Name: Student Number: University of Manitoba - Department of Chemistry CEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 13, 2014; 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 (4 Marks) 3. Mechanism (4 Marks) 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 9 Feb 13, 2014 1. (8 MARKS) An example of the Ritter Reaction, a general process for the formation of amides by the acid-catalyzed reaction of nitriles with alkenes, is shown below. Write a stepwise mechanism for this reaction.
CEM 2220 Test #1 Page 3 of 9 Feb 13, 2014 2. (4 MARKS) 1-Bromobicyclo[2.2.2]octane (A) does not undergo an E2 elimination when treated with a strong base. Briefly explain why not. 3. (4 MARKS) Write a stepwise mechanism to explain the following rearrangement.
CEM 2220 Test #1 Page 4 of 9 Feb 13, 2014 4. (8 MARKS) When compound A was treated with Br in C 2 Cl 2 and with complete exclusion of light, bromocyclohexane product B was formed. owever, when the reaction was conducted in the presence of peroxides, product C was obtained instead. Write out stepwise mechanisms for these two products and briefly summarize why the different results are obtained.
CEM 2220 Test #1 Page 5 of 9 Feb 13, 2014 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 relative product stereochemistry (wedge and dash bonds) where appropriate if a racemic product is formed, simply indicate +/ or racemic. a. (2 Marks) b. (2 Marks) c. (2 Marks) d. (2 Marks)
CEM 2220 Test #1 Page 6 of 9 Feb 13, 2014 e. (4 Marks) f. (2 Marks) g. (2 Marks) h. (4 Marks)
CEM 2220 Test #1 Page 7 of 9 Feb 13, 2014 6. (6 MARKS Total) The IR and NMR spectra of an unknown organic compound A having the formula C 6 13 NO are shown on the next page. Based on these data, answer the following questions about compound A. a. (1 Mark) What is the degree of unsaturation in compound A? b. (1 Mark) What functional group is present in A? c. (1 Mark) What is the significance of the 3-proton singlet at δ1.98 ppm in the 1 NMR spectrum? d. (3 Marks) What is the structure of compound A?
CEM 2220 Test #1 Page 8 of 9 Feb 13, 2014 IR C 6 13 NO 13 C NMR C 6 13 NO 1 NMR C 6 13 NO s, 1 tr, 2 s, 3 m, m, 2 2 tr, 3
Page 9 of 9 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 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 13, 2014; 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 (4 Marks) 3. Mechanism (4 Marks) 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 9 Feb 13, 2014 1. (8 MARKS) An example of the Ritter Reaction, a general process for the formation of amides by the acid-catalyzed reaction of nitriles with alkenes, is shown below. Write a stepwise mechanism for this reaction. You did not have to show all the resonance structures in your answer. A correct mechanism can be written using only one of the possible canonical forms.
CEM 2220 Test #1 ANSWERS Page 3 of 9 Feb 13, 2014 2. (4 MARKS) 1-Bromobicyclo[2.2.2]octane (A) does not undergo an E2 elimination when treated with a strong base. Briefly explain why not. E2 elimination requires that the C- and C-X bonds adopt an antiperiplanar arrangement, which permits orbital overlap in the transition state. The ring bonds in this molecule cannot rotate, and the C-Br bond is nearly orthogonal to the adjacent C- bonds. Thus, the necessary geometry is impossible, and E2 elimination cannot occur. This is Klein problem 8.73. It can also be noted that the product would be impossibly strained, since if such an alkene could form the two ends would be rotated almost 90 from one another! Note the image on the right, which shows a view similar to a Newman projection. The C-Br is in the background, and the C2 is in the foreground. Clearly the C- and C-Br bonds are not antiperiplanar. 3. (4 MARKS) Write a stepwise mechanism to explain the following rearrangement. This is Groutas #21.
