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2007 key - Name SISD Chemistry 35 , Exam 2 Answers —...

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Unformatted text preview: Name SISD Chemistry 35 , Exam 2 Answers — April 9, 2007 Problem I . (14 points) Provide the products for the reactions shown below. If stereochemistry is important, be sure to indicate stereochemistry clearly. If no reaction would occur, write NO REACTION. A) (2 points) No partial credit 0 O o Intramolecular Mitsunobu reaction. B) (2 points) No partial credit MUST PROVIDE EXPLANATION!!! N N SEA/\OMS a 3 NO REACTION F’h DMF This is similar to a neopentyl system. Beta branching does not allow an SN2 reaction to occur. C) (2 points) No partial credit A(E) CB oisfl or; The base is very strong, with a pKa around 35. However, it is a poor nucleophile because it is very sterically hindered due to the two tBu groups. This base ' deprotonates the thiol to generate a thiolate, which is a strong nucleophile- The thiolate initiates an intramolecular 8N2 reaction With the benzyl bromide to give the cyclic thioether. Name SISD (C) Snl reaction — good leaving group, resulting carbocation has a resonance structure with complete (9 octets on every atom. NOV RACEMIC PRODUCT — 2pts (1 pt each enantiomer or 2 pts for squiggly line) 0 O 0 NC CN 4 QtS total [or ert C (d) H H a Q) H Me Me—S Na H S H Bl’ ___> 1 I \Me THF 20% Me Me Me S R S I I toriabeled4 R Slow Sn2 — inversion of configuration, Sn2 at cyclohexane requires axial bromide (WHY ? when the Br is equatorial, the axial hydrogens on the C-Br backside block backside access. When the Br is axial, the molecule acts as a normal 2° halide. ) “CORRECT” — 2pts 01' “N0 REACTION” w 2 pts if student states that this reaction required extensive heating in lab (No points if student does not give explanation) 6 gts total [or Qart D 10 it Name SISD Problem 6. (18 points) Provide the missing starting materials (left box), reagents (box around arrow), product of first reaction (middle box) or product of second reaction (right box) that complete the following reaction sequences. (a) 6 pts total H2304 1 pt - aoetic acid 1 pt — sulfuric acid or Ts ll (catalyst) (HCI, HBr, HI not acoep able 33:35 as ea a ys since e X- reacts TsCl I pyridine @ 0° C NaOH or MsCI I pyridine @ 0° 2 pts D 2H 2 pts either Sulfonate 1 pt if mt at 0 C 2 pts - must be retention Best - 2 pts Ms-Cl pyridine @ 0 followed by Br“ of I‘ NaCN I DMSO 2 pts (may give some SM) 2 pts - substitution of OH group must occur with inversion HCI I ZnCl2, HBr/HZSO4. or Hl - greatest chance for mixed Sn2 / SM 2 pts 0 pts for H2804 + NaCN 11 Name SISD D) (2 points) 2 points for either answer, but 2 points total. No partial credit. 0 Ph - Ph 0 Ph AOH WM OR /l§/\ JJ\ N O 0 Ph 0 Ph OTs A \f This is an SNl reaction. The tosylate leaves to generate a carbocation. Resonance distributes the positive charge over two different carbon atoms, either of which can be attacked by acetic acid to generate the indicated products. E) (2 points) No partial credit. OH 1 equivalent of NaH OH HOAQ—J Me04©—/ “I equivalent of Mel NaI-I deprotonates the most acidic position, which is the phenolic OH rather than the benzylic OH. The resulting phenoxide ion does an 8N2 reaction with Mel (Williamson ether synthesis) to generate the methyl ether product. F) (2 points) No partial credit. 0K to show the Na salt of the product. Q‘ 00 5‘ NaN3 Q‘s/OH O N /\/\/ o CHSCN 3 o The starting material is an internal sulfonic ester, and behaves just like a tosylate or mesylate. Azide is a strong nucleophile, and it attacks the sulfonic ester (8N2 reaction) to give the ring opened product. G) (2 points) No partial credit OH PPhg, DEAD —» NO REACTION N NHZ The Mitsunobu reaction only works with nucleophiles that have a conjugate acid with a pKa between 3 and 12. The primary amine in this reaction has a pKa of approximately 35, which is outside of this range. Name SISD Problem 2 (14 points) A) (12 points) Provide a detailed, step—by—step mechanism for the following reaction. In this question, we provide the reactant and reaction conditions. YOUR JOB IS TO: 1) Show the individual chemical steps and all intermediates formed in the process of converting the reactant to the product. 2) Use curved arrows to show all changes in bonding and lone pair electrons. 