FINAL EXAM ANSWER KEY Winter Session 2011R
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1 ANSWE KEY Page 1 of 11 CEM 2220 rganic Chemistry II: eactivity and Synthesis Prof. P.G. ultin, Prof. T. egmann, Dr.. Luong FINAL EXAM ANSWE KEY Winter Session 2011 Monday April 25, :00 am 12:00 pm Frank Kennedy Gold Gym Students are permitted to bring into the exam room NE SEET of 8½ x 11 paper with any ANDWITTEN notes they wish (both sides). Molecular model kits and calculators (no text or graphics memory!) are permitted but no other aids may be used. Question 1 eactions and Products Question 2 Synthesis Question 3 Mechanism Question 4 Mechanism Question 5 Mechanism Question 6 Laboratory Question 7 Spectroscopy TTAL: (30 Marks) (10 Marks) (10 Marks) (15 Marks) (15 Marks) (15 Marks) (5 Marks) (100 Marks)
2 CEM 2220 Winter 2011 ANSWE KEY Page 2 of (30 MAKS) eactions and Products. Supply the missing molecular structure or reagent/solvent/reaction conditions to correctly complete the following reactions. 1. Cr 3, 2 S 4 aq. (Jones' eagent) (a) 2. C 3 C 3, + cat. (b) (c) (d) (e) (f) (4 Marks) (g)
3 CEM 2220 Winter 2011 ANSWE KEY Page 3 of 11 (h) LDA, TF, -78 C Then add Et (i) Et Et (j) (k) (l) (m) (n)
4 CEM 2220 Winter 2011 ANSWE KEY Page 4 of (10 MAKS) Propose a sequence of reactions to the target compound 1 below starting from cyclopentanone. You may use any of the organic compounds shown in the cupboard in order to build the exo-cyclic chain up this will require more than one step! In addition you have any reagents or solvents you may need to perform your reactions. For each reaction in your sequence show the starting material, all necessary reagents/solvents, and the expected product.
5 CEM 2220 Winter 2011 ANSWE KEY Page 5 of (10 MAKS) The unusual compound emycin F is produced by a strain of Streptomyces cellulosae along with several other related compounds. ne of these, emycin E was found to be formed from emycin F by an acid-catalyzed reaction. Write a mechanism to explain the formation of emycin E from emycin F. 3 * emycin F emycin E *a mixture of stereoisomers at this carbon The fact that the position marked * is formed as a mixture of stereoisomers suggests that this centre becomes a cation at some point during the reaction,whichcanbeattackedbyoxygenfromeitherface. * *
6 CEM 2220 Winter 2011 ANSWE KEY Page 6 of (15 MAKS Total) Alkenes can be isomerized using the two-step sequence shown below. The process is stereospecific: E-alkenes are converted to Z-alkenes, and vice versa. (a) (5 MAKS) What product is formed in the first step, the treatment of an alkene with a peroxycarboxylic acid? Write a detailed mechanism for this step. Thereactionofanalkenewithaperoxycarboxylicacidformsanepoxidewithretentionofthe conf iguration of the alkene. C 3 + C 2 (b) (10 MAKS) Provide a detailed stepwise mechanism for the second step, in which the product of step 1 is heated with triphenylphosphine. This reaction has some similarities to the Wittig reaction which we discussed in class. Clearly show how your mechanism forces the stereochemical inversion shown (Newman projections may be helpful). Using the epoxide formed from a trans alkene as the example: P P P 180 o rotation of C-C bond P P P +3 P Note that without the 180 o rotation, the oxygen cannot attack the phosphonium ion. It is this rotation that forces the trans/cis or cis/trans inversion.
7 CEM 2220 Winter 2011 ANSWE KEY Page 7 of (15 MAKS TTAL) Treatment of dimethyl 3-oxopentanedionate (1) first with one molar equivalent of sodium hydride (a non-nucleophilic strong base) and then with methyl but-2-ynoate (2) at low temperature forms the symmetrical resorcinol (3). In contrast, at room temperature or above the unsymmetrical resorcinol isomer (4) is obtained. (a) (7 MAKS) Provide a detailed, stepwise mechanism for the formation of 3 at low temperature. 3 C C 3 1 Na (1 equiv.), dry TF then add 3 C 2 C 3 low temperature (0 o C) C 3 C 3 C 3 3 Note that you have only 1 equivalent of Na, and this strong base deprotonates irreversibly. 3 C C 3 Na 3 C C 3 3 C C (g) 3 C C 3 C 3 3 C C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 C 3 Product
8 CEM 2220 Winter 2011 ANSWE KEY Page 8 of 11 (b) (8 Marks) Write a mechanism for the formation of 4 at higher temperatures. (INT: the reaction initially follows the same path as the cold process, but diverges part way along).
