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exam2key - SF 350 Organic Chemistry 2nd Exam NAME Spring...

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Unformatted text preview: SF: 350 Organic Chemistry 2nd Exam NAME Spring 2011 TR 2:10-3:25 Prof. Bartmess (C ey 8) 1) (14) - This is a 75 minute test. - Please think your answers through before committing 2) (06) them to paper. - Be legible; if we can't read it, it's wrong. 3) (55) - Don't put any answers on the back of the pages (although you can use the backs for scratch paper); 4) _(08) material on the back of pages will not be graded. _ - Please put your name on the back of the last page by the staple, as well 5) _(31) as above in the blank. 6)— (24) 3’) _(12) 1.(14) Give complete IUPAC names for the following compounds, including BIZ stereochemistly. H OH / H Br ’1 H... ’ H .0, ME Cl‘ 5. a: ’ npr/ \c=c/ /C\. /C\ H’ \ H30 32 CHZCH3 2. (06) Circle the side favored at equilibrium (higher concentration): q —h. 3 6 H202 + MCO' “— H00- '1' MCOH H: + H20 NH3 + rho+ 9w 3.(55) Give the missing reagent(s), reactant(s) or product(s). Include stereochemistry and regioehemistry where pertinent. If more than one step or set of reagents are necessary, be sure to g—Bv 13 make each step clear. a. CH3CH2CH=CH2 b. Et E C. H Me (1. Me H e. Me ./ H“\|-C—C Et/ \ f. ICHZCHZCH2C1 g. H Me 30:03 Wm”, CH3CH2CH2CH2B I‘ \ 1.12 c67— a—mkd It —I- “1% --l ”rang Pd H L“ stow Cog MN ) —\ Mm 550’“ Bl'z H20 03 ——.—-.—————’-._._._._+ -78C, CHZCIZ l eqv. HS' -—--——--—-—--+ DMSO RC03H Jr fiWcLM C?“ drew ‘5 g Bsz H202, NaOH ———-—-——)-—)-r WET L-#\W' H Et20 H20 Et ; 0T5 OH é . T/sYiZl -- 2_ \A ' MD pyridine M Et 31.2 w M M143“? ‘ \ ET ‘u j \ .--- 1.4 Walk. \Et CH2012 H Bf at LOW SW “/1 "” 4.(08) The following conversion is not accomplished by a single set of reagents, but by more than one consecutive reaction. What reagents, and in what order, are needed to carry out this process? 1'31‘ CH3CH2CH2CH2BI' '——"* CH3CH2CHCH3 $2.0” W ,. fl MW: star A M _ M Lag 5.(31) There are six different dichloro-cyclobutanes C4H6C12(including structural and stercoisomers) that exist. a. Draw the structures of these six isomers below, next to the I, II, III, IV, V, VI. b. Circle the designator (I to VI) of the ones that if pure would be optically active. 0 For each horizontal pair (I & II, III & IV, V & VI) indicate their relationship: structural isomers, diastereomers, enantiomers) on the line between the pair I. Q08 _5—_”\~::i- HC/ELj/LQ v. l LQ ‘ . v1. . DK Big: {E IEKCQ 6.(24) a. Give a complete curved-arrow mechanism for the following reaction, leading to both products shown (and o_nly those products). Show all intermediate structures and steps. Circle the straight arrow of the rate-determining step. \L’f ”H lTf / (CH3)3C- -CH-CH3 .———+ \c=c + N—E' b. Label the chirality center in the reactant with an asterisk. Circle the intermediate that is the 5 product of the step where chirality is lost. + 2, Jr 1 Lt c. What IS the valence bond structure of the -0Tf group? What is its name? 4 2, P692? . "CPL \a-h- “FIL- \(‘O 0 +1" . ”CWLg-AQC a3(12) For each pair of reactions, (a) give the likely mechanism (E1,E2,SN1,SN2) 0n the “mech” line, (b) state which is faster, A or B, on the “faster” line, and (c) underneath each pair of reactions, briefly indicate the chemical reason for that rate difference. Note that I don’t want the non-answer “because structure X is different (or faster, or slower) than structure Y”, but rather the chemical reason why a structural difference affects that mechanism. I’ll state that both A and B go by the same mechanism in each pair /_\ A. CH3CH2CHZI +NaF in DMF -> CH3CHZCH2F Mech; 5w 1 {a ( B. CH3CHch-I21 +NaF in MeOI—I —> CH3CH2CH2F Faster: IQ: 1‘ * 0*?”‘3—6“ 5,660ij Wham Nut. 4L A. tBuBrinMeOH -> MeZC=CH2 +HBr . Mech: E l + I B. tBuBr in iPrOH _> MeZC=CH2+HBr Faster: A ’t l VPoicM 5mm stale; [Ly—a “TS +I “H A. EtOTs+KClinMeOH .> EtCl Mech: (SM D, Pt- I B. EtOTs +KI in MeOH —> EtI ' Faster:___ B J? l o Ufli‘c) 0494 EM W S [ (135:;me Bing km: ...
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