Important Note: We will NOT accept papers written in pencil back for re-marking after they have been returned to you. Please do not ask!

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1 Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2 Thursday, March 15, 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 (5 Marks) 2. Mechanism (5 Marks) 3. Mechanism (8 Marks) 4. Reactions and Products (22 Marks) 5. Reactions and Products (5 Marks) 6. Spectra and Structures (5 Marks) TOTAL (50 Marks) MARKS Important Note: We will NOT accept papers written in pencil back for re-marking after they have been returned to you. Please do not ask!

2 CHEM 2220 Test #2 Page 2 of 9 March 15, (5 MARKS) Draw a reasonable stepwise mechanism for the following reaction. 2. (5 MARKS) When aniline (aminobenzene) is treated with fuming sulfuric acid, electrophilic aromatic substitution occurs at the meta position even though we expect an amino group to be an ortho, para director. Briefly explain why this is so.

3 CHEM 2220 Test #2 Page 3 of 9 March 15, (8 MARKS) The Benzoin Addition Reaction was discovered in the 1830s, but its mechanism was only determined in In its classic form, two equivalents of benzaldehyde are heated with aqueous potassium cyanide to form benzoin (2-hydroxy-2-phenylacetophenone). Suggest a stepwise mechanism for this reaction. HINT: the CN group in a nitrile behaves as a good electronwithdrawing group.

4 CHEM 2220 Test #2 Page 4 of 9 March 15, (22 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Give product stereochemistry where appropriate. a. (2 Marks) b. (2 Marks) c. (2 Marks) d. (6 Marks)

5 CHEM 2220 Test #2 Page 5 of 9 March 15, 2012 e. (2 Marks) f. (4 Marks) g. (4 Marks)

6 CHEM 2220 Test #2 Page 6 of 9 March 15, (5 MARKS) We have seen that acetylene is a very useful starting material for organic synthesis. Using acetylene as the starting compound, propose a sequence of 3 reactions to produce (E)-oct-3-ene-1-ol. For each reaction, give the necessary reagent(s) and conditions, as well as the product formed. You may use any additional organic compound having four or fewer carbon atoms, as well as any reagents or solvents you may require.

7 CHEM 2220 Test #2 Page 7 of 9 March 15, (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula C9H7NO2 are shown on the next page. Answer the following questions about this compound. a. (1 MARK) What is the unsaturation number for compound A? b. (1 MARK) What functional group(s) is(are) present? Indicate the specific signal(s) in the spectra that support your answer. c. (1 MARK) What does the 13 C NMR spectrum tell you about compound A? Be specific. d. (2 MARKS) What is the structure of compound A?

8 CHEM 2220 Test #2 Page 8 of 9 March 15, 2012 IR 13 C NMR 1 H NMR

9 CHEM 2220 Test #2 Page 9 of 9 March 15, 2012 Spectroscopy Crib Sheet for 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 O C C C H C H O O H OH C OH (solvent dependent) (solvent dependent) 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 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 IR Typical Functional Group Absorption Bands 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 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

10 ANSWER KEY University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 2 Thursday, March 15, 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 (5 Marks) 2. Mechanism (5 Marks) 3. Mechanism (8 Marks) 4. Reactions and Products (22 Marks) 5. Reactions and Products (5 Marks) 6. Spectra and Structures (5 Marks) TOTAL (50 Marks) MARKS Important Note: We will NOT accept papers written in pencil back for re-marking after they have been returned to you. Please do not ask!

11 CHEM 2220 Test #2 ANSWERS Page 2 of 9 March 15, (5 MARKS) Draw a reasonable stepwise mechanism for the following reaction. This is problem 20.65a from Klein. Note that students could just write H + instead of showing H 3 O +. It was not required to draw water explicitly as the base when losing a proton, and students could draw forward arrows instead of equilibrium arrows if they wanted. 2. (5 MARKS) When aniline (aminobenzene) is treated with fuming sulfuric acid, electrophilic aromatic substitution occurs at the meta position even though we expect an amino group to be an ortho, para director. Briefly explain why this is so. When aniline is put into strong acid, the amino group is immediately protonated. Ammonium ions are strong electron withdrawing groups, and since the lone pair is tied up now, they cannot stabilize the Wheland complexes formed from ortho or para addition. The ammonium group behaves as a deactivating meta director, similar to a nitro group. This is based on material on page 882 and Table 19.1 in Klein, and is similar to problem