3) Show all formal charges and all contributing resonance structures. Remember, the sequence of steps you propose must convert the reactant into the product using only the starting material provided- You may not use additional acids or bases, since they are not present in the reaction mixture. The solvent for the reaction is THF, the structure of which is provided below. / OTs A THF solvent HO O 0 >003 H0 Hicé Hid Hid. —a. —. m + TsOH (969 O H we OTs + TsOH Arrows (4 points): 1 point for each of the four arrows shown in the 1St , 4th , and 5“1 structures. The arrows must clearly start and end in the proper place (bond or atom). Formal charge (4 points): 1 point for each of the + formal charges shown in the 2nd, 3rd, 4th and 5th structures. Resonance structures (3 points): 1 point for each resonance structure that is completely correct. They do not have to ShOW the arrows in the 2ncl and 3rd structures that show movement of electrons to interconvert resonance structures. Deprotonation (1 point): 1 point if the deprotonation step is drawn completely right: deprotonation occurs AFTER the alcohol reacts with the carbocation. No point if the student shows deprotonation of the alcohol BEFORE reaction with the carbocation. They do not have to show the lone pair on oxygen. It is OK if they use THF rather than tosylate as the base during the deprotonation. Name SISD B) (2 points) Based on your mechanism, predict one other ether that might be formed as a product during this reaction. m OR ,I/ Q 2 points for either answer, but 2 points total. No partial credit. Problem 3. (14 points) For each set of reactions shown below, circle the reaction that is fastest, and cross out the reaction that is slowest. Grading: 1 point for each correct circled reaction, and 1 point for each correct crossed out reaction. A) (2 points) Am NaSMe A» B SM F MeOH solvent J“ \R MeOH : Br cameo/H am We M NaOMe AW 8 Chile I” MeOH solvent B) (2 points) fifer EtOH, A dOEt Name SISD F) (2 points) Br DMSCL/ < > NMez /'7 /\/Br / NaBr, DMSO —————+ /‘\/Br NaBr, DMSO —-— /\/B" Ozmzo 0:0}:0 Ozmzo i <2 G) (2 points) The products for these three reactions are intentionally not provided. MeBr, DMSO MeNH2 MeBr, DMSO IVIePH2 ————» Me—CENfiwé/y Name C) (2 points) D) (2 points) E) (2 points) SISD OTs NaCN. DMF CN 69 \NaSMe 0132186“ \ / H, u N c: MeaN + MeZS 69 NaSIVIe, DMSO “(I)/ Go; MeZO + Me28 EB NaSMe, DMSO \ / 90] ———-——_— M628 + MeZS is the NH solvent omkxflaIE/r/x N3 © / Q4 -/”F120 solvent Name SISD Problem 4. Me\ (a) a points Me'gfof Meg? \tSMe Starting Material Final Product Requires - substitution (MeS‘) with inversion at less substituted epoxide carbon, producing alkoxicle at more substituted epoxide carbon (4 points properly completed) - ether formation (Me—X electrophile) at 3° alkoxide (or alcohol) (4 points properly completed) 9 Me\ Me“ 0 - + Me, O ‘SMe Me, 0 0 “‘95 Na 9 ~‘ Me-X (X=Cl, Br, I, 0M3. 0T3) " ~“SMe THF or Me Me, 0 - Me, 0” t 1-Mes Na+ c‘SMe 1. NaH (THF or other polar aprotic solvent) Me] O 8M9 mm o" \\\ 2- d" H30+ 2. Me-X (X=C|, Br, l, 0M5, 0T5) O Name (b) 12 points Mex/f0? Me’O’OMf eNSMe Starting Material Final Product Requires - substitution with inversion (MeS‘) at less substituted epoxide carbon - substitution with inversion (MeO‘ ??) at more substituted epoxide carbon SUBSTITUTFON WITH lNVERSiON AT 3° epoxide carbon only possible using H2804 I MeOH Me}, O Mao, Me OH Need to convert to a good leaving group wlo inversion, followed ’ H2804 " by Sn2 with inversion using MeS'. ——+ MeOH MsCI I pyridine or TsCi I pyridine Me \Meol" OMs or OTs Me,OzMe SMe 4 points proper epoxide ring opening, . AND 4 pts proper OMs or OTs, 4 points proper MeS' 0R 8 points proper Mitsunobu silty errors H2304 IMeO' 2 pts - reaction works but MeO' is protonated by H2804 MsCl lTsCI without pyridine 2 pts - pyridine prevents formation of HCI which will open epoxide 0 points if use SOCi2 or PBr3 to convert alcohol into good leaving group — inversion, not retention \J Name SISD Problem 5. (20 points) For the following reactions, (1) Give the absolute configuration, (R) or (S), of all stereocenters in the reactants and products PROVIDED BY US. (2) If the reaction is 100% correct as written, write “CORRECT” in the box. If the reaction product(s) provided are not entirely correct, draw ALL THE CORRECT PRODUCTS in the box and SPECIFY THE SUBSTITUTION MECHANISM(S). If no reaction occurs, write “NO REACTION” in the box. (a) 8 N 6) Me, Br (3/0 8 Men O'Ph AA© THF 00 c Sn2 will not occur at a tertiary carbon. The solution temperature (0° C) greatly slows Snl carbocation formation. If one waits long enough, Snl racemic product forms. NO REACTION - 2 pts (must explain that Snl is slow because THF is not very polar) or RACEMIC PRODUCT Zpts ( 1 pt each enantiomer or 2 pts squiggly line) 4 21‘s total {or part (A) (1)) Me 6 MeO, I I MeOH ——-----I- 50° C 1 pt each center ,Br \\ Me MeO,, 3 (S) MeQ, \Me Mao, \,0Me lpt (R) 1 pt Snl reaction: obtain both absolute configurations at reacting C (NOT RACEMIC — left stereocenter (8)) (squiggly line okay for right stereocenter) ts total or art B Na® 6) CN THF 50° C ...
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2007 key - Name SISD Chemistry 35 , Exam 2 Answers —...

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