Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2
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1 Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2 Thursday, March 12, 2015; 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 (6 Marks) 2. Mechanism (8 Marks) 3. Mechanism (6 Marks) 4. Reactions and Products (18 Marks) 5. Reactions and Products (7 Marks) 6. Spectra and Structures (5 Marks) TOTAL (50 Marks) MARKS
2 CHEM 2220 Test #2 Page 2 of 10 March 12, (6 MARKS TOTAL) There are not very many reagents that can directly introduce an amino (NH2) group by electrophilic aromatic substitution. George Olah reported a method (J. Org. Chem. 1989, 54, ), involving a mixture of hydrazoic acid HN3 and triflic acid HO3SCF3. a. (2 Marks) Draw two resonance structures for HN3. Note that the azide anion (its conjugate base) is linear. b. (4 Marks) Based on your structures from part a, draw mechanisms showing how HN3 reacts with triflic acid to form a functional equivalent of H2N + and how this reacts with toluene to form p-aminotoluene. Hint: the byproduct is N2 (g).
3 CHEM 2220 Test #2 Page 3 of 10 March 12, (8 MARKS TOTAL) When we discussed reductive amination in class we focused on intermolecular processes, but it is also possible to perform intramolecular reductive aminations. Suggest a stepwise mechanism for the following reaction in which an aminosugar is converted into a pyrrolidine product by reductive amination.
4 CHEM 2220 Test #2 Page 4 of 10 March 12, (6 MARKS) Provide a stepwise mechanism for the following reaction.
5 CHEM 2220 Test #2 Page 5 of 10 March 12, (18 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Show product stereochemistry where appropriate, using the wedge-and-dash formalism. If the product formed is racemic, indicate this by writing racemic or ± below the structure. a. (2 Marks) b. (2 Marks) c. (2 Marks) d. (2 Marks)
6 CHEM 2220 Test #2 Page 6 of 10 March 12, 2015 e. (2 Marks) f. (2 Marks) g. (2 Marks) h. (4 Marks)
7 CHEM 2220 Test #2 Page 7 of 10 March 12, (7 MARKS) The following 4-step sequence was employed during a synthesis of an inhibitor of a key enzyme in the replication of HIV-1. Fill in the necessary reagents and intermediate products to complete the sequence. HINT: treat this as a synthesis planning problem. Each step accomplishes one specific task in the overall process. Figure out the intermediates first, and then supply the necessary reagents.
8 CHEM 2220 Test #2 Page 8 of 10 March 12, (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula C9H13N 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) What do the IR bands at 3430 and 3351 cm -1 tell us about the compound? c. (1 MARK) What conclusion about the structure of compound A can be drawn from the splitting patterns combined with the integration of the 1 H NMR signals at ppm and ppm? d. (1 MARK) What conclusion about the structure of compound A can be drawn from the signals between 6.5 and 7.1 ppm in the 1 H NMR? e. (1 MARK) Draw the structure of compound A
9 CHEM 2220 Test #2 Page 9 of 10 March 12, 2015 IR 13 C NMR 1 H NMR 2 H 2 H 2 H 1 H 6 H
10 CHEM 2220 Test #2 Page 10 of 10 March 12, 2015 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 13 C NMR Typical Chemical Shift Ranges High Field Ester IR Typical Functional Group Absorption Bands CH x -Y Alkene Y = O, N Aryl Ketone, Aldehyde Amide Acid RC N CR 3 -CH 2 -CR 3 CH x -C=O RC CR CH 3 -CR 3 Group Frequency (cm -1 ) Intensity Group Frequency (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 R2N 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
11 ANSWER KEY University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2 Thursday, March 12, 2015; 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 (6 Marks) 2. Mechanism (8 Marks) 3. Mechanism (6 Marks) 4. Reactions and Products (18 Marks) 5. Reactions and Products (7 Marks) 6. Spectra and Structures (5 Marks) TOTAL (50 Marks) MARKS
12 CHEM 2220 Test #2 Answers Page 2 of 10 March 12, (6 MARKS TOTAL) There are not very many reagents that can directly introduce an amino (NH2) group by electrophilic aromatic substitution. George Olah reported a method (J. Org. Chem. 1989, 54, ), involving a mixture of hydrazoic acid HN3 and triflic acid HO3SCF3. a. (2 Marks) Draw two resonance structures for HN3. Note that the azide anion (its conjugate base) is linear. b. (4 Marks) Based on your structures from part a, draw mechanisms showing how HN3 reacts with triflic acid to form a functional equivalent of H2N + and how this reacts with toluene to form p-aminotoluene. Hint: the byproduct is N2 (g).
13 CHEM 2220 Test #2 Answers Page 3 of 10 March 12, (8 MARKS TOTAL) When we discussed reductive amination in class we focused on intermolecular processes, but it is also possible to perform intramolecular reductive aminations. Suggest a stepwise mechanism for the following reaction in which an aminosugar is converted into a pyrrolidine product by reductive amination.
14 CHEM 2220 Test #2 Answers Page 4 of 10 March 12, (6 MARKS) Provide a stepwise mechanism for the following reaction.
15 CHEM 2220 Test #2 Answers Page 5 of 10 March 12, (18 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Show product stereochemistry where appropriate, using the wedge-and-dash formalism. If the product formed is racemic, indicate this by writing racemic or ± below the structure. a. (2 Marks) b. (2 Marks) c. (2 Marks) d. (2 Marks)
16 CHEM 2220 Test #2 Answers Page 6 of 10 March 12, 2015 e. (2 Marks) f. (2 Marks) g. (2 Marks) h. (4 Marks)
17 CHEM 2220 Test #2 Answers Page 7 of 10 March 12, (7 MARKS) The following 4-step sequence was employed during a synthesis of an inhibitor of a key enzyme in the replication of HIV-1. Fill in the necessary reagents and intermediate products to complete the sequence. HINT: treat this as a synthesis planning problem. Each step accomplishes one specific task in the overall process. Figure out the intermediates first, and then supply the necessary reagents.
18 CHEM 2220 Test #2 Answers Page 8 of 10 March 12, (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula C9H13N 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 = 4 b. (1 MARK) What do the IR bands at 3430 and 3351 cm -1 tell us about the compound? These bands are N-H stretches; because there are two of them you might also conclude this is probably due to an NH2 group (in-phase and out-ofphase stretching). c. (1 MARK) What conclusion about the structure of compound A can be drawn from the splitting patterns combined with the integration of the 1 H NMR signals at ppm and ppm? Signal at integrates for 6 H and is a doublet. Signal at integrates for 1 H and is a septet. Thus, it has six identical neighbours. The structure is therefore an isopropyl group CH(CH3)2. d. (1 MARK) What conclusion about the structure of compound A can be drawn from the signals between 6.5 and 7.1 ppm in the 1 H NMR? There are four protons total in the aromatic region of the spectrum. They are grouped in two pairs, with a dominant doublet-like splitting pattern. Thus, the aromatic ring is para disubstituted. e. (1 MARK) Draw the structure of compound A
19 CHEM 2220 Test #2 Answers Page 9 of 10 March 12, 2015 IR 13 C NMR 1 H NMR 2 H 2 H 2 H 1 H 6 H
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