Name: Student No: Page 1 of 15

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1 Name: Student No: Page 1 of 15 CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong FINAL EXAM Winter Session 2014R Tuesday April 22, :00 pm 9:00 pm Frank Kennedy Gold Gym Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN 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 (12 Marks) Question 4 Mechanism grab-bag (32 Marks) Question 5 Laboratory (10 Marks) Question 6 Spectroscopy (6 Marks) TOTAL: (100 Marks)

2 CHEM 2220 Final Exam 2014R Page 2 of (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. Simple aqueous acid or base workups can be assumed but any special workup conditions should be specified. (a) (b) (c) (3 Marks) (d) (3 Marks) (e) (f) (g)

3 CHEM 2220 Final Exam 2014R Page 3 of 15 (h) (i) (j) (k) (4 Marks) (l) (4 Marks)

4 CHEM 2220 Final Exam 2014R Page 4 of (10 MARKS) Propose a synthetic route (sequence of reactions) to prepare 3-phenylheptanoic acid starting with diethyl malonate and benzaldehyde, plus any other reagents, solvents, or catalysts necessary. This can be accomplished in 3 or 4 steps using reactions covered in class. Be sure to write out each reaction in your route showing the starting compound, product and specific reaction conditions.

5 CHEM 2220 Final Exam 2014R Page 5 of (12 MARKS TOTAL) In appropriate molecules, a series of Michael reactions can occur in sequence during a single reaction to form quite complicated carbon skeletons. One such cascade reaction was reported by Fukumoto and his co-workers in (a) Draw the structure of the enolate formed when the starting material was treated with LDA. Show both resonance structures and include the positive counter ion. (b) (3 Marks) In the structures below, certain carbons in the starting material have been labeled. i) Indicate which carbons in the product correspond to these designated atoms, by applying the same labels to the structure provided. ii) Label the carbon in the product that corresponds to the nucleophilic site of the enolate you drew in part (a). iii) Label the bonds in the product that were formed during the reaction.

6 CHEM 2220 Final Exam 2014R Page 6 of 15 (c) (7 Marks) reaction. Based on your analysis in parts (a) and (b), write a detailed stepwise mechanism for this

7 CHEM 2220 Final Exam 2014R Page 7 of (32 MARKS TOTAL) Mechanism grab-bag! (a) (4 Marks) Reaction of α-chloroketones with sodium borohydride produces epoxides. Write out a mechanism for the following reaction. You do not have to show the fate of the boron during workup. (b) (4 Marks) Draw a plausible mechanism for the following transformation.

8 CHEM 2220 Final Exam 2014R Page 8 of 15 (c) (4 Marks) Write a mechanism for the following variation on a reaction we discussed in class. (d) (4 Marks) Alkaline hydrogen peroxide can promote a reaction similar to the Baeyer-Villiger oxidation. Write a mechanism for the following reaction that accounts for the stereochemistry of the product.

9 CHEM 2220 Final Exam 2014R Page 9 of 15 (e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide) hydrolyzes to form HCl extremely rapidly in biological fluids, in contrast to ordinary alkyl chlorides that are essentially inert under these conditions. Thus, it incapacitated soldiers who breathed it or who got it into their eyes by causing acid burns. i) (4 Marks) Write a mechanism and briefly explain why mustard gas undergoes hydrolysis so much faster than does an analogous primary chloride.

10 CHEM 2220 Final Exam 2014R Page 10 of 15 ii) (6 Marks) Those soldiers who survived exposure to mustard gas were often horribly maimed, but they also tended to have a high incidence of cancer later in life. It was discovered that this was because of alkylation and cross-linking of DNA by the mustard agent. Draw a plausible structure for the mustard gas cross-link between the DNA bases adenosine and thymidine, and based on your mechanism for hydrolysis write a mechanism for the formation of this cross-link. (f) (6 Marks) Like alkenes, simple cyclopropanes do not react with nucleophilic reagents. However, certain types of substituents on the cyclopropyl ring permit nucleophilic ring opening to occur. Based on the examples shown, draw a generalized mechanism and briefly explain this observation. (N.B. phenyl selenide is a reasonably strong nucleophile.)

