Name: Student No: Page 1 of 14 CEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. ultin, Dr.. Luong FINAL EXAM Winter Session 2012R Friday April 20, 20122 1:30 pm 4:30 pm Frank Kennedy Brown Gym Students are permitted to bring into the exam room ONE SEET of 8½ x 11 paper with any ANDWRITTEN 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 Reactions and Products Question 2 Synthesis Question 3 Mechanism Question 4 Mechanism Question 5 Mechanism grab-bag Question 6 Laboratory Question 7 Spectroscopy (30 Marks) (10 Marks) (8 Marks) (12 Marks) (20 Marks) (15 Marks) (5 Marks) TOTAL: (100 Marks)
CEM 2220 Final Exam 2012R Page 2 of 14 1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction conditions to correctly complete the following reactions. Show stereochemistry when necessary. (a) (b) (c) (d) (e) O PBr 3 Ph 3 P Product is ionic - show both cation and anion parts! nbuli, TF then add O (f) (6 Marks)
CEM 2220 Final Exam 2012R Page 3 of 14 (g) (h) (i) (j) (k) (6 Marks)
CEM 2220 Final Exam 2012R Page 4 of 14 2. (a) (7 MARKS) Synthesis. 1-(ydroxymethyl)cyclohexene can be prepared in four steps starting from diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary reactions, as well as the intermediate products formed at each step. EtO O O OEt Reaction 1 Reaction 2 Reaction 3 Reaction 4 O (b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required starting materials can be obtained, briefly explain why or why not.
CEM 2220 Final Exam 2012R Page 5 of 14 3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that this process occurs at an unusually low temperature for this type of reaction. C 2 Cl 2 /MeO 25 o C O O O O (racemic)
CEM 2220 Final Exam 2012R Page 6 of 14 4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes. This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17). (a) (5 MARKS) Johnson s team studied the following reaction as a model process. Suggest a stepwise mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)
CEM 2220 Final Exam 2012R Page 7 of 14 (b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of them. Write a mechanism for the formation of the two products shown.
CEM 2220 Final Exam 2012R Page 8 of 14 5. (20 MARKS TOTAL) Mechanism grab-bag! (a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CEM 2220. Write a mechanism for this reaction. (b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction. Write a stepwise mechanism to explain this transformation.
CEM 2220 Final Exam 2012R Page 9 of 14 (c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure. (d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).
CEM 2220 Final Exam 2012R Page 10 of 14 6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAl 4 ) is a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville s proposed synthetic procedure is shown below. Most data and procedures in this question are fictitious and should not be used for experimental purposes. Acid A Alcohol B LiAl 4 Diethyl ether MW (g/mol) 136.15 122.14 38.39 74.12 MP ( o C) 52 20 N/A -116 BP ( o C) 120 195 N/A 35 Solubilities (g/ml) Water 0.3 Ethanol - 6 Diethyl ether - 3 Water 0.05 Ethanol - 10 Diethyl ether - 20 Water N/A Ethanol N/A Diethyl ether 10 Densities 1.76 0.89 N/A 0.88 Neville s Proposed Procedure Water - 0.03 Ethanol miscible Dissolve Acid A (10 g) in diethyl ether (100 ml). Cool the solution on an ice bath. Slowly add LiAl 4 (8 g). After 30 minutes of stirring, slowly add ice cold water (50 ml) to the solution. Remove and discard the top (aqueous) layer. The organic layer is extracted three times with 50 ml aliquots of 1 M Cl. The solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from the distillation flask (round bottom flask). (a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with Neville s procedure as you can, and indicate how you would modify the procedure to make it work. Your goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield. State all assumptions.
CEM 2220 Final Exam 2012R Page 11 of 14 (b) (3 MARKS) After implementing your recommendations, Neville collected his product and then brought you the TLC plate shown in Figure 1. Unfortunately Neville had used up all the starting material so he could not use it as a reference spot. owever, he found a sample of pure Alcohol B and used that as the reference. elp Neville interpret this TLC plate by explaining whether the product has been made and comment on the purity of the product. Solvent front Origin Alcohol B cospot Product from the reaction Eluent: ethyl acetate:hexanes (1:9) Silica plate Figure 1: TLC of Neville's first product. (c) (3 MARKS) Neville performed the reaction again and examined his product by TLC on silica gel. The plate is shown on the left in Figure 2. Using the blank TLC plate template on the right in Figure 2, draw a diagram of how the plate should look if the plate substrate (absorbent) was modified to the LESS POLAR cellulose but using the same eluent. The spots from left to right are starting material, co-spot, and product. Figure 2: TLC for Neville's second attempt.
