Massachusetts Institute of Technology Dr. Kimberly L. Berkowski 5.13, Fall 2006 Organic Chemistry II EXAM #3 EXTRA PROBLEMS KEY What to expect on Exam #3: 1. ~1 Labeling experiment 2. ~2 Mechanisms 3. ~2 Syntheses 4. ~5 transformations – supply missing product 5. ~5 transformations – supply missing reagents 6. ~3 General questions 1. (4 points each, 8 points total) In the boxes, please provide the reagents for the illustrated transformations. More than one step may be required. (a) 1. KCN 2. H3O i-Pr or + 1. Mg ,ether 2. CO2 + 3. H3O /workup Br O i-Pr OH (b) 1. SOCl2 2. LiAl(OtBu)3H 3. workup O i-Pr or 1. LiAlH4 (XS) 2. H2O 3. PCC O i-Pr OH H Figure by MIT OCW. 2. (2 points each, 8 points total) Please provide the products of the following reactions. If no reaction is expected, write “NR”. (a) O Et (b) 1. Excess Na BH4 Cl O Et (c) O (d) O Et NR or O 2. Workup 1. Excess MeLi OMe Et 2. Workup O HO Me Et Me 2. Workup 1. Excess LiAlH4 NMe2 OH 2. Workup 1. Excess MeMgBr OH Et Et Et NMe2 Figure by MIT OCW. 3. (2 points each, 16 points total) Please provide the requested products or reagents. If no reaction is expected, write “NR”. (a) Br2, NaOH O Me NH2 Me Me NH2 H2O2, ∆ Me (b) O Me H2N Me OH N-OH 1. LiAlH4 + cat. H NH2 2. workup (c) POCl3 O n-Bu n-Bu NH2 CN H3O + H2O O nBu OH (d) NH2 NaNO2 HCl N2 CuBr Br Figure by MIT OCW. 4. (11 points) Please provide a detailed mechanism for the illustrated conversion of acetic acid (A) to acetyl chloride (B). O Me OH + Cl O S O Cl Me O S Me + HCl Cl Cl OH O S Cl O O Me + SO2 B A O Cl S Cl O OH Cl OH + Cl Me Cl O H Me + SO2 Cl Figure by MIT OCW. 5. (11 points each, 22 points total) Please provide syntheses for only two of the three indicated compounds. All the carbon atoms should be derived from the allowed starting materials. You may use any common reagents. Allowed Starting Materials: Me Me CO2 Me Me OH Me A Pick Two: N C Me B C HO Me H N Synthesis # 1: D A O OH 2 O Me 1.O3 2.MeS 3.Na2Cr2O7 H2SO4 SOCl2 Cl NH3 B Me Me Me KMnO4 Me OH 1.BH3, THF 2. H2O2,−OH or H3O+/H2O O NH2 OH POCl3 1.PBr3 C N MgBr O 1.CO2 OH 2. H + 2.Mg SOCl2 OH (excess) 1.MeMgBr 2.H2O O Cl Figure by MIT OCW. 5. (Continued) Allowed Starting Materials: Me Me Me CO2 Me A Pick Two: B Me C N OH Me Me-OH C HO Me Me Me H N Me Me O 1. Synthesis #2: C MeOH OH O 2. H2NNH2 PBr3 or TsCl, Pyridine Br O NH2 PCC H H N K NH2 or 1. NaN3 2. LiAlH4 + 3. H3O or XS NH3 Cat H N 1. LiAlH4 2. H2O H N Figure by MIT OCW. 6. (11 points) Provide a synthesis that will selectively convert A to B. Show all the key intermediates and furnish all the important reagents. This is not a one-step process. O Me A O Me HNO3 H2SO4 O 2N O HO Me B O Me H2, Pd H2N N2 O Me NaNO2, HCl O Me H2SO4, H2O HO O Me Figure by MIT OCW. 7. Methyl acetimidate (A) is hydrolyzed in aqueous sodium hydroxide to give mainly acetamide and methanol (eq 1). In aqueous acid, A hydrolyzes to give primarily methyl acetate and ammonium ion (eq 2). a) Provide a detailed mechanism for the illustrated process. Please show all arrow pushing. HO- NH Me N-H Me OMe A O H2O OMe Me NH2 PT Me OH NH2 O OMe Me O MeOH + H OH + MeO (1) MeOH + OH NH2 Figure by MIT OCW. b) Provide a detailed mechanism for the illustrated process. Please show