Ian R. Gould
the exam cover sheets look kind of like this
PRINTED
FIRST NAME
PRINTED
LAST NAME
ASU ID or
Posting ID
THIS IS A CHM 233 PRACTICE EXAM
Person on your LEFT (or Aisle)
Person on your RIGHT (or Aisle)
1__________/14 ........
FINAL EXAM
3__________/6 ........
PRACTICE TEST #1
4__________/22 ........
• PRINT YOUR NAME ON EACH PAGE!
2__________/27 ........
• READ THE DIRECTIONS CAREFULLY!
• USE BLANK PAGES AS SCRATCH PAPER
work on blank pages will not be graded...
5__________/14 ........
•WRITE CLEARLY!
6__________/44 ........
• MOLECULAR MODELS ARE ALLOWED
7__________/24 ........
• DO NOT USE RED INK
8__________/24 ........
• DON'T CHEAT, USE COMMON SENSE!
Total (incl Extra)________/175+5
Extra Credit_____/5
H
He
Li Be
B
N
O
F
Ne
Na Mg
Al Si P
S
Cl
Ar
Ga Ge As Se Br
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
small range
range of values
broad peak
Re Os Ir Pt Au Hg
O H
C N
N H
C O
C
~1.0
Kr
H/Me
In Sn Sb Te I
Xe
Me/Me
Tl Pb Bi Po At
Rn
Me/Et
H
C
H
N H
C
N
N
11
220 O
10
200
R C OH
H
~10
~8
H
1600
O
NR2
H
O
C CH3
–H2C NR2
7
140
6
120
5
100
R2C
Aromatic
CR2
C CH
4
80
3
60
RC
CR
Alkyl
3Y > 2Y > 1Y
2
40
–OCH2–
R C N
H
~15
C C
1500
C CH2
8
160
~2
H
C
–OCH2–
NMR Correlation Charts
H
H
1650
2000
~2
C C
O
Aromatic Ar H
mainly 8 - 6.5
O
C
~7
H
C C
1735
CH
2500
9
180
H H
H
–H2C X
(δ, ppm)
Approximate Coupling
Constants, J (Hz), for
1H NMR Spectra
OR
1710
3000
O
C H
~2.7
1680
C
2200
amine R NH2 variable and condition
alcohol R OH dependent, ca. 2 - 6 δ
O
R C OH
t-Bu/Me
C C
C
O
C O H
3500
~2.9
O
broad ~3000
(cm-1)
~1.1
H
C
2850–2960
broad ~3300
~0.95
i-Pr/Me
C
1600–1660
H
broad with spikes ~3300
O H
Et/Me
~2.6
C
2200
C
~1.4
O
2720–2820
2 peaks
3000–
3100
~0.9
Infrared Correlation Chart
H
C
3300
Me/Me
usually
strong
C
Gauche
Eclipsing
H/H
O
C H
Interaction Energies, kcal/mol
1
20
0
0
Alkyl 3Y > 2Y > 1Y
C X
C NR2
CHM 233 Final Exam : Practice Exam #1
-2-
NAME
Question 1. Give the IUPAC name for the following. Specify stereochemistry as appropriate.
H
a)
Br
R
1
2
H
3
R Br
b)
(1R,3R)-dibromocyclohexane
(5S)-bromo-(3S)-methylcyclohexene
Br
Question 2 On the following structures and for the orbitals indicated, roughly sketch the
shape of the wavefunctions ON TOP OF THE STRUCTURES, taking care to make the
locations of any nodes clear, and to explicitely state the A.O.'s used to make any M.O.'s.
H3C
C - sp3
Br - p
the C-Br s* anti-bonding orbital (where the
electrons "go to", THE LUMO in an SN2 reaction)
Br more electronegative, therefore anti-bonding M.O. larger on C
Br
C
H
H
H
C
H
H
H
H
C - sp3
H - 1s
C
H
one of the C–H s-bonding orbitals (i.e., where the
electrons are that determine the shapes of alkanes)
Question 3. Draw all reasonable resonance contributors, identify the major contributor and
draw the "actual" structure (use the δ symbol rather than trying to calculate accurate
fractional charges)
••
••
••
••
O
S O d
"actual" O d
••
••
••
••
••
••
O
all 3 are equal major
S O
••
••
••
••
the tosylate anion (–OTs)
S O
••
••
••
••
O
••
••
••
S O
O d
O
••
••
••
••
O
••
••
O
-3-
CHM 233 Final Exam : Practice Exam #1
Question 4. For the following two reactions A and B, give the full curved arrow-pushing
mechanisms, indicate the Lewis acid/base and Bronsted acid/base at each step as
appropriate and state which reaction would be faster and explain using a reaction energy
diagram that has BOTH reactions on the SAME diagram
Br
HBr
A
LB/BB
Br LB
SLOWER
Br
CCl4
H LA/BA
HBr
B
CCl4
Br
FASTER
LB/BB
H
Br
LA/BA
Br LB
LA
H
H LA
energy
tep
‡
HBr
‡
ed
etn
.s
rat
Ea(B)
Ea(A)
The rate determining step is the initial
protonation. Protonation of both alkenes
give the same cationic intermediate. The
alkene in B is disubstituted and thus
"starts" higher in energy than the
tetrasubstituted alkene in A. Ea(B) is thus
smaller than Ea(A).
