CHEM 233, Fall 2015
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FIRST NAME
Final Exam
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Ian R. Gould
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H
Interaction Energies, kcal/mol
He
Li Be
B
N
O
F
Ne
Na Mg
Al Si P
S
Cl
Ar
H/H
~1.0
Me/Me
~0.9
Ga Ge As Se Br
Kr
H/Me
~1.4
Et/Me
~0.95
In Sn Sb Te I
Xe
Me/Me
~2.6
i-Pr/Me
~1.1
Tl Pb Bi Po At
Rn
Me/Et
~2.9
t-Bu/Me
~2.7
K
Ca
Sc Ti V
Cr Mn Fe Co Ni Cu Zn
Rb Sr
Y
Zr Nb Mo Tc Ru Rh Pd Ag Cd
Cs Ba
Lu Hf Ta W
Re Os Ir Pt Au Hg
small range
range of values
broad peak
C
O H
C N
N H
C O
C
3300
C
O
N
C
R
R
C
O
CH
NH2 variable and condition dependent,
ca. 2 - 6 δ
OH
O
C
2200
11
220
O
10
200
R C OH
X
–H2C
8
160
7
140
6
120
R C
O
R2C
C
Aromatic
1650 1500
NR2
O
C CH3
–H2C NR2
C CH2
9
180
C
1710
–OCH2–
NMR Correlation Charts
R C OH
(δ, ppm)
O
2000
2500
Aromatic Ar H
mainly 8 - 6.5
H
1600
C
3000
O
1680
1735
2850–2960
O
C O H
3500
OR
2200
H
broad ~3000
(cm-1)
N
C
C
broad with spikes ~3300
broad ~3300
1600–1660
O
2720–2820
2 peaks
C
O H
C
C
H
H
H
3000–
3100
N H
Infrared Correlation Chart
usually
strong
O
C H
Gauche
Eclipsing
CR2
5
100
N
4
80
Alkyl
C CH
3
60
–OCH2–
3° > 2° > 1°
2
40
1
20
Alkyl
RC
CR
C
C NR2
X
3° > 2° > 1°
0
0
-2-
CHEMISTRY 233, Fall 2015 FINAL EXAM
NAME
Question 1 ( 13 pts.) Give the IUPAC name for the following structure. Specify
stereochemistry as appropriate.
(E)
(S)
H
6
7
5
4
8
3
1
(6S)-ethyl-3,7-dimethyl-(3E)-octene
2
longest chain that contains the most substituents
and functional groups
Question 2 (24 pts.) For the following substituted cylohexane, draw both chair conformations
and indicate which chair would allow an E2 elimination reaction, and which is the lowest energy
chair (these may be the same or they may be different), AND, give the alkene product expected
for E2 elimination (assume a strong bulky base).
Et
Me
Br
Et
Me
Br
Me
Et
Br
reactive chair
Et
Me
lowest energy
chair 2
chair 1
alkene product
Extra Credit (5 pts) The hole in the ozone layer has been attributed to the atmospheric
chemistry of which kind of molecule?
halides
alkanes
alkanes
alcohols
-3-
CHEMISTRY 233, Fall 2015, FINAL EXAM
Question 3 (42 pts.)
a) Give a curved arrow mechanism for BOTH the SN1 and the E1 reactions for the provided
bromide in hot methanol. Label the Lewis acid and Lewis base in each step as appropriate, and
whether they are also Brønsted acids and bases. Indicate the number of sets of intermediates and
transition states for each mechanism
clearly indicate the rate determining step for BOTH reactions!!
OMe
MeOH
Br
heat
rate
detn
step
Me LB
O H
Me
O H
Me
O H
H
H
LB/BB
3 sets of intermediates
4 transition states
LA/BA
LA
MeOH
Br
heat
rate
detn
step
Me
H
H
LB/BB
2 sets of intermediates
3 transition states
O H
LA/BA
b) Draw a properly labelled reaction energy diagram for BOTH of these reactions on the SAME
diagram, clearly indicate which reaction is which. Assume that both reactions are exothermic
under the conditions, but your diagram should clearly show which reaction you expect to be more
exothermic, do not draw any transition states, but INDICATE THEIR POSITIONS
Clearly indicate the activation energy for the rate determining step and the exothermicity for
each reaction.
