exam i su08 - university of illinois urbana-champaign · chemistry 332 exam i,8 summer 08 5 20) [4...

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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 exam. 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 exam. Answers must be submitted via ACE Organic 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 O F Si S Cl Br I

104 99 93 111 135 76 83 103 87 71 H

83a 73b 86c 116d 72 65 81 68 52 C 39 53e 65 46 N 47 45 108 52 48 56 O 37 135 F

aC=C 146, C≡C 200 53 91 74 56 Si 60 61 52 S

bC=N 147, C≡N 213 dIn CF4 58 Cl 46 Br

cC=O 176 (aldehydes) e In nitrites and nitrates 36 I 179 (ketones)

CHEMISTRY 332 Exam I, Summer 08

1

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).

2) [4 pts.] Based on the MO images in Figure 1, place a check in each box that corresponds to an image of an antibonding MO. 3) [4 pts.] Using the information provided in Figure 1, match the numbers in the energy level diagram to the letters of the MO 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?

Figure 1 shows nitrogen fluoride molecular orbital images and its MO 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

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?

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 QUESTION 9 is pi* not the LUMO (check all that apply)? 11) [4 pts.] On the basis of the MO diagrams, rank the cations in QUESTION 9 from most to least electrophilic (most = 1 and least = 5). 12) [4 pts.] For each of the cations in QUESTION 9, identify the most electrophilic atom in that structure by placing the number “1” next to this atom. The MO diagrams in Figure 2 can help you. To add the number "1", 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 → M1 (or Off). 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 Off).

Figure 2 shows energy diagrams of atomic p orbitals and their resulting pi/pi* MOs for the series of cations containing double or triple bonds in QUESTION 9. 1 2 3 4 5


3

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. Org. Chem., 73, 4342, 2008).

O

CH3

H

O

CH3

Cl I

OCH3

H

O

CH3

I

O

CH3

H

O

CH3

I

O

O

CH3

I

CH3

H

O

O

CH3

I

CH3

H

OH

H

O

O

CH3

I

CH3

QUESTION 14

QUESTION 15

QUESTION 16

QUESTION 17

QUESTION 18

alkynyl-2,3-epoxy alcohol

iodopyran-4-one

H3O+

Cl–

14) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled “QUESTION 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 “QUESTION 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.


4

16) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in the mechanism step labeled “QUESTION 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 “QUESTION 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 “QUESTION 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 DOI: 10.1039/b805446k). Use the information in this figure to answer QUESTION 19.

Structure of "cavitand" molecular capsules

side view cavitand interior

OH

HO

estradiol

complex

complex

Keqestradiol

cavitand

19) [4 pts.] Check all of the boxes that are true based on your interpretation of Figure 4. The reported Keq for estradiol at room temperature in aqueous solution is about 1 × 108 M-2.


5

20) [4 pts.] Use the table of bond energies to estimate the enthalpy change for the following reaction (Ref. Org. Lett., 2008, 10, 2569).

O

O

O O O

O

O

O

O

+ H2O

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 Off). 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

0

5

10

15

20

25

92949698100102104106108110112114116118120122

<C-As-C> Valence Angle (Degrees)

Fre

qu

en

cy


6

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

92

94

96

98100102104106108110112114116118120122

<C-N-C> Valence Angles (Degrees)

A


92

94

96

98100102104106108110112114116118120122


B


92

94

96

98100102104106108110112114116118120122


C


92

94

96

98100102104106108110112114116118120122


D


7

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. One 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 –CH3 groups on nitrogen (use the bond tool and click on the N atom to add a methyl group).

Cl

N

CH3

H3C

N

CH3

H3C

fragmentationproduct

N

CH3

H3C

+

N

CH3

H3C

mixture of E2 elimination products

EtOH, Et3N, H2O

Cl–

trans-isomer


8

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. Org. Chem., 2008, 73, 4753).

O

O

P

OCH3

OO

R

NH2

k1

k-1

SN

O

O

P

OCH3

OO

H2N

R

I

+

k2

P

OCH3

O O

HO

NHR

O

+

P 26) [4 pts.] The rate coefficient that governs the rate expression in QUESTION 25 is directly related to the pKa of the amine’s conjugate acid (the larger the pKa, the larger the rate coefficient). Rank these amines from largest rate coefficient to smallest (1 = largest rate coefficient, 5 = smallest rate coefficient). 27) [2 pts.] Assume that the [I] follows steady-state behavior; also assume that the values of k-1 and k2 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.

ACE Organic homework http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable...

1 of 8 6/25/08 7:29 AM

Exam 1 - Summer 08 Maximum allowed tries per question: Unlimited

(1) QUESTION 1) [4pts.] Follow theinstructions in the exam.

(Question -1118)


(Question -1119)

Option 1. aOption 2. bOption 3. cOption 4. dOption 5. eOption 6. f


(Question -1120)

Item 1. aItem 2. bItem 3. cItem 4. dItem 5. eItem 6. f


(Question -1121)

Option 1. A nitrogen 2p atomic orbital.Option 2. A nitrogen 2s atomic orbital.

Option 3. A fluorine 2p atomic orbital.Option 4. A fluorine 2s atomic orbital.


(Question -1122)

Option 1. A nitrogen 2p atomic orbital.Option 2. A nitrogen 2s atomic orbital.