+
Br–
Br
reaction coordinate
Question 5. In WORDS, describe the differences and similarities between a nucleophile and a
Lewis base. Give one example of a strong base that is NOT a strong nucleophile.
All nucleophiles are Lewis bases, both provide the electrons to make a new bond in a reaction
with a Lewis acid/electrophile
The definition of basicity is thermodynamic, stronger bases undergo more exothermic reactions
The definition of nucleophilicity is kinetic, stronger nucleophiles undergo faster reactions
the tertiary butoxide anion (t-BuO–) is a strong base but a weak nucleophile for steric reasons
CHM 233 Final Exam : Practice Exam #1
-4-
NAME
Question 6. Classify the two alkenes shown as enantiomers, diastereomers or the same
structure drawn a different way. Classify the provided reactions as addition, substitution,
elimination or rearrangement. Which reaction is more exothermic? Give a BRIEF explanation.
H
H2
addition
A
H
Pd/C
H
H
B
H2
addition
Pd/C
A is a trans-stereoisomer, B is a cis-stereoisomer, they are not teh same structure, B is
the higher energy isomer, thus reaction B is more exothermic
both reactions have the same product, thus any differences in exothermicity must be
in the reactants
A is trans-, B is cis-, B is the higher energy isomer, thus reaction B starts higher in
energy and is thus more exothermic
Question 7. Assign absolute configurations to all asymmetric centers in the following two
structures and determine whether they are enantiomers, diastereomers or the same structure
drawn differently. Identify any meso compounds.
Br
H3C
C
R
H
C
Br
meso
H
S
H
S
CH3
H3C
diastereomers
Br
C
C
CH3
S
Br
H
CHM 233 Final Exam : Practice Exam #1
-5-
NAME
Question 8. For the following reactions:
a) Give the missing major ORGANIC PRODUCT
b) Properly describe all stereoisomeric products if any and identify any MESO compounds
c) Explain whether a solution of the product(s) would be optically active and why or why not
d) Assign each reaction to addition, elimination, rearrangement or substitution
Ph
a)
achiral
CCl4
Ph
Br
Ph
=
addition
not optically active,
achiral meso compound
H2
b)
Ph
addition
not
optically
active,
*
*
*
*
achiral
meso
Br
Br
Ph
Br
compound
overall anti-addition (via backside attack on the bromonium ion)
Ph
Br2
Pd/C
H H
(CH3)3O– +Na
c)
Br
E2
acetonitrile
Na+ –CN
d)
Br
CH3
acetonitrile
SN2
NC
(±)
H2O
substitution
optically active
single enantiomer
chiral product
addition
not optically active,
racemic mixture formed
Br
Br2
e)
CH3
elimination
not optically active
achiral product
OH
1. Hg(OAc)2 / H2O
f)
addition
not optically active,
achiral product formed
OH
2. NaBH4
Ph
g)
Ph
1. Hg(OAc)2 / H2O
2. NaBH4
Ph
Ph
(±)
HO
H
addition
not optically active,
racemic mixture formed
CHM 233 Final Exam : Practice Exam #1
-6-
NAME
Question 9. Draw a reaction energy diagram for both reactions A and B ON THE
SAME DIAGRAM with properly labelled axes (normalize your energy diagrams at the
transition states), include a drawing of the transition state for the rate determining step
(it is the same for both reactions). Indicate the activation energy for both reactions and
indicate which is faster (A or B) and give a brief explanation for your choice.
CLASSIFY EACH REACTION as addition, elimination, substitution or rearrangement.
Cl
CH3O– +Na
elimination
–
+
CH
A
3O-H + Cl
CH3OH
SLOWER
Cl
elimination
CH3O– +Na
+ CH3O-H + Cl–
B
DMF
FASTER
Strong base, must be E2. The polar protic solvent in A strongly solvates (stabilizes) the CH3O–,
lowering its energy compared to the polar aprotic solvent in B, which does not solvate the base
as strongly. The negative charge in the transition state is "spread out" and is not strongly solvated
in either solvent. More energy is required to reach the transition state in A than in B, the products
are also lower in energy in the protic solvent since the chloride anion is also more strongly
solvated in the protic solvent
‡
Cl
Relative
Energy
OCH3
Reaction B
H
EaB
EaA
(normalized at ‡ to illustrate the fact
that the energy differences are
largest in the reactant and product
ions)
Reaction A
Reaction Coordinate
Question 10. Draw minor resonance contributors that show which of the carbon atoms A or B will have the
SMALLEST chemical shift in a carbon NMR spectrum, include curved arrow pushing and resonance
arrows and brackets. GIve a brief explanation.