11 CHEM 2220 Final Exam 2014R Page 11 of Lab Questions (10 MARKS Total) After a long day of filling potholes in Winnipeg, Filmia Crater (a fictional city worker who lives next door to you) came home with a terrible headache. As the kind neighbour that you are, you check your medicine cabinet and there s no acetaminophen (para-acetamidophenol, Compound C) to be found. However, in the organic laboratory you have access to all the necessary materials to synthesize it. Assume you want to prepare about 15 g of Compound C in the most pure form possible (it s a bad headache!). To start off, you have a protected aminophenol (Compound A) to ensure that there s reactivity only at the amine. To remove the protecting group will only require heating with acid for a period of time. In addition to Compound A, you are to use only the reagents and chemicals typically available in the CHEM 2220 laboratories. Data Table: Most of the data are fictitious, and should NOT be used for real experimental purposes! MW (g/mol) BP/MP ( o C) Solubility (g/l) Compound A /130 Compound B /20 Compound C /168 Acetic Anhydride / 73 In planning your synthesis you will need to make the following assumptions: H 2 O: 1 Ethyl acetate: 50 Ethanol: 80 H 2 O: 1 Ethyl acetate: 50 Ethanol: 80 H 2 O: 0.1 Ethyl acetate: 20 Ethanol: 10 H 2 O: 100 Ethyl acetate: 100 Ethanol: After the first step, there will be residual starting material (limiting reagent AND those in excess). 2. You need to isolate Compound B free of contaminants before proceeding to remove the protecting group. 3. After the second step, there will be residual Compound B in the crude product mixture. 4. Since the reactions are incomplete, assume the isolated yield after each step will be 50% There is some math involved but we have chosen numbers that can easily be approximated.

12 CHEM 2220 Final Exam 2014R Page 12 of 15 a) (7 Marks) Write out the procedure you will use to make acetaminophen. b) Product Analysis You spotted a sample of the reaction mixture in Step 1 on a TLC plate towards the end of the reaction period, using Compound A as a reference. Draw what this TLC plate would look like after you eluted it with ethyl acetate: hexanes (1:1). Make sure to label aspects of your plate and spots. c) (1 Mark) In many of the laboratory experiments, you used brine to wash an organic layer. Why is washing with brine necessary would washing with plain distilled water have the same effect?

13 CHEM 2220 Final Exam 2014R Page 13 of (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C 7 H 14 O 2. The IR, 13 C NMR and 1 H NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (0.5 MARK) What is the unsaturation number for this compound? (b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for your conclusion. (c) (1 MARK) What can you conclude from the number of 13 C NMR signals? (d) (1 MARK) What can you conclude from the integral and splitting of the 1 H NMR signal at 2.5 ppm? (e) (2 MARKS) Draw the structure of this compound in the box below. Structure for C 7 H 14 O 2

14 CHEM 2220 Final Exam 2014R Page 14 of 15 Spectra for Question 6 IR 13 C NMR NB: all signals are single lines. 1 H NMR 2H tr 1H heptet 2H m 6H d 3H tr

15 Spectroscopy Crib Sheet for CHEM 2220 Introductory Organic Chemistry II 1 H NMR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C CH C C H C CH 2 C C C C H C C H O H O OH C OH (solvent dependent) (solvent dependent) O O C H H Aryl C H Cl C H H Br C H Aryl H I C H RCO 2 H Aromatic, heteroaromatic X C H X = O, N, S, halide R 3 C H Aliphatic, alicyclic Y = O, NR, S Y H H Y H Y = O, NR, S Low Field High Field Alkene Aryl Ketone, Aldehyde Ester Amide Acid RC N CH x -Y Y = O, N CR 3 -CH 2 -CR 3 CH x -C=O RC CR CH 3 -CR C NMR Typical Chemical Shift Ranges IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity C H Medium RO H Strong, broad C=C H Medium C O Strong C=C Medium C=O Strong C C H Strong R 2 N H Medium, broad R C C R Medium (R R ) C N 1230, 1030 Medium Aryl H Medium C N Medium Aryl C=C 1600, 1500 Strong RNO Strong

16 ANSWER KEY Page 1 of 15 CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong FINAL EXAM Winter Session 2014R Tuesday April 22, :00 pm 9:00 pm Frank Kennedy Gold Gym Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN 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 (12 Marks) Question 4 Mechanism grab-bag (32 Marks) Question 5 Laboratory (10 Marks) Question 6 Spectroscopy (6 Marks) TOTAL: (100 Marks)

17 CHEM 2220 Final Exam 2014R ANSWERS Page 2 of (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. Simple aqueous acid or base workups can be assumed but any special workup conditions should be specified. See Klein section 9.11 and Skill Builder 9.8. (a) See Klein pp (b) See Klein section 23.11, and SkillBuilder 23.6 (c) (3 Marks) (d) The Strecker Synthesis was discussed in class, see also Klein pp (3 Marks) Baeyer-Villiger oxidation, Klein section (e) (f) Alkylation of a kinetic enolate: Klein section Similar to Klein problem d.