‡
Energy
‡
Ea RDS
same for both
‡
SN1
‡
‡
SN1
exothermicity
E1
E1
exothermicity
Reaction Coordinate
-4-
CHEMISTRY 233, Fall 2015 FINAL EXAM
NAME
Question 4 (14 pts). For the reaction below.
a) Use the sum of the bond dissociation energies provided to determine the reaction enthalpy, ΔH
b) Classify the reaction as addition, elimination, substitution or rearrangement
c) Is the reaction exothermic or endothermic, give a BRIEF EXPLANATION
O
O H
C-O (π) BDE = 87 kcal/mol
C-C (π) BDE = 65 kcal/mol
H
C-H BDE = 98 kcal/mol
O-H BDE = 110 kcal/mol
Sum = 185 kcal/mol
Sum = 175 kcal/mol
ΔH =
-10 kcal/mol
the reaction is
rearrangement
the reaction is exothermic, the energy cost associated with breaking the bonds is lower than the
energy release by making the bonds
Question 5 (14 pts.) Give the hybridization of the OXYGEN atom in the provided structure, and
show that you understand the meaning of the hybridization assignment by making a small table
that summarizes all of the valence hybrid (and any unhybridized) atomic orbitals associated with
this oxygen atom, and state how they are used (e.g. used to make a specific kind of bond to a
specific atom, etc.)
H
O
sp2 - s-bond to hydrogen
sp2 - s-bond to carbon
C
sp2 - non-bonding electrons
H3C
CH3
p - π-bond to carbon
the oxygen is sp2 hybridized
Question 6 (24 pts.) Give the major organic product of the following reaction of halide A
Na+ –OCH3
acetone
Cl
CH3O
A
Comparing A with the halides B - D, rank all four halides in order of increasing rate of reaction with
Na+ –OCH3 in acetone. Justify your choice with a brief explanation that includes an assignment of
the reaction mechanism.
Br
Cl
slowest
D
<
A
<
B
Br
C
B
<
C
D
fastest
This is an SN2 reaction. Substitution will be slowest for D because the bromine is bonded to
an sp2 carbon, and fastest for C because the bromine is in an allylic position. Both A and B
will be slower than C because chloride is a poorer leaving group than bromide, and B will be
faster than A because the chlorine is allylic.
CHEMISTRY 233, Fall 2015 FINAL EXAM
-5-
NAME
Question 7 (80 pts.) Give the missing major ORGANIC PRODUCT for each reaction
a) Show all stereochemistry as appropriate, identify any MESO compounds
b) Briefly explain whether and why a solution of the product would be optically active or not
c) assign each reaction as addition, elimination, substitution or rearrangement
OH
1. Hg(OAc)2/H2O
(±) addition
a)
2. NaBH4
not optically active, racemc
Br
Br
1 Equiv. Na+ –OH
b)
Br
substitution
optically active
single enantiomer
formed
DMSO
OH
1. BH3 . THF
c)
Ph
OH
not optically active, racemc
Ph
Br2
d)
Ph
CCl4
addition
(±)
2. –OH, H2O2
addition
Br Br Ph
not optically active, meso
Br
HBr
e)
addition
CCl4
not optically active, achiral
Cl
f)
Br
Cl
1 Equivalent
K+ –O-t-Bu
optically active
single enantiomer
formed
DMF
elimination
H2
g)
not optically active,
meso
addition
Pd/C
Br2
h)
hν
Br
not optically active,
achiral
substitution
-6-
CHEMISTRY 233, Fall 2015, FINAL EXAM
NAME
Question 8 (20 pts.) For each reaction, decide whether the mechanism would be SN1/SN2/E1 or
E2 and give a brief explanation. Draw the product and show stereochemistry where appropriate
and state whether a solution of the product would be optically active or not AND give a
BRIEF explanation.
OEt
Br
EtOH
a)
(±)
heat
NOT optically active, racemic mixture
allylic (actualy benzylic), weak nucleophile/base, polar aprotic solvent, E1 not
possible, must be SN1
Me
Br
1 Equiv. t-BuO– +K
b)
DMF
Cl
Cl
optically active, single enantiomer
1 Equiv, Br better leaving group, bulky base with tertiary halide gives AntiSayetzeff alkene product, still has one chiral center
Question 9 (42 pts.)
