CH2420, C , Organicc Chemisstry II, Exam 1 2:00–3:000 PM, Februaary 8th, 20100 (total 200 points) p ur name CL LEARLY heere: Pleaase print you Last, TEST T FORM First A Makke sure you have h chosen the form of test that is different d from m those chossen by the stuudents next to you; theree are three forms, A, B and C,, which are printed p on paapers with diifferent coloors. The quesstions are thee same but they and the answ wers to them m are arrangeed in differennt orders. The exam has tw wo parts. Parrt one containns 20 multipple choice quuestions. Thee total pointss for this parrt are 120 (6 Y need to mark m your annswers of theese questionns with a × 200). Each quesstion has onlly one correcct answer. You penccil on your annswer sheet.. This part will w be gradedd using a scaanner. As a result, r answeers marked on o the exam will NOT be graaded. In addiition to answ wers, please also a fill the following f infformation onn your answer sheet: 1. yourr name; 2. I. D. number, print it in thhe form and then t mark thhe corresponnding numbeers; 3. Test foorm number, whicch is “A” forr you. The second part of the exam m contains 6 questions q thaat need shortt written ansswers. The tootal points foor this part are 80. 8 Answers to these queestions need to be put in the space prrovided in thhe exam. Thiis part will be b graded manuually. The total points your earned will be enteered into WeebCT. The annswer sheet for the multiple choice questions q may not be returrned to you. The whole exams e will be b placed in front f of the organic o lab on o the 6th flooor of Chem mistry Buildding after thee points are entered e into WebCT. Thhe keys will be b posted in WebCT. I will w send an emaiil to the class when read dy. The exam ms that are noot picked up will remainn in my office for severall weeks and will then be desttroyed. 1 Part 1: Multiple choice questions. 1. Arrange the following compounds in order of decreasing chemical shift for the underlined hydrogens (largest δ value first, smallest value last). a) CH3OCH2CH3; b) CH3CH2CH3; c) Cl2CHCH2CH3; d) ClCH2CH2CH3 A) B) C) D) b>c>a>d c>b>a>d a>c>d>b c>b>d>a 2. How many nonequivalent protons does the following structure have? A) B) C) D) 12 6 4 8 3. How many signals are expected in a 13C NMR spectrum of methylcyclohexane? A) B) C) D) 4 5 6 7 4. Give the number of lines in the coupling pattern for each type of hydrogen. A) B) C) D) a =2, b=8,c=8, d=2 a=2, b=8, c=2, d=1 a=2, b=7, c=7, d=2 a=2, b=7, c=1, d=1 2 5. The 1H NMR spectrum of a compound is shown below. What is the structure of the compound? A) O B) O C) O D) O 3 6. Rank the following alkyl radicals in order of increasing stability (least < < <most). Ph 1 A) B) C) D) 2 3 4 4<2<1<3 3<1<2<4 1<3<4<2 2<4<3<1 7. What is the major product of the following reaction? A) Br B) Br Br C) Br 4 D) Br 8. Consider the reaction of 2-methylpropane with a halogen. With which halogen will the product be almost exclusively 2-halo-2-methylpropane? A) B) C) D) F2 Cl2 Br2 I2 9. Which of the following elementary reactions corresponds to a propagation step in the formation of chloroethane from ethane and chlorine? CH3CH3 + CH2CH2Cl Cl2 + HCl A) CH3CH2 + CH3CH3 CH3CH3 + CH3CH2 B) hv Cl2 2 Cl C) CH3CH2 + Cl CH3CH2Cl D) CH3CH2 + Cl2 CH3CH2Cl + Cl 5 10. Identify the reagent(s) that will promote the following transformation. A) B) C) D) HBr, tetrahydrofuran HBr, peroxides Br2, CCl4 Br2, H2O 11. Which of the following is the MOST stable carbocation? A) B) C) D) 12. Which of the following resonance structures contributes the most to the resonance hybrid? A) OMe 6 The thermodynamic product. 1-bromo-2-butene. Formation of the two products arises from a common resonance stabilized carbocation intermediate.4-conjugate addition of HBr to 1. The kinetic product is favored by carrying out the reaction at high temperature. arises from the more stable carbocation intermediate. 14. Which of the following statements regarding 1.2. 3-bromo-1-butene.or 1.3-butadiene is FALSE? A) B) C) D) The kinetic product. is the more stable product. What is the major product of the following reaction? 7 .B) OMe C) OMe D) OMe 13. A) CHO CHO B) CHO CHO C) CHO CHO D) CHO CHO 15. What are the reactants needed to accomplish the following reaction? 8 . Which of the following compounds are aromatic? A) B) C) D) a. b. and d 17. d. How many uncharged resonance structures are there for azulene? 9 .A) + O B) O + C) + O D) O + 16. and e a. c. and d a. b. b. c. and d a. A) B) C) D) 1 2 3 4 18. c only c. CH3CH2OCH2CH3. a d. b only a. CH3CH2CH3. a. d. d 19. b. Which of the following alkyl halides can be prepared in good yield by radical halogenation of an alkane? Cl Br b a A) B) C) D) Cl c Br d a b c d 20. b. b. c. Rank the following compounds in the order of increasing λmax a A) B) C) D) c b d c. c. d only 10 . c a. c. b. CH3OCH2CH2OCH3 A) B) C) D) a. d b. c. a. Which compounds give a 1H NMR spectrum with two signals in a ratio of 2:3? a. CH3CH2Cl. d. d only a. Please use appropriate curved arrows to convert the first one to second one and so forth. Draw five resonance structures for cyclopentadienyl cation. However. the arrows must be appropriate. Missing or incorrectly drawing each resonance structure will reduce 2 points. 11 .Part 2: Questions that need short answers: 1. (10 points. missing or incorrectly drawing each arrow will reduce 1 point until 0 point is reached) Other reasonable versions may also be correct Pushing electrons two or more times to get the next resonance forms are also OK as long as 5 resonance forms are drawn. Draw a stepwise mechanism for the following reaction. Make sure to use appropriate curved arrows to indicate movement of all electrons that moved during the reaction.2. The arrows for converting one resonance form to another one may be omitted) D D D D 1) NaH + + 2) H2O D omitting the sodium cation is OK Na O H D D H D D H O with two dots here is OK O H H D H H D D H omitting this H is OK D + HO NaOH may be omitted here NaOH is also good 12 . (15 points. missing or incorrectly drawing each arrow will reduce 1 point until 0 point is reached. Indicate how each of the products is formed. Missing or incorrectly drawing each structure including that of small molecules will reduce 2 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. it is even better. Draw products of the following two Diels Alder reactions. (20 points.3. Missing or incorrectly drawing a structure will reduce 8 points. But this is not required for the test 13 . missing or incorrectly drawing each arrow will reduce 3 points until 0 point is reached) MeO O Heat + O OMe MeO O O OMe O OMe OMe O O OMe Heat O OMe O OMe omitting re-draw the starting material and drawing the arrows on the orignial one is also OK If stereochemistry of the product is given. (15 points. Use appropriate curved arrows to indicate the movement of all electrons. missing or incorrectly drawing an arrow will reduce 1 point until 0 point is reached) omitting this will reduce 2 points Initiation hv Br Br Br + All arrows should be single-headed instead of double headed Propagation H C H Br H Br + H H H C HBr H H H C H + + Br Br H C H + Br Br H Termination Br + Br H H Br2 H C + C H H H H H H C H H H C C H H H H + Br H C Br H Drawing electrons aroung Br is OK. Write a mechanism for the reaction of CH4 with Br2 to form CH3Br and HBr under photo conditions. Omitting the words such as initiation is OK 14 .4. Missing or incorrectly drawing a structure will reduce 2 points. Draw a stepwise mechanism for this reaction. Ethers are oxidized with O2 to form hydroperoxides that decompose violently when heated. Missing or incorrectly drawing each structure including that of small molecules will reduce 2 points. (10 points. missing or incorrectly drawing each arrow will reduce 1 point until 0 point is reached) + O2 OOH O O H + O + O O H O OH O O All arrows should be single-headed instead of double headed O O OH O Omitting lone pairs on oxygen is OK Omitting any upaired electrons will be considered wrong H O O + O OH O 15 . Make sure to use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction.5. Devise a synthesis of compound B from compound A. You may use any other required organic or inorganic reagents. You do not need to provide mechanism however. Missing or incorrectly drawing a structure including that of small molecules that are required for the transformations will reduce 2 points until 0 point is reached) OH B A Br2 KOC(CH3)3 H2O Br hv OH H2SO4 Using thermo conditions for halogenation is OK Using other strong base for elimination is OK Uisng other strong acid for hydration is OK Drawing the minor elimination product instead of the major one will not reduce points 16 .6. (10 points. Major reagents and conditions should be provided. The second part of the exam m contains 3 questions q thaat need shortt written ansswers.. 2010 (total 200 points) ur name CL LEARLY heere: Pleaase print you Last. B and C. The total points your earned will be enteered into WeebCT. Test foorm number. Each quesstion has onlly one correcct answer. D. I. theree are three forms. I willl send an emaiil to the class when read dy. This part will w be gradedd using a scaanner. Thee total pointss for this parrt are 120 (8 Y need to mark m your annswers of theese questionns with a × 15). In addiition to answ wers. You penccil on your annswer sheet. A. please also a fill the following f infformation onn your answer sheet: 1. Marchh 17th. Thiis part will be b graded manuually. Exam 2 2:00–3:000 PM. 1 . whicch is “A” forr you. The exam has tw wo parts.CH2420. The quesstions are thee same but they and the answ wers to them m are arrangeed in differennt orders. 