Option 3. A fluorine 2p atomic orbital.Option 4. A fluorine 2s atomic orbital.


(Question -1123)

Option 1. sigma-typeOption 2. pi-type

Option 3. constructiveOption 4. destructive

Option 5. fluorine 2sOption 6. fluorine 2p

Option 7. nitrogen 2sOption 8. nitrogen 2p

(7) QUESTION 7) [4pts.] Follow theinstructions in the

Option 1. The nitrogen 2s orbital makes a significant contribution to thisMO.


2 of 8 6/25/08 7:29 AM

exam.

(Question -1124)

Option 2. Coaxial interaction of a pair of p-orbitals provides less overlapthan side-by-side overlap.Option 3. s/p mixing decreases the energy of the pi MO levels.

Option 4. All of the above.


(Question -1125)

Option 1. 1.0

Option 2. 1.5Option 3. 2.0

Option 4. 2.5Option 5. 3.0


(Question -1126)

Item 1.

Item 2.

Item 3.

Item 4.

Item 5.


3 of 8 6/25/08 7:29 AM


(Question -1127)

Option 1.

Option 2.

Option 3.

Option 4.

Option 5.

(11) QUESTION 11) [4pts.] Follow theinstructions in the

Item 1.


4 of 8 6/25/08 7:29 AM

exam.

(Question -1128)

Item 2.

Item 3.

Item 4.

Item 5.


(Question -1129)

Fig. 1 of 1


Fig. 1 of 1


5 of 8 6/25/08 7:29 AM

(Question -1141)


(Question -1132)

Option 1. sigma → a

Option 2. sigma → pi*Option 3. sigma → sigma*

Option 4. pi → aOption 5. pi → pi*

Option 6. pi → sigma*Option 7. n → a

Option 8. n → pi*Option 9. n → sigma*

Option 10. coaxial (sigma-type) interactionOption 11. side-by-side (pi-type) interaction


(Question -1133)








(Question -1134)







6 of 8 6/25/08 7:29 AM



(Question -1135)

Option 1. sigma → aOption 2. sigma → pi*

Option 3. sigma → sigma*Option 4. pi → a

Option 5. pi → pi*Option 6. pi → sigma*

Option 7. n → aOption 8. n → pi*

Option 9. n → sigma*Option 10. coaxial (sigma-type) interaction

Option 11. side-by-side (pi-type) interaction


(Question -1136)

Option 1. sigma → aOption 2. sigma → pi*

Option 3. sigma → sigma*Option 4. pi → a

Option 5. pi → pi*Option 6. pi → sigma*

Option 7. n → aOption 8. n → pi*

Option 9. n → sigma*Option 10. coaxial (sigma-type) interaction

Option 11. side-by-side (pi-type) interaction


(Question -1137)

Option 1. As shown in Figure 4, the stoichiometry of the cavitand:estradiolcomplex is 1:1.

Option 2. In the absence of solvent, ΔS for complex formation is positive.

Option 3. Ionization of the cavitand’s carboxylic acid groups will helppromote solubility of the cavitand in water.?

Option 4. In aqueous solution without added estradiol, the cavitand’sinterior is filled with ordered water molecules.?

Option 5. In aqueous solution, complex formation is entropically favoredbecause estradiol binding liberates water molecules from the cavitand’sinterior.?

Option 6. Higher temperature will bring about a decrease in Keq.


(Question -1138)

Option 1. +258 kcal/molOption 2. +36 kcal/mol

Option 3. +4 kcal/molOption 4. -12 kcal/mol

Option 5. -18 kcal/mol


7 of 8 6/25/08 7:29 AM

Option 6. -162 kcal/mol


(Question -1142)

Fig. 1 of 1


(Question -1143)

Option 1. pyramidal

Option 2. trigonal planarOption 3. linear

Option 4. bent


(Question -1144)

Option 1. AOption 2. BOption 3. COption 4. D


(Question -1145)

Fig. 1 of 1


(Question -1146)

Option 1. rate = k1[N][S]

Option 2. rate = k-1[I]

Option 3. rate = k2[I]

Option 4. rate = k1[N][S] - k2[I]

Option 5. rate = (k-1 - k2)[I]


(Question -1147)

Item 1.

Item 2.


8 of 8 6/25/08 7:29 AM

Item 3.

Item 4.

Item 5.

(27) QUESTION 27) [2pts.] Follow the instructions in the exam.

(Question -1149)

Option 1. [I] = {k1/(k-1 + k2)}[N][S]

Option 2. [I] = {k1/(k-1 - k2)}[N][S]

Option 3. [I] = {k1k2/(k-1 - k2)}[N][S]

Option 4. [I] = {k1k2/(k-1 + k2)}[N][S]

(28) QUESTION 28) [4pts.] Follow the instructions in the exam.

(Question -1148)

Option 1. The most basic amino group will react fastest in a competitivesituation as long as the pH of the reaction solution is above the pKa of that of the amine.Option 2. The most basic amino group will react slowest in a competitivesituation as long as the pH of the reaction solution is above the pKa of that of the amine.Option 3. The rate of the reaction will be independent of pH.

Option 4. In addition to the rate coefficient and concentrations of thereactants, the relative amount of free vs protonated amine determines the overall rate of amide formation.

exam i su08 - university of illinois urbana-champaign · chemistry 332 exam i,8 summer 08 5 20) [4...

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