a) draw a Lewis structure with wedged/dashed bonds for (1S)-chloro-(1,2S)-diphenylpropane
b) for the Lewis structure that you drew, draw a 3D/sawhorse structure AND a Newman
projection of the lowest energy conformation for rotation around the C1-C2 bond,
LOOKING FROM C1 to C2
c) for the Lewis structure that you drew, draw a 3D/sawhorse structure AND a Newman
projection of the conformation the molecule must achieve in order to undergo an E2
elimination, LOOKING FROM C1 to C2
d) give the alkene product of E2 elimination for the conformation you drew in part c)
H
a)
b)
C
Me
Cl
C
c)
H
Ph
b)
Cl
Ph
Me
H
sawhorse/3D structure
in the conformation required
for E2 elimination
H
Newman projection
in lowest energy conformation for
rotation around the C1-C2
Ph
c)
C
Cl
Ph
sawhorse/3D structure
in lowest energy conformation for
rotation around the C1-C2
Lewis
structure
Me
H
Ph
H
H Ph
Me
C
Cl
Ph
Ph
H
Ph
Cl
H
Me
Ph
Ph
d)
H
Ph
H
Newman projection
in the conformation required for
E2 elimination
E2 elimination
alkene product
-7-
CHEMISTRY 233, Fall 2015 FINAL EXAM
NAME
Question 10 (22 pts.) Give a curved arrow mechanism for the following reaction. Label the
Lewis acid and Lewis base in each step as appropriate, and whether they are also Brønsted
acids and bases. Indicate the number of sets of intermediates and transition states for your
mechanism but do not draw any transition states.
O
H
O
S
O
H
O
LA/BA
HO
LB/BB
conc. H2SO4
heat
O
O
O
LA/BA
H
H
H2O
S
H
O
LB/BB
LA/BA
number of sets of intermediates
number of transition states
3
4
Question 11 (16 pts) The 3D structure of (+)-hydroxycitric acid is shown below (non-bonding
electrons omitted for clarity).
a) Indicate the positions of all chiral/asymmetric carbons with the * symbol
b) Give the absolute configuration for all chiral/asymmetric carbons
c) Draw a 3D structure for (–)-hydroxycitric acid and give the absolute configuration for ALL
chiral/asymmetric carbons
HO
HO2C
*
R
H
OH
CO2H
*R
CO2H
HO
HO2C
*
S
H
OH
CO2H
*S
CO2H
-8-
CHEMISTRY 233, Fall 2015 FINAL EXAM
NAME
Question 12 (20 pts.) For the following two reactions, indicate which would be faster and draw
a reaction energy diagram for both on THE SAME DIAGRAM (normalize them at the reactants)
and give a BRIEF explanation that includes the type of mechanism that is operating
(SN1/SN2/E1/E2 etc.). Your discussion MUST mention the Hammond postulate. Indicate the
positions of the transition states and teh activation energies for both reactions and draw
the transition state for reaction A only!
Br Na+ –OMe
Br Na+ –SMe
OMe
SMe
A
B
DMF
DMF
‡
Relative
Energy
‡
EaA
‡
OMe
B
EaB
Br
A
reaction coordinate
these are SN2 reactions, strong nucleophiles and a polar aprotic solvent, reaction B is
slower because nucleophlicity and basicity parallel in an aprotic solvent and O is
smaller and more elctronegative than S (makes stronger bonds), -OMe is a stronger
base and stronger nucleophile, the -OMe reaction is more exothermic, faster and
according to the Hammond postulate has an earlier transition state
Question 13 (20 pts) For the C-Ha and the C-Hb bonds in the structure below, add the curved arrow
pushing and give the products of homolytic bond dissociation, include all resonance contributors
as appropriate. Give an explanation for which would have the LARGER and which the SMALLER
bond dissociation energy, C-Ha or C-Hb, that includes the term "energy of the electron(s)"
Hb
Ha
Hb
Hb
Ha
break C-Ha
Hb
Ha
break C-Hb
Hb
Ha
Ha
Ha
dissociation of C-Hb generates a radical that is more stabilized by resonance, the energy of the
non-bonding electron is lower in the radical from Hb, it takes less energy to form, the BDE C-Hb
is lower than that for C-Ha
-9-
CHM 233, Fall 2015, Final Exam
NAME
Question 14 (24 pts) Provided are spectra for a compound with molecular formula C6H12O2
degree of unsaturation
a) Give the degrees of unsaturation 1________________
b) On the infrared spectrum, indicate the peaks that identify the functional groups in the molecule
(including C(sp3)-H). Indicate BOTH the functional group, and where appropriate, the specific
BOND in the functional that corresponds to the peak.
2301
2707
3442
1663
2826
sp3
H C
1470
2936
O
2969
1362
1370
OR
1064
1728
1225
c) draw the structure and clearly indicate which hydrogens correspond to which
signals in the proton nmr spectrum (only)
a
3H d
a
O
d
H3C
CH3 b
CH
C
O
a
CH3
c
CH2
note compressed
scale here
c
2H
b
1H
(multiplet)
6H