2. The exam ms that are noot picked up will remainn in my office for severall weeks and will then be desttroyed. Thhe keys will be b posted in ARIS. 3. C . Your entire exam will be placed in front f of the organic o lab on o the 6th flooor of Chem mistry Buildding after thee points are entered e into WebCT. number. which are printed p on paapers with diifferent coloors. print it in thhe form and then t mark thhe corresponnding numbeers. yourr name. Organicc Chemisstry II.. 8 Answers to these queestions need to be put in the space prrovided in thhe exam. The annswer sheet for the multiple choice questions q may not be returrned to you. TEST T FORM First A Makke sure you have h chosen the form of test that is different d from m those chossen by the stuudents next to you. As a result. Parrt one containns 15 multipple choice quuestions. r answeers marked on o the exam will NOT be graaded. The tootal points foor this part are 80. Part 1: Multiple choice questions. The 1H NMR spectrum of a compound is shown below. C. 1. D. The intermediate is less stable and the product is less stable. The intermediate is more stable and the product is more stable. What is the structure of the compound? A) O B) O C) 2 . D. The intermediate is less stable and the product is more stable. B. 3. A methyl radical A primary radical A secondary radical A tertiary radical 2. C. In comparing kinetic control to thermodynamic control for the reaction of butadiene and HCl. which of the following statements represents kinetic control? A. B. The intermediate is more stable and the product is less stable. Which of the following is the MOST stable radical? A. 6. Predict the major product of the following reaction.O D) O 4. sp-xylene. CHO HNO3 H2SO4 A. C. C. SO3 NO3 NO2+ NO+ 7. B. What is the electrophile in the electrophilic substitution reaction of benzene using HNO3 and H2SO4? A. B. Electrophilic addition of 1-chloropropane. The name of the following compound is H3C CH3 A. Nucleophilic aromatic substitution using propyl Grignard. The best sequence to use to form propylbenzene from benzene is A. Friedel–Crafts alkylation using 1-chloropropane and aluminum trichloride. p-xylene. D. m-xylene. D. D. o-nitrobenzaldehyde and p-nitrobenzaldehyde 3 . C. B. Friedel–Crafts acylation followed by reduction. 5. o-xylene. CH3OH. m-formylbenzenesulfonic acid 8. FeBr3 2) Br2. D. C. D.B. o-formylbenzenesulfonic acid and p-formylbenzenesulfonic acid D. HF A. C. B. H2O 4) H2CrO4 ? O A. B. B. H2SO4. Which of the following compounds is the most acidic? CH3COOH. CH3COOH H2SO4 CH3OH HF 10. 4 . Y OCH3 Z X<Y<Z Z<Y<X Y<X<Z X<Z<Y 9. m-nitrobenzaldehyde C. D. What is the product of the following synthesis? 1) MeBr. light 3) NaOH. tert-Butylbenzene Ethylbenzene Toluene Styrene 11. Which of the following compounds does NOT give benzoic acid when heated with alkaline potassium permanganate? A. Rank the following compounds in order of increasing reactivity in electrophilic aromatic substitution Cl X A. C. B. C. B. Which of the following products is formed in good yield by the reduction of methyl 4-oxo-hexanoate (M) with NaBH4? O O M A. D. C. X<Y<Z Z<Y<X X<Z<Y Y<X<Z 13. O H D. 12. D. O Br C.O OH B.5-Hexanediol 4-Hydroxyhexanal 5 . OH OH OH NO2 Br NO2 NO2 X Y Z A. O Methyl 4-hydroxyhexanoate 6-Hydroxy-3-hexanone 1. Rank the following compounds in order of increasing acidity. Alkyllithium reagents (RLi) add to the carbonyl group of aldehydes and ketones. Part 2: Questions that need short answers: 1. C. Grignard reagents (RMgBr) add to the carbonyl group of aldehydes and ketones. X and Y undergo substitution. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Y and Z undergo addition. But this is not required for the test 6 . X and Y undergo addition. X. C. CH3CH2CH2COCl Y. D. W and Z undergo substitution. (20 points. Y and Z undergo substitution. Draw product of the following Diels-Alder reaction. W and Z undergo addition. stereochemistry is not required) O OMe Heat O OMe O OMe omitting re-draw the starting material and drawing the arrows on the orignial one is also OK If stereochemistry of the product is given. W and X undergo substitution. C6H5CHO W and X undergo addition. B. (CH3)2C=O A. Which compounds undergo nucleophilic addition and which undergo substitution? W. B. The structure of the product worth 11 points. each arrow worth 3 points. Grignard reagents are prepared in ether or tetrahydrofuran (THF). D. Which of the following statements about organometallic compounds is FALSE? A. Alkyllithium and Grignard reagents do not add to esters. 15.14. it is even better. CH3COOCH3 Z. Please do not omit any important steps (30 points.2. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. in this case.) AlCl3 + + HCl Cl H AlCl3 Cl Cl AlCl3 + AlCl4 Product + H H H H H H H H H H Each structure highlighted in red and each arrow highlighted in blue worth 3 points until the maximum 30 points are reached Writing side products as HCl and AlCl3 is OK. Draw a stepwise mechanism for the following reaction. the mechanism is slightly different and is more close to the writing in the book 7 . Each arrow and intermediates worth 3 points until 30 points are reached. 7 points) 1) NaNH2 H H 2) CH3I 1) NaNH2 A 2) CH3CH2I A H B 1) O3 2) H2O C + D B CH3 C or D CH3CH2COOH H3CH2C CH3 C or D CH3COOH 8 . C. B. and D. 7 points.3. 8 points. A. Identify the lettered compounds in the following reaction sequence (30 points. 8 points. Test foorm number.. The total points for the final will be enteered into WeebCT. B and C. The tootal points foor this part are 100.CH H2420. Finaal Exam 3:00–5:000 PM. I. In addiition to answ wers. Aprill 26th. As a result. The exam ms that are not picked up p will remainn in my officce for severaal weeks andd will then bee destroyed. TEST T FORM First A Makke sure you have h chosen the form of test that is different d from m those chossen by the stuudents next to you. yourr name. 2. Organic O C Chemistr ry II. whicch is “A” forr you. Thhis part will be graded manuually. You may picck up your exam s be abble to finish grading g befoore Thursdayy April 29th. D. You needd to mark youur answers of o these quesstions with a penccil on your annswer sheet. The exam has tw wo parts. theree are three forms. which are printed p on paapers with diifferent coloors. may not be returrned to you. print it in thhe form and then t mark thhe corresponnding numbeers. This part will w be gradedd using a scaanner. Thee total pointss for this parrt are 200 (10 × 20). r answeers marked on o the exam will NOT be graaded. 3. please also a fill the following f infformation onn your answer sheet: 1. The second part of the exam m contains 6 questions q thaat need shortt written ansswers. The keys will be postted in ARIS and sent to you y by emaiil. 1 Answerrs to these qu uestions needd to be put inn the space provided p in the t exam. The quesstions are thee same but they and the answ wers to them m are arrangeed in differennt orders. A. 1 . number. Each question q hass only one coorrect answeer. You might want w to checck the accuraacy of your We should total score in WeebCT beforee the grades are a submitteed to the Reggistrar's Officce (I will sennd an email to remind you.. Parrt one containns 20 multipple choice quuestions. and will givve you one day d to do thiss). The annswer sheet for the multiiple choice questions q e in my office o after the t scores arre entered intto WebCT. 2010 (total ( 300 pooints) ur name CL LEARLY heere: Pleaase print you Last. b) CH3CH2CH3. the energy that is released (the heat of hydrogenation) is A) B) C) D) the same as for cyclohexatriene. the same as for 1. 3. The diene must be s-cis and the major product is the exo product.3-cyclohexadiene. The diene must be s-trans and the major product is the endo product.4-cyclohexadiene. Arrange the following compounds in order of decreasing chemical shift for the underlined hydrogens (largest δ value first. Which of the following aromatic compounds undergoes Friedel–Crafts alkylation with methyl chloride and aluminum chloride? A) B) C) D) benzoic acid nitrobenzene aniline toluene 5. 1. the same as for 1. Which of the following compounds does NOT give benzoic acid when heated with alkaline potassium permanganate? A) B) C) D) tert-butylbenzene ethylbenzene toluene styrene 2 .Part 1: Multiple choice questions. smallest value last). Which of the following statements regarding the Diels–Alder reaction is TRUE? A) B) C) D) The diene must be s-cis and the major product is usually the endo product. less than for 1. c) Cl2CHCH2CH3. a) CH3OCH2CH3.3-cyclohexadiene. When benzene undergoes hydrogenation in the presence of a catalyst. 4. d) ClCH2CH2CH3 A) B) C) D) b>c>a>d c>b>a>d a>c>d>b c>b>d>a 2. The diene must be s-trans and the major product is the exo product. 