Chemistry 216 Second Examination December 11, 2007 Name _____KEY__________ Please Print Dr. Andrew Karatjas (2 hrs, 100 points) Signature _______________________ Student ID # _____________________ Please CHECK OFF your lab section. ____section #_________GSI_____________ _____130 Thomas Sundberg _____131 Peter Mai _____132 Peter Ung _____133 Therese Dorau _____134 Candice Paulsen _____135 Ahleah Rohr _____136 Justin Lomont _____150 Kapil Karki _____151 Ben Thompson _____152 Thomas Sundberg _____153 Candice Paulsen _____154 Peter Ung _____155 Joseph Jankolovits _____156 Ahleah Rohr This exam has 15 pages including this cover page. The last 4 pages include a periodic table, tables of characteristic IR absorption frequencies of common chemical bonds, and representative 1 H and 13 C NMR chemical shifts. ____section #____GSI____________ _____170 Kapil Karki _____171 Ben Thompson _____172 Peter Mai _____173 Therese Dorau _____174 Joseph Jankolovits _____175 Christopher Taylor _____176 Christopher Taylor I 15 II 9 III 16 IV 8 V 12 VI 7 VII 3 VIII 10 IX 10 X 10 Total 100 I Draw the structures for the organic products containing a phenyl group which would be expected upon treatment of each of the following chemicals with PhMgBr followed by acidic workup (15 points). O Ph H dry ice CO 2 H 2 O O OH + HO Ph H Ph OH O O OH II The attempted Grignard reaction of 6-bromohexanal with benzophenone did not proceed to give the desired product. Provide in the box an explanation for the failure of this reaction on the basis of its mechanism (9 points). Br H O Mg, Et 2 O O HO O H + X Br H O BrMg H O Thealdehydeisanelectrophile, the Grignard will react intramolecularly as soon as it is formed. OMgBr OH H 3 O + III Fill in the box(es) with the expected outcome of the reaction or with the correct starting material(s). For each, also give the name of the reaction (16 points). O O O +(a) O O O Diels-Alder (b) O Ph 3 P CH O MeO + O O Wittig O O O +(c) This diene is too hindered to adopt the needed s-cis conformation for the Diels-Alder reaction. No reaction will take place. O O 1. NaOH/CH 3 CH 2 OH H 2 O, reflux 2. H 3 O + workup (d) CO 2 H OH benzilic acid rearrangement IV For the following reaction, draw a complete step-by-step mechanism (8 points). O O Ph Ph NaOH, CH 3 OH Ph HO Ph CO 2 H HO - O O Ph Ph HO O Ph Ph O - Ph - O Ph CO 2 H Ph HO Ph CO 2 - Ph HO Ph CO 2 H H 3 O + V For each of the following compounds, predict the number of peaks you would expect to observe in both the 1 H NMR spectra. Do not count the peaks for the solvent or for TMS (12 points). OH OH H 2 N OHOHOO O 1 8 5 6 4 10 VI Draw the expected 1 H NMR spectrum for the following compound. Both correct chemical shift and splitting are needed. Do not draw in integrals (7 points). 012345678910 PPM s d d d d q t O O H VII The proton NMR of a mixture of acetonitrile (CH 3 CN), benzene (C 6 H 6 ), and cyclohexane (C 6 H 12 ) is integrated and results in heights of 18, 30, and 12 for the peaks at 2.10, 7.26, and 1.43 ppm respectively. In what mole ratio are the three substances present. No explanation is required. No partial credit will be given for this problem (3 points). :: acetonitrile : benzene : cyclohexane 651 In questions VIII ? X you will be given the molecular formula, IR, 13 C NMR and 1 H NMR spectra of a compound. You are required to deduce its molecular structure and give your answer as part d of each question. If you answer part d completely correctly you will receive full credit for the question. The answers to parts a, b, and c will help you deduce the structure. You will receive partial credit by answering these questions if your answer in part d is not correct. Note: 13 C data refer to whether the carbon signal is expected to be a singlet (s), doublet (d), triplet (t) or quartet (q). The spectra you are given were recorded in ?decoupled? mode (all the carbon peaks appear as singlets). However, the multiplicity (when recorded in ?coupled? mode) is indicated above each peak by these letters. The numbers below each peak in the 1 H NMR spectrum correspond to the relative integration for the peak. VIII Compound A has molecular formula: C 5 H 8 (10 points) VIIIa The number of double bond equivalents for compound A is: VIIIb The functional group or groups present in compound A is/are: (write one functional group in each box, you may not need both boxes) VIIIc The substructures or fragments (containing more than one carbon atom) that make up compound A are: (write one substructure in each box, you may not need all the boxes) VIIId The structure of compound A is: (draw your answer in the box) IX Compound B has molecular formula: C 8 H 14 O 4 (10 points) IXa The number of double bond equivalents for compound B is: IXb The functional group or groups present in compound B is/are: (write one functional group in each box, you may not need both boxes) IXc The substructures or fragments (containing more than one carbon atom) that make up compound B are: (write one substructure in each box, you may not need all the boxes) IXd The structure of compound B is: (draw your answer in the box) O O O O X Compound C has molecular formula: C 11 H 11 N O 4 (10 points) Xa The number of double bond equivalents for compound C is: Xb The functional group or groups present in compound C is/are: (write one functional group in each box, you may not need both boxes) Xc The substructures or fragments (containing more than one carbon atom) that make up compound C are: (write one substructure in each box, you may not need all the boxes) Xd The structure of compound C is: (draw your answer in the box) NO 2 O O Andy Microsoft Word - Chemistry 216 - Exam 2-KEY.doc
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