Name
ID No
Chem 340, Organic Chemistry I, Spring 2003
Final Exam
This is a closed-book exam. No aid is to be given to or received from another
student. Model set and calculator may be used. Please write your name and WSU ID
number on each page (including this page) at the top right corner. The exam time is 2
hours. Hand in your exam papers to the proctor at the end of the test. Good luck and
have fun.
Thank you for taking Chem 340.
For grader only:
Checked:
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Total
Page 1 of 11
Name
ID No
Question 1. (11 points)
The structure of methionine, an essential amino acid, is shown below.
HO2C
S
Methionine
NH2
a.
Draw the structures for the S-isomer. Clearly show the absolute chiral configuration.
1
H2N
4
H
3
S
2CO2H
b.
In term of stereochemistry, how are R- and S-methionine related to each other?
Enantiomer to each other.
c.
There are two bottles of methionine in your pharmacy, one contains the racemic
methionine and the other one contains the S-isomer. Unfortunately, they are not labeled.
Please devise a test to unambiguously identify which bottle contains the S-isomer.
Measure optical rotation. The S-isomer is optical active, but the racemic methionine is
not.
d.
Among the following properties, which are the same for the two samples in part c?
Properties
Same
Different
Melting point
X
Density
X
Molecular weight
X
Taste
Probably different
Page 2 of 11
Probably different
Name
ID No
Question 2 (11 points)
Constitutional isomers are compounds that have the same molecular formula, but differ in the
way the atoms are connected in three dimensional space. Please draw the structures for all
constitutional isomers that have the molecule formula of C3H6Br2. Name each structure
according to the IUPAC nomenclature system.
Br
Br
Br Br
Br
Br
Br
Br
1,1-dibromopropane
1,2-dibromopropane
2,2-dibromopropane
1,3-dibromopropane
13C: 3 signals
13C: 3 signals
13C: 2 signals
13C: 2 signals
Br
Br Br
2
2
2 1
1
Br 1
2
1
3
Br
Br
1H: 3 signals
1
Br
H
HH
Br
H
H
H
1H: 3 signals
1
H
H
2
3
3
Br
H
H
Br H H
2
3
1
Br
1
Br
1H: 1 signals
1H: 2 signals
Br Br
2
H H 1
1
H
H
H
H
Br Br
H
HH
H
HH
1H integration:
1H integration:
1H integration:
1H integration:
1, 2, 3 protons
1, 2, 3 protons
6 protons
2, 4 protons
1H: Signal from H3 is
1H: Signal from H3 is
a triplet
a doublet
In addition, the chemical shift values are different.
NMR analysis can be employed to distinguish the possible structures. Describe the expected
results from such experiments.
See above.
Page 3 of 11
Name
ID No
Question 3 (11 points)
Provide necessary reagents to complete the following transformation. Multiple steps may be
required.
Starting
Reagents and Conditions
Desired
Materials
Product as
the Major
Product
+
mCPBA
or peracid
+
OH
OH +
OH
OH
OH
OH
O + NaOH
H2O
OsO4
or
KMnO4
TsCl
(Tosyl chloride)
OH
OH
OH
OH
O Ts
Pd/C
or
Pt/C
Page 4 of 11
H2 +
+ t-BuOK
Name
ID No
Question 4 (11 points)
Please devise a synthetic route for isopropyl propyl ether shown below. All of the carbon atoms
in the final product should be derived from propene. Specify the reagents used. For each step of
your proposed synthesis, the material leading to the final product should be the major reaction
product for that step.
O
Many routes can lead to the desired product. The last step is the formation of ether bond
between an alcohol (nucleophile) and an electrophile. The key transformations are regioselective formations of terminal alcohol and secondary alcohol or halide. Only some examples
are shown. As long as your route leads to the desired ether as the major product, fully credit is
given.
+ HBr
Ethyl ether
1. BH3/THF
2. NaOH, HOOH
Br
Base
such as tBuOK
HO
O
Page 5 of 11
KO
Name
ID No
Question 5 (11 points)
Epoxides can undergo ring-opening reaction under either basic or acidic conditions. However,
different regioselectivities are observed under these two different conditions. Two examples are
shown below. Explain the regioselectivity. Use arrow formalism to illustrate your proposed
mechanism.
Hints: The reactions are substitution reactions. Epoxides are strained ethers, in essence.
O
+ NaOCH2CH3
O
+ CH3CH2OH
EtOH
OH
H2SO4 (trace)
30 min, 25 C
1.