4-dimethylcyclohexanone 4-heptanone 7. NaOEt. A) B) C) D) 1-butanol 2-butanol 1-butene butanoic acid 8.6. H3O . Which of the following compounds does NOT give a tertiary alcohol upon reaction with excess methylmagnesium bromide? A) B) C) D) 3-methylpentanal ethyl benzoate 4. What is the product formed when benzonitrile is treated with phenylmagnesium bromide followed by acid hydrolysis? A) B) C) D) benzophenone benzyl alcohol benzaldehyde triphenylmethyl amine 11. heat A) 3-hexanone B) 3-heptanone 3 . Predict the product of the following reaction. OEt O O 1. What ylide is needed to make 3-ethyl-3-heptene from 3-pentanone in a Wittig reaction? A) B) C) D) Ph3P=CH-CH2-CH3 Ph3P=CH(CH2CH3)2 Ph3P=CH-CH2CH2CH3 Ph3P=CH2 9. Predict which of the following compounds will have a lower boiling point than 1-butanal. Which of the following compounds will react most rapidly with water? A) B) C) D) acetone acetyl chloride acetic anhydride ethyl acetate 10. PrBr 3. EtOH 2. C) ethyl pentanoate D) 2-butanone 12. What is the major product of the following reaction? NaOH O O Heat ? 4 . reacts with methyl iodide? O O a b A) B) C) D) OH c d a b c d 13. Which compound is formed when the enolate of 2-methylcyclohexanone. H2O OH O 2X 2X a A) B) C) D) O O 2X c b 2X d a b c d 14. What is the starting material used in the following reaction? O Ph OH ? 10% NaOH. formed from lithium diisopropylamide (LDA). O O a A) B) C) D) O b c O d a b c d 15. Which of the following compounds form the highest percentage of gem-diol at equilibrium? A) B) C) D) CH3CH2CH2CHO CH3CH2COCH3 CH3CF2CHO CH3CH2CHO 5 . Predict the product of the following reaction. What is the name of the following compound? N Li A) B) C) D) lithium diisopropylamine lithium diisopropylamide lithium diisopropylammonium diisopropylamine anion 16. Benzenediazonium chloride + CuCN → A) B) C) D) Ph2CuCl Ph–CN Ph2CuLi Ph–N=N–CN 17. NH2 NH2 b c H H N H a A) B) C) D) a<b<c b<a<c c<b<a b<c<a 6 . (CH3CH2CO)2O (Z) A) B) C) D) W<X<Y<Z Z<X<Y<W W<Y<Z<X W<Y<X<Z 19. Hz < Hy < Hx. In compound 1. Rank the following compounds in order of increasing reactivity in nucleophilic acyl substitution. CH3CH2CH2COOCH2CH2CH3 (Y). In compound 1. CH3CH2CH2CONH2 (W). Hx < Hy < Hz. In compound 2. Hx < Hy < Hz. Hy < Hz < Hx. CH3CH2CH2COCl (X). Hz < Hx < Hy. Hx < Hz < Hy. In compound 2. Hy O Hx O O O 1 A) B) C) D) Hx Hz Hy Hz 2 In compound 1. Hx < Hy < Hz. In compound 2. Hy < Hy < Hx. In compound 2. In compound 1.18. 20. Rank the following three compounds in the order of increasing basicity. Rank the labeled protons in each compound in order of increasing acidity. The total may not exceed 15 points. each structure and curved arrow in blue worth 1 point.Part 2: Questions that need short answers: 1. 7 . each curved arrow and side product worth 1 point until 15 points are reached.) O COOH HO H2SO4 O + H2O H H O HO O OH H H H HO HO O O O H H H H O HO HO O H OH O O O H2O H O O Each structure in red worth 2 points. Drawing the equlibrium reaction arrow as irreversible reaction arrow will not reduce points. Other appropriate drawings may be considered correct. Please do not omit any important steps (15 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Each intermediate structure worth 2 points. Draw a stepwise mechanism for the following reaction. each curved arrow and side product worth 1 point until 20 points are reached). each structure and curved arrow in blue worth 1 point.2. Each intermediate worth 2 points. Other appropriate drawings may also be considered correct. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Draw a stepwise mechanism for the following reaction. 1) NaOEt 2) CO2Et O O CH2(CO2Et)2 O 3) H3O O O O EtO O OEt H H O O EtO O EtO OEt EtO O O O OEt H HOEt OEt O OEt EtO O O O O O H3O O O O O H OEt OEt OEt H HOEt EtO Each structure in red worth 2 points. Please do not omit any important steps (20 points. 8 . The total may not exceed 20 points. Writing a name instead of a structure or drawing a structure instead of a name will also be considered correct. omitting reagents that are highlighted in red will not reduce points. NH4OH C 1) F O D A O Si 1) BH3 2) H2O2.3. imidazole F G Si O OH 2) H2O Ag2O. HgSO4 MeLi TBAF or Bu4NF Each anwser worth 2 points. please do not forget to label them with A-G. more than one step/reagent may be needed. OH HO CHO COOH HO HO H B O C HO 1) A G O 2) H2O HO Li E H B D E TBDMS-Cl. H2SO4. 9 . For some letter. Fill in the lettered reagents (A-G) in the following reaction scheme (draw structures or write the names of the reagents in the space below the scheme. Total 14 points). -OH H2O. other suitable reagents are also OK. 1) LDA O (1) OH 2) CH3CH2CHO 3) H2O O A O (2) Ph NaOCH3 Ph + O CH3OH O B CHO (3) O O HO + Ph H2O Ph C Drawing the Z isomer is also OK For A. 10 . No mechanism is required. Each strucutre worth 7 points. if the aldol is given. 5 points will be given. elimination product will also be considered correct. For B. The total is 21 points. just need the structure of the major final organic products (total 21 points).4. Draw the organic products formed in the following three reactions. O NaBH4 H N H2O + CH3OH NH2 H H OH O H N H H H H H O H N N NH2 H2O H Each structure in red worth 4 points.5. each structure and curved arrow in blue worth 1 point. Other appropriate drawings may also be considered correct. Each intermediate worth 4 points and each curved arrow and side product worth 1 point until 15 points are reached). Draw a stepwise mechanism for the following reaction. N 11 . Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Please do not omit any important steps (15 points. The total may not exceed 15 points. 5 point. (15 points. Each intermediate worth 3 points and each curved arrow and side product worth 1 point until 15 points are reached) + O2 OOH O O H + O + O O H O O O O O O O O H O O + O OH O Each structure in red worth 3 points. Omiting lone pairs on oxygen atoms is OK. Draw a stepwise mechanism for this reaction. The total may not exceed 15 points. but the unpaired electrons cannot be omitted.6. 12 . Drawing the single-headed arrow as double headed arrow will only give 0. each structure and curved arrow in blue worth 1 point. Other appropriate drawings may be considered correct. Ethers are oxidized with O2 to form hydroperoxides that decompose violently when heated. Make sure to use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. February 2nd. The answer sheet for the multiple choice questions may not be returned to you. Part one contains 10 multiple choice questions. 3. There are three forms. number. Each question has only one correct answer. which are printed on papers with different colors. which is “A” for you. As a result. You can pick up your exam in the Chemistry Learning Center after the points are entered into WebCT (see syllabus for details). Organic Chemistry II. The total points for this part are 40 (20 × 2). 2011 (total 200 points) Please print your name CLEARLY here: Last. You need to mark your answers of these questions with a pencil on your answer sheet. B and C. print it in the form and then mark the corresponding numbers. A. D. I. This part will be graded using a scanner. your name. This part will be graded manually.CH2420. In addition to answers. The keys will be posted in WebCT. The second part of the exam contains 2 questions that need short written answers. The total points for this part are 160 (16 × 10). I will send an email to the class when ready. 2. please also fill the following information on your answer sheet: 1. 1 . Answers to these questions need to be put in the space provided in the exam. Exam 1 2:00–3:00 PM. answers marked on the exam will NOT be graded. TEST FORM First A Make sure you have chosen the form of test that is different from those chosen by the students next to you. The total points your earned will be entered into WebCT. The exam has two parts. Test form number. The questions are the same but they and the answers to them are arranged in different orders. (16 × 10 = 160 points) 1. What is the organic product of the following reaction? CH2CH2I + OCH(CH3)2 A. trans-Cycloheptanol 2.Part 1: Multiple choice questions. The following internal alkene D. The following terminal alkene C. Give the IUPAC name for the following compound: CH 3 OH A. cis-2-Methylcyclohexanol C. cis-1-Hydroxymethylcyclohexane D. The following unsymmetrical ether CH2CH2OCH(CH3)2 B. trans-6-Methylcyclohexanol B. The following unsymmetrical ether O 2 . Compound B. B is least reactive. D. Compound A. C is least reactive. A is most reactive. C is the major product. 5. B. 3 . and A in between.3. All are alcohols and therefore they have the same reactivity. B is the major product. and C in between. D. and B in between. A is least reactive. A is the major product. Rank the following alcohols in order of increasing reactivity when dehydrated with H2SO4. O H3C H 1) H H H H CH 3 OH 2) H2O HO H HO H H H H A H H 3C B H 3C H H H3C H H H OH C D A. C. What is the major product formed when alcohol X is treated with HCl? Cl Cl OH Cl X A B C A. CH3 OH OH HO CH3 A B C A. B. B is most reactive. Compound D. None of the above is correct. B. 4. C. D. C is most reactive. Compound C. C. What is the product of the following reaction? Pay attention to the stereochemistry of the reaction. Its name is (1Z. 4S)-1. the products are: HBr Br Br H Br Br H A B C D A. C. Select the appropriate reagents to complete the following multistep synthesis. Step 1: KOCH3. Step 1: KO(CH3)3. A and D. 4R)-1.4-dimethylcyclodecene.6. 8. 4S)-1. Its name is (1E. A is the S enantiomer and D is the R enantiomer. Cl Br OH A. B.4-dimethylcyclodecene. 7. H2O. Its name is (1E. C.4-dimethylcyclodecene. D. Its name is (1Z. 4R)-1. A. A and B. step 2: HCl. C. A is R enantiomer and B is the S enantiomer.4-dimethylcyclodecene is: A B C D A. Step 1: KO(CH3)3. step 2: Cl2. step 2: Cl2. 