CEM 2220 Final Exam 2012R Page 12 of 14 7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C 7 16 O. The IR, 1 NMR and 13 C NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (1 MARK) What is the unsaturation number for this compound? (b) (1 MARK) your answer? What functional group(s) is(are) present? What specific spectroscopic evidence supports (c) (1 MARK) Briefly explain the signals in the 1 NMR spectrum between 0.80 0.88 ppm. (d) (2 MARKS) Draw the structure of this compound in the box below. Structure for C 7 16 O
CEM 2220 Final Exam 2012R Page 13 of 14 Spectra for Question 7 IR 13 C NMR NB: all signals are single lines. 1 NMR 0.999 9.009 2.993 2.004 1.000
Spectroscopy Crib Sheet for CEM 2220 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 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) O 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 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 13 C NMR Typical Chemical Shift Ranges IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (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 Page 1 of 15 CEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. ultin, Dr.. Luong FINAL EXAM Winter Session 2012R Friday April 20, 2012 1:30 pm 4:30 pm Frank Kennedy Brown Gym Students are permitted to bring into the exam room ONE SEET of 8½ x 11 paper with any ANDWRITTEN 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 Reactions and Products (30 Marks) Question 2 Synthesis (10 Marks) Question 3 Mechanism (8 Marks) Question 4 Mechanism (12 Marks) Question 5 Mechanism grab-bag (20 Marks) Question 6 Laboratory (15 Marks) Question 7 Spectroscopy (5 Marks) TOTAL: (100 Marks)
CEM 2220 Final Exam 2012R ANSWER KEY Page 2 of 15 1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction conditions to correctly complete the following reactions. Show stereochemistry when necessary. (a) (b) (c) (d) (e) (f) (6 Marks)
CEM 2220 Final Exam 2012R ANSWER KEY Page 3 of 15 (g) (h) (i) (j) (k) (6 Marks)
CEM 2220 Final Exam 2012R ANSWER KEY Page 4 of 15 2. (a) (7 MARKS) Synthesis. 1-(ydroxymethyl)cyclohexene can be prepared in four steps starting from diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary reactions, as well as the intermediate products formed at each step.
CEM 2220 Final Exam 2012R ANSWER KEY Page 5 of 15 (b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required starting materials can be obtained, briefly explain why or why not. The Diels-Alder reaction generally requires an electron-withdrawing group on the dienophile. The cyclohexene product will have the double bond between the carbons that were C2 and C3 of the diene. The diene and dienophile needed to make 1-(hydroxymethyl)cyclohexene in the simplest way do not fit these requirements, so the DA reaction is probably not appropriate.
CEM 2220 Final Exam 2012R ANSWER KEY Page 6 of 15 3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that this process occurs at an unusually low temperature for this type of reaction.
CEM 2220 Final Exam 2012R ANSWER KEY Page 7 of 15 4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes. This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17). (a) (5 MARKS) Johnson s team studied the following reaction as a model process. Suggest a stepwise mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)
CEM 2220 Final Exam 2012R ANSWER KEY Page 8 of 15 (b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of them. Write a mechanism for the formation of the two products shown.
CEM 2220 Final Exam 2012R ANSWER KEY Page 9 of 15 5. (20 MARKS TOTAL) Mechanism grab-bag! (a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CEM 2220. Write a mechanism for this reaction. (b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction. Write a stepwise mechanism to explain this transformation.
CEM 2220 Final Exam 2012R ANSWER KEY Page 10 of 15 (c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure. (d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).
