Name: Student Number: University of Manitoba - Department of Chemistry CEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 16, 2012; 7-9 PM This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions. Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached at the end of the exam. QUESTION 1. Mechanism (8 Marks) 2. Mechanism 3. Mechanism 4. Mechanism (8 Marks) 5. Reactions and Products (20 Marks) 6. Spectra and Structures (6 Marks) TOTAL (50 Marks) MARKS
CEM 2220 Test #1 Page 2 of 8 Feb 16, 2012 1. (8 MARKS) When a solution of the diol 1 in toluene solvent was treated with concentrated phosphoric acid ( 3 PO 4 ) and boiled for 10 hours, the product 2 was formed. Provide a detailed stepwise mechanism for this reaction.
CEM 2220 Test #1 Page 3 of 8 Feb 16, 2012 2. (4 MARKS) The reactions of two alkyl bromides with NaOC 2 C 3 in C 3 C 2 O solvent at 55 C were studied, and the results are shown below. Briefly explain why the product ratios are different in these two reactions. 3. (4 MARKS) When trans-2-butene is treated with dibromine the product is exclusively formed by an anti addition process. In contrast, under the same conditions trans-1-phenylpropene forms mostly the product of anti addition but some of the isomer arising from syn addition is also formed. Briefly explain why the stereoselectivity is lower in the reaction of 1-phenylpropene.
CEM 2220 Test #1 Page 4 of 8 Feb 16, 2012 4. (8 MARKS) When 2-pentyne is treated with aqueous sulfuric acid and mercuric sulfate, two new compounds (B and C) are formed, both of which have the formula C 5 10 O. The IR spectrum of each of these products shows a strong band in the 1715-1720 cm -1 region. Draw a detailed stepwise mechanism for this reaction that shows what the two products are and how they are formed.
CEM 2220 Test #1 Page 5 of 8 Feb 16, 2012 5. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Show product stereochemistry (wedge and dash bonds) where appropriate. a. b. c. d.
CEM 2220 Test #1 Page 6 of 8 Feb 16, 2012 e. f. g. h.
CEM 2220 Test #1 Page 7 of 8 Feb 16, 2012 6. (6 MARKS) The spectra of an unknown organic compound A having the formula C 6 12 O 2 are shown below. What was the structure of compound A? NOTE: putting your rough work in the left margin can earn part marks. IR 4000 3000 2000 1500 1000 500 13 C NMR 1 NMR Exchangeable Structure of Compound A
Page 8 of 8 Spectroscopy Crib Sheet for 2.222 Introductory Organic Chemistry II 1 NMR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C C 3 0.7 1.3 C C 2.5 3.1 C C 2 C 1.2 1.4 C O C C C C 1.4 1.7 O O O 1.5 2.5 C O 9.5 10.0 10.0 12.0 (solvent dependent) 1.0 6.0 (solvent dependent) 2.1 2.6 O C 3.3 4.0 Aryl C 2.2 2.7 Cl C 3.0 4.0 4.5 6.5 Br C 2.5 4.0 Aryl 6.0 9.0 I C 2.0 4.0 RCO 2 Aromatic, heteroaromatic X C X = O, N, S, halide R 3 C Aliphatic, alicyclic Y = O, NR, S Y Y Y = O, NR, S 12 11 10 9 8 7 6 5 4 3 2 1 0 Low Field 13 C NMR Typical Chemical Shift Ranges igh Field Alkene Aryl Ketone, Aldehyde Ester Amide Acid RC N C x -Y Y = O, N CR 3 -C 2 -CR 3 C x -C=O RC CR C 3 -CR 3 220 200 180 160 140 120 100 80 60 40 20 0 IR Typical Functional Group Absorption Bands Group Frequency (cm -1 ) Intensity Group Frequency (cm -1 ) Intensity C 2960 2850 Medium RO 3650 3400 Strong, broad C=C 3100 3020 Medium C O 1150 1050 Strong C=C 1680 1620 Medium C=O 1780 1640 Strong C C 3350 3300 Strong R 2 N 3500 3300 Medium, broad R C C R 2260 2100 Medium (R R ) C N 1230, 1030 Medium Aryl 3030 3000 Medium C N 2260 2210 Medium Aryl C=C 1600, 1500 Strong RNO 2 1540 Strong
ANSWER KEY University of Manitoba - Department of Chemistry CEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 16, 2012; 7-9 PM This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions. Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached at the end of the exam. QUESTION 1. Mechanism (8 Marks) 2. Mechanism 3. Mechanism 4. Mechanism (8 Marks) 5. Reactions and Products (20 Marks) 6. Spectra and Structures (6 Marks) TOTAL (50 Marks) MARKS
CEM 2220 Test #1 ANSWERS Page 2 of 8 Feb 16, 2012 1. (8 MARKS) When a solution of the diol 1 in toluene solvent was treated with concentrated phosphoric acid ( 3 PO 4 ) and boiled for 10 hours, the product 2 was formed. Provide a detailed stepwise mechanism for this reaction. COMMENTS: This is Groutas problem # 7. Although all the steps are in fact reversible, students did not need to show equilibrium arrows for full marks. It was not necessary to explicitly show the phosphoric acid molecule as the proton source; students could just indicate +. Likewise, it was not necessary to explicitly indicate the 2 PO 4 as the base in the final step. Note that because this reaction was conducted in the nonpolar solvent toluene, the strong acid is not dissociated as it would be if the reaction had been done in water.