18 CHEM 2220 Final Exam 2014R ANSWERS Page 3 of 15 Crossed aldol addition under irreversible conditions. Discussed in class, also see Klein p (g) Wittig reaction. Klein pp (h) Diels-Alder reaction. This is Klein problem a. (i) Acid-catalyzed hydrolysis of an acetal and an ester forms a β-ketoacid, which decarboxylates. See Klein sections 20.5 and 21.11, and also p (j) 1) Hydroboration is the only method we know for anti-markovnikov hydration of alkenes. 2) Chromic acid oxidation of primary alcohols forms carboxylic acids. 3) Formation of acid chloride using thionyl chloride. 4) Amide formation: excess amine plus acid chloride. (k) (4 Marks) (l) (4 Marks) Spectroscopic data indicate 3 kinds of aromatic hydrogens and 2 alkene hydrogens. Also note 2 CH3 groups, and 2H doublet at 2.21 for CH2 next to alkene.

19 CHEM 2220 Final Exam 2014R ANSWERS Page 4 of (10 MARKS) Propose a synthetic route (sequence of reactions) to prepare 3-phenylheptanoic acid starting with diethyl malonate and benzaldehyde, plus any other reagents, solvents, or catalysts necessary. This can be accomplished in 3 or 4 steps using reactions covered in class. Be sure to write out each reaction in your route showing the starting compound, product and specific reaction conditions. There are numerous ways in which the required synthesis could be accomplished. Any answer was judged on its own merits.

20 CHEM 2220 Final Exam 2014R ANSWERS Page 5 of (12 MARKS TOTAL) In appropriate molecules, a series of Michael reactions can occur in sequence during a single reaction to form quite complicated carbon skeletons. One such cascade reaction was reported by Fukumoto and his co-workers in (a) Draw the structure of the enolate formed when the starting material was treated with LDA. Show both resonance structures and include the positive counter ion. (b) (3 Marks) In the structures below, certain carbons in the starting material have been labeled. i) Indicate which carbons in the product correspond to these designated atoms, by applying the same labels to the structure provided. ii) Label the carbon in the product that corresponds to the nucleophilic site of the enolate you drew in part (a). iii) Label the bonds in the product that were formed during the reaction. These initial questions guide you towards the correct mechanism. There is only one possible enolate that can be formed so this should have been straightforward. Identifying the labeled atoms in the product is important because this locates where structural changes must occur. Look at the functional groups and where the labels are relative to them in the starting material. Then, spot the functional groups in the product and they will tell you where the labels probably are.

21 CHEM 2220 Final Exam 2014R ANSWERS Page 6 of 15 (c) (7 Marks) reaction. Based on your analysis in parts (a) and (b), write a detailed stepwise mechanism for this As you can see, the mechanism is actually extremely simple once you know WHAT structural changes have to be accomplished. From part b-iii of the question we know we must make C-C bonds from the enolate α-position to A, and from the ester α-position to C. Since we have a nucleophile in the enolate ready to go, making this bond first is reasonable. We notice that the result of doing this is to make the enolate of our ester. This permits us to make the second bond to carbon C, leading to yet another enolate. There is no other structural change needed, so we can assume that this enolate remains until the acid workup when it is protonated.

22 CHEM 2220 Final Exam 2014R ANSWERS Page 7 of (32 MARKS TOTAL) Mechanism grab-bag! (a) (4 Marks) Reaction of α-chloroketones with sodium borohydride produces epoxides. Write out a mechanism for the following reaction. You do not have to show the fate of the boron during workup. This is Klein problem 14.57, and it is similar to Groutas #51. Basically, you know that borohydride reduction of ketones forms alkoxide anion intermediates. You also know that epoxides can be formed by treating vicinal halohydrins with base, and that this proceeds via the alkoxide. So, this reaction is simply the combination of the two pathways. (b) (4 Marks) Draw a plausible mechanism for the following transformation. This is Klein problem The important features of a correct mechanism are seen in Klein Mechanism 20.5 (page 925): 1) protonation on O followed by opening to give O-stabilized tertiary cation; 2) hemiacetal intermediate; 3) protonation on O and collapse to release diol plus acetone. Then, the rest of the mechanism is standard acid-catalyzed esterification (see Klein SkillBuilder 21.1 page 989).