CEM 2220 Final Exam 2012R ANSWER KEY Page 11 of 15 6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAl 4 ) is a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville s proposed synthetic procedure is shown below. Most data and procedures in this question are fictitious and should not be used for experimental purposes. Acid A Alcohol B LiAl 4 Diethyl ether MW (g/mol) 136.15 122.14 38.39 74.12 MP ( o C) 52 20 N/A -116 BP ( o C) 120 195 N/A 35 Solubilities (g/ml) Water 0.3 Ethanol - 6 Diethyl ether - 3 Water 0.05 Ethanol - 10 Diethyl ether - 20 Water N/A Ethanol N/A Diethyl ether 10 Densities 1.76 0.89 N/A 0.88 Neville s Proposed Procedure Water - 0.03 Ethanol miscible Dissolve Acid A (10 g) in diethyl ether (100 ml). Cool the solution on an ice bath. Slowly add LiAl 4 (8 g). After 30 minutes of stirring, slowly add ice cold water (50 ml) to the solution. Remove and discard the top (aqueous) layer. The organic layer is extracted three times with 50 ml aliquots of 1 M Cl. The solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from the distillation flask (round bottom flask). (a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with Neville s procedure as you can, and indicate how you would modify the procedure to make it work. Your goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield. State all assumptions. ere are some key points that could be part of your answer: 1) The amount of Acid A is incorrect. Since we want 122 g of Alcohol A (1 mole) and we know that the reaction produces product at 50% yield, then we require 2 moles of starting material, or 272 g of Acid A. 2) To dissolve Acid A, notice that the solubility in diethyl ether is 3 g/ml, we ll require a minimum of 90 ml to dissolve all of Acid A. 3) Reduction of an acid requires 3 hydride equivalents: 1 to neutralize the acidic proton plus 2 more to add to the carbonyl. A minimum of 28.8 g of LiAl 4 is required (0.75 moles) assuming the reagent delivers all four hydrides. In practice one would probably add at least 1 mole, and probably more just to be sure. 4) As mentioned by Dr.ultin in class, the LiAl 4 reduction reaction will have to be refluxed (note that TF is usually used as the solvent since it has a higher boiling point than diethyl ether) and the reaction should be monitored by TLC to determine its completion. 5) After water is added to quench the reaction, the ether will be the top layer and not the bottom. 6) A WASING should be done and not an EXTRACTION using aliquots of 1 M NaO to remove the residual acid. Note that the product is in the ether layer. 7) Dry the ether layer with a drying agent potassium carbonate would be a good choice to remove any final traces of acid. 8) The product can be isolated by distilling off the solvent and saving the liquid in the round bottom
CEM 2220 Final Exam 2012R ANSWER KEY Page 12 of 15 (b) (3 MARKS) After implementing your recommendations, Neville collected his product and then brought you the TLC plate shown in Figure 1. Unfortunately Neville had used up all the starting material so he could not use it as a reference spot. owever, he found a sample of pure Alcohol B and used that as the reference. elp Neville interpret this TLC plate by explaining whether the product has been made and comment on the purity of the product. There are two spots in the product lane. The one with the higher Rf corresponds to the product while the bigger spot with the lower Rf corresponds to the starting material. Therefore the TLC suggests that very little product was made in the reaction. The product, be it the acid or alcohol is considered impure since there are two spots in the TLC. Figure 1: TLC of Neville's first product. (c) (3 MARKS) Neville performed the reaction again and examined his product by TLC on silica gel. The plate is shown on the left in Figure 2. Using the blank TLC plate template on the right in Figure 2, draw a diagram of how the plate should look if the plate substrate (absorbent) was modified to the LESS POLAR cellulose but using the same eluent. The spots from left to right are starting material, co-spot, and product. Figure 2: TLC for Neville's second attempt. The samples will travel to higher Rf values since the plate is less polar. Alcohol A will still have a higher Rf than Acid A.
CEM 2220 Final Exam 2012R ANSWER KEY Page 13 of 15 7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C 7 16 O. The IR, 1 NMR and 13 C NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (1 MARK) What is the unsaturation number for this compound? Unsaturation number is ZERO. (b) (1 MARK) your answer? What functional group(s) is(are) present? What specific spectroscopic evidence supports There is an alcohol because the IR shows a broad band at around 3400 cm -1. (c) (1 MARK) Briefly explain the signals in the 1 NMR spectrum between 0.80 0.88 ppm. There are 9 protons in this region. One signal is a 3-proton singlet suggesting a C3 group with no neighbour protons. The other signal integrates for 6 protons and at first might appear to be a triplet. owever, the peaks of a triplet must be in a 1:2:1 ratio and this is not the case. It looks more like this is actually two different signals on top of one another, perhaps a 3-proton triplet with a 3-proton singlet on top. (d) (2 MARKS) Draw the structure of this compound in the box below. Structure for C 7 16 O The 1-proton quartet at about 3.5 ppm tells us that the C next to the O group is also next to a C3 and nothing else. We see that C3 as the doublet at 1.1 ppm. Our analysis for part c above also tells us that there must be two methyls with no neighbours, and one methyl with a C2 next to it.
CEM 2220 Final Exam 2012R ANSWER KEY Page 14 of 15 Spectra for Question 7 IR 13 C NMR NB: all signals are single lines. 1 NMR
Spectroscopy Crib Sheet for CEM 2220 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 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) O 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 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 13 C NMR Typical Chemical Shift Ranges IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (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