CEM 2220 Test #1 ANSWERS Page 3 of 8 Feb 16, 2012 2. (4 MARKS) The reactions of two alkyl bromides with NaOC 2 C 3 in C 3 C 2 O solvent at 55 C were studied, and the results are shown below. Briefly explain why the product ratios are different in these two reactions. In the first reaction, the S N 2 product is the major and the E 2 product is the minor, but in the second reaction this is reversed. In the first case, the starting compound is a primary bromide, which has negligible steric hindrance, so the S N 2 is favoured. In the second case, the reactant is a secondary bromide which is much more hindered, slowing the S N 2 process relative to the E 2 process. 3. (4 MARKS) When trans-2-butene is treated with dibromine the product is exclusively formed by an anti addition process. In contrast, under the same conditions trans-1-phenylpropene forms mostly the product of anti addition but some of the isomer arising from syn addition is also formed. Briefly explain why the stereoselectivity is lower in the reaction of 1-phenylpropene. The bromonium ion formed from 2-butene has no special cation-stabilizing groups, so it does not open to form a discrete carbocation and must be attacked in a backside fashion to make the anti dibromide. The bromonium ion obtained from 1-phenylpropene is much less rigid because the phenyl ring can stabilize a carbocation. Thus, the bridged ion can open to the discrete cation, which can be attacked by bromide in an S N 1-like fashion from either side, leading to the small proportion of syn dibromide. COMMENT: students did not have to draw a mechanism to explain this. The mechanism is shown here only for clarity.
CEM 2220 Test #1 ANSWERS Page 4 of 8 Feb 16, 2012 4. (8 MARKS) When 2-pentyne is treated with aqueous sulfuric acid and mercuric sulfate, two new compounds (B and C) are formed, both of which have the formula C 5 10 O. The IR spectrum of each of these products shows a strong band in the 1715-1720 cm -1 region. Draw a detailed stepwise mechanism for this reaction that shows what the two products are and how they are formed. COMMENT: The mechanisms must show how the gso4 catalyzes the process that is, how it adds and how it is removed from the organic molecule. It was not necessary to write out the steps twice if students just wrote out one of the pathways in detail and then said the same kind of steps leading to the correct other product that would be fine too. It obviously doesn t matter which product is B and which is C.
CEM 2220 Test #1 ANSWERS Page 5 of 8 Feb 16, 2012 5. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Show product stereochemistry (wedge and dash bonds) where appropriate. a. b. c. d.
CEM 2220 Test #1 ANSWERS Page 6 of 8 Feb 16, 2012 e. f. g. h.
CEM 2220 Test #1 ANSWERS Page 7 of 8 Feb 16, 2012 6. (6 MARKS) The spectra of an unknown organic compound A having the formula C 6 12 O 2 are shown below. What was the structure of compound A? NOTE: putting your rough work in the left margin can earn part marks. IR 4000 3000 2000 1500 1000 500 13 C NMR Formula gives 1 degree of unsaturation. IR shows alcohol plus carbonyl. 13 C shows carbonyl is a ketone. 1 shows nothing but singlets, ratio 1:2:3:6. The 6-proton singlet is obviously two identical methyls with no neighbors. The chemical shifts of the 3-proton singlet and 2-proton singlet suggest that they are next to the ketone. There remains only one place for the O group. 1 NMR Exchangeable Structure of Compound A