First Three Letters of Last Name NAME Network ID CHEMISTRY 332 SUMMER 08 EXAM I June 26-28, 2008 The following materials are permissible during the exam: molecular model kits, course notes (printed, electronic, or hand written), textbook, calculator, Internet browser. You are not permitted to receive assistance, in any way, from another person during the Receiving aid on an exam from any person, by any means of communication, is considered academic misconduct and can result in a failing grade for that Answers must be submitted via ACE rganic during the scheduled examination period. The paper copy of your exam will be collected, but not graded. Your goal is to answer every question correctly, in as few attempts as possible. Average Bond Energies (kcal mole -1 ) H C N F Si S Cl Br I 104 99 93 111 135 76 83 103 87 71 H 83 a 73 b 86 c 116 d 72 65 81 68 52 C 39 53 e 65 46 N 47 45 108 52 48 56 37 135 F a C=C 146, C C 200 53 91 74 56 Si b C=N 147, C N 213 c C= 176 (aldehydes) 60 61 52 S d In CF 4 58 Cl 46 Br e In nitrites and nitrates 36 I 179 (ketones)
CHEMISTRY 332 Exam I, Summer 08 1) [4 pts.] Draw the Lewis structure of the diatomic molecule, nitrogen fluoride (NF). You are expected to follow the standard practice showing an electron configuration that satisfies the octet rule for all atoms and which contains no unpaired electrons (i.e., radicals). Be sure to indicate formal charges (the overall charge on the molecule is neutral). Figure 1 shows nitrogen fluoride molecular orbital images and its M energy diagram. In the images, fluorine is on the left side and nitrogen is on the right. In the energy diagram, electrons have been not been included. 2) [4 pts.] Based on the M images in Figure 1, place a check in each box that corresponds to an image of an antibonding M. 3) [4 pts.] Using the information provided in Figure 1, match the numbers in the energy level diagram to the letters of the M images. 4) [2 pts.] Which atomic orbital makes the greatest contribution to the molecular orbital designated c? (select one box only) 5) [2 pts.] Which atomic orbital makes the greatest contribution to pi*? (select one box only) 6) [6 pts.] Check 4 boxes for this problem. The molecular orbital designated b in Figure 1 is best described as a (sigma-type or pi-type) interaction resulting from the (constructive / destructive) combination of the (fluorine 2s, fluorine 2p) and (nitrogen 2s, nitrogen 2p) atomic orbitals? 1
CHEMISTRY 332 Exam I, Summer 08 7) [4 pts.] Which statement explains why level 4 in the energy diagram of Figure 1 is higher than level 3? 8) [2 pts.] What is the bond order of nitrogen fluoride? Figure 2 shows energy diagrams of atomic p orbitals and their resulting pi/pi* Ms for the series of cations containing double or triple bonds in QUESTIN 9. 1 2 3 4 5 9) [4 pts.] Match the numbers associated with the energy diagrams in Figure 2 to the corresponding molecular structure of the cations. 10) [4 pts.] For which of the cations in QUESTIN 9 is pi* not the LUM (check all that apply)? 11) [4 pts.] n the basis of the M diagrams, rank the cations in QUESTIN 9 from most to least electrophilic (most = 1 and least = 5). 12) [4 pts.] For each of the cations in QUESTIN 9, identify the most electrophilic atom in that structure by placing the number 1 next to this atom. The M diagrams in Figure 2 can help you. To add the number "1", right-click (Safari and Netscape for Mac users: control- or optionclick) on the atom that needs to be labeled (or unlabeled), and choose Map M1 (or ff). 13) [4 pts.] The structure shown in ACE was recently suggested as an intermediate in the mechanism for an acyl transfer reaction (J. Am. Chem. Soc. 2006, 128, 4556). There are 3 nitrogen atoms in the structure of this molecule. Label the nitrogen atoms 1 through 3 to indicate the most to least basic sites (assign 1 to the most basic nitrogen, 2 to the second most basic nitrogen, and 3 to the least basic nitrogen). To add a number, right-click (Safari and Netscape for Mac users: control- or option-click) on the atom that needs to be labeled (or unlabeled), and choose Map number (or ff). 2