A
B
NH2
NH2
carbon atom A will carry a partial negative charge as a result of resonance
delocalization, atoms are deshielded by removal of electron density, thus A is less
deshieled and thus has the SMALLER chemical shift
CHM 233 Final Exam : Practice Exam #1
NAME
-7-
Question 11 (22 pts.) For the following Bronsted acid/base equilibrium
a) Draw the curved-arrows that describes the bond making and breaking in BOTH directions
b) Label the STRONGER acid/base and the WEAKER acid/base on EACH side
c) Indicate which reaction would be faster, left to right or right to left
d) Indicate on which side the equilibrium will lie
e) Indicate which acid has the smaller and which the larger pKa
f) Give a BRIEF explanation for your choice of stronger/weaker Bronsted acids/bases that includes
drawings of ALL relevant resonance contributors
H
N
O
+
O
O
H
faster
H
N
N
slower
H
O
+
O
N
O
H
stronger base
stronger acid
smaller pKa
H
N
weaker acid
larger pKa
O
Equilibrium on
THIS side
O
weaker base
the stronger acid has the lower energy electrons in the conjugate base, the base
on the right hand side is more stable due to resonance stabilization
g) Draw a reaction energy diagram and include the activation energy is BOTH directions and a
drawing of the transition state
H
Energy
N
N
EaL>R
H
N
O
O
H
O
EaR>L
H
N
H
H ‡
H
O
O
O
‡
N
O
O
reaction coordinate
H
O
N
H
CHM 233 Final Exam : Practice Exam #1
-8-
NAME
Question 12.
a) Give a full curved-arrow pushing mechanism for the provided reaction, indicate the Lewis and
Bronsted acids/bases as appropriate and indicate the rate determining step.
H
LB/BB OH
O H
O
H2O
H
HCl (cat.)
LA/BA
rate
determining
step
H
H
OH
H LB/BB
O
H
O
LA/BA
b) For your mechanism:
2
give the number of sets of intermediates ______________
3
give the number of transition states ______________
c) Briefly describe your reasoning for your choice of rate determining step in your mechanism
the rate determining step is the slowest, it is the most endothermic, it is endothermic because 2
bonds are broken in this step and only one bond is formed, this requires energy
d) Draw a properly labelled reaction energy diagram that corresponds to the mechanism that you
drew. On the diagram, clearly indicate the activation energy for the rate determining step, and the
overall reaction exo- or endothermicity. Indicate the positions of the transition states but do not
draw them
‡
Energy
‡
‡
Eards
exothermicity
reaction coordinate
-9-
NAME
CHM 233 Final Exam : Practice Exam #1
Question 13. For BOTH reactions A and B
a) Give the products AND THE mechanisms for ELIMINATION of HCl for the provided
reagents/conditions, indicate the Lewis and Bronsted acids/bases at each BIMOLECULAR
step as appropriate
b) For each mechanism, give the number of sets of intermediates and transition states for each
H
Cl
H
CH3OH
A
O
Me LB/BB
H
heat
LA/BA
2
number of sets of intermediates ______________
3
number of transition states ______________
H
B
Cl
O
CH3OH
Me
LB/BB
H
heat
LA/BA
1
number of sets of intermediates ______________
2
number of transition states ______________
c) Draw a reaction energy diagram ON THE SAME DIAGRAM for each mechanism, clearly
indicate which reaction refers to which diagram, indicate the rate determining steps for each
reaction (you do not need to indicate or draw transition states)
Energy
reaction A
EaB
EaA
reaction B
Reaction Corodinate
d) Which reaction, A or B, would be faster? Give a BRIEF explanation.
this is an E1 reaction, the rate determining step is heterolysis of the C-Cl bond in the polar
protic solvent, this will be faster for B because it forms a 3° cation initially, whereas A forms an
initial 2° cation
CHM 233 Final Exam : Practice Exam #1
- 10 -
NAME
Question 14 Provided are spectra for a compound with molecular formula C9H10O
degrees of unsaturation
a) Give the degrees of unsaturation 5________________
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.
3412
3392
3051
3044
2730
O
2826
H
sp2
2899
H
H
2981
sp3
O
776
1529
1376
2991
1110
817
1198
1681
1598
c) draw the structure and clearly indicate which hydrogens correspond to which
signals in the proton nmr spectrum (only)
e
3H
triplet
b
b
c
O
H
a
b
c
CH2
d CH
e
a
1H
singlet
c
2H
doublet 2H
doublet
3
d
2H
quartet
-2-
CHM 233 Final Exam : START
OF Practice Exam #2
NAME
Question 1. Give the IUPAC name for the following. Specify stereochemistry as appropriate.
H
a)
H3C
C C
3 2
C C
CH3
5 4
1
H3CH2CH2C
CH3
8 7 6
1
b)
H
4,5-dimethyl-(2Z,4E)-octadiene
Cl
2
3
4
5
7
3-chloro-1-cyclohexyl-(2S,4)-dimethyl-(3Z)-heptene
6
H
Question 2.
a) Rank the following in order of increasing basicity in a polar APROTIC solvent. Give a
BRIEF explanation.