(~80% yield)
O
5 h, 80 C
OH
(~80% yield)
O
Under basic condition, the epoxide oxygen (alkoxide) is a poor leaving group, so longer
reaction time and harsher condition is required. The reaction goes through an SN2 mechanism.
The observed regio-selectivity can be attributed to steric effects, as the terminal carbon is less
sterically hindered.
O
EtOH
+ :OCH2CH3
2.
O
5 h, 80 C
(~80% yield)
OH
Under acidic condition, the epoxide oxygen can be protonated to form an oxonium ion,
now the epoxide oxygen (an alcohol) is a very good leaving group, so shorter reaction time and
mild condition is required. The reaction goes through an SN1 mechanism. The observed regioselectivity can be attributed to higher stability of a tertiary carbocation compared to a secondary
carbocation. The oxonium ion is analogous to a bromonium ion in bromination reaction.
H+
O
H2SO4 (trace)
O
H
O
CH3CH2OH
Page 6 of 11
OH
OH
Name
ID No
Question 6 (11 points)
Allyl carbocation is considerably more stable than other primary carbocation, such as n-propyl
carbocation.
a.
Explain.
Resonance stabilization.
b.
Devise a mechanistic pathway that accounts for the following transformation. Be sure to
count carbons and use arrow formalisms when drawing the mechanism.
Allyl carbocation is very stable and iodide is a good leaving group. In addition, methanol
is a poor necleophile. For these reasons, it is very easy to form a allyl carbocation
intermediate via an SN1 mechanism. The resonance forms of allyl carbocation indicate
that two carbon atoms bear positive charges, thus both are electrophilic. As a result,
methanol oxygen can attack either position.
I
+ MeOH
Page 7 of 11
O
Name
ID No
Question 7 (12 points)
When the following chloride undergoes an E2 reaction, the major product formed is the less
substituted olefin (compound A), instead of the more substituted and stable olefin (compound B).
Based on stereochemistry and conformational analyses, explain the observation.
CH3
+ t-BuOK
CH3
t-BuOH
CH3
Cl
Compound A
(major product)
Compound B
(not detected)
E2 mechanism require the leaving group, Cl anion, and the proton being abstracted are either 0
or 180 to each other.
H
H
H
Cl
CH
Cl 3
H
none of the hydrogen is of
the right geometry for E2 reaction
H
H
H
H
CH3
the circled hydrogen, but not the boxed one, is of
the right geometry for E2 reaction
Page 8 of 11
Name
ID No
Question 8 (11 points)
The following bromide can under go both elimination and substitution reactions. Please show all
possible products for the mechanisms specified. Show the intermediate(s) if they exist.
Br
Reaction
Intermediates (if exist)
Major product
Condition
Minor
product(s)
and
Mechanism
hydride shift
form more stable
carbocation
in MeOH
SN1
O
O
H
NaI in
acetone
I
SN2
hydride shift
form more stable
carbocation
Et3N in EtOH
E1
H
t-BuOK in
EtOH
E2
Page 9 of 11
Name
ID No
Question 9 (11 points)
Thiolate is considered as an excellent nucleophile. As such, alkylation of thiolate can be carried
out in a nucleophilic solvent (the solvent reacts with the alkylation reagents as well). The
relative nucleophilicity of thiolate (RS-) and water is 126,000 to 1. When methyl iodide (0.1
mM) is incubated with sodium methanethiolate (NaSCH3, 2 mM) in water, two major products
are formed.
a.
Please draw the structures of the two products in the boxes.
CH3I + NaSCH3
H2O
S
Compound A
b.
+
OH
Compound B
Calculate the relative ratio of the two possible products. Since the nucleophiles are in
large excess to the alkylation reagents, the nucleophile concentrations are considered
unchanged during the course of the reaction.
Numbers you may need for the calculation: 1 mM equals 0.001 M. Water’s molecular
weight is 18 and density is 1 g/mL.
First, calculate molar concentration of water
1 L of water weights 1000 g
[water] = {1000g/(18 g/mole)}/1 L = 55.5 M
rate for A formation = k(RS-) [CH3I] [CH3S-]
rate for B formation = k(water) [CH3I] [water]
methyl iodide concentration changes during the reaction, but at any moment, identical amount is
available to react with the thiolate and water.
Ratio of [CH3SCH3]/[CH3OH] = {k(RS-) [CH3I] [CH3S-]}/{ k(water) [CH3I] [water]}
= {k(RS-) [CH3S-]}/{ k(water) [water]}= [126,000 x 0.002 M]/[1 x 55.5 M] = 4.61
Page 10 of 11
Name
Page 11 of 11
ID No