23 CHEM 2220 Final Exam 2014R ANSWERS Page 8 of 15 (c) (4 Marks) Write a mechanism for the following variation on a reaction we discussed in class. This is Klein problem It is simply a Diels-Alder reaction between the diene and the alkyne, followed by the reverse of a Diels-Alder reaction to release CO 2. (d) (4 Marks) Alkaline hydrogen peroxide can promote a reaction similar to the Baeyer-Villiger oxidation. Write a mechanism for the following reaction that accounts for the stereochemistry of the product. The mechanism of a Baeyer-Villiger reaction involving a peroxyacid like mcpba is shown in Klein page 954, and we covered it in class. We discussed the behaviour of hydrogen peroxide in the presence of NaOH when we talked about hydroboration reactions (see Klein Mechanism 9.3 page 415), and we saw that oxyanions can attack ketones (Mechanism 20.3 page 924). The mechanism here is directly derived from consideration of the Baeyer-Villiger mechanism and the oxidative workup in hydroboration/oxidation.

24 CHEM 2220 Final Exam 2014R ANSWERS Page 9 of 15 (e) During World War I, mustard gas was developed as a weapon. Mustard gas (bis(2-chloroethyl)sulfide) hydrolyzes to form HCl extremely rapidly in biological fluids, in contrast to ordinary alkyl chlorides that are essentially inert under these conditions. Thus, it incapacitated soldiers who breathed it or who got it into their eyes by causing acid burns. i) (4 Marks) Write a mechanism and briefly explain why mustard gas undergoes hydrolysis so much faster than does an analogous primary chloride. We saw something similar in the mechanism of forming an epoxide from a halohydrin (Klein page 648), and we know that sulfides RSR are good nucleophiles (Klein Fig and page 655). Part marks were given for suggesting that the electron-withdrawing nature of sulfur increased the reactivity of the primary halide towards SN2 displacement.

25 CHEM 2220 Final Exam 2014R ANSWERS Page 10 of 15 ii) (6 Marks) Those soldiers who survived exposure to mustard gas were often horribly maimed, but they also tended to have a high incidence of cancer later in life. It was discovered that this was because of alkylation and cross-linking of DNA by the mustard agent. Draw a plausible structure for the mustard gas cross-link between the DNA bases adenosine and thymidine, and based on your mechanism for hydrolysis write a mechanism for the formation of this cross-link. Since mustard reacts with water by an SN2 mechanism, it makes sense that it would react in a similar way with other nucleophiles. Note that reaction with ring nitrogens in adenosine would either make cationic products or disrupt the aromaticity of the ring so these are much less likely. (f) (6 Marks) Like alkenes, simple cyclopropanes do not react with nucleophilic reagents. However, certain types of substituents on the cyclopropyl ring permit nucleophilic ring opening to occur. Based on the examples shown, draw a generalized mechanism and briefly explain this observation. (N.B. phenyl selenide is a reasonably strong nucleophile.) This is a slightly modified version of Groutas #172.

26 CHEM 2220 Final Exam 2014R ANSWERS Page 11 of Lab Questions (10 MARKS Total) After a long day of filling potholes in Winnipeg, Filmia Crater (a fictional city worker who lives next door to you) came home with a terrible headache. As the kind neighbour that you are, you check your medicine cabinet and there s no acetaminophen (para-acetamidophenol, Compound C) to be found. However, in the organic laboratory you have access to all the necessary materials to synthesize it. Assume you want to prepare about 15 g of Compound C in the most pure form possible (it s a bad headache!). To start off, you have a protected aminophenol (Compound A) to ensure that there s reactivity only at the amine. To remove the protecting group will only require heating with acid for a period of time. In addition to Compound A, you are to use only the reagents and chemicals typically available in the CHEM 2220 laboratories. Data Table: Most of the data are fictitious, and should NOT be used for real experimental purposes! MW (g/mol) BP/MP ( o C) Solubility (g/l) Compound A /130 Compound B /20 Compound C /168 Acetic Anhydride / 73 In planning your synthesis you will need to make the following assumptions: H 2 O: 1 Ethyl acetate: 50 Ethanol: 80 H 2 O: 1 Ethyl acetate: 50 Ethanol: 80 H 2 O: 0.1 Ethyl acetate: 20 Ethanol: 10 H 2 O: 100 Ethyl acetate: 100 Ethanol: After the first step, there will be residual starting material (limiting reagent AND those in excess). 2. You need to isolate Compound B free of contaminants before proceeding to remove the protecting group. 3. After the second step, there will be residual Compound B in the crude product mixture. 4. Since the reactions are incomplete, assume the isolated yield after each step will be 50% There is some math involved but we have chosen numbers that can easily be approximated.