F
Cl
B
A
weakest
D
<
C
<
A
Br
I
C
D
<
B
strongest
I– forms weakest bonds, F– forms strongest bonds (note that nucleophilicity
depends upon solvent, but basicity much less so)
b) How would nucleophilicity vary for the same anions in a polar APROTIC solvent? Give a
VERY brief explanation.
basicity and nucleophilicity follow the same trends in a polar aprotic solvent,
iodide is the weakest nucleophile because it forms the weakest bonds
c) How would nucleophilicity vary for the same anions in a polar PROTIC solvent? Give a
VERY brief explanation.
basicity and nucleophilicity follow OPPOSITE trends in a polar aprotic solvent,
iodide is still the weakest base because it forms the weakest bonds, however,
in the protic solvent it is less solavted becvasue it is large, AND it can form
stronger partial bonds in the transition state because it is large
CHM 233 Final Exam : Practice Exam #2
-3-
Question 3) Give a curved arrow mechanism and a reaction energy diagram for the following
reaction. IIndicate LB, LA, BB, BA as usual and state the number of sets of intermediates and
transition states. Indicate the positions of the transition states on the diagram but do not draw them.
This mechanism is NOT IN THE NOTES, you need to figure it out using what you know about
Lewis/Bronsted acid base and other reactions that you have learned. The first step is given for you.
LA/BA
H
O
LB/BB
Cl
HCl
OH
EtOH
OEt
H
LA
O
OH
HO Et
LB
‡
H
‡
energy
Et
LB/BB
O
Et
LA/BA
OH
2 sets of
intermediates and 3
transition states
‡
reaction coordinate
Question 4) Below is an intermediate in an electropohilic aromatic substitution reaction we will
study next semester that gives a minor product. Draw all reasonable resonance contributors for
this ion and also an "actual" structure (use the δ notation for partial charge, do nt attempt to
assign absolute fractional charges). Indicate the major resonance contributor, if any, and give a
brief explanation.
H
H
H
Br
δ
δ
OH
OH
H
Br
Br
Br
δ
OH
"actual"
OH
all 3 are equal major, there is no reasonable way of getting the non-bonding electrons on the
oxygen involved in any resonance contributors
-4CHM 233 Final Exam : Practice Exam #2
NAME
Question 4)
a) Give a full curved-arrow pushing mechanism for the provided reaction, indicate the
Lewis and Bronsted acids/bases as appropriate and indicate the rate determining step.
b) Draw a properly labelled reaction energy diagram that corresponds to the mechanism
that you drew. On the diagram, clearly indicate the activation energy for the rate
determining step, and the overall reaction exo- or endothermicity. Indicate the positions
of the transition states but do not draw them. INDICATE THE POSITIONS OF THE
VARIOUS SETS OF INTERMEDIATES ON YOUR DIAGRAM.
LA/BA H
H O+
LB/BB
Ph
HCl cat./heat
H
LB
••O
H
HO
Ph
••
H
H2O
H LB/BB
••
O
+
Ph
H
energy
Ph
‡
+
LA/BA
H
H
LA
H
O
+
Ph
‡
‡
‡
reaction coordinate
c) For your mechanism:
2
give the number of sets of intermediates ______________
3
give the number of transition states ______________
d) Briefly describe your reasoning for your choice of rate determining step in your
mechanism
the rate determining step is the slowest, it is the most endothermic, it is endothermic
because 2 bonds are broken in this step and only one bond is formed, this requires
energy
CHM 233 Final Exam : Practice Exam #2
-5-
NAME
Question 5. For the following reactions:
a) Give the missing major ORGANIC PRODUCT
b) Properly describe all stereoisomeric products if any and identify any MESO compounds
c) Explain whether a solution of the product(s) would be optically active and why or why not
d) Assign each reaction to addition, elimination, rearrangement or substitution
OH
H2SO4
a)
H
addition
not optically
active achiral
product
H2O/heat
will rearrange
Br
substitution
NBS
b)
(±)
hν
strong Nucleophile/weaker
base - SN2
substitution
CH2CH2CH2OCH3
CH3S– +Na
optically active single
C H
enantiomer formed
CH3S
acetone
CH3
CH2CH2CH2OCH3
c)
H C Br
H3C
d)
D
not optically
active, racemic
mixture formed
D
Br2
CCl4
D
*
*
Br
Br
OR
(±)
D
Br
*
*
D
(±)
Br
D
Na+ –OH
e)
elimination
Br
DMF
non-bulky base gives
Sayetzeff alkene
1. Hg(OAc)2/H2O
f)
addition
not optically
active, racemic
mixture formed
not optically
active achiral
product
addition
not optically
active, racemic
mixture formed
*
(±)
OH
no rearrangment occurs here due to ABSENCE of carbocation intermediate
2. NaBH4
g)
*
*
H2/Pd/C
*
*
addition
OPTICALLY
ACTIVE, single
enantiomer reactant,
single enantiomer
product
CHM 233 Final Exam : Practice Exam #2
-6-
NAME
Question 6) For (2R,3S)-dibromobutane (meso-1,2-dibromobutane)
a) Draw a 3-D structure using wedged/dashed bonds showing the stereochemistry at the two
asymmetric centers, IN THE REACTIVE CONFORMATION FOR AN E2 ELIMINATION
Me
H
H
H
H
Br
R
S
R
S
OR
C*
C*
*C
*C
Me
Me
Br
Me
Br
Br
b) Draw a Newman projection for the reactive conformation for E2 elimination, looking FROM
carbon #2 TO carbon #3.