12 CHEM 2220 Test #2 ANSWERS Page 3 of 9 March 15, (8 MARKS) The Benzoin Addition Reaction was discovered in the 1830s, but its mechanism was only determined in In its classic form, two equivalents of benzaldehyde are heated with aqueous potassium cyanide to form benzoin (2-hydroxy-2-phenylacetophenone). Suggest a stepwise mechanism for this reaction. HINT: the CN group in a nitrile behaves as a good electronwithdrawing group. This is based on the formation of cyanohydrins. Because the CN group is a good EWG, the adjacent C-H bond becomes much more acidic than usual and can be deprotonated (reversibly) by hydroxide ion. The resulting carbanion then attacks another molecule of benzaldehyde. Note that only relatively small amounts of the anion are present at any given time, as the equilibrium constants for the first few steps do not favour the products. However, the later steps do favour their products, and the final equilibrium is extremely favourable, so the overall reaction proceeds smoothly. Students did not have to give any explanatory details: the question only required a mechanism.

13 CHEM 2220 Test #2 ANSWERS Page 4 of 9 March 15, (22 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NOT required. Give product stereochemistry where appropriate. a. (2 Marks) b. (2 Marks) c. (2 Marks) d. (6 Marks) The first step must form the Hoffmann product, so a sterically bulky base for E2 elimination is the key point. In the second step, it might be tempting to think of ozonolysis but note that the number of carbons in the product is the same as in the starting material. There is no cleavage therefore. Anti-Markovnikov hydration followed by PCC oxidation is thus the only answer.

14 CHEM 2220 Test #2 ANSWERS Page 5 of 9 March 15, 2012 e. The molecule might look scary but this is just Grignard addition to a ketone. This reaction is actually part of the synthesis of the estrogen-modulating drug Raloxifene, which is prescribed for treatment of osteoporosis in post-menopausal womene. (2 Marks) f. (4 Marks) g. (4 Marks)

15 CHEM 2220 Test #2 ANSWERS Page 6 of 9 March 15, (5 MARKS) We have seen that acetylene is a very useful starting material for organic synthesis. Using acetylene as the starting compound, propose a sequence of 3 reactions to produce (E)-oct-3-ene-1-ol. For each reaction, give the necessary reagent(s) and conditions, as well as the product formed. You may use any additional organic compound having four or fewer carbon atoms, as well as any reagents or solvents you may require. This is similar to many examples in Chapter 10 (Alkynes) and Chapter 12 (Synthesis) in Klein.

16 CHEM 2220 Test #2 ANSWERS Page 7 of 9 March 15, (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula C9H7NO2 are shown on the next page. Answer the following questions about this compound. a. (1 MARK) What is the unsaturation number for compound A? Unsaturation number is SEVEN! b. (1 MARK) What functional group(s) is(are) present? Indicate the specific signal(s) in the spectra that support your answer. There is a C=O group present (IR, ~1710 cm -1 ). From the 13 C NMR (see below) it is clearly a carboxyl-family carbonyl, but the lack of any IR signal for an OH shows that it is NOT a carboxylic acid. The 1 H NMR has no signals around 4 ppm so it is not an ester. If it were a nitrile there would be an IR band at around 2250 cm -1, which there is not. It is probably some kind of amide therefore. c. (1 MARK) What does the 13 C NMR spectrum tell you about compound A? Be specific. The signal at ~168 ppm is a carboxyl-family carbonyl carbon. The signal at ~24 ppm is likely a CH 3 group. There are only 3 other signals, which are in the aromatic region. This also tells us there are no alkenes or alkynes present. Since there are 9 carbons in total, this molecule is highly symmetrical! The aromatic ring accounts for 4 unsaturations, leaving 3 more which must be two carbonyls plus one more ring. d. (2 MARKS) What is the structure of compound A? Note integral in 1 H NMR is 4:3. Four aryl protons means disubstituted aromatic ring. The methyl group at about 3 ppm is consistent with N-CH 3. Solving this problem was all about symmetry. The 13 C NMR showed only 1 kind of carbonyl and 1 kind of aliphatic, with 3 kinds of aryl carbons. The key was to ask yourself how many ways can 9 carbons be sorted into 5 kinds? Furthermore, since the 1 H NMR told you that there was only 1 methyl, you could deduce that it had to lie on the symmetry plane. Finally, the symmetry strongly suggested that there were actually 2 carbonyls, both identical. These points all should have suggested an orthodisubstituted aromatic compound as shown above.

17 CHEM 2220 Test #2 ANSWERS Page 8 of 9 March 15, 2012 IR 13 C NMR 1 H NMR