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Final exam 2013, questions
Introductory Organic Chemistry 1: Structure and Function (University of Manitoba)
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CHEM 2210 Final Exam, December 9, 2013
CHEM 2210 Introductory Organic Chemistry I: Structure and Function
Dr. J. L. Sorensen
Total time = 180 minutes
Total Pages in this Exam Booklet = 13 (including spec. problem)
First Name:
.
LAST Name:
(Print Clearly)
.
(Print Clearly)
Student Number:
.
Instructions: Attempt all questions. The marks available for each question are indicated on the
question. You may use either pencil or pen but please write legibly. You may use the back of the
pages and attached data sheets for rough work. Only answers drawn in the boxed area will be
marked for credit. All long answer questions will be eligible for part marks. You may use the back of
the data sheet for scrap paper. Model Kits are allowed. Calculators are not allowed.
Question
Score
Question 1 (Stereochemistry)
(Max = 5 Marks)
Question 2 (Spectroscopy I)
(Max = 4 Marks)
Question 3 (Conformations I)
(Max = 5 Marks)
Question 4 (Conformations II)
(Max = 5 Marks)
Question 5 (Energy Diagrams)
(Max = 5 Marks)
Question 6 (Mechanism I)
(Max = 5 Marks)
Question 7 (Mechanism II)
(Max = 5 Marks)
Question 8 (Synthesis I)
(Max = 5 Marks)
Question 9 (Synthesis II)
(Max = 5 Marks)
Question 10 (Predict Product I) (Max = 5 Marks)
Question 11 (Predict Product II) (Max = 6 Marks)
Question 12 (Spec Problem)
TOTAL
(Max = 5 Marks)
(Max = 60 Marks)
Good Luck!
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CHEM 2210 Final Exam, December 9, 2013
Question 1: Stereochemistry (5 Marks total)
Question 1A (2 Marks):Lovastatin is a cholesterol-lowering drug that is produced by fermentation
cultures of the fungus Aspergillus terreus. In the box provided please assign the absolute
configuration for each of the indicated stereogenic centres. Each stereogenic center is worth 0.5
marks.
HO
O
O
O
O
Question 1B (1 Mark): Shown below is the Fisher Projection for D-galactose. Please assign the
absolute configuration for the indicated stereogenic centers. Each stereogenic center is worth 0.5
marks.
H
HO
HO
H
CHO
OH
H
H
OH
CH2OH
Question 1C (2 Marks): In the space below please draw the wedge and dash representation of
the following Newman projection. Be certain to indicate the correct stereochemistry.
H
H
CH 3
H
CH2CH 3
CH2OH
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CHEM 2210 Final Exam, December 9, 2013
Question 2: Spectroscopy I (4 Marks total) Please examine the spectroscopy questions below
and place your answers in the spaces provided. There is no need to show your work.
Question 2A:(2 Marks)
Molecular formula: C10H10O2
H – NMR: δ 2.82 (6H, s), 8.13 (4H, s)
13
C – NMR: δ 197.0, 141.1, 128.7, 26.6
IR: 1681 cm–1, no OH stretch
(HINT: This question is from Sapling)
1
Question 2B: (2 Marks)
Molecular formula: C7H16O
H – NMR: δ 1.5 (q), 1.4 (broad singlet, exchanges with D2O), 0.9 (t)
13
C – NMR: δ 80.3, 31.8, 10.0
IR: broad OH stretch
(HINT: This question is from Sapling)
1
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CHEM 2210 Final Exam, December 9, 2013
Question 3 Conformations I (5 Marks): Please draw both chair conformations of cis–
1,4-diethyl-cyclohexane. Indicate in the appropriate boxes the energy difference
between chairs, and estimate the value of the equilibrium constant (Keq). Be certain to
draw clearly and neatly – poorly drawn chairs or substituents that are not clearly axial
or equatorial will not be graded.
Energy Difference
between Chairs
Equilibrium constant
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CHEM 2210 Final Exam, December 9, 2013
Question 4 Conformations II (5 Marks): Please use the space below to explain the results
shown in the box. Please provide an explanation in the second box below. Please be certain to
draw clearly and neatly – poorly drawn chairs or substituents that are not clearly axial or
equatorial will not be graded.
OH
Br
OH
NaH
O
SLOW
t-Bu
NaH
O
Br
Very FAST
Explanation:
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CHEM 2210 Final Exam, December 9, 2013
Question 5 Energy Diagram (5 Marks) The
H
Br
H
Br
bromination of benzene proceeds via the
Br
mechanism shown below. In the first step a
positively charged bromine atom is attacked by
the electrons of the benzene ring to form a
carbocation intermediate. In the final step a
A
B
C
proton is lost to restore the aromatic ring.
Please calculate the overall enthalpy change for this reaction given that the C–Br B.D.E. is 330
kJ/mole. In the space provided below please show an energy vs reaction co-ordinate diagram for
the reaction. Feel free to use the letters beneath each structure to indicate their position on the
energy diagram.
Enthalpy 1 Mark
ΔH° =
Energy Diagram = 4 Marks
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CHEM 2210 Final Exam, December 9, 2013
O
Question 6 Mechanism I (5 marks):
N H
C
In the space provided below please draw the detailed
NH 2
H 2O
reaction mechanism that shows all of the steps in the
hydrolysis of a nitrile to an amide. Be certain to clearly indicate all necessary protonation and
deprotonation steps. You must use good curved arrow notation to receive full credit.
(HINT: all of the steps of this reaction are consistent with the mechanisms that we have
previously covered in class)
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CHEM 2210 Final Exam, December 9, 2013
O
Question 7 Mechanism II (5 marks):
In the space provided below please draw the detailed
reaction mechanism that shows all of the steps in the
reaction indicated below. Be certain to indicate all
protonation and deprotonation steps. You must use
good curved arrow notation to receive full credit.
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OH
H
H 2O
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CHEM 2210 Final Exam, December 9, 2013
Question 8 Synthesis I (5 marks):
OH
In the space provided below please show
the sequence of substitution reactions that
are necessary to accomplish the synthesis
shown. A detailed mechanism is not
necessary. However, you must show all
required reagents and solvents for each synthetic transformation.
Question 9 Synthesis II (5 marks):
In the space provided below please show
the sequence of substitution reactions
O
that are necessary to accomplish the
synthesis shown. A detailed mechanism is
not necessary. However, you must show
all required reagents and solvents for each synthetic transformation.
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OH
CH 3
CN
CN
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CHEM 2210 Final Exam, December 9, 2013
Question 10 Predict Product I (5 Marks – 1 Marks for Each Box):
For the questions below please indicate the expected major product for reaction of each of the
molecules with a single molecule of ethyl bromide.
NH 2
Br
DMSO
N
H
Br
OH
H 2N
DMSO
NH 2
NH 2
Br
DMSO
O
Br
N
H
OH
DMSO
OH
Br
DMSO
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CHEM 2210 Final Exam, December 9, 2013
Question 11 Predict product (6 Marks Total – 2 marks each):
Mg
Br
then
1.
H 2O
2. TsCl/py
3. NaCN/DMSO
O
1. TsCl/py
2. O DMSO
OH
CH 3
O
3. dilute HCl
OH
HBr
H 2O
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CHEM 2210 Final Exam, December 9, 2013
Question 12 (5 marks):
In the space provided below please provide a structure that is consistent with the spectroscopic
data that is provided on the attached page
The End!
Happy Holidays
Everyone!
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