H
H
H
Br
Me
H
OR
Me
Br
Me
Br
Me
Br
c) Give the product of E2 elimination to give the Sayetzeff (Zaitsev) alkene product
Me
Me
C
C
Br
H
Question 7) For the two SN2 reactions shown, A and B
a) Give the curved arrow pushing and indicate the Lewis acid and base (LA/LB) and ALSO the
nucleophile (Nuc) and electrophile (Elec)
b) Draw reaction energy diagrams on the SAME DIAGRAM (normalize at the reactants),
indicate the activation energy for both reactions and indicate the positions of the transition states
c) Indicate the faster reaction and give a brief explanation that includes Hammond Postulate
LB/Nuc
LA/Elec
Na NH2
NH2
Br
+ Br
A
CH3CN
LA/Elec
B
relative
energy
Br
Na
OH LB/Nuc
CH3CN
OH
+
Br
‡
‡
EaB
EaA
B
A
reaction coordinate
reaction A is faster, the -NH2 anion is a stronger Lewis base and nucleophile because the energy
of the non-bonding electrons is higher, they are more reactive, because N is less electronegative
than O, reaction A is more exothermic, the Hammond Postulate states that the more exothermoic
reaction will have teh smaller activation energy, be faster and have the earlier transitions state
CHM 233 Final Exam : Practice Exam #2
NAME
-7-
Question 7) Draw the conjugate base anion that results from deprotonation of the N-H bond in A,
B and C below. Include ALL resonance contributors where appropriate. AND, rank A B and C in
order of increasing Brønsted acidity. Give a BRIEF explanation.
H
N
A
N
B
H
C
N
H
A
<
B
weakest
O
C
<
strongest
deprotonation of C gives a resonance stabilized anion with negative charges on N and O,
which makes the most stable anion, deprotonation of B puts the negative charge on N and C
which is thus higher in energy than that from C, deprotonation of A gives an anion that is not
resonance stabilized, thus highest in energy and thus its conjugate acid is the weakest
-H+
H
N
H
N
H
N
-H+
-H+
N
N
N
N
N
O
O
O
Question 8) On the following structures and for the orbitals indicated, roughly sketch the shape of
the wavefunctions ON TOP OF THE STRUCTURES, taking care to make the locations of any
nodes clear, and to explicitely state the A.O.'s used to make any M.O.'s.
node
C-p
O-p
H
C
O
node the C–O π* antibonding orbital (i.e., where the electrons will "go", THE
LUMO, when a C=O acts as an electrophile in second semester)
H3C
O more electronegative, therefore anti-bonding M.O. larger on C
H
H
O
H - 1s
O - sp3
the H-O σ* antibonding orbital (i.e., where the electrons are
"given to", THE LUMO, when H3O+ reacts with a base)
H
node node
O more electronegative and positively charged, anti-bonding M.O. larger on H
-8-
NAME
CHM 233 Final Exam : Practice Exam #2
Question 9) For the following acid/base reaction, show the curved-arrow pushing, identify the
Lewis acid/base and whether they are also Brønsted acids/bases, draw a reaction energy
diagram and the structure of the transition state. Indicate the activation energy and the reaction
exothermicity or endothermicity as appropriate. Briefly explain why you think the reaction is
endo- or exothermic.
O
H3C
+
Li+
H2C
LB
Li+
O
CH2
H3C CH2
CH2
LA
O
‡
ΔG
‡
Ea
H3C Li
O
H2C
H3C CH2 CH2
note that the transition state has a formal
negative charge IF you do not include the Li+, if
you include the Li+ then the ‡ would have no
overall charge, either is OK
exothermicity
CH2
‡
O
H3C CH2
CH2
reaction coordinate
the reaction is exothermic, the negative charge moves from the less electrinegative carbon
to the more electronegative oxygen, and ring strain is released
Question 10) For (2R)-bromo-2,(3R)-diphenylbutane
a) Draw a 3-D structure in the reactive conformation for an E2 elimination using wedged/dashed
bonds to indicate stereochemistry
Ph
Br
R
*C
R
Ph
Me
Me
C*
H
b) Draw a Newman projection in the reactive conformation for E2 elimination
Br
Ph
Me
Ph
Me
H
c) Give the alkene product obtained upon E2 elimination using a non-bulky base
Ph
Me
Ph
Ph
Me
Me
≡
Ph
Me
-9-
NAME
CHM 233 Final Exam : Practice Exam #2
Question 11) Give a full curved arrow mechanism for the following reaction, indicate
LB/AL/BB/BA as appropriate, and give a reaction energy diagram. State the number of sets of
intermediates and transition states, do not draw transition state structures but indicate the
positions of the transition states on the diagram and how these positions correspond to the
mechanism that you drew. Include all reasonable resonance contributors as appropriate.