4R)-1. step 2: HCl. D. The enantiomer of (1E. For the following reaction. D.4-dimethylcyclodecene. B is the S enantiomer and C is the R enantiomer. B is the R enantiomer and C is the S enantiomer. B. D. Step 1: KO(CH3)3. B and C. H2O. B. B and C. B. C. 4 . C. both methods give a single product. OH A. Only acid-catalyzed addition of water gives a single product. (1) Use KOC(CH3)3 to convert the starting material to A. (2) use Br2/H2O to convert A to D. (1) Use KOC(CH3)3 to convert the starting material to C. Yes. 10. (1) Use KOC(CH3)3 to convert the starting material to A. (3) use an acid to convert D to the product. B. (2) use Br2/H2O to convert C to D. For the following multistep synthesis: Br O OH Br A Br O OH B C D A. D. Can the following alcohol be prepared as a single product by hydroboration-oxidation of an alkene? Can it be prepared as a single product be the acid-catalyzed addition of water to an alkene? Select the single best answer. both methods give product mixtures. Only hydroboration-oxidation gives a single product. (3) use a base to convert D to the product. B. (1) Use KOC(CH3)3 to convert the starting material to A. (2) use Br2/H2O to convert A to D. (3) use a base to convert D to the product.9. 5 . D. C. No. (2) use Br2/H2O to convert A to B. (3) use a base to convert B to the product. Part 2: Questions that need short answers. Each intermediate structure worth 4 points. (20 × 2 = 40 points) 1. each curved arrow and structure in green worth 1 point until 20 points are reached. Other correct drawings will also get points. 6 . Each structure in red worth 4 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. each curved arrow and side product worth 1 point until 20 points are reached). O OH + N S Cl Cl Cl Cl O OH O S S Cl O Cl Cl H N Cl O S O Cl N H Cl + O S O + Cl Total 20 points. Draw a stepwise mechanism for the following reaction. you do not need to be completely correct to get full points. So. Please do not omit any important steps (20 points. H3C CH 3 H C C CH 3 H3C H2SO4 CH3 CH3 C H3C O C CH3 H H3C CH 3 H C C CH 3 CH3 H3C O H H C C CH 3 O O S CH3 H O H O CH 3 + O S OH O H OH H2O O H3C CH3 H C C CH3 CH3 H CH 3 H3C C C CH3 CH3 H3C CH3 C H3C C CH3 H H2C H CH 3 H C C CH3 CH3 H Total 20 points. each curved arrow and structure in green worth 1 point until 20 points are reached. each curved arrow and side product worth 1 point until 20 points are reached). Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Each structure in red worth 4 points. Next page has a short survey. So. Other correct drawings will also get points. 7 . Each intermediate structure worth 4 points.2. Please do not omit any important steps (20 points. you do not need to be completely correct to get full points. Draw a stepwise mechanism for the following reaction. This is a voluntary anonymous survey. How about the pace of the lectures? Can be somewhat faster? Or it is OK or it should be slower. I believe that both you and I can benefit from them. detach this page from the exam and put in the box located at the doors. Instead. Please do not submit with your exam. 1. The goal is to help to improve teaching. Are the numbers of questions in the homework OK for you? More or less is better? 5. Any other comments are also appreciated. Overall. Can you hear me well at the back of the classroom? Are the slide show and project clear enough? 3. Please do not let your frustrations from COONECT to affect your answers. 8 . Although I cannot follow all of your suggestions (they can be contradictory to each other too). 2. Do you like the grading policy? Do you know what materials you should review before an exam? 4. You can answer all or part of the questions. how do you rank the course? If ranking from 1 to 5 with 1 being the least satisfactory and 5 being the most satisfactory. which do you choose? 6. print it in the form and then mark the corresponding numbers. The answer sheet for the multiple choice questions may not be returned to you. You need to mark your answers of these questions with a pencil on your answer sheet. This part will be graded manually. Answers to these questions need to be put in the space provided in the exam. Each question has only one correct answer. The exam has two parts. answers marked on the exam will NOT be graded. which is “A” for you. This part will be graded using a scanner. your name. which are printed on papers with different colors. please also fill the following information on your answer sheet: 1. I. The total points you earned will be entered into WebCT. In addition to answers. B and C. 3. 2011 (total 200 points) Please print your name CLEARLY here: Last. Test form number. There are three forms. The second part of the exam contains 2 questions that need short written answers. The total points for this part are 140 (7 × 20). Part one contains 20 multiple choice questions. You can pick up your exam in the Chemistry Learning Center after the points are entered into WebCT (see syllabus for details). Exam 2 2:00–3:00 PM. 1 . The total points for this part are 60 (30 × 2). Organic Chemistry II. A. As a result. D. 2. number.CH2420. TEST FORM First A Make sure you have chosen the form of test that is different from those chosen by the students next to you. I will send an email to the class when ready. The questions are the same but they and the answers to them are arranged in different orders. March 16th. The keys will be posted in WebCT. (7 × 20 = 140 points) 1. 4R)-1.4-dimethylcyclodecene. Compound B has the highest boiling point. Its name is (1E.4-dimethylcyclodecene. Its name is (1Z.4-dimethylcyclodecene. D. 4R)-1. Compound A. 3. Its name is (1Z. 2. compound A has the lowest boiling point. Compound A has the highest boiling point. 4S)-1. B. 4R)-1. D. Compound D. The enantiomer of (1E. Compound C. A. C. Compound A: CH3CH2OCH3 Compound B: CH3CH2CH2OH Compound C: (CH3)2CHOH A. Rank the following compounds in order of increasing boiling point. Compound B has the highest boiling point. C. O H3C H 1) H H H H CH 3 OH 2) H2O HO H HO H H H H A H H 3C H 3C B H H H3C H H H OH D C A. What is the product of the following reaction? Pay attention to the stereochemistry of the reaction. Compound C has the highest boiling point. D. C. Its name is (1E. C.Part 1: Multiple choice questions. B.4-dimethylcyclodecene is: A B C D A. 2 .4-dimethylcyclodecene. Compound B. B. compound A has the lowest boiling point. compound C has the lowest boiling point. compound B has the lowest boiling point. D. 4S)-1. B. The major organic product of the following two reactions are: H2O.4. HBr. HOB O OH O OH W X Y Z A. BH3. 3 . B. Product A is Z. B. POCl3 in pyridine. B. product B is Z. H2O2. Compound C. HgSO4 A BH3. D. What reagents are needed to prepare CH3CH2CH2CH2C≡CH from CH3CH2CH2CH2CH=CH2? A. then H2O2. 5. KOC(CH3)3 (2 equivalents) in DMSO. H2SO4. C. product B is X. product B is Y. NaH. Product A is X. What is the major organic product of the following reaction? A. C. Compound D. Product A is W. Compound B. D. C. D. KOC(CH3)3 (2 equivalents) in DMSO. 6. product B is W. HO-. Product A is Y. Compound A. Cl2. Which bases can deprotonate acetylene.(pKa of conjugate acid is 16).(pKa of conjugate acid is 44) and CH3NH.(pKa of conjugate acid is 44). Compound A is W. A. Compound A is Z. which has a pKa of 25. Compound A is W. Identify the compounds in the following multi-step reaction.(pKa of conjugate acid is 10. 4 . Compound A is X. Compound C is Z. CH3CH2O.5) and CH2=CH. CH3CH2S. 8. Compound C is W.(pKa of conjugate acid is 40). C. B.7. Compound C is W. D.(pKa of conjugate acid is 40). C. CH3CH2S. CH2=CH. Compound B is X.(pKa of conjugate acid is 16) and CH3NH. Compound C is Z. D. Compound B is Y. Compound B is Y.(pKa of conjugate acid is 10. A. Compound B is Y. B.5) and CH3CH2O. 6−dimethyloctane when treated with H2 and a Pd catalyst. Ozonolysis of oximene forms (CH3)2C=O. Compound X. Oximene and myrcene. Compound Y. myrcene is Y. 10. two hydrocarbons isolated from alfalfa that have the molecular formula C10H16. D. Identify the structures of oximene and myrcene. and HCOCH2CH2COCHO. D. both yield 2. and CH3COCHO. CH2=O (two equiv). myrcene is X. Oximene is Y. C. myrcene is Z. C. B. Oximene is Y. CH2(CHO)2. 5 . myrcene is W. Compound Z. Oximene is X. B. Compound W. Oximene is X. CH2=O. Ozonolysis of myrcene yields (CH3)2C=O.9. A. What is the major organic product of the following reaction? A. Compound Z. compound Z is least reactive. B. Compound X is most reactive. D. What product is formed. Compound X. including stereochemistry. when CH3OCH2CH2C≡CH2CH(CH3)2 is treated with Na in liquid NH3? A.11. A. 13. They have the same reactivity. Pd-C. C. From the list below. Compound Y. B. 12. O OH T S W X U Y V Z 6 . Compound Z is most reactive. compound Y is least reactive. Rank the following alkenes in order of increasing rate of reaction with H2. C. select all of the structures for a hydrocarbon with a molecular ion at m/z = 112 in mass spectra. D. Compound W. compound Z is least reactive. Compound Y is most reactive. Compounds S. B. One is at ~2950 cm-1. Compounds W and Z D. V: 1650 cm-1. Y. Compounds X and Y C. Y. Z. C. V. Z: 3500 cm-1. U. T. W. Y. Z. Compounds V and W B. X. X. A. W. Y. W. Y: 2250 cm-1. Y. D. X. 7 . D. in the functional group region. X: 3000-2850 cm-1. V. V. The structure of the compound is most likely to be: O O OH O OH V U W O X Y Z A. B. 14. V. What are the major IR absorptions in the functional group region for the following compound? U: 3300 cm-1. Can only be compound U. Compounds U. W. Compounds S. X. T. In the IR spectrum of an organic compound. Z. C. 15. Compounds T. there are only two absorptions.A. the other is at ~1750 cm-1. W: 3150-3000 cm-1. W. Compound Y. What is the compound? A.86 ppm (doublet.55 ppm. and 2120 cm-1. Compound X. The 1H NMR spectrum of 1. D.16. 