Chem. 310N, Spring 2008, Professor Krische Final Exam KeyChem. 310N, Spring 2008, Professor Krische...
Transcript of Chem. 310N, Spring 2008, Professor Krische Final Exam KeyChem. 310N, Spring 2008, Professor Krische...
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Chem. 310N, Spring 2008, Professor Krische
Final Exam Key Last Name:
First Name: Problem
1. (30 points)
2. (12 points)
3. (10 points)
4. (14 points)
5. (15 points)
6. (10 points)
7. (24 points)
8. (20 points)
9. (15 points)
Total Points: /150
Letter Grade T-Score A+ 100-96 A 95-92 A- 91-90 B+ 89-86 B 85-82 B- 81-80 C+ 79-76 C 75-72 C- 71-70 D+ 69-66 D 65-62 D- 61-60 F 60-0
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1. (30 Points) For the following compounds, circle the one that embodies the indicated property. (2 pts each) a. Retains deuterium upon elimination
N
CH3
CH3
CH3N
CH3
O
CH3CH3H3C
HD D
H
CH3H3C
b. Aromatic Stabilization
HNNH
c. Faster Rate of Reaction with Br2/NaOH
O
CH3
O
CH2
Br
d. Faster Rate of Reaction with Br2/HBr
O
CH3
O
CH2
Br
e. Anomeric Hydroxyl in α-configuration
OOH
HOHO
HO
OOH
H3COH3CO
H3CO
OH
OH
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f. Faster Rate of Reaction with CH3OH through an SN1 Mechanism
Br Br
CH3 CH3
g. Greater Dipole Moment
O
O
CH3
O
O
H3C
h. Faster Rate of Reaction with HNO3/H2SO4
Br O O
CH3
i. Faster Rate of Reaction with NaOCH3/HOCH3
F F
MeO2C
MeO2C
NO2
NO2 j. Greater Acidity
OHHO
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k. Faster Rate of Reaction in Diels-Alder Cycloaddition with cyclohexenone
CH3OMe j. Greater Concentration of Hydrated Form at Equilibrium in water
O
CH3
O
H
h. Greater Basicity of the Carbonyl Oxygen
O
NH3C
CH3
H3C
CH3
CH3
O
OH3C
CH3
H3C
CH3
i. Greater Number of Signals in the 13C NMR
CH3
CH3
CH3
CH3
j. Faster Rate of Reaction with CH3NH2 through an Addition-Elimination Mechanism
O O O O O
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2. (12 points) At low temperature the following Diels-Alder reaction provides mainly the endo product, while at high temperature the exo product predominates. A. (8 Points) Draw the endo- and exo products.
Exo ProductMajor Product at
High Temperature
Endo ProductMajor Product at Low Temperature
O
O
O
O
HH
O
OH
H
OO
O
B. (4 Points) Draw an energy versus reaction progress diagram that accounts for the observed change in product distribution as a function of temperature.
Ener
gy
Reaction Coordinate
Exo
Endo
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3. (10 points) Consider the following hexoses. The hexoses indicated below are drawn as Kekulé type structures. Convert these hexoses to Hayworth projections and, finally, Fischer projections (in their acyclic forms). Answer the questions by circling the correct response. (5 pts each)
Glucose
O
OHH
CH2OH
OHH
H
HO OH
HH
CHO
OH
HHO
OHH
OHH
CH2OH
Gulose
O
H
CH2OH
H
OHOH
OH
H H
OHH
CHO
OH
OHH
HHO
OHH
CH2OH
HO
HO
OH
H
H
HOHH OH
OH
Glucose depicted as the... a-anomer or β-anomer.
The compound above is... D-Glucose or L-Glucose.
Gulose is depicted as the... α-anomer or β-anomer.
The compound above is... D-gulose or L-gulose.
HOO
H
H
HO
H
OHOHH H
OH
OHO
OHH
D-Glyceraldehyde
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4. (14 points) Consider the hydrochlorination of the following alkene. Write separate mechanisms to account for the formation of each product, indicating any relevant resonance structures. Based on a comparison of the two mechanisms, predict which product is expected to be the major product. Circle the major product and justify your prediction in writing.
ORH-Cl
CH3 CH3Cl CH3
Cl
Product A Product B Mechanism for formation of Product A: (5 pts)
H-Cl
CH3 CH3 CH3
Etc.