Br
these are NOT
2
intermediates,
this is ONE
intermediate
drawn 2
different ways
‡1
H3C
H
‡3
O
H3C
O
H
‡1
LB/BB
CH3
LA/BA
‡2
LA
this is NOT
a step in
the
mechanism
MeO
MeOH
heat
H
LB
O
2 sets of
intermediates and 3
transition states
‡2
energy
‡3
reaction coordinate
Question 12) For each of the following reactions, give the major organic product and assign the
mechanism as E1, E2, SN1 or SN2 and briefly explain the reasoning for your choice. Indetify all
stereoisomeric products as appropriate.
a)
Na+ –OH
Cl
DMF
E2, hydroxide is a strong base and strong nucleophile, but SN2 is not possible for a tertiary
halide, hydroxide is a non-bulky base therefore the product is Sayetzeff (Zaitsef)
CH3OH
b)
Cl
heat
(±) SN1
OCH3
SN1, the halide is benzylic, which does not help to distinguish mechanisms, but MeOH is a weak
nucleophile and weak base and a polar protic solvent, which suggests SN1 or E1, eliminatioin is
not possible for this strcture, however, therefore can only be SN1
CHM 233 Final Exam : Practice Exam #2
- 10 -
NAME
Question 14 Provided are spectra for a compound with molecular formula C10H14
degrees of unsaturation
a) Give the degrees of unsaturation 4________________
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.
1947
1704
1604
C
sp2
1399
1422
H 3097
3027
1498
2883
2907
H C
2945
sp3
c) draw the structure and clearly indicate which hydrogens correspond to which signals in the
proton nmr spectrum ONLY
6H
doublet
A
H
H
A
CH3
C
CH2
CH B
H
H
H
D
CH3
A
D
5H
multiplet
C
2H
doublet
B
1H
multiplet
CHM 233 Final Exam : START
OF Practice Exam #3
-2-
NAME
Question 1. Give the IUPAC name for the following. Specify stereochemistry as appropriate.
Br
H
Ph
1
Br =
2 H
a)
Ph
(1S)-bromo-(2R)-phenylcyclohexane
because all other things are equal, and ONLY because all other things are equal,
number alphabetically to give the carbon with the Br #1 in these structures
b)
6
Br
4
H
5
3
2
Br
1
1,6-dibromo-(3R)-methylhexane
compared to l) switched
the position of 2
substituents at the chiral
center (Me and H), thus
simply the enantiomer
Question 2. Glucose exists mainly in the two cyclic forms shown below (the non-bonding
electrons are omitted for clarity).
a) For each structure, identify ALL chiral/asymmetric carbins using the symbol *
b) Assign the absolute configuration of the carbon indicated as C1 in each structure
c) Assign the two structures as enantiomers, diastereomers and give a reason for your choice
C1 (R)
C1 (S)
O
HO
*
*
HO
*
OH
*
*
H
OH
OH
α-D-glucopyranose
O
OH
*
*
H
* * *
HO
OH
OH
β-D-glucopyranose
HO
the structures are stereoisomers, but do not have opposite configuration at ALL asymmetric
centers, therefore they are DIASTEREOMERS
Question 3. Draw the requested wavefunction or wavefunction squared. Indicate the positions
of all nodes for wavefunctions, and the places where electron density is zero for wavefunctions
squared. INCLUDE THE RELEVANT ATOMIC ORBITALS YOU USED TO BUILD THE
MOLECULAR ORBITALS
H3C
CF3
C
H3C
C
CF3
the Ψ for the π-bonding M.O. that is the HOMO when an
C - p alkene reacts with any electrophile, e.g. HBr, etc. (take the
C-p
effect of the electronegative elements into account)
in the wavefunction squared, there is no phase information
H
C
H3C
O
2
C - sp2 the Ψ for the C–O σ bonding orbital (i.e. where the
electrons are that don't get involved in reactions!!)
O - sp2
O more electronegative, therefore bonding M.O. larger on O
-3-
CHM 233 Final Exam : Practice Exam #3
Question 4) Give a curved arrow mechanism for the following reaction. Indicate LB, LA, BB, BA as
usual and state the number of sets of intermediates and transition states. This mechanism is NOT IN
THE NOTES, you need to figure it out using what you know about Lewis/Bronsted acid base and
other reactions that you have learned. The first step is given for you. INCLUDE A DRAWING OF
ALL OF THE TRANSITION STATES
H
LA/BA
H O
H
LB/BB
O
O H
H O / HCl cat.
2
H O
H ‡
H
O
‡
H
H
H
O
O
LA
O
O
O
O LB LA/BA
H
H
‡ for step 1
H
H
H
H
O
H
O
H
LB/BB
H
H
O
H
‡ for step 3
H
‡
O
the mechanism has
3 steps and
2 sets of intermediates
‡ for step 2
O
H
H
Question 4) For (1S)-t-butyl-(2S)-methyl-cyclohexane, draw both chair conformations and determine
the energy difference using the tables on the front page of this test. Indicate the lower energy chair
and give the energy difference.