all arrow pushing. NH Me H Me N A H Me OMe OMe NH2 OMe excess H H 2O PT OH2 O + Me Me OMe NH4 + NH3 OMe OH H O Me (2) OMe + NH3 H2O O Me OMe + NH4 Figure by MIT OCW. c) Briefly explain why the two reactions provide different products. Basic conditions: NH2 worse L.G. than OMe . Elimination favors amide Acidic conditions: Acid/base equilibrium favors protonation of ntrogen making it a good L.G. Also, NH4 is not nucleophilic + formation of ester is reversible Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II 8. Synthesize the indicated compounds from the allowed starting materials shown below. All of the carbons of the target compounds should be derived from the allowed starting materials. 8. b) TsCl MeOH NaCN MeOTs MeCN N 1) LAH NH2 2) H2O O O HBr EtOH EtBr Mg 1) H, O 2) H3O+ EtMgBr O Na2Cr2O7 OH EtNH2, H+ PCC O N T H O O c) Me OH Me OH Me OH KMnO4, H+ PBr3 2 Br Me Cl 1 Br 2 O Me Me OH Me PCC CN SOCl2 Me Me 3 H C N LiAlH4 O Me 4 H NH2 Me Me Me NH Me H 3 LiAlH4 O 1 Cl Me Me Me N T O Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II 1 from part c O Cl (d) Me CN OH O Me H-CN Me AlCl3 LiAlH4 OH NH2 Me NO2 H2SO4 (e) NH2 H2/Ni cat Br Br2(excess) FeBr3 HNO3 Br NH2 Br NaNO2 HCl Br Br O H2O, H OH Br Br Br Br C N N N CaCN Br Br Br SOCl2 Br Br O Cl Br Br + H 2N O N H Me 4 from part c Br Me Br Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II 9. Provide the best stepwise mechanism for the illustrated process. Please show all arrow pushing. (9) O O H NH2 O H+ H O H H H NH2 O O O H H N H H 2O H H H N H + H H OH N H H H OH NaBH3CN "H " O O H H H H N H H N H HO N H H H H H NaBH3CN N H N "H " H N H2O H O H N H Target Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II 10. (a) Provide the best mechanism. Please show all arrow pushing. (a) O N benzylic cation H H C Me O OH CH3 N C CH3 C N OH OH OH OH etc N C CH3 O H or workup H Me N O Target Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II (b) Provide the best mechanism. Please show all arrow pushing. Me N H N H O OH2 O H 2O H H Me N O H O Me H N H Me O H O H H NH2 O NH3 H O O O Me Me + H NH3 O H Me O H O H NH3 O O H Me OH2 H NH3 O H Me O OH H OH2 NH3 OH O H HO Me workup NH2 OH Figure by MIT OCW. Massachusetts Institute of Technology 5.13: Organic Chemistry II 11. Consider the labeling experiments outlined below: 11.Start with the mechanisms: O H NHMe H2O HO k1 PT O H2 OH H 2O O k1 NHMe NHMe NHMe PT PT HO OH k2 O H OH OH NH2Me Proton transfer are very fast. O H Acid/base equilibria favor the protonated N (not O). Therefore, once the tetrahedral intermediate forms, loss of NH2Me (k2) is faster than loss of OH (k1).Very little O is incorporated into the unreacted starting material. H O O OH OH O H OMe H2O k1 HO O H2 PT H2O OH O k1 OMe OMe OMe PT PT HO All of the oxygens in the tetrahedral intermediate are roughly equally basic. Therefore, each protonated form 13 present in the same concentration ,and k2 ~= k1. as a result, you would expect more O incorporation into the ester starting material than you would into the corresponding amide. OH OMe H k2 O H O H OH OH H O O OH OH Figure by MIT OCW.

Organic Chemistry 5.13, Solutions to Exam 3 Extra Problems

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