1.
Rank the following compounds in the trend requested. (15 points each)
a.
Rank by energy of the lowest unoccupied molecular orbital (LUMO). The molecule with
the LUMO that is lowest in energy is 1, while the compound with the LUMO that is highest in
energy is 5.
NO2
3
1
NH2
O
4
2
5
b.
Rank by rate of SN1 reaction. The compound that will react the fastest in a SN1 reaction
is 1, while the compound that reacts the slowest in a SN1 reaction is 5.
Br
Br
Br
Br CF3
3
4
Br
1
5
2
c.
Rank by rate of electrophilic aromatic substitution reaction. The compound that reacts
the fastest in an electrophilic aromatic substitution reaction is 1, while the compound that reacts
the slowest is 5.
O
NH
NH2
O
O
O
5
2
3
4
1
2.
Consider the molecule cytosine shown below which is one of the four main bases found
in RNA and DNA structures. Using the numbering shown, answer the following questions.
N1
NH2
N2
N
N
H
N3
a.(6)
O
What is the hybridization of each nitrogen atom?
__sp2__
N1
__sp2__
N2
__sp2__
N3
b.(6) Each nitrogen atom has a lone pair of electrons. Which orbital is each lone pair of
electrons held?
__p__
N1
__sp2__
N2
__p__
N3
c.(4)
How many electrons are conjugated in the cyclic ring?
6
The ring has a large resonance contributor to allow aromatic stability.
NH2
N
N
H
O
d.(6) Rank by ability to protonate the lone pair of electrons in acidic conditions. The nitrogen
that is easiest to protonate is 1, while the nitrogen that is hardest to protonate is 3.
__2__
N1
__1__
N2
__3__
N3
N3 is hardest to protonate as this would make ring nonaromatic and lose aromatic stability.
Protonation of N1 would lose resonance stability with lone pair with ring. N2 does not lose any
extra stability by protonation and thus is easiest to protonate.
3.
Consider the molecule 1,3,5-hexatriene.
a.(6) Below are shown the six molecular orbitals for the π system of hexatriene. Write a
number corresponding to the energy level for each molecular orbital. Write a 1 under the MO
that is lowest in energy and a 6 under the MO that is highest in energy.
2
5
1
6
4
3
HOMO
HOMO
HOMO
HOMO
HOMO
HOMO
LUMO
LUMO
LUMO
LUMO
LUMO
LUMO
As the number of nodes increases, the energy increases.
b.(6) In the space above, circle HOMO under the orbital that corresponds to the highest
occupied molecular orbital and LUMO under the orbital that corresponds to the lowest
unoccupied molecule orbital.
There are 6 π electrons, thus MO3 is the HOMO and MO4 is the LUMO.
c.(5) Would hexatriene react preferentially in a pericyclic reaction with ethylene or butadiene
under thermal conditions? The HOMO and LUMO molecular orbitals for ethylene and
butadiene are shown. Circle the reaction that would occur thermally.
!
!
LUMO
LUMO
HOMO
HOMO
The HOMO of hexatriene is symmetric, thus the LUMO of the reacting pair must also be
symmetric for a symmetry allowed reaction.
d.(5) What conditions would be required for hexatriene to react in a pericyclic reaction with
the other reagent (either ethylene or butadiene which did not react in part c)?
Need to change symmetry of HOMO of hexatriene, thus if compound is photolyzed an electron
is promoted to change the symmetry of the HOMO.
4.
Consider the reaction of 3-nitrotoluene with Br2/FeBr3.
CH3
Br2, FeBr3
NO2
5-position
4-position
a.(6) Draw the arenium ion (sigma complex) that result when the reaction occurs at the 4position. Include all relevant resonance structures.
CH3
CH3
CH3
NO2
NO2
NO2
H Br
H Br
H Br
HIGHEST
b.(6) Draw the arenium ion (sigma complex) that result when the reaction occurs at the 5position. Include all relevant resonance structures.
CH3
CH3
H
Br
c.(5)
CH3
H
NO2
Br
H
NO2
Br
NO2
Circle the resonance form indicated in either part a or part b which is lowest in energy.
Resonance form is the only one that places positive charge adjacent to electron donating methyl
d.(5) Write the word HIGHEST under the resonance form in either part a or part b which is
highest in energy.
Resonance form is the only one that places positive charge adjacent to electron withdrawing nitro
e.(4) Should the reaction shown be faster or slower if 4-nitrotoluene was reacted instead of 3nitrotoluene? Circle the starting material (3-nitrotoluene or 4-nitrotolune) that reacts faster.
3-nitrotoluene
4-nitrotoluene
Arenium ion intermediate for 4-nitrotoluene would not place positive charge adjacent to nitro
5.
Indicate the preferred product for the following reactions. Assume proper work-up for
each step. (7 points each)
a.
KMnO4
HO2C
NaOH
b.
NBS, h!
Br
c.
HO
1) PhCO3H
2) CH3NH2
NHCH3
d.
NO2
NO2
e.
CN
CN
f.
N
S
Br
S
B(OH)2
Pd(PPh3)4
K2CO3
N
g.
OCH3
OCH3
HNO3, H2SO4
NO2
h.
Br
Br2, FeBr3
N
N
i.
Cl
O
O
Cl2, AlCl3
O
O
O
O
O
O
j.
Br
Br
F
NaOCH3
OCH3
O2N
O2N
k.
O
1) Br2, FeBr3
2)
O
Br
Cl
AlCl3
l.
O
HI
!
OH
I
6.
Indicate a method to synthesize the products requested in the following examples. You
may use any organic or inorganic reagents you desire, but you must begin with the starting
material given and synthesize the product requested. More than one step will be required. (15
points each)
a.
Synthesize paracetamol (also called acetaminophen) starting with phenol
OH
H
N
?
O
HO
OH
HNO3
H2SO4
b.
OH
OH
Fe, HCl
O
Cl
O 2N
H2N
N
H
Starting with benzene, synthesize the compound shown
Br
?
O
O
Cl
AlCl3
O
Br2, FeBr3
Br
OH
O
NH2NH2
KOH
Br
7.
Benzene was reacted as shown in the scheme below. The 1H NMR of intermediate A is
shown with overlapping peaks between 7.0-7.5 ppm, which in total integrate to 5. The 13C NMR
for intermediate B displays 6 peaks at ppm values of 150, 131, 127, 119, 34 and 31. Indicate the
structures for compounds A, B and C. Assume proper work-up for each step. (7 points each)
Br2
Cl
A
KMnO4
B
AlCl3
C
FeBr3
NaOH
CO2H
Br
__________
A
Br
__________
B
__________
C
3
5
3
2
1