5
5
4
1
6
3
2
Me
4
Me
5
6
t-Bu
3
2
1 x t-Bu/Me gauche = 2.7
Total Energy = 2.7 kcal/mol
1
3
1
2
t-Bu
lower E
6
4
2 x t-Bu/Me gauche = 5.4
2 x Me/Me gauche = 1.8
Total Energy = 7.2 kcal/mol
t-Bu
priority = 1
C has 3 x C attached
Me
Energy difference = 4.5 kcal/mol
t-butyl comes before methyl "alphabetically" in a IUPAC name because the "tert" part of the t-butyl
name is ignored because it is hyphenated, unlike isopropyl, where "iso" is not ignored because it is
not hyphenated, so B for butyl comes before m for methyl
this also determkines thge numbering in this case, because ALL ELSE is equal (and only because
all else is equal) the numbering of the ring is alphabetical
-4-
CHM 233 Final Exam : Practice Exam #3
NAME
Question 5) Here is a reaction we will not do until the end of second semester organic. It is a
Hofmann E2 type elimination. It is unusual in that the NON-Sayetzeff product as the major one.
Indicate which reaction (A and B) is which in the provided reaction energy diagram, give a
BRIEF explanation. Draw the transition state for reaction A only. The reactions are slightly
exothermic. Indicate which reaction you would expect to be less exothermic and why.
H
OH
H3C CH CH CH3
H3C CH CH CH3
minor
NMe3
HO
A
H
H2C CH CH2 CH3
H2C CH CH2 CH3
B
major
‡
NMe3
‡
energy
H
A
OH
H3C CH CH CH3
‡
NMe3
B
(no overall charge)
reaction coordinate
The alkene in B is less stable, but is the major product, but its alkene is less substituted, which means
that its energy diagram must be the least exothermic. Because it is the major product it must be formed
fastest, must have the smallest activation energy and also be the highest energy product.
(the reason that B is faster is related to a steric effect due to the large -NMe3+ group)
Question 6) For the following Bronsted acid/base equilibrium, add the curved arrow-pushing that
describes the bond making/breaking in both directions, identify the acids and the bases and
decide which acid/base pair is the stronger and give a BRIEF explanation. You will need to add
the important hydrogen atoms that are missing from the line-angle structures to do the
proper curved arrow-pushing. Draw all reasonable resonance contributors for any species that is
involved in the reaction as appropriate. Which reaction is faster and why?
H
+
stronger base
H
faster
H
slower
+ H
weaker acid
stronger acid
the base on the right has resonance stabilized non-bonding
electrons that are less reactive, it is thus the weaker base, the
base on the left must be stronger, the stronger acid deprotonates
more easily because its base is more stable, reaction left ot right
is faster because lower energy electrons are formed
H
weaker base
CHM 233 Final Exam : Practice Exam #3
-5-
NAME
Question 5. For the following reactions:
a) Give the missing major ORGANIC PRODUCT or reagents/conditions
b) Properly describe all stereoisomeric products if any and identify any MESO compounds
c) Explain whether a solution of the product(s) would be optically active and why or why not
d) Assign each reaction to addition, elimination, rearrangement or substitution
Cl
(CH3)3O– +K
a)
3° halide, strong bulky base, E2
to give Anti-Sayetzeff, even
though in polar protic solvent
not optically active, achiral ptroduct
E2
elimination
(CH3)3OH
no SN1 at a primary carbon
b)
substitution
CH3OH
Br
Br
SN1
Br
MeO
boil
(give substitution product(s) only, no elimination)
1. Hg(OAc)2/H2O
c)
OH
(±)
2. NaBH4
can't use H2O/H2SO4, this would result in rearrangement
I
acetone
NC
(give substitution product)
Cl
addition
not optically active
racemic mixture
formed
substitution
optically active,
single enantiomer
formed
1 Equiv. Na+ –CN
d)
not optically
active, achiral
product
CH3
better leaving group
addition
1) BH3 . THF
e)
OH
2) H2O2, –OH
(±)
H
not optically active
racemic mixture
formed
Cl
addition
HCl
f)
not optically active, achiral product
CCl4
note carbocation rearrangement here!