CHEMISTRY 332 Exam I, Summer 08 Figure 3 shows a mechanism for the electrophile-induced cyclization of an alkynyl-2,3- epoxy alcohol into an iodopyran-4-one. Use this figure to answer questions 14-18 (ref. J. rg. Chem., 73, 4342, 2008). Cl H I QUESTIN 14 H I Cl alkynyl-2,3-epoxy alcohol QUESTIN 15 H I QUESTIN 16 H I QUESTIN 17 H H H I H 3 + I QUESTIN 18 iodopyran-4-one 14) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled QUESTIN 14 of Figure 3. First, indicate the frontier orbitals involved; your answer should be one of the nine possible combinations of filled empty orbital pairs (e.g., π π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type (side-by-side) interaction. 15) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled QUESTIN 15 of Figure 3. First, indicate the frontier orbitals involved; your answer should be one of the nine possible combinations of filled empty orbital pairs (e.g., π π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type (side-by-side) interaction. 3
CHEMISTRY 332 Exam I, Summer 08 16) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled QUESTIN 16 of Figure 3. First, indicate the frontier orbitals involved; your answer should be one of the nine possible combinations of filled empty orbital pairs (e.g., π π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type (side-by-side) interaction. 17) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the resonance interaction labeled QUESTIN 17. First, indicate the frontier orbitals involved; your answer should be one of the nine possible combinations of filled empty orbital pairs (e.g., π π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type (side-by-side) interaction. 18) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled QUESTIN 18 of Figure 3. First, indicate the frontier orbitals involved; your answer should be one of the nine possible combinations of filled empty orbital pairs (e.g., π π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type (side-by-side) interaction. Figure 4 shows a reversible chemical reaction in which the hydrophobic effect drives complex formation (Ref. Chem. Commun., 2008 DI: 10.1039/b805446k). Use the information in this figure to answer QUESTIN 19. complex H estradiol K eq cavitand complex H estradiol Structure of "cavitand" molecular capsules side view cavitand interior 19) [4 pts.] Check all of the boxes that are true based on your interpretation of Figure 4. The reported K eq for estradiol at room temperature in aqueous solution is about 1 10 8 M -2. 4
CHEMISTRY 332 Exam I, Summer 08 20) [4 pts.] Use the table of bond energies to estimate the enthalpy change for the following reaction (Ref. rg. Lett., 2008, 10, 2569). + H 2 21) [4 pts.] Three structures taken from the CSD are shown in ACE. For each structure, you are to indicate the most likely geometry of the nitrogen or phosphorus atom. Specifically, label each nitrogen or phosphorus with a number to indicate the expected geometry (1 = linear, 2 = trigonal planar, 3 = pyramidal, or 4 = bent). To add a number in ACE, right-click (Safari and Netscape for Mac users: control- or option-click) on the atom that needs to be labeled (or unlabeled), and choose Map number (or ff). 22) [2 pts.] A CSD search was performed similar to Quiz 5, only arsenic was used instead of nitrogen or phosphorus. The search structure and histogram are provided. Given this data, what geometry would you predict for the arsenic atom? Histogram of <C-As-C> Valence Angles 25 20 Frequency 15 10 5 92 94 96 98 0 100 102 104 106 108 110 112 114 116 118 120 <C-As-C> Valence Angle (Degrees) 122 5
CHEMISTRY 332 Exam I, Summer 08 23) [2 pts.] A search of the CSD was performed similar to Quiz 5 only the carbons were constrained to having a coordination number of 4. Which of the following histograms would you expect as your result? Histogram of <C-N-C> Valence Angles Histogram of <C-N-C> Valence Angles A B 92 94 96 98 100 102 104 106 108 110 112 114 116 118 <C-N-C> Valence Angles (Degrees) 120 122 92 94 96 98 100 102 104 106 108 110 112 114 116 <C-N-C> Valence Angles (Degrees) 118 120 122 C Histogram of <C-N-C> Valence Angles D Histogram of <C-N-C> Valence Angles 92 94 96 98 100 102 104 106 108 110 112 114 116 <C-N-C> Valence Angles (Degrees) 118 120 122 92 94 96 98 100 102 104 106 108 110 112 114 116 <C-N-C> Valence Angles (Degrees) 118 120 122 6