27 CHEM 2220 Final Exam 2014R ANSWERS Page 12 of 15 a) (7 Marks) Write out the procedure you will use to make acetaminophen. Dissolve about 66 g of Compound A in acetic anhydride (50 ml minimum) to acetylate the amine group. This reaction should not require heating or a catalyst. The reaction can be monitored by TLC. At the end of the reaction you have a mixture of Compound A and B and residual acetic anhydride and acetic acid. Add water to quench the anhydride. Compound B is a liquid and insoluble in water and therefore you can also wash with a basic solution to remove the acetic acid. Compound A is removed only by an acidic wash. Drying Compound B would be chemically wasteful since the next stage we re heating with acid (aqueous) anyway. As Compound C is produced, we expect that a solid precipitates out (since it is insoluble in water, and the phenol group can t be deprotonated under acidic conditions to make it more water soluble). Since there is Compound B present (as a liquid) at the end of the reaction, just perform a suction filtration to isolate Compound C and then wash the crystals with cold ethanol to remove Compound B contaminant. Compound C can be further purified by recrystallization with ethanol. b) Product Analysis You spotted a sample of the reaction mixture in Step 1 on a TLC plate towards the end of the reaction period, using Compound A as a reference. Draw what this TLC plate would look like after you eluted it with ethyl acetate: hexanes (1:1). Make sure to label aspects of your plate and spots. c) (1 Mark) In many of the laboratory experiments, you used brine to wash an organic layer. Why is washing with brine necessary would washing with plain distilled water have the same effect? Brine is used after washing with another aqueous solution. Its purpose is to crudely remove water before drying the organic layer over drying agent. Replacing the brine with only water would not have the same effect.

28 CHEM 2220 Final Exam 2014R ANSWERS Page 13 of (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C 7 H 14 O 2. The IR, 13 C NMR and 1 H NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (0.5 MARK) What is the unsaturation number for this compound? Unsaturation = 1 (b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for your conclusion. Contains an ester. IR ~1740 cm -1, no OH band; 13 C NMR ~178 ppm; 1 H NMR 2H triplet ~ 4 ppm suggests OCH 2 (c) (1 MARK) What can you conclude from the number of 13 C NMR signals? 6 13 C NMR signals but 7 carbons in the formula: one signal represents 2 symmetrical carbons. (d) (1 MARK) What can you conclude from the integral and splitting of the 1 H NMR signal at 2.5 ppm? A 1-proton heptet (7 lines) means this proton has 6 identical nearest neighbour protons. Since we also see a 6-proton doublet at ~1.2 ppm this suggests an isopropyl group CH(CH 3 ) 2. (e) (2 MARKS) Draw the structure of this compound in the box below. Structure for C 7 H 14 O 2 1. Chemical shift of the 1-H heptet ~2.5 ppm suggests it is next to the carbonyl of the ester H triplet at ~4 ppm means OCH 2 CH H triplet at ~0.9 ppm means methyl with 2 neighbours only possibility is OCH 2 CH 2 CH 3 Note multiplet at ~1.6 ppm is the middle CH 2, split by 2H of one kind and 3H of another. 4. Only one way to put these pieces together consistently.

29 CHEM 2220 Final Exam 2014R ANSWERS Page 14 of 15 Spectra for Question 6 IR 13 C NMR NB: all signals are single lines. 1 H NMR 2H tr 1H heptet 2H m 6H d 3H tr

30 Spectroscopy Crib Sheet for CHEM 2220 Introductory Organic Chemistry II 1 H NMR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C CH C C H C CH 2 C C C C H C C H O H O OH C OH (solvent dependent) (solvent dependent) O O C H H Aryl C H Cl C H H Br C H Aryl H I C H RCO 2 H Aromatic, heteroaromatic X C H X = O, N, S, halide R 3 C H Aliphatic, alicyclic Y = O, NR, S Y H H Y H Y = O, NR, S Low Field High Field Alkene Aryl Ketone, Aldehyde Ester Amide Acid RC N CH x -Y Y = O, N CR 3 -CH 2 -CR 3 CH x -C=O RC CR CH 3 -CR C NMR Typical Chemical Shift Ranges IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity C H Medium RO H Strong, broad C=C H Medium C O Strong C=C Medium C=O Strong C C H Strong R 2 N H Medium, broad R C C R Medium (R R ) C N 1230, 1030 Medium Aryl H Medium C N Medium Aryl C=C 1600, 1500 Strong RNO Strong