Cl
CH3Cl
Mechanism for formation of Product B: (5 pts)
H-Cl
CH3 CH3
Cl
CH3
Cl
Justification (One sentence): (4 pts) Resonance stabilization of the carbocation from the aromatic ring causes A to be favored.
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5. (15 points) Propose mechanisms to account for the following transformations. For each intermediate, draw all lone pair electrons, any formal charges, and all important resonance structures. Use the arrow pushing formalism to interconvert intermediates. (5 pts each) A.
OOH
H
N
CH3
CH3 N
CH3
CH3
O
OOH
H
N
CH3
CH3 N
CH3
CH3
O OH
H
N
CH3
CH3
O
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B.
H2SO4CH3
CH3
H3C CH3
CH3
CH3
H3C CH3
S
O
O
O OH
H
CH3
CH3
CH3
CH3
H
S
O
O
O OH
H3C CH3
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C.
O
+H-Cl
H2O
D.S. TrapHS SH
SS
O H Cl OH
OH
Protonation Activates CarbonylToward Nucleophilic Addition
SOH
H Cl
Cl
H2O
SSH
SS
HCl
HCl
Cl
Cl
Cl
SHHSS
H
H
SOH
SH
SO
HS
H
HS
SH
S
SH
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6. (10 points) Consider the proposed pericyclic reactions. A. (5 Points) Based on your understanding of transition state aromaticity, predict which of the two reactions should be successful. Circle the reaction that you expect to proceed most readily. Explain your answer in one sentence. (4 pts)
Explanation: (2 pts) Aromatic stabilization of transition state. B. Draw the final product of the following Cope rearrangement. (4 pts)
ΔOH
O
OH
or
-1 pt
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7. (24 points) Draw the major product(s) expected when the indicated starting materials are subjected to the following transformations. Be sure to indicate stereochemistry when relevant. A.
KOH (aq)
O
H3C CH3
O
HHeat
CH3
CH3
CH3
CH3
CH3
CH3
H3C
O
B.
1) NaN3, DMFCl2) LiAlH4 then H2O
NH2
C.
NH2
1) NaNO2, H3O+
2) CuBr, HBrMeO BrMeO
D.
N
H3Cthen H2O
LiAlH4
O
N
H3C
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E.
O
Then H2O
BrMg
OH
CH3
F.
1) AlCl3
2) H2N-NH2, KOHH20, 100oC
Cl
O
H3CH3C H3C
G.
CH2CH3
O
then H2O
BuLi, then
S S
HH
S S
OH
H.
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CH3
CH3HCl
O
H3C
H3C
HN CH3
D. S. Trap
CH3
CH3
N
H3C
H3C
H3C CH3
I.
CH3H3CO
O NaOCH3, HOCH3
Then HCl H3CO
O
CH3
O
J.
150oC
H3C
O
H
CH3
O
H
K.
1) I-CH3 (Excess)
H3C
NH2
2) AgOH, heat
L.
1) LDA, -78oC
CH3
O
2)
Then H2OH CH3
O
O OH
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8. (20 points) Devise a synthetic route to accomplish the following transformations. Clearly number each step of your proposed sequence and Draw all Intermediates. (Hint: Try to work backwards).
A. (5 pts)
?H3CH3C
OH
HNO3, H2SO4
H3C
NO2
H2, PtH3C
NH2
NaNO2, HClH3C
N2
H2O
B. (5 pts)
?OCH3
O2N
Br
HNO3, H2SO4
Br
O2N
NaOMe
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C. (5pts)
H3C?OH
H3COH
PBr3
H3CBr Mg
H3CMgBr
O
Then H2O
D. (5 pts)
?H3COCH3
OO
H3C
O
NaOCH3
OCH3
OO
H3C
then Br
1. NaOH, H2O2. HCl, H2O3. Heat
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9. (15 points) The molecular formulas of three compounds determined by combustion analysis are given below. By considering the 1H NMR spectra of these compounds (found on the following page), formulate structural assignments. Indicate your answer by drawing the compound in the appropriate box. (5 pts each) Compound A) C3H7Cl
Cl
Compound B) C3H6Br2
Br Br
Compound C) C3H7I
I
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A. C3H7Cl
(4.2 ppm, 1H, septet), (1.5 ppm, 6H, doublet)
B. C3H6Br2
(3.55 ppm, 4H, triplet), (2.34 ppm, 2H, pentet)
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C. C3H7I
(3.14 ppm, 2H, triplet), (1.82 ppm, 2H, sextet), (0.94 ppm, 3H, triplet)
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