CHEMISTRY 332 Exam I, Summer 08 24) [4 pts.] When exposed to the reaction conditions indicated below, the trans isomer results in both the fragmentation product and a mixture of E2 elimination products. The fragmentation pathway requires that the sigma bond undergoing fragmentation overlap in a pi-type fashion with sigma* of the C Cl bond. ne specific ring-flipped chair conformation of the trans-isomer is ideally suited for fragmentation and the other ring-flipped chair conformation is ideally suited for E2 elimination. Follow steps (i) (iv) below to draw the ring-flipped chair conformation that is best suited to undergo the fragmentation pathway. i. To add a nitrogen or chlorine substituent to the ring, press the appropriate atom button. ii. Click on an H atom in the structure to replace that H atom with the atom or group at the tip of the cursor. iii. Do not delete any of the atoms or bonds in the ring or directly attached to the ring with the Erase button, and do not move atoms or groups directly attached to the ring by dragging them. iv. Don t forget to draw the groups on nitrogen (use the bond tool and click on the N atom to add a methyl group). H 3 C N H 3 C N Cl fragmentation product Cl EtH, Et 3 N, H 2 trans-isomer H 3 C N H 3 C N + mixture of E2 elimination products 7
CHEMISTRY 332 Exam I, Summer 08 25) [2 pts.] The mechanism for an amine (N) reacting with an acyl phosphate monoester (S) is provided below. The rate-determining step is decomposition of the tetrahedral intermediate (I). Which equation determines the overall rate of the reaction? (Ref. J. rg. Chem., 2008, 73, 4753). S + R P NH 2 N k 1 k- 1 H 2 N I 26) [4 pts.] The rate coefficient that governs the rate expression in QUESTIN 25 is directly related to the pk a of the amine s conjugate acid (the larger the pk a, the larger the rate coefficient). Rank these amines from largest rate coefficient to smallest (1 = largest rate coefficient, 5 = smallest rate coefficient). R P k 2 P + NHR H P 27) [2 pts.] Assume that the [I] follows steady-state behavior; also assume that the values of k -1 and k 2 have comparable magnitudes (unlike the case in your lecture notes). Given these assumptions, derive an expression for the steady-state concentration of [I] in terms of [N] and [S]. Select the equation that you derived. 28) [4 pts.] Based on the above information, check all the statements that are true. 8
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... Exam 1 - Summer 08 Maximum allowed tries per question: Unlimited (1) QUESTIN 1) [4 (Question -1118) (2) QUESTIN 2) [4 (Question -1119) ption 1. a ption 2. b ption 3. c ption 4. d ption 5. e ption 6. f (3) QUESTIN 3) [4 (Question -1120) Item 1. a Item 2. b Item 3. c Item 4. d Item 5. e Item 6. f (4) QUESTIN 4) [2 (Question -1121) ption 1. A nitrogen 2p atomic orbital. ption 2. A nitrogen 2s atomic orbital. ption 3. A fluorine 2p atomic orbital. ption 4. A fluorine 2s atomic orbital. (5) QUESTIN 5) [2 (Question -1122) ption 1. A nitrogen 2p atomic orbital. ption 2. A nitrogen 2s atomic orbital. ption 3. A fluorine 2p atomic orbital. ption 4. A fluorine 2s atomic orbital. (6) QUESTIN 6) [6 (Question -1123) ption 1. sigma-type ption 2. pi-type ption 3. constructive ption 4. destructive ption 5. fluorine 2s ption 6. fluorine 2p ption 7. nitrogen 2s ption 8. nitrogen 2p (7) QUESTIN 7) [4 ption 1. The nitrogen 2s orbital makes a significant contribution to this M. 1 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... (Question -1124) ption 2. Coaxial interaction of a pair of p-orbitals provides less overlap than side-by-side overlap. ption 3. s/p mixing decreases the energy of the pi M levels. ption 4. All of the above. (8) QUESTIN 8) [2 (Question -1125) ption 1. 1.0 ption 2. 1.5 ption 3. 2.0 ption 4. 2.5 ption 5. 3.0 (9) QUESTIN 9) [4 Item 1. (Question -1126) Item 2. Item 3. Item 4. Item 5. 2 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... (10) QUESTIN 10) [4 ption 1. (Question -1127) ption 2. ption 3. ption 4. ption 5. (11) QUESTIN 11) [4 Item 1. 