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CHEMISTRY 343
05/15/2007
Final Exam
D. LEE
PRINT NAME and STUDENT ID: _______________KEY____________________
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Problem 1
(10 points)
_________
Problem 2
(10 points)
_________
Problem 3
(15 points)
_________
Problem 4
(10 points)
_________
Problem 5
(15 points)
_________
Problem 6
(15 points)
_________
Problem 7
(40 points)
_________
Problem 8
(25 points)
_________
Problem 9
(15 points)
_________
Problem 10
(20 points)
_________
Problem 11
(10 points)
_________
Problem 12
(15 points)
_________
(200 points)
_________
Total
1. (a) (4 points) Diazomethane (CH2N2) is a source of a reactive entity (:CH2) called
carbene by extruding N2, which has only six electrons on the carbon and thus reacts
with an alkene to form a cyclopropane. Provide the best Lewis structure of
1
diazomethane by adding appropriate atoms, bonds (-bonds if any), lone pair
electrons, and appropriate formal charges at all atoms. Write another resonance
structure based on the structure you have completed.
H
H
H
C N N
C N N
C N N
H
H
H
very minor contributor
(b) (3 points) Choose the resonance structure that can easily explain the cleavage
of C–N bond to generate a carbene in (a) and briefly explain your choice.
H
H
C
C N N
+
This is resonance structure that can undergo
C–N single bond cleavage form to generate
carbene (:CH2) and N2 more easily than C=N
double bond in the other resonace .
N N
H
H
(c) (3 points) Use two resonance structures you have in (a), and indicate the
hybridization state of all atoms in both resonance structures. Considering the
hybridization, predict the shape of diazomethane (planar, linear, tetrahedral, bent,
trigonal pyramidal and etc.) and justify your prediction.
H
H
C N N
H
SP3
Because the central nitrogen is
sp-hybridized in both resonance
forms, diazomethane should be
linear molecule.
C N N
H
SP
SP2
SP2 SP
2. (10 points) From erythromycin A below, name the indicated functional groups.
Ketone
O
H3 C
CH3
OH
3° Amine(Amino group)
OH
H3 C
OH
H3C
H3 C
C
H2
Ester
H3 C
CH3
O
N
HO
CH3
O
O
CH2
O
O
OCH3
CH3
CH3
Ether
OH
O
2° Alcohol (Hydroxyl group)
CH3
3. (15 points) There are many isomers with the formula C7H10O2. Draw specific
examples that incorporate the following restrictions:
(a) (5 points) What is the Index of hydrogen deficiency? < 3 >
2
(b) (5 points) An ester with a quaternary carbon.
O
O
O
or
O
or
O
O
O
or
O
O
O
O
or
O
or
etc.
(c) (5 points) An unsaturated carboxylic acid.
O
O
O
or
or
OH
OH
OH
etc.
4. (10 points) Number the following compounds in order of increasing acidity (1 being
least acidic).
O
H
OH
H
3
N
HCl
H
H
H
5
1
OH
2
4
5. (15 points) Draw the structure of the following compounds. Use perspective
drawings
(i.e. for cyclohexanes show the conformation clearly.)
(a) (5 points) 1,4-dibromocyclohexane having molecular dipole moment.
Br
Br
Br
Br
Br
Br
(b) (5 points) the most stable conformation of trans-1,2-dimethylcyclohexane and its
ring-flipped structure
ring flip
(c) (5 points) What is the energy difference between the two conformations in (b)?
There is “three gauche interaction difference” between the two
conformations, thus the energy difference is 3 x 3.8 kj/mol = 11.4 kj/mol.
6.
(15
points)
(S,R)-1-Bromo-2-methylcyclohexane
is
subjected
to
dehydrohalogenation reaction. Predict the product(s) by drawing a reasonable transition
state. If there is more than one product please indicate which is the major and the minor
isomer.
3
CH3
Ha elimination
Major
Br
CH3
Br
CH3
Ha
Hb
Minor
Hb elimination
7. (40 points) Starting from 1-methyl-1-cyclohexene, a tertiary alcohol, a secondary
alcohol, and diols can be synthesized. With the indicated specific reagents and the
reaction conditions, draw unambiguously the product of each transformation (the
stereochemistry needs to be specified for full credit).
Hg(OAc)2, H2O-THF
then NaBH4, HO-
OH
BH3:THF
H
then H2O2, OH-
OH
OsO4, pyridine
OH
then NaHSO3, H2O
OH
+ enantomer
OH
MMPP (or mCPBA)
then H3
+ enantomer
+ enantomer
O+
OH
8. (25 points) The ketone shown below can be made from bromobenzene and
ethylene. Please show your synthesis using Grignard reagents as the key
component (hint: different types of transformations including epoxidation and two
different Grignard reagents are required).
4
BH3
ethylene
PBr3
HO
Mg
Br
ether
H2O2, OH-
BrMg
ether
MMPP
bromobenzene
Mg, ether
O
H2SO4
MgBr
heat
MMPP
O
or PBr3 then
O
then PCC
OH
NaOEt
BH3
ethylene
PBr3
HO
Br
ether
H2O2, OH-
Mg
BrMg
ether
O
bromobenzene
Mg, ether
MMPP
Jones or
PCC
MgBr
PCC
O
OH
O
OH
9. (15 points) Stereospecific reactions are important tools in organic chemistry. To
show this point, please specify reaction conditions for the conversion of the given
chiral primary alcohol to two different enantiomers of the ether shown below.
NaH, EtBr
H
D
R
O
H
D
R
OH
1.
TsCl, pyr
D
H
S
O
2.
NaO
10. (20 points) Allylic substitution reaction with NBS (N-bromosuccinimide) is very
powerful synthetic tool because it can increase the oxidation state of hydrocarbons
under mild conditions. From the reaction below, four major monobrominated products
(two pairs of diastereomers) are expected. Draw these structures with unambiguous
indication of stereochemistry.
5
O
Br
Br
N Br
O
(NBS)
peroxide
60 °C
Br
Br
11. (10 points) The transformation above involves a facile formation of allylic radical.
Sketch the molecular orbitals of the parent allyl radical. Indicate the energy levels
of the bonding, nonbonding and antibonding molecular orbitals relative to an isolated
p-orbital on carbon with appropriate number of electrons(s).
Node
energy
Antibonding MO
H
H
C
H
C
C
H
Three isolated
p orbitals
Node
H
Nonbonding MO
Allyl Radical
Bonding MO
12. (a) (10 points) Provide an expected structure with the stereochemistry of an endo
Diels-Alder product.
6
CO2Me
H
110 °C
H
+
CO2Me
toluene
CO2Me
CO2Me
(b) (5 points) Use -molecular orbitals (HOMO of a diene and LUMO of a
dienophile) to show how the above Diels-Alder reaction proceeds to give the
expected product.
HOMO of diene
H
H
MeO2C
MeO2C
H
H
MeO2C
H
MeO2C
H
LUMO of diene
7
CO2Me
CO2Me