3.91 ppm (multiplet. B. 2H).93 ppm (doublet.33 ppm (quartet.55 ppm. 3 H). 8 . Compounds S and X. Compounds U and Y. Which of the following compounds are consistent with the following set of data: a hydrocarbon with a molecular ion at m/z = 68. C.2-dimethoxyethane recorded on a 300 MHz spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. Compound Z. 3000-2850 cm-1. B.55 ppm. 18. B.39 ppm and 2. Compounds W and Z. 4.39 ppm and 4. 2H).55 ppm. 1. D. 1. C. 2. What are the chemical shifts of the two absorptions? A. A molecule has the molecular formula of C7H14O2. 17. Compound W. and IR absorptions at 3310 cm-1.15 ppm (triplet. 2. 1 H). A.39 ppm and 3. C. 3. Compounds V and T. 6 H). D.39 ppm and 1. Its 1H NMR data are 0. 1. Z. X. C. Rank the indicated carbon atoms in order of increasing chemical shift. carbon Y has the largest chemical shift. Z. 20. carbon X has the largest chemical shift. D. U. U. Carbon X has the smallest chemical shift. Carbon Y has the smallest chemical shift. A. W. Compounds S. Z. carbon Z has the largest chemical shift.19. Carbon Z has the smallest chemical shift. Compounds T. D. 9 . U. B. V. Compounds S. C. Carbon X has the smallest chemical shift. carbon X has the largest chemical shift. B. Compounds T. W. Which of the following compounds give one singlet in 1H NMR spectra? O Cl Cl Br Br Br Br S O O Br T O Y W V U Z X A. Y. X. each curved arrow and structure in green worth 2 points until 30 points are reached. Draw a stepwise mechanism for the following reaction. So. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. each curved arrow and other items worth 2 points until 30 points are reached). Each structure in red worth 4 points. Other correct drawings will also get points.Part 2: Questions that need short answers. Please do not omit any important steps (30 points. (30 × 2 = 60 points) 1. 10 . Each intermediate structure worth 4 points. you do not need to be completely correct to get full points. H H2O H+ H H H3C CH3 H 3C + + O H+ H CH 3 H 3C CH3 Either E or Z is OK H H O H3C H H CH3 + H H3C H + O CH3 H H 3C H H +O CH3 H+ H H3C H O CH3 Total 30 points. you do not need to be completely correct to get full points. Na. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Please do not omit any important steps (30 points. each curved arrow and other items worth 2 points until 30 points are reached). Each structure in red worth 4 points.2. So. Each intermediate structure worth 4 points. Draw a stepwise mechanism for the following reaction. Other correct drawings will also get points. NH3 (liquid) R H R R H R H H N R R C H H R C R C C R Na Na + + e e- - H H H 2N- R + H R C R R + H 2N- R C - H H N H Total 30 points. 11 . each curved arrow and structure in green worth 2 points until 30 points are reached. The questions are the same but they and the answers to them are arranged in different orders. please see Syllabus. In addition to answers. This part will be graded manually. The total points for this part are 70 (20 + 25 + 25). The exam has three parts. please also fill the following information on your answer sheet: 1. Each question has only one correct answer. I. which are printed on papers of different colors. D. Answers to these questions need to be put in the space provided in the exam. For grading policy. Organic Chemistry II. 1 . The keys will be posted in WebCT. There are three forms. 2. A. Answers to these questions need to be put in the space provided in the exam. As a result. number. which is “A” for you. The total points for this part are 180 (9 × 20). Exam Final 3:00–5:00 PM. The total points you earned will be entered into WebCT. I will send an email to the class when ready. This part will be graded using a scanner. The third part contains 3 questions that need you to provide stepwise mechanism of organic reactions. your name. April 25th. 2011 (total 300 points) Please print your name CLEARLY here: Last. B and C. This part will be graded manually. print it in the form and then mark the corresponding numbers. 3. The total points for this part are 50 (10 × 5). You need to mark your answers of these questions with a pencil on your answer sheet. TEST FORM First A Make sure you have chosen the form of test that is different from those chosen by the students next to you. The second part contains 5 questions that need you to draw the structure of organic reaction products.CH2420. Test form number. The first part contains 20 multiple choice questions. answers marked on the exam will NOT be graded. D. respectively. X. What reagent(s) is/are needed to convert 1−ethylcyclohexene into1. Compounds U. X. ROOR C. V. 11. 1. B. 12 and 13. Compounds U. 12 and 13. W. 4. 10. X. W. 1. respectively. respectively. V. Pd−C B. HBr D. Y and Z are 2. respectively. H2.2−dibromo−1−ethylcyclohexane? A. 6. HBr. Y and Z are 2. Y and Z are 2. (9 × 20 = 180 points) 1. 3. CH3CH2Cl Cl2 U AlCl3 FeCl3 Br2 Heat HNO3 V mCPBA KOtBu X [+ o isomer] W H2SO4 Y Z Cl Cl Cl 1 4 2 Cl NO2 NO2 3 Br 5 Cl Cl O Cl NO2 Cl NO2 7 6 Cl NO2 Br NO2 9 8 O2N 10 Br O O Cl Cl NO2 11 NO2 12 Cl 13 NO2 Br NO2 14 Br NO2 15 A. Fill in the intermediate products in the following multiple step synthesis. 9.Part 1: Multiple Choice Questions. Y and Z are 2. 5. V. Br2 2 . V. Compounds U. 11. 14 and 15. 10. 2. 7 and 8. 3. W. C. X. Compounds U. W. D. U. V. 4. Compounds W. and spectrum F is from compound X. C. and spectrum F is from compound Y. B. Y. V only. T. X. U.3. All are aromatic. 3 . Which of the following compounds are aromatic? H N N O N S O N N N N T U O V S O W X Y Z A. Compounds S. Spectrum E is from compound X. Spectrum E is from compound Z. Spectrum E is from compound W. Compounds S. X. and spectrum F is from compound Z. B. D. Z only. C. and spectrum F is from compound X. Z only. O O OH O OH W O O X Y O Z A. Spectrum E is from compound X. Match 1H NMR spectra with compounds. T. W. Hc is least acidic because its conjugate base is antiaromatic. mCPBA. HNO3/H2SO4. HNO3/H2SO4.5−cycloheptatriene (Z). Consider both 5−methyl−1. Hb is least acidic because its conjugate base is antiaromatic. respectively. Br2/FeBr3. W. Br2/FeBr3. W. C. 7. Ha is least acidic because its conjugate base is aromatic. X < Y < Z. X. respectively. B. Y and Z are CH3CH2COCl/AlCl3. Z < X < Y. 6. Br2/FeBr3. D. A. D. Zn(Hg)/HCl. Y and Z are CH3CH2COCl. respectively. Ha is most acidic because its conjugate base is antiaromatic. HNO3/H2SO4. Hd is most acidic because its conjugate base is aromatic. SO3/H2SO4. Y < X < Z. Add reagents for the following multiple step synthesis. Rank the following three compounds in order of increasing reactivity in electrophilic aromatic substitution reactions. A. X. B. Hd is least acidic because its conjugate base is aromatic. Zn(Hg)/HCl.3−cyclopentadiene (Y) and 7−methyl−1. W. C. Y and Z are CH3CH2COCl/AlCl3. X. Z < Y < X.3. Hc is most acidic because its conjugate base is antiaromatic. respectively. D. which labeled H atom is most acidic? Which labeled H atom is least acidic? A. Hb is most acidic because its conjugate base is aromatic. B.5. X. 4 . Zn(Hg)/HCl. C. Br2/heat. Y and Z are CH3CH2COCl/AlCl3. intramolecular hydrogen bonding stabilizes the conjugate base. A. 5 . Select all statements that explain this result. respectively. Match the following structures to the appropriate pKa values.86. W.8. 0. Y. C. Y. Select the single best answer. the OH group donates electron density by its resonance effect A. A. the OH group destabilizes the conjugate base. X.8.6. Statement Z: In the p form. the OH group withdraws electron density by its resonance effect Statement V: In the p form. 1. The pKa of compounds V. Statement W: In the o form.3 and 2.3 and 0.5.6. Statements U. The pKa of compounds V.2. X.8. Z. The 2° alkyl halide is too crowded for nucleophilic substitution. Y. C. D. 4.8.8. X.86.3 and 2. W. intramolecular hydrogen bonding destabilizes the conjugate base Statement X: In the p form. Statement U: In the p form. Y and Z are 2. 9. and internal alkynes cannot be made using acetylide anions. respectively. 4. 4. W. p −Hydroxybenzoic acid is less acidic than benzoic acid.86. The 3° alkyl halide is too crowded for nucleophilic substitution. The pKa of compounds V. X. the OH group stabilizes the conjugate base. 0. 1. Statement Y: In the o form. B. It has an internal alkyne. respectively. W. B.8. Explain why 2. 1. Statements X.3. Y and Z are 2. V.8 and 2. Y and Z are 2. W. W.6.8. Select all that apply. X. respectively. 10. D. 0.6.5−tetramethyl−3−hexyne can't be made using acetylide anions. o −hydroxybenzoic acid is slightly more acidic than benzoic acid.8.86. 1. 2. 