CEM 2220 Test #1 ANSWERS Page 4 of 9 Feb 13, 2014 4. (8 MARKS) When compound A was treated with Br in C 2 Cl 2 and with complete exclusion of light, bromocyclohexane product B was formed. owever, when the reaction was conducted in the presence of peroxides, product C was obtained instead. Write out stepwise mechanisms for these two products and briefly summarize why the different results are obtained. The cationic cyclization is a straightforward extension of electrophilic addition to alkenes. Similar reactions are found in Groutas, for example #29. The conditions clearly indicate a radical mechanism. By analogy with the cationic mechanism, the initial radical can attack the remaining alkene. This mechanism is a chain in which the propagation requires a few steps. It is written here simply as a linear sequence but obviously the final step produces bromine radical to re-enter the first step.
CEM 2220 Test #1 ANSWERS Page 5 of 9 Feb 13, 2014 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 relative product stereochemistry (wedge and dash bonds) where appropriate if a racemic product is formed, simply indicate +/ or racemic. Any hydroboration reagent could be used here. a. (2 Marks) b. (2 Marks) This reaction cannot form the Zaitsev product, since the tertiary C- cannot align antiperiplanar to the C-OTs. See Klein pp. 359-360, and SkillBuilder 8.7. c. (2 Marks) d. (2 Marks)
CEM 2220 Test #1 ANSWERS Page 6 of 9 Feb 13, 2014 e. (4 Marks) Klein suggests NaO as the base. You could also have used the non-nucleophilic Na in TF solution, or several other relatively strong bases like NaOMe or NaOEt. This is very similar to problem 3 from Sapling omework #4. f. (2 Marks) Recall Klein Figure 8.25 on page 375. Sulfur compounds RS are nucleophiles but not bases, so an S N 2 reaction is the only possibility here (Klein section 8.13). g. (2 Marks) Note that there is no water present during the mercuration, so the only nucleophile available is the alcohol group. h. (4 Marks) Alkylation of terminal alkynes was first presented in CEM 2210, see Klein section 10.3. This reaction has featured in a few Sapling problems this year. You could also have suggested forming the ketone by simple acid hydration but this would require heating as well as aqueous acid. Remember that this process is slow due to a termolecular rate-limiting step.
CEM 2220 Test #1 ANSWERS Page 7 of 9 Feb 13, 2014 6. (6 MARKS Total) The IR and NMR spectra of an unknown organic compound A having the formula C 6 13 NO are shown on the next page. Based on these data, answer the following questions about compound A. a. (1 Mark) What is the degree of unsaturation in compound A? Unsaturation = 1 2 6 + 2 = 14; 14 13 = 1; 1 + 1 (for N) = 2; 2/2 = 1. b. (1 Mark) What functional group is present in A? Strong IR bands at ~1675 and ~3280 cm -1, plus 13 C NMR at ~176 ppm, plus N in the formula = AMIDE If you just put carbonyl, ½ Mark. c. (1 Mark) What is the significance of the 3-proton singlet at δ1.98 ppm in the 1 NMR spectrum? A 3-proton singlet strongly suggests a methyl with no nearest neighbors; the chemical shift of 1.98 ppm says it is next to the carbonyl of the amide (i.e. we have C 3 C(O)-N). d. (3 Marks) What is the structure of compound A? You were not required to assign the signals to specific parts of your structure. For future reference: 13 C NMR: 170.54 2 39.47 3 31.73 4 23.06 1 20.19 5 13.73 6 1 NMR: 7.05 (1, broad s, N) 3.21 (2, tr, C 2-3) 1.98 (3, s, C 3-1) 1.49 (2, m, C 2-4) 1.35 (2, m, C 2-5) 0.92 (3, tr, C 3-6) IR: (notice that detailed analysis can get complicated!) 3289 (amide N- stretch) 3088 (related to an overtone of 1558) 2960, 2934, 2874 (C- stretches) 1654 (amide I mostly C=O stretch) 1558 (amide II mostly N- bend)
CEM 2220 Test #1 ANSWERS Page 8 of 9 Feb 13, 2014 IR C 6 13 NO 13 C NMR C 6 13 NO 1 NMR C 6 13 NO s, 1 tr, 2 s, 3 m, m, 2 2 tr, 3
Page 9 of 9 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 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