9 CEM 2220 Winter 2011 ANSWE KEY Page 9 of Lab Questions (15 MAKS total) (a) (12 MAKS) Propose an experimental procedure starting from Aldehyde A to make about 2.5 g of Ester D. You have access to any chemicals required for the synthesis. Assume that the reaction will give 50% yield and be sure to include a purification step. Indicate the approximate amount of starting materials and solvents needed (you should be able to perform the math without the aid of a calculator). Please rationalize each step in your procedure and state any necessary assumptions. Below are physical constants of some chemicals that you may wish to use or make. Many of the physical constants below are fictional and should be used NLY for the purposes of this examination. Chemical Name Aldehyde A osphorane B osphine xide C Ester D M.W. (g/mol) Boiling Point ( o C) Melting Point ( o C) room temperature (amount of solute/amount of solvent) C 3 (1 g/1ml) Pentane (1 g/1ml) Water (2g/10L) C 2 Cl 2 (1g/10L) C 2 Cl 2 1.5g/1mL Pentane 1g/1mL Water 3g/10L C 3 2g/10L C 2 Cl 2 1g/1mL C 3 0.9g/1mL Pentane 0.8g/1mL Water 0.5g/10L C 3 1 g/1ml C 2 Cl 2 1 g/1ml Pentane 0.01 g/1l Water 0.01 g/1l Common Solvents Name C 2 Cl 2 Pentane C 3 Boiling Point ( o C) Dissolve Aldehyde A (3.6 g, 0.02 mol, 1 mark for correct quantity) in about 20 ml of pentane (1 mark for appropriate volume and 1 mark for solvent choice). Cool the solution down in an ice bath (1 mark). After a sufficient cooling period, phosphorane B (7.2 g, 0.02 mol, 1 mark for correct quantity) is added. As the reaction proceeds, the product will precipitate out because of its low solubility in pentane (2 marks for recognizing this). Conveniently, the phosphine oxide C byproduct is pentane soluble, so it will remain in solution (2 marks for recognizing this). When the reaction is complete, the product can be suction filtered (1 mark). To purify the product, a recrystallization can be performed using a mixture of methanol/water (Ester B has low solubility in water and high solubility in methanol). This mixture was used because the solvents are miscible in each other. (2 marks) Note that the physical data you were given indicates that C2Cl2, the solvent you used in your experiment in the CEM 2220 lab, would not be a wise choice. Further examination of the solubility data should convince you that PENTANE is the best solvent to use, since it dissolves both starting material and reagent. Furthermore, the product is insoluble in pentane, but the phosphine oxide byproduct is fairly soluble (again, the opposite was true in your experiment) so you would expect that the product would precipitate as it forms and can be collected by simple filtration.
10 CEM 2220 Winter 2011 ANSWE KEY Page 10 of 11 (b) (3 Marks) Suppose at the end of the reactions that you proposed you examined the isolated product by TLC. You developed one silica TLC plate with ethyl acetate while the other was developed with petroleum ether. Based on the TLC plate observations: i) Was Ester D made? What is your evidence for your answer? Ester B was made and this is evident from the second spot in the product lane when petroleum ether was used as the eluent. This conclusion cannot be drawn when examining the plate developed with ethyl acetate. (2 marks) ii) Comment on the purity of the product. What is the evidence for your evaluation? Since there is more than one spot in the product lane, it is safe to conclude that the product is impure. (1 mark)
11 CEM 2220 Winter 2011 ANSWE KEY Page 11 of (5 MAKS) The molecular formula of the unknown organic compound whose spectra are shown is C What is its structure? Show your work in the margin for partial credit. d q m m t t Formula gives 1 degree of unsaturation. I shows clear C= at ~1740 cm -1. N present. 13 C NM shows 7 signals total = some symmetry. Carbonyl C at ~175 says this must be an ester. Note 2 13 C signals around 10 ppm, one larger than the other, suggests 2 kinds of C 3 groups, and perhaps 2 of one kind plus 1 of the other. 1 NM shows 6 distinct signals in ratio 2:2:1:4:3:6. Two-proton doublet at 4 ppm is C 2 C of ester. Two-proton quartet at 2.3 ppm plus 3-proton triplet at 1.1 ppm are C 3 C 2 C=. Note 6-proton triplet at 0.9 ppm is 2 identical C 3 groups each having a C 2 next door. The C 2 groups are the 4-proton multiplet at 1.4 ppm. The remaining 1-proton multiplet is the C at the branch point, with all the C 2 groups next to it. Structure for C
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