3 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... (Question -1128) Item 2. Item 3. Item 4. Item 5. (12) QUESTIN 12) [4 Fig. 1 of 1 (Question -1129) (13) QUESTIN 13) [4 Fig. 1 of 1 4 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... (Question -1141) (14) QUESTIN 14) [4 (Question -1132) ption 1. sigma a ption 2. sigma pi* ption 3. sigma sigma* ption 4. pi a ption 5. pi pi* ption 6. pi sigma* ption 7. n a ption 8. n pi* ption 9. n sigma* ption 10. coaxial (sigma-type) interaction ption 11. side-by-side (pi-type) interaction (15) QUESTIN 15) [4 (Question -1133) ption 1. sigma a ption 2. sigma pi* ption 3. sigma sigma* ption 4. pi a ption 5. pi pi* ption 6. pi sigma* ption 7. n a ption 8. n pi* ption 9. n sigma* ption 10. coaxial (sigma-type) interaction ption 11. side-by-side (pi-type) interaction (16) QUESTIN 16) [4 (Question -1134) ption 1. sigma a ption 2. sigma pi* ption 3. sigma sigma* ption 4. pi a ption 5. pi pi* ption 6. pi sigma* ption 7. n a ption 8. n pi* ption 9. n sigma* 5 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... ption 10. coaxial (sigma-type) interaction ption 11. side-by-side (pi-type) interaction (17) QUESTIN 17) [4 (Question -1135) ption 1. sigma a ption 2. sigma pi* ption 3. sigma sigma* ption 4. pi a ption 5. pi pi* ption 6. pi sigma* ption 7. n a ption 8. n pi* ption 9. n sigma* ption 10. coaxial (sigma-type) interaction ption 11. side-by-side (pi-type) interaction (18) QUESTIN 18) [4 (Question -1136) ption 1. sigma a ption 2. sigma pi* ption 3. sigma sigma* ption 4. pi a ption 5. pi pi* ption 6. pi sigma* ption 7. n a ption 8. n pi* ption 9. n sigma* ption 10. coaxial (sigma-type) interaction ption 11. side-by-side (pi-type) interaction (19) QUESTIN 19) [4 (Question -1137) ption 1. As shown in Figure 4, the stoichiometry of the cavitand:estradiol complex is 1:1. ption 2. In the absence of solvent, ΔS for complex formation is positive. ption 3. Ionization of the cavitand s carboxylic acid groups will help promote solubility of the cavitand in water.? ption 4. In aqueous solution without added estradiol, the cavitand s interior is filled with ordered water molecules.? ption 5. In aqueous solution, complex formation is entropically favored because estradiol binding liberates water molecules from the cavitand s interior.? ption 6. Higher temperature will bring about a decrease in K eq. (20) QUESTIN 20) [4 (Question -1138) ption 1. +258 kcal/mol ption 2. +36 kcal/mol ption 3. +4 kcal/mol ption 4. -12 kcal/mol ption 5. -18 kcal/mol 6 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... ption 6. -162 kcal/mol (21) QUESTIN 21) [4 Fig. 1 of 1 (Question -1142) (22) QUESTIN 22) [2 (Question -1143) ption 1. pyramidal ption 2. trigonal planar ption 3. linear ption 4. bent (23) QUESTIN 23) [2 (Question -1144) (24) QUESTIN 24) [4 ption 1. A ption 2. B ption 3. C ption 4. D Fig. 1 of 1 (Question -1145) (25) QUESTIN 25) [2 (Question -1146) ption 1. rate = k 1 [N][S] ption 2. rate = k -1 [I] ption 3. rate = k 2 [I] ption 4. rate = k 1 [N][S] - k 2 [I] ption 5. rate = (k -1 - k 2 )[I] (26) QUESTIN 26) [4 Item 1. (Question -1147) Item 2. 7 of 8 6/25/08 7:29 AM
ACE rganic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable... Item 3. Item 4. Item 5. (27) QUESTIN 27) [2 (Question -1149) ption 1. [I] = {k 1 /(k -1 + k 2 )}[N][S] ption 2. [I] = {k 1 /(k -1 - k 2 )}[N][S] ption 3. [I] = {k 1 k 2 /(k -1 - k 2 )}[N][S] ption 4. [I] = {k 1 k 2 /(k -1 + k 2 )}[N][S] (28) QUESTIN 28) [4 (Question -1148) ption 1. The most basic amino group will react fastest in a competitive situation as long as the ph of the reaction solution is above the pk a of that of the amine. ption 2. The most basic amino group will react slowest in a competitive situation as long as the ph of the reaction solution is above the pk a of that of the amine. ption 3. The rate of the reaction will be independent of ph. ption 4. In addition to the rate coefficient and concentrations of the reactants, the relative amount of free vs protonated amine determines the overall rate of amide formation. 8 of 8 6/25/08 7:29 AM