2. Statements W. The 1° alkyl halide is too crowded for nucleophilic substitution. B. D. The pKa of compounds V. Y and Z are 4. Statements U. C. C. then HBr.11. First HBr. First Br2/heat. First mCPBA. B. W and Y. B. then KOtBu. Which of the following reagents/conditions are appropriate to convert compound Y to compound Z? Y Z A. D. X and Z. 6 . D. then H2/Lindlar catalyst. Only X. V. Which of the following compounds matches or match the IR spectrum shown? OH OH W V U O O O O X Z Y A. Only U. 12. First Br2/heat. C. then PCC. U. respectively. A. 32 and 10 ppm. B. 2 and 3 are from compounds X. respectively. C3 and C4 appear at 70. C2. respectively. Mass spectra 1. Assign carbon-13 NMR signals to each carbon in the compound shown. Z and X. 2 and 3 are from compounds X. 70. C2. Y and Z. Mass spectra 1. 7 . respectively. C2. 32 and 10 ppm. B. C1. respectively. Match structures to mass spectra. C. respectively. Mass spectra 1. D. respectively. C2. 70. C1. Mass spectra 1. 70. C3 and C4 appear at 23. respectively. C3 and C4 appear at 32. C3 and C4 appear at 10. Y and Z. Y and Z. 14. 32 and 23 ppm.13. C. 23. D. C1. 2 and 3 are from compounds X. C1. 23 and 10 ppm. Cl O O Br O X Y Z A. 2 and 3 are from compounds Y. Statement Y: The product is a carboxylic acid. Only statements T. Y are correct. Y. D. B. Only statements W.15. W are correct. Compound X. V are correct. Reagent X is Cl2 and water. Alkyne S is treated with O3 followed by water. Reagent X is Cl2. X. reagent Y is NaH. D. T. reagent Z is first NaC≡CH. Reagent X is Cl2 and water. Statement V: One product is an alcohol. U. reagent Y is HCl. reagent Y is NaH. Z are correct. reagent Z is first water. Statement U: The only product is benzoic acid. Which of the following statements are correct? S Statement T: One product is an epoxide. 8 . Statement Z: One product is an aldehyde. C. Provide reagents for the following multistep synthesis. Compound W. Compound Y. D. then NaC≡CH. Only statements U. reagent Z is first NaC≡CH. 16. then water. Statement X: Only one organic product will be formed. then water. Compound Z. Only statements W. Which compound absorbs UV light at the longest wave length? A. Statement W: More than one organic product will be formed. X Cl OH Z Y O OH + enantiomer A. reagent Y is NaH. C. A. 17. Reagent X is first Cl2. reagent Z is NaC≡CH and water. B. C. B. then water. X < Y < Z. D. 19. Carbocation X. X < Z < Y. B. 9 . Z < Y < X. C.18. B. Y < X < Z.2-shift product of carbocation V? A. What is the most stable 1. Rank the following dienophiles in order of increasing reactivity in Diels Alder reactions. C. Carbocation Y. D. Carbocation Z. Carbocation W. A. 20. All of the above bases are strong enough. C. Only bases Y and W are strong enough. Only bases Z and V are strong enough.8)? A. B. Z and X are strong enough. Using the pKa values given. D. Only bases V. 10 . which of the following bases are strong enough to deprotonate CH3COOH (pKa = 4. Reaction 2: HBr 40 oC (Hint: There are two products. The reaction occurs in an ionic media and under thermodynamic control) 11 . (10 × 5 = 50 points) Draw the structure of the organic products of the following reactions. Try your best to draw accurately as partial points may not be given due to the difficulty to be fair in that case.Part 2: Reaction Products. please draw both and indicate which is the major product and which is the minor one. 12 . 13 . So.Part 3: Mechanism of Organic Reactions. 14 . you do not need to be completely correct to get full points. each curved arrow and structure in green worth 1 point until 20 points are reached. (20 + 25 +25 = 70 points) 1. Please do not omit any important steps (20 points. Each intermediate structure worth 3 points. Other correct drawings will also get points. each curved arrow and other items worth 1 point until 20 points are reached). Each structure in red worth 3 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Draw a stepwise mechanism for the following reaction. H H N R R C H H R C R C C R Na Na + + e H H2N- R + H R C R R H2N- e- - H + R C - H H N H Total 20 points. So. Each intermediate structure worth 5 points. Other correct drawings will also get points. 15 . each curved arrow and structure in green worth 1 point until 25 points are reached. AlCl4 H3C C C H Cl CH3 H CH3 H CH3 H H3C CH3 C C H Cl AlCl3 CH3 H3C C C H CH3 CH3 H CH3 H H Cl HCl + CH3 AlCl3 H3C CH3 C H CH3 H H3C C H3C C C CH3 H H H AlCl3 AlCl3 Total 25 points. Each structure in red worth 5 points. you do not need to be completely correct to get full points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. each curved arrow and other items worth 1 point until 25 points are reached). Please do not omit any important steps (25 points.2. Draw a stepwise mechanism for the following reaction. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. So. Draw a stepwise mechanism for the following reaction. each curved arrow and structure in green worth 1 point until 25 points are reached. Please do not omit any important steps (25 points.3. 16 . Each intermediate structure worth 10 points. each curved arrow and other items worth 1 point until 25 points are reached). R R C O O O R Cr H C R O O C Cr O O O OH O H Cr H O R H H R H O H O H O OH O Cr + O H H O + R R H Omitting the lone pair is OK Total 25 points. you do not need to be completely correct to get full points. Other correct drawings will also get points. Each structure in red worth 10 points. CH2420. 2. I will send an email to the class when ready. The total points for this part are 150 (15 × 10). your name. In addition to answers. Part one contains 15 multiple choice questions. Answers to these questions need to be put in the space provided in the exam. The total points for this part are 50 (25 × 2). print it in the form and then mark the corresponding numbers. The questions are the same but they and the answers to them are arranged in different orders. The answer sheet for the multiple choice questions may not be returned to you. Each question has only one correct answer. Test form number. which is “A” for you. As a result. There are three forms. please also fill the following information on your answer sheet: 1. B and C. This part will be graded using a scanner. which are printed on papers with different colors. 1 . Organic Chemistry II. I. 2012 (total 200 points) Please print your name CLEARLY here: Last. answers marked on the exam will NOT be graded. A. First TEST FORM A Make sure you have chosen the form of test that is different from those chosen by the students next to you. The total points you earned will be entered into Blackboard. The keys will be posted in Blackboard. February 1st. D. Exam 1 2:00–3:00 PM. 3. You can pick up your exam in the Chemistry Learning Center after the points are entered into Blackboard (see syllabus for details). The second part of the exam contains 2 questions that need short written answers. number. This part will be graded manually. The exam has two parts. You need to mark your answers of these questions with a pencil on your answer sheet. D. NH3 Y Z Z Reactions (a).Part 1: Multiple choice questions. H2SO4 HgSO4 Na. (e). What is the product of the following reaction? O 1) LiAlH4 2) H2O OH OH H H OH OH W A. Devise a synthesis of CH3CH2CH2CHO (Z) from two-carbon starting materials. OH- X (b) H H X (d) H H CH3CH2Br (e) X (g) X (i) A. B. D. and (l) 2. (a) H H (c) H H NaH 1) BH3 2) H2O2. H2SO4 HgSO4 H2O. OHBrCH2CHO Z (j) Z (l) Y Y Br2 X HBr X BrCH2CH2OH Y H2O. and (i) Reactions (d). X Y OH Z The product is W The product is X The product is Y The product is Z 2 . (e). and (j) Reactions (b). B. X Y (h) X NaNH2 (2 equiv) Y (k) Y (f) Y (1) BH3 (2) H2O2. C. (g). (15 × 10 = 150 points) 1. and (k) Reactions (a). C. (f). Select reactions that convert cis-2-butene to trans-2-butene. and secondary alcohols to ketones. (d) and (e) 5. D. T and U The products are Y and Z The products are W and X 4. C. Which of the following compounds does NOT give an aldehyde upon reaction with PCC? A.2-Dimethyl-1-butanol 3-Methyl-3-pentanol 3 . D. B. what are/is the product(s)? R O O O O T S O O O OH H H Z Y X W V U O OH A. D. O O The only product is V The products are S. Pyridinium chlorochromate (PCC) is a commercially available reagent that provides Cr6+ (as CrO3). B. It is useful for oxidizing primary alcohols to aldehydes. NH3 Y NaNH2 (2 equiv) X Y H2. (c) (e) A. B. C.3. (c) and (e) Reactions (b). (c) and (f) Reactions (b). C. The alkyne R is treated with O3 followed by water. and does not contain water. Benzyl alcohol Cyclopentylmethanol 2. Br2 HBr (a) X CH3CO3H X Y (b) (d) X Na. Lindlar catalyst (f) Y Reactions (a). (d) and (f) Reactions (a). H2O H 1) CH3CO3H Ph 2) OH-. H X 1) OsO4 Ph OH H 1) OsO4 Ph 2) NaHSO3. H2O (b) 2) NaHSO3. What is the product of the following reaction? Pay attention to the stereochemistry of the reaction. Compound C D. C. D. A. H2O Ph H 1) CH3CO3H 2) OH-. Compound D 4 . H2O H (d) Ph Reactions (a) and (c) Reactions (b) and (d) Reactions (a) and (d) Reactions (b) and (c) 7. Compound A B.6. B. Which of the following reactions can be used for the synthesis of diol X and its enantiomer? H HO H Ph H Ph (a) Ph H H Ph (c) Ph A. Compound B C. Product A is X. product B is W C. Predict the product of the reaction of 1-pentyne and excess Br2.2. product B is Z D. 1-Butyne C.2-Tetrabromopentane D.1-Dibromo-1-pentene B. 1.2-Dibromo-1-pentene C. Butane 10. Propyne B. 2-Chloropropene D.2-Dibromo-1-pentene 5 . The major product of the reaction of 2. The major organic products of the following two reactions are: A. 1. 1. Product A is W. Product A is Y. 2-Butyne B. 1.2-Propadiene C.1.8. product B is Y B. product B is X 9.2-dichloropropane and excess NaNH2 is A. A. Which of the following is the MOST acidic? A. 1-Butene D. Product A is Z. 2. 