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1 Page 1 of 5 Name: Student Number: l University of Manitoba - Department of Chemistry CEM Introductory rganic Chemistry II - Term Test 1 Thursday, February 14, 2008 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. 1. (10 MARKS) In the reactions of cis- and trans-1-bromo-4-tert-butylcyclohexane, the cis isomer shows a noticeably higher reaction rate (rel. rate > 500) in comparison to the trans-isomer (rel. rate = 1). Show the mechanisms of the two elimination reaction (use conformational structures), and briefly explain the difference in reactivity. C(C 3 ) 3 C(C 3 ) 3 C(C 3 ) 3 K t Bu, t Bu K t Bu, t Bu rel. rate = 1 rel. rate >500

2 Page 2 of 5 2. (8 MARKS) Acid-promoted addition to alkenes sometimes forms rearranged products. Write a stepwise mechanism to explain the following reaction. (int: only a few bonds need to move!) 3 C C 3 3 C C 3 3 C 3 C 3. (7 MARKS) Suggest a mechanism to explain how the addition of sodium bicarbonate to this epoxidation reaction changes the outcome. mcpba C 2 2 mcpba C 2 2 NaC 3

3 Page 3 of 5 4. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NT required. Give product stereochemistry where appropriate. NaN 2 a. N 2 excess C 3 I Ag 2, 2 heat b. (4 Marks) + heat C 3 c. d. + NaCN Et 2 S 4 (cat.) 2, heat e.

4 Page 4 of 5 1) 2) f. (two successive steps) g., ether -20 o C C 3 Zn, C 2 I 2 h. i.

5 Page 5 of 5 5. (5 MARKS) The spectra of an unknown organic compound A having the formula C are shown below. What was the structure of compound A? IR 1 NMR 13 C NMR Structure: A

6 Page 6 of 5 Spectroscopy Crib Sheet for CEM 2220 Introductory rganic Chemistry II 1 NMR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C C C C C C 2 C C C C C C C (solvent dependent) (solvent dependent) C Aryl C C C Aryl I C RC 2 Aromatic, heteroaromatic X C X =, N, S, halide R 3 C Aliphatic, alicyclic Y =, NR, S Y Y Y =, NR, S Low Field δ igh Field 13 C NMR Typical Chemical Shift Ranges Alkene Aryl Ketone, Aldehyde Ester Amide C x -Y Y =, N CR 3 -C 2 -CR 3 C x -C= C 3 -CR IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity C Medium R Strong, broad C=C Medium C Strong C=C Medium C= Strong C C Strong R 2 N Medium, broad R C C R Medium (R R ) C N 1230, 1030 Medium Aryl Medium C N Medium Aryl C=C 1600, 1500 Strong RN Strong δ

7 ANSWER KEY Page 1 of 5 ANSWER KEYl University of Manitoba - Department of Chemistry CEM Introductory rganic Chemistry II - Term Test 1 Thursday, February 14, 2008 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. 1. (10 MARKS) In the reactions of cis- and trans-1-bromo-4-tert-butylcyclohexane, the cis isomer shows a noticeably higher reaction rate (rel. rate > 500) in comparison to the trans-isomer (rel. rate = 1). Show the mechanisms of the two elimination reaction (use conformational structures), and briefly explain the difference in reactivity. C(C 3 ) 3 C(C 3 ) 3 C(C 3 ) 3 K t Bu, t Bu K t Bu, t Bu Consult Fox and Whitesell pp for discussion of the geometric requirements of E2 eliminations. rel. rate = 1 rel. rate >500 The slow reaction minor conformation due to axial position of bulky group major conformation in solution The E2 elimination requires an antiperiplanar alignment of the C- and C- bonds which only exists in the disfavoured conformation. Because this is present only in trace amounts, elimination is very slow. The fast reaction major conformation minor conformation due to axial in solution position of bulky group In this case, the preferred conformation has the necessary antiperiplanar arrangement. The reaction is faster.

8 ANSWER KEY Page 2 of 5 2. (8 MARKS) Acid-promoted addition to alkenes sometimes forms rearranged products. Write a stepwise mechanism to explain the following reaction. (int: only a few bonds need to move!) 3 C C 3 3 C C 3 You did not have to explain why this rearrangement occurs, but you can see that since it converts a 4- membered ring in the starting material to a much more favourable 5-membered ring, the result is energetically downhill. 3 C 3 C 1,2-shift In fact, the attack of chloride is concerted with the 1,2-shift, which is why only one stereoisomer is obtained (similar to an SN2 process in some ways). owever, we will give full marks for a stepwise process as well. 3. (7 MARKS) Suggest a mechanism to explain how the addition of sodium bicarbonate to this epoxidation reaction changes the outcome. mcpba C 2 2 mcpba C 2 2 NaC 3 N BASE Ar Ar Ar NaC 3 Acid catalysis is necessary to assist in nucleophilic opening of the epoxide. Adding bicarbonate neutralizes the m-chlorobenzoic acid formed during epoxidation, preventing ring-opening. For discussion about epoxidation by peroxyacids and the nucleophilic opening of epoxides, see Fox and Whitesell pp Note especially the importance of acid catalysis in ring-opening as indicated at the bottom of page 496. This is also shown in Groutas on page 15 in Example 4, and features in Groutas problems 15, 21 and 34.

9 ANSWER KEY Page 3 of 5 4. (20 MARKS) Provide the necessary reagents/solvents or starting materials or major products to correctly complete the following reactions. Mechanisms are NT required. Give product stereochemistry where appropriate. NaN 2 N 2 a. b. N 2 excess C 3 I 3 C C 3 N C 3 I Ag 2, 2 heat ofmann elimination! (4 Marks) + C 3 heat C 3 c Zn/Ac or KMn 4 / 3 + d. e. NaCN Et C N 2 S 4 (cat.) 2, heat or carbox. acid N 2

10 ANSWER KEY Page 4 of 5 1) B 2 6 or B 3 2) 2 2, Na (aq.) f. (two successive steps) g., ether -20 o C or Low temperature leads to kinetic control which favours 1,2-addition. Note however, that the two double bonds are not equal in this case. We gave full marks for either 1,2- adduct, but most likely the kinetic product would be formed from the more electron-rich trisubstituted alkene. C 3 Zn, C 2 I 2 2 C C 3 h. i. s 4 (cat.) K 3 Fe(CN) 6 2

11 ANSWER KEY Page 5 of 5 5. (5 MARKS) The spectra of an unknown organic compound A having the formula C are shown below. What was the structure of compound A? IR Unsaturation = 1 IR shows C=, no. 13 C shows 4 signals thus must have 2 identical C atoms in molecule; note C= at ca. 178 plus no means must be ester! 1 NMR singlet at 3.6 ppm for 3 = C 3 : Must be a methyl ester. Multiplet at 2.5 ppm for 1, doublet at 1.1 ppm for 6 indicates isopropyl group. NB: C must be adjacent to carbonyl based on chemical shift. 1 NMR 13 C NMR Note: If the methyl group was next to the carbonyl and the isopropyl group was attached to oxygen, the 3 singlet would have been at about 2.1 ppm and the 1 multiplet would have been around 3.7 or so! Structure: 3 C C 3 C 3 A

Name: Student Number: University of Manitoba - Department of Chemistry CHEM Introductory Organic Chemistry II - Term Test 1

Name: Student Number: University of Manitoba - Department of Chemistry CHEM 2220 - Introductory Organic Chemistry II - Term Test 1 Thursday, February 11, 2016; 7-9 PM This is a 2-hour test, marked out