water contains an alkyl bromide (RBr) at a concentration of 0.010 M, and potassium hydroxide at a concentration of 0.010 M. The nucleophilic substitution reaction of RBr ----> ROH proceeds by an 5N1 mechanism with a rate constant of 0.050 s.1 at 500°C. The initial rate of appearance of ROH at 500°C will be (circle answer) rate = 5.0 M/S 0.050 M/s 0.00050 M/s 0.0000050 M/S Chemistry 322a / 325a -6— Name Exam No. 3 (4) (10) When (S)-1-bromo-1-phenylethane (I) (100% ee) is heated in 1-butanol, the ether product II is produced as a racemic form (0% ee). CH3qHBr heat CH3§ZHOCH2CH2CH2CH3 C6H5 CH3CH2CH2CHZOH c6H5 I (S) (100% ee) II racemic form (0% ee) (A) (2) In the box to the right, complete the stereostructure for (S)-1-bromo-1-phenylethane. (B) (2) Based on the product produced, what notation describes the mechanism of this reaction. S A] l (C) (6) Provide the detailed mechanism of this reaction below, starting with the stereostructure of (S)-1-bromo-1-phenylethane, as a series of chemical reactions with detailed structural formulas that show all key bond-making and bond—breaking steps. In the box below, draw the stereostructure of the key intermediate in this reaction. Identify the rate-determ' ' step and be sure your mechanism clearly shows how a racemic roduced. \_\ Q“: r DS = ' L AW W “$1 1* r Br 3 Emmi c g Catriona-t, Que “ 5 Hdu—oH'itS admit-rd TM; Q¢\'\\T‘L\ <““‘°°<°_H "K (“m-‘1' V‘MQ ‘\ m‘l’fl key intermediate erL'tS uk't‘k 4-le "I\\.¢.\zo Pk} \( C(-—hQ-\’0.N\.\ N\*\r\ (Rum.\ _—————————1 Chemistry 322a/ 325a -7- Name Exam No. 3 (5) (10) When a racemic form of 1-bromo-1-phenylethane (0% ee) is heated in (R)-2- methyl-l-butanol (100% ee), two separable products (A and B) are formed that have the same constitutional structure (shown below). ' CH3C;HBr + CH3CHZQHCHZOH CH3§HOCHZQHCH2CH3 C6H5 CH3 C6H5 CH3 I racemic form (R) (100% cc) (00/0 ee) (A) (2) Products A and B may be separated b fractional distillation. What is the stereochemical relationship of A and B? d“: 9,5;(1 n o Mt. VS (B) (6) Complete the stereostructures below for the two 3 parable products of this reaction, and indicate the configuration of ea (as a mixture of A and B) stereostructures for products A and B, and assign R / S labels to the \“ numbered carbo s on the lines below the structures . (C) (2) Will each of the above products (A and B) be optically active? __\[;s_ Chemistry 322a / 325a -8- Name , Exam No. 3 (6) (15) A chemical reaction of A ------ > C proceeds by way of the intermediate B. Draw the free energy diagram below for this multi-step reaction by incorporating the following energetics: (i) The standard free energy change (AGO) for A——--—> C is AGO = -2.0 kcal/mol. (ii) The intermediate B is 15.0 kcal/mol higher in free energy than A. (iii) The free energy of activation (A61) for step 1 is AG1= 20.0 kcal/mol. (iv)The free energy of activation (AG1) for step 2 is AGi = 2.0 kcal/mol. Your diagram must be drawn to scale and it must clearly show the starting, intermediate and final states, properly labeled. It also must show all the energy changes Free Energy Diagram for A kcal/mol 24 _ 6) 22—- 20 —~ ~~ 18— 7.. W“ A Q is In cu : u cu ca 3-. H-t q; > .— dd a 2 progress of reac tion Chemistry 322a / 325a -9- Name Exam No. 3 (7) (15) Answer the questions below. (A) (3) Briefly explain why a racemic form of a chiral product is always produced, if the mechanism of the reaction involves an achiral stage at any point. vax excl-nymk $41139 mace—n3 Jerl-vk gr; MTFer‘ (mgal- ReceAfib'K ?ou‘r\a\3 Vigil-“5 +. 'k—ke‘ We QMmq—dfiomu—s o";- t LK\rg\' Pn4g¢+.1—qu~ ?Q~\r‘,\g bet¢¢d wfik Qfiklk‘ ?T€>\¢§E:\f-\— lt-Q- Filth—\I SD +kL "(9.90 Q'r\¢—v\'\":‘b~\r\-S avg [Jr-e» sluc ¢¢ \"V\ Qqu cL\ qMou—vc’rs —— °~ T‘°~(9Mt‘q—CQ1~N\. (B) (3) According to transition state theory, as reactants proceed to products, they pass through a "transition state" that is higher in energy than both the reactant and product states. What is the source of the additional energy required for the reactants to reach the higher energy transition state? Th‘ \fivrxz-‘nq ngvq\{ r? <1 Ntufi‘owvi* \5 Cbmwex—‘Ntd \K‘V’Q *km Q\\q M\<L\ ?o+'lz—~,\"€\¢c\ K—h-Rr37 mi“? NA +0 “KC—k +lN=L +rn—AST-H0’IN Shite. (C) (9) Provide a statement of the Hammond-Leffler Postulate. How does it explain the reactivity order of alkyl halides (RX) in nucleophilic substitution reactions that proceed by the 5N1 mechanism? The. \—\—L eta—mica era»: +¥s+ v“ K; EyL'N‘Xififi.“ O’N‘s -_. Flacr‘ciomS {—Le‘ +r6msl+Fom 34-654 Cop-\zs \o.+~<7\ Q\°%§ -———\__J '\"\N7_ mack-10x Coorsl‘t’pxgfi Q'KA +l~g Mmt+t b’v\ fiatl‘x S+TVQ+° (‘1 FLSLmlolzs Jrllxk eroslwsj’ 3"?“3“ 6“ (xhrmzd‘xdtk s+~<£fk t-t— (3 qp'pT-chM? mg l’“ +l\z Sail (can‘t—Row» Jrka. \“DS \5 key-mi \xl—‘l‘f0\\[$[5 ?T‘OAU¢“ M: Q QC;?£bQC.—‘—:\O’K \x'vcrm \S k\coL\\ I,“ QNV$7 SD ‘90“; khr‘ol\/3KS L's a l\\3kl7 Q'KéfiK‘SD’YxTQ ?T0 ¢¢<;S_ Bx/ +\.‘Q H‘L ©6$4Ybl¢~lk( +\¢— TS \zaAim3% “‘3'le5 \Mhrlmfiéfa—‘l‘g Ra; com3]A¢Tcxl1\K Con—\DOQCJFTOM Ckorocl‘lt‘. The caeta‘ttvrrlj ovégfi c?— R14 Lowvooq/«lg LS. SKA “aim?“ 73 $0 > 20 >10 >w+h7|~ W73 orcler mglet‘bs +\\L S‘l‘a—lei‘bfi arele— 0‘? :leDDCmHDMS. [\(COT‘ é§m$ l7 +‘lVL S’fi’o\\=T\\‘~)—3 £3 0'? +\’\K \S S—k-mg-t-LAPLS «\Sb mglt—Q‘C‘l +\r\T$ 2%533‘\ i Omk+k<ze \IcAucs a? jig Ag? G + ““90“”‘83 est 2% 2% {04 Mrstkvrl‘ ko‘ ”33 4;“ R7< ...
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