2-Chloropropane 11. C. B. D. D.4. What 10-carbon compound will undergo ozonolysis to give 2. A. Indicate the product resulting from the hydrogenation (excess H2/Pd on carbon) of 2-hexyne.4-pentadione as the only product? A.4-cyclooctadiene 15. C.4-cyclohexadiene 2.2-Dimethylcyclopropene 1.4-Dimethyl-1. B. B. Which starting materials can be used to make the alkyne S? A.5-Tetramethyl-1. W and Y U. Anti A mixture of syn and anti Neither syn or anti Syn 13. 1.2. X and Z 6 . D. V and Z W. V and Y T.12. (Z)-2-Hexene (E)-2-Hexene Hexane 1-Hexene 14. D. C. The metal-catalyzed addition of hydrogen to an alkene occurs in what fashion? A.6-Cyclodecadiene 1. T. B. C. each curved arrow and structure in green worth 2 points until 25 points are reached. 7 . (25 × 2 = 50 points) 1. Please do not omit any important steps (25 points. Other correct drawings should be graded appropriately. Draw a stepwise mechanism for the following reaction. H3C H H2O H+ H H CH3 H3C + + O H+ H CH3 H3C CH3 Either E or Z is OK H H H O H3C CH3 H+ H H3C H + O CH3 H H3C H H +O CH3 H+ H H3C H O CH3 Total 25 points. each curved arrow and other items worth 2 points until 25 points are reached). Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. So. Each structure in red worth 3 points.Part 2: Questions that need short answers. the students do not need to be completely correct to get full points. Each intermediate structure worth 3 points. Each intermediate structure worth 9 points. the students do not need to be completely correct to get full points. Draw a stepwise mechanism for the following reaction. 8 . The next page (page 9) has a voluntary anonymous survey. each curved arrow and structure in green worth 1 point until 25 points are reached. Each structure in red worth 9 points. So. R R C H + O H O O O Cr R O R R C O Cr H O R C O R O O Cr H R H H R H O R O C H O H OH O Cr O O O H OH O Cr O + O- H O H + R R H Omitting the lone pair is OK Total 25 points. Please do not omit any important steps (25 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction.2. Other correct drawings will also get points. each curved arrow and other items worth 1 point until 25 points are reached). You can answer all or part of the questions. Even though I cannot follow all of your suggestions (they can be contradictory to each other too). 2. Instead. how do you rank the course? If ranking from 1 to 5 with 1 being the least satisfactory and 5 being the most satisfactory. Can you hear me well at the back of the classroom? Are the slide show and projector clear enough? 3. 1. How about the pace of the lectures? Can be somewhat faster? Or it is OK or it should be slower. From your experience so far. detach this page from the exam and give it to a TA separately. I believe that both you and I will benefit from information you give. Any other comments are also appreciated although I may not be able to follow. 9 . which do you choose? 4. Please do not submit with your exam. The goal is to help to improve teaching.This is a voluntary anonymous survey. The points you earned will be entered into Blackboard. 3. This part will be graded manually. please also fill the following information on your answer sheet: 1. There are three forms. Exam 2 2:00–3:00 PM. The total points for this part are 50 (25 × 2). I will send an email to the class when ready. This part will be graded using a scanner. The total points for this part are 150 (15 × 10). answers marked on the exam will NOT be graded.CH2420. You can pick up your exam in the Chemistry Learning Center after the points are entered into Blackboard (see syllabus for details). Test form number. Each question has only one correct answer. B and C. The keys will be posted in Blackboard. which is “A” for you. Organic Chemistry II. TEST FORM First A Make sure you have chosen the form of test that is different from those chosen by the students next to you. 2012 (total 200 points) Please print your name CLEARLY here: Last. The questions are the same but they and the answers to them are arranged in different orders. your name. Answers to these questions need to be put in the space provided in the exam. The second part of the exam contains 2 questions that need short written answers. The exam has two parts. print it in the form and then mark the corresponding numbers. 2. number. 1 . A. You need to mark your answers of these questions with a pencil on your answer sheet. The answer sheet for the multiple choice questions may not be returned to you. Part one contains 15 multiple choice questions. I. D. which are printed on papers with different colors. In addition to answers. As a result. March 21st. Part 1: Multiple choice questions. (15 × 10 = 150 points) 1. The major organic products of the following two reactions are: A. Product A is X; product B is Y B. Product A is Z; product B is W C. Product A is W; product B is Z D. Product A is Y; product B is X 2. Why is the following compound not aromatic? A. It has 4n electrons B. It isn't planar C. It has 4n+2 electrons D. The electron system is not continuous 3. What are the major IR absorptions in the functional group region for the following compound? U: 3300 cm-1; V: 1650 cm-1; W: 3150-3000 cm-1; X: 3000-2850 cm-1; Y: 2250 cm-1; Z: 3500 cm-1. A. U, W, X B. W, X, V C. Y, W, Z D. V, W, Y 2 4. Which of the following compounds are aromatic? A. Compounds S, T, U, V only B. Compounds W, X, Y, Z only C. Compounds S, T, U, V, X, Z only D. All are aromatic 5. What 10-carbon compound will undergo ozonolysis to give 2,4-pentadione as the only product? A. B. C. D. 1,2-Dimethylcyclopropene 1,6-Cyclodecadiene 1,2,4,5-Tetramethyl-1,4-cyclohexadiene 2,4-Dimethyl-1,4-cyclooctadiene 6. The 1H NMR spectrum of 1,2-dimethoxyethane recorded on a 300 MHz spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. What are the chemical shifts of the two absorptions? A. 4.39 ppm and 4.55 ppm B. 1.39 ppm and 1.55 ppm C. 2.39 ppm and 2.55 ppm D. 3.39 ppm and 3.55 ppm 7. Which of the following statements is true about free radical halogenation of alkanes? A. The first of the chain-propagating steps is rate-determining B. The reaction proceeds by way of a flat sp2 hybridized free radical C. The chain-initiating step involves cleavage of a carbon-hydrogen bond to afford a carbon radical and a hydrogen atom D. Both statements A and B are true 3 8. Which of the following compounds give one singlet in 1H NMR spectra? O Cl Cl Br Br Br Br T S O W O Br V U O Y Z X A. Compounds S, U, X, Z B. Compounds T, V, X, Y C. Compounds T, U, W, Z D. Compounds S, U, W, Z 9. Consider both 5−methyl−1,3−cyclopentadiene (Y) and 7−methyl−1,3,5−cycloheptatriene (Z), which labeled H atom is most acidic? Which labeled H atom is least acidic? A. Hc is most acidic because its conjugate base is antiaromatic; Hd is least acidic because its conjugate base is aromatic B. Hb is most acidic because its conjugate base is aromatic; Hc is least acidic because its conjugate base is antiaromatic C. Hd is most acidic because its conjugate base is aromatic; Hb is least acidic because its conjugate base is antiaromatic D. Ha is most acidic because its conjugate base is antiaromatic; Ha is least acidic because its conjugate base is aromatic 10. Which molecule absorbs UV light at the longest wavelength? A. B. C. D. 1,3-heptadiene 1,4,7-octatriene 1,4-pentadiene 1,4-cyclohexadiene 4 Y only C. X only B.33 ppm (quartet. Compound Z 13. Its 1H NMR data are 0. 3.91 ppm (multiplet. Z only D. A molecule has the molecular formula of C7H14O2. 2.11. then HBr 12. 2H).15 ppm (triplet. then PCC B.93 ppm (doublet. What is the compound? A. What will the product(s) be when compound W is reacted with NBS and UV light? A. then KOtBu C. Compound X C. 1. Which of the following reagents/conditions are appropriate to convert compound Y to compound Z? A. 2H). 3 H). Y and Z 5 . Compound W B. 1. 6 H). 1 H). First mCPBA.86 ppm (doublet. First HBr. First Br2/heat. Compound Y D. then H2/Lindlar catalyst D. First Br2/heat. C. B.14. 2900 cm-1. 3300 cm-1 (weak). The reaction is stereospecific B. 3040 cm-1. 3400 cm-1 (weak) A. The diene must be in the s-cis conformation in order to react C. Which of the following statements about the Diels-Alder reaction is false? A. The dienophile usually contains an electron withdrawing group for the reaction to be synthetically useful D. Which of the following molecules would give an IR spectrum with the following peaks: 1650 cm-1. The diene and dienophile line up so that the exo product is kinetically favored 15. D. Compound W Compound X Compound Y Compound Z 6 . each curved arrow worth 1 point until 25 points are reached). Please do not omit any important steps (25 points. 7 . Each intermediate structure worth 1 point. Part 2: Questions that need short answers. (25 × 2 = 50 points) 1. Draw a stepwise mechanism for the following reaction.E. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Please do not omit any important steps (25 points. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. each curved arrow and other items worth 2 points until 25 points are reached). Each intermediate structure worth 5 or 2 points. 8 . Draw a stepwise mechanism for the following reaction.2. The total points for this part are 60 (20 × 3). This part will be graded manually. The keys will be posted in WebCT. A. D. 2012 (total 300 points) Please print your name CLEARLY here: Last. In addition to answers. As a result. I will send an email to the class when ready. Test form number. please see Syllabus. The exam has two parts. 