please print clearly please print clearly ID No PRINTED Name Chemistry 210 Third Examination Signature December 6 2011 1 5 hr 140 points PUT AN X ON THE LINE BELOW INDICATING WHICH CHEM 210 LECTURE SECTION YOU ATTEND Section 100 Pomerantz Section 200 Wolfe Section 300 Montgomery Section 400 Loy Please read each question carefully and answer it completely and clearly Precision in drawing is an important skill Double check any three dimensional representations to ensure you are implying an unequivocal direction of bonding Individual point values are given in the corner of each answer space The exam has 7 pages including this cover page and the pKa table on the last page Score Prob Pts I 28 II 24 36 III IV 21 31 V Total 140 GSI Page 1 Name I 28 points A a Provide the major product 2 formed in the reaction between 1 and K OC CH3 3 no partial H H Br OC CH3 3 K HOC CH3 3 solvent H3C H 1 H3C H Product 2 5 b Major product 2 is formed as circle one A single chiral molecule A single achiral molecule A mixture of enantiomers A mixture of diastereomers c The transformation above is an example of what type of reaction circle the one best description SN1 reaction reduction reaction electrophilic addition reaction Bronsted acid base reaction complexation decomplexation reaction SN2 reaction E1 reaction E2 reaction B a Provide the major product 5 formed in the reaction between 3 and 4 H Br 3 O 4 O O K OH acetic acid solvent connectivity 2 stereo 3 O O S H 5 Product 5 b Draw the structure of the transition state for the reaction of 3 4 5 Be sure to clearly illustrate the position of nonbonded electrons and show dotted lines to illustrate the position of electrons in bonds being formed or broken H O O Br 2 if missing or incorrect partial charges or nbe s 5 transition state for 3 4 5 c Which of the following will increase the rate of the reaction of 3 with 4 circle all that apply note the stereochemistry label does not need to be drawn this is an aid to the graders HBr no partial DMSO solvent instead of acetic acid increased concentration of 4 3 d Will the reaction of 6 with 4 be faster or slower than the reaction of 3 with 4 You may assume the two reactions proceed through the same mechanism Briefly explain your answer increased temperature decreased temperature 2 2 O O K 4 acetic acid solvent H Br 6 Product 7 The reaction of 6 with 4 will be faster slower than the reaction of 3 with 4 circle one The SN2 reaction of 6 with 4 will be slower because 6 is more sterically hindered The large t butyl group will impede attack by the nucleophile thus slowing the rate of the bimolecular reaction brief explanation 6 Page 2 II 24 points A Provide the reagents or structures as necessary to complete the following sequences of reactions If more than one synthetic step is required number them sequentially You may use abbreviations for reagents e g TsCl but not acronyms e g mCPBA draw the structure instead If any reaction produces a mixture of stereoisomers follow the specific instructions to draw the single indicated stereoisomer and write enantiomer or diastereomer in the box Name a Draw the product that does not have an S stereocenter H3C H H3C H 1 BH3 2 NaOH H2O2 Br2 HOCH3 b Draw the product that has only one R stereocenter H3C H c O H H C C CH3 H2 Pd BaSO4 or Pd CaCO3 4 PbO or quinoline no partial H3C H H connectivity 2 then stereo 2 diastereomer 1 diastereomer OH product with no S stereocenters 5 connectivity 2 then stereo 2 diastereomer 1 diastereomer Br H CH3 OCH3 5 product with only one R stereocenter CH3 C H O H H C H N O HO2C connectivity 2 then stereo 3 O Cl OH O HO2C diastereomer H O H3C OH O H OH OsO4 CH3 3Si Ph OH CO2CH3 diastereomer N H O O H H3C OH connectivity 2 then stereo 3 CH3 3Si Ph CO2CH3 5 5 d e Page 3 III 36 points A A recent paper described a new reaction that causes anti Markovnikov addition of a hydrogen atom and a PPh3 group to an alkene This reaction provides an alkylphosphonium salt as the product Given this information draw the structure of the cationic product formed in the reaction shown below Name H PPh3 BF4 PPh3 light CH2Cl2 Ph3P no partial cationic product anti Markovnikov addition 5 B A seemingly simple acid catalyzed addition of methanol to compound 7 was carried out in an attempt to make product 8 However molecule 9 was formed as the major product instead of 8 Molecule 9 results from a single favorable rearrangement step Draw the structure of product 9 H3C H3C H 7 CH3OH cat H2SO4 H CH3 H3CO H3C H3C H minor product 8 no partial CH3 H3C H3C OCH3 H NMR 4 peaks 6 4 3 3 major product 9 5 C Provide the reagents or structures as necessary to complete the following reaction scheme If more than one synthetic step is required number them sequentially You may use abbreviations for reagents e g TsCl but not acronyms e g mCPBA draw the structure instead CH3 H C C 1 NaNH2 2 CH3 CH3 C C no partial 5 H2 Pd C no partial H3C H3C Br no partial 5 HBr H3C OH 1 Li NH3 OR Na NH3 2 Br2 no partial 4 Br H H Br 4 CH3 CH3 CH3 D Using lines dashes and wedges to indicate stereochemistry if when appropriate draw the structure of 2Z 4E 3 bromoocta 2 4 dien 6 yne H3C Br 2 points per error 8 Page 4 IV 21 points A Molecule 11 below was recently used as a key intermediate in the synthesis of a pharmaceutal that is being examined for the treatment of respiratory diseases Molecule 11 was prepared via treatment of molecule 10 with a catalytic amount of H2SO4 Provide a complete curved arrow mechanism for the most straightforward conversion of 10 to 11 You may abbreviate any Bronsted acid you use for the mechanism as H B and any Bronsted base as B Name O N H3CO OCH3 O cat H2SO4 polar solvent 70 C Cl H B Cl H3CO CH3 O O O N CH3 H O O H3CO Cl 10 O N 10 each intermediate 5pts each set of arrows 2 pts nbe are optional B Provide the uncharged product formed in the reaction shown below Cl N O 1 2 NaOH H2O Cl NH2 CH3OH Cl Cl 11 Cl O N H3CO O N H3CO O H H O B Cl 11 O N H3CO O 16 N O N H no partial 5 Page 5 V 31 points A Treatment of molecule 12 with a strong base leads to an E2 reaction that provides a single alkene stereoisomer a Draw the structure …