1 . Exam Final 3:00–5:00 PM. 3. which is “A” for you. TEST FORM First A Make sure you have chosen the form of test that is different from those chosen by the students next to you. print it in the form and then mark the corresponding numbers. April 23rd. Organic Chemistry II. your name. The second part contains 3 questions that need you to provide stepwise mechanism of organic reactions. Answers to these questions need to be put in the space provided in the exam. I. number. answers marked on the exam will NOT be graded. There are three forms. 2. The total points for this part are 240 (8 × 30). This part will be graded using a scanner. For grading policy. The questions are the same but they and the answers to them are arranged in different orders. which are printed on papers of different colors.CH2420. You need to mark your answers of these questions with a pencil on your answer sheet. B and C. please also fill the following information on your answer sheet: 1. The total points you earned will be entered into WebCT. Each question has only one correct answer. The first part contains 30 multiple choice questions. 4-pentadione as the only product? A) B) C) D) 1.2-Dimethylcyclopropene 1.4-Dimethyl-1. A) A ketone or an aldehyde and a secondary amine B) A ketone or an aldehyde and a primary amine C) An ylide and a primary amine D) An ylide and a secondary amine 3.4-cyclooctadiene 4.6-Cyclodecadiene 1. 2 .2. Identify the reactants required for the formation of an enamine.5-Tetramethyl-1. What 10-carbon compound will undergo ozonolysis to give 2. (8 × 30 = 240 points) 1. What is the product of the following reaction sequence? A) Compound A B) Compound B C) Compound C D) Compound D 2.Part 1: Multiple Choice Questions. Determine the product of the following reaction.4.4-cyclohexadiene 2. O O OH O OH W O O X Y O Z A) Spectrum E is from compound W. and spectrum F is from compound X B) Spectrum E is from compound X. Which reagent would be used to reduce the alkene in cyclopentenone? A) NaBH4 B) LiAlH4 C) 1 equivalent H2 and catalytic amount of Pd/C D) DIBAL-H 3 . and spectrum F is from compound X 6. and spectrum F is from compound Z D) Spectrum E is from compound Z.R Br NH2 T S Br U A) Compound R B) Compound S C) Compound T D) Compound U 5. and spectrum F is from compound Y C) Spectrum E is from compound X. Match 1H NMR spectra with compounds. 3400 cm-1 (weak) A) B) C) D) Compound W Compound X Compound Y Compound Z 8. Identify the hemiacetal carbon atom in this structure. 2900 cm1 . The structure below is -D-galactose. A) Compound A B) Compound B C) Compound C D) Compound D 4 . Which of the following molecules would give an IR spectrum with the following peaks: 1650 cm-1. A) Carbon A B) Carbon B C) Carbon C D) Carbons D and E 9. 3040 cm-1. 3300 cm-1 (weak). Select the product of the following reaction.7. X. X. mCPBA.1 1. Y and Z are CH3CH2COCl. Y and Z are CH3CH2COCl/AlCl3. HNO3/H2SO4. Br2/FeBr3. Add reagents for the following multiple step synthesis.5 3. Y and Z are CH3CH2COCl/AlCl3. X. HNO3/H2SO4. An unknown compound X has the molecular formula C6H14O. Br2/FeBr3. Br2/FeBr3. respectively 12. X. Zn(Hg)/HCl. Y and Z are CH3CH2COCl/AlCl3. Zn(Hg)/HCl. Zn(Hg)/HCl. HNO3/H2SO4. The 1H NMR spectral data of X are given below. respectively B) W.0 1. Which of the following alkenes undergo allylic bromination to form a single monobrominated product (not including stereoisomers)? A) Alkene A B) Alkene B C) Alkene C D) Alkene D 5 . What is the most likely structure of X? absorption triplet singlet quartet singlet 1. A) W. X shows a strong peak in its IR spectrum at 3000 cm–1. respectively D) W.10. SO3/H2SO4. Br2/heat.5 H ratio 3 6 2 3 A) Compound A B) Compound B C) Compound C D) Compound D 11. respectively C) W. Z B) Compounds T. W. X. Select the structure of the product in the following reaction. Y C) Compounds T. A) Product A B) Product B C) Product C D) Product D 14. W. U. U. Which of the following compounds give one singlet in 1H NMR spectra? O Cl Cl Br Br Br Br S O O Br T O Y W V U Z X A) Compounds S. Which of the following reaction starting materials would yield the compound below? 6 . X. Z 15. V.13. Z D) Compounds S. U. CO2 Me A. CO2Me + + D. Which of the following compounds matches or match the IR spectrum shown? OH OH V U O W O O O X Y Z 7 . CO2Me + + B. MeO2C CO2Me A) Starting materials A B) Starting materials B C) Starting materials C D) Starting materials D 16. MeO2C CO2Me CO2Me C. What is the first step in the general mechanism for electrophilic aromatic substitution? A) Protonation of the aromatic ring B) Deprotonation of the aromatic ring C) Addition of the electrophile to the aromatic ring D) Loss of the electrophile from the aromatic ring 8 . Name the following compound. Rank the following in terms of increasing reactivity for an SN1 reaction. then PCC B) First Br2/heat. Which of the following reagents/conditions are appropriate to convert compound Y to compound Z? A) First HBr.o-Dibromonitrobenzene 18.A) U. then KOtBu C) First mCPBA. A) 2. then HBr 20. X and Z B) V. A) X<Y<Z B) Z<Y<X C) Z<X<Y D) Z<Y=X 19. then H2/Lindlar catalyst D) First Br2/heat.3-Dibromoaniline B) 2-Bromo-5-aminobromobenzene C) 2-Bromo-3-aminobromobenzene D) m. W and Y C) Only U D) Only X 17. Why is the nitro group a meta director? A) Because it is sterically very large B) Because it adds electron density to the meta position. 9 . C3 and C4 appear at 32. respectively D) C1. 70. 32 and 10 ppm. respectively 24. 70. C2. then extract with an organic solvent C) Make an acidic aqueous solution. How could you separate the following compounds? A) React with KMnO4 B) Make a weakly basic aqueous solution. thus activating it C) Because it stabilizes the intermediate cation D) Because it removes more electron density from the ortho and para positions than the meta position. C2. 32 and 23 ppm. C3 and C4 appear at 70. 32 and 10 ppm. C3 and C4 appear at 23.21. respectively C) C1. Rank the following from most to least acidic. Assign carbon-13 NMR signals to each carbon in the compound shown. C2. C2. then extract with an organic solvent D) Hydrogenate 23. A) C1. C3 and C4 appear at 10. thus deactivating the meta position less 22. 23 and 10 ppm. 23. respectively B) C1. 70. What is the missing reagent in the reaction below? A) Compound A B) Compound B C) Compound C D) Compound D 26.A) Compound A > Compound B > Compound C B) Compound B > Compound A > Compound C C) Compound A > Compound C > Compound B D) Compound B > Compound C > Compound A 25. What is the product of the following reaction? A) Compound A B) Compound B C) Compound C D) Compound D 10 . CCl4 29. then NaC≡CH D) Reagent X is Cl2. reagent Y is NaH. reagent Z is first NaC≡CH. Which reagent would bring about the following transformation? A) LiAlH4 C) Br2. reagent Z is first water. What reagent(s) is (are) needed to add NO2 to benzene? A) HNO3. An electron donating group attached to a benzoic acid will: A) Make the benzoic acid less acidic B) Make the benzoic acid more acidic C) Deactivate the benzene ring D) Not change the acidity of the benzoic acid 30. A) Reagent X is Cl2 and water. reagent Z is NaC≡CH and water C) Reagent X is Cl2 and water. light B) FeCl3 D) Basic KMnO4 28.27. then water 11 . then water B) Reagent X is first Cl2. then water. reagent Y is NaH. reagent Y is NaH. reagent Z is first NaC≡CH. reagent Y is HCl. HCl B) HNO3. Provide reagents for the following multistep synthesis. H2SO4 C) NO2(g) D) NO2. Each intermediate structure worth 3 points. each curved arrow and other items worth 1 point until 20 points are reached).Part 2: Mechanism of Organic Reactions. Use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. O CH2 Ph3P + CH3Br Ph3PCH3 BuLi Ph3P CH2 Br 12 . (20 × 3 = 60 points) 1. Please do not omit any important steps (20 points. Draw stepwise mechanisms for all the following reactions that are used to synthesize the alkene product. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Draw a stepwise mechanism for the following reaction. Other correct drawings will also get points. Please do not omit any important steps (20 points. AlCl4 H3C C C H Cl CH3 H CH3 H CH3 H H3C CH3 C C H Cl AlCl3 CH3 H3C C C H CH3 CH3 H CH3 H H Cl HCl + CH3 AlCl3 H3C CH3 C H CH3 H H3C C H3C C C CH3 H H H AlCl3 AlCl3 Total 20 points. Each intermediate structure worth 4 points. each curved arrow and structure in green worth 1 point until 20 points are reached.2. each curved arrow and other items worth 1 point until 20 points are reached). Each structure in red worth 4 points. 13 . You do not need to be completely correct to get full points. You do not need to be completely correct to get full points. Each intermediate structure and curved arrow worth 1 point until 20 points are reached). 14 . H H Al H H Al H O H O + H H H + AlH3 N N H H H H H Al O AlH3 N O H N H H AlH3 H O H Al H H H AlH3 H H H Al H H H H H O H N N H H AlH3 OH H H H N H Total 20 points. Draw a stepwise mechanism for the following reaction. Please use appropriate curved arrows to indicate the movement of all electrons that moved during the reaction. Other reasonable drawings will also get points. Please do not omit any important steps (20 points.3. Each curved arrow in green and each structure in red worth 1 point until 20 points are reached.