Ph
g)
Ph
H2
Pd/C
achiral
Ph
Ph
*
*
H
H
syn-addition
addition
not optically active
achiral meso product
CHM 233 Final Exam : Practice Exam #3
-6-
NAME
Question 6) For the following two reactions
a) Add the curved arrow-pushing showing bond making/breaking, assigen them both as
addition, elimination, substitution or rearrangement
b) Draw a reaction energy diagram for both reactions ON THE SAME DIAGRAM,
normalize your diagrams at the transition states and show BOTH activation energies
c) Indicate which reaction would be faster and give a brief explanation
N
LA/Elec
C N
C
+ I
A
I Na
acetonitrile
LB/Nuc
LA/Elec
B
C N
I Na
methanol
C
N +
I
‡
relative
energy
EaA
A
EaB
B
reaction coordinate
Reaction A is faster, because the anion LB/Nuc is less solvated in the polar APROTIC solvent,
resulting in a smaller activation energy and faster rate, the solvation of the large anionic transition
states are similar in both solvents, the solvation of the small anion leaving group will be greater in the
protic solvent B, which is why the energy of the products are lower in B compared to A, note that
although iodine is a large atom, it is still considered to be a small anion compared to the transition state
Question 7) Draw the base anions that results from deprotonation of the most acidic proton in A
B and C, include ALL resonance contributors as appropriate, rank A B and C in order of
increasing Brønsted acidity. Give a BRIEF explanation.
HC
HB
HA
CF3
CF3
F3C
A < C < B
A
C
B
weakest
strongest
F3C
CF3
F3C
To answer this question, you FIRST need to identify WHICH is the most acidic proton on each
structure.
deprotonation of B gives a resonance stabilized anion in which the -CF3 group directly stabilizes
the negative charge in both structures, deprotonation of C has direct -CF3 stabilization in 1
structure, deprotonation of A results in only "indirect stabilization by two nearby -CF3
H
F3C
F3C
F3C
H
H
F3C
F3C
CF3
F3C
F3C
F3C
CF3
F3C
CF3
CF3
CF3
CF3
CF3
CF3
CF3
CHM 233 Final Exam : Practice Exam #3
NAME
-7-
Question 8) Write a curved arrow mechanism for the following reaction. Indicate LB, LA, BB, BA
as usual and State the number of sets of intermediates and transition states. AT EACH STEP IN
THE MECHANISM, PROPERLY IDENTIFY THE RELATIVE STEREOCHEMISTRY OF THE
SUBSTITUENTS USING WEDGED/DASHED BONDS, AND ALSO INDICATE THE PRESENCE
OF ANY RACEMIC MIXTURES IN THE INTERMEDIATES USING THE (±) NOTATION AND
IDENTIFY ANY MESO STRUCTURES, EVEN IF THEY ARE INTERMEDIATES
Br
Br
LA
LB
Br
Br2
(±)
EtOH
OEt
at THIS point
we have a pair
of enantiomers
Br
Br
the
bromonium ion
is a meso
structure
Et
LA
Et
O LB
H
(±)
LA/BA
there is actually no really good way to do the
curved arrow pushing for this first step to
generate the bromonium ion, sometimes it is
done this way.....
E
t
O
O
H
Br
H
Br
Question 9) Below is the structure of (+)-xylose. Identify ALL of the chiral/asymmetric
centers and indicate with the * sykmbol, and assign the absolute configuration in each
case (non-bonding electrons omitted for clarity). Draw a structure for (-)-xylose.
OH
H
(S)
* *
OH
*
OHC (R)
(R)
H
OH
H
HO
OH
HO
OH
CHO
OH
(-)-xylose
we simply need to draw the enantiomer, which
has the opposite configuration at all 3 chiral
centers, or more simply, just the mirror image. An
enantiomer is simply the mirror image no matter
how many chiral centers there are
LB/BB
-8-
NAME
CHM 233 Final Exam : Practice Exam #3
Question 10) Give a curved-arrow pushing mechanism for the following reaction, indicate the
Lewis acid and base at each step and whether they are also Bronsted acids/bases. Assign the
reaction as addition, elimination, substitution or rearrangement.
LA/BA
H
Cl
LB/BB
HCl cat.
O
H
H
O
H
HO
LA
addition
O
O
LB
LA/BA
HO
O
O
O
H
LB/BB
Cl
Quetsion 11) Rank the following in order of increasing rate of solvolysis in methanol. Give
a BRIEF explanation that includes drawings of he primary intermediate cations for each
structure (you do not need to draw ALL resonance contributors).
Cl
A
Cl
B
O
Cl
C
C
slowest
<
A
<
B
fastest
O
O
Solvolysis is an SN1 reaction. The rate determining step involves heterolytic cleavage of the
C-Cl bond to form an intermediate cation. The cations from A and B are resonance stabilized,
that from C is not. That from B has an extra resonance form compared to that from A, and
thus is particularly stabilized, formstion of cation from B will be fastest.
-9-
NAME
CHM 233 Final Exam : Practice Exam #4
Question 11) For the acid/base reaction shown below
a) add the curved arrow pushing that shows bond making/breaking IN BOTH DIRECTIONS
b) identify the stronger and weaker acids and stronger and weaker bases, and give a brief
explanation for your choice
c) state which reaction would have the larger rate constant (be faster) left to right or right to left
d) identify on which side the equilibrium would lie
e) draw a reaction energy diagram that includes a drawing of the transition state
if your explanation uses resonance arguments, draw all important resonance contributors
NH
H
C
slower
(smaller k)
O
H
C
+
H
H
H
C
H
weaker acid
NH
C
H
H
faster
(larger k)
+
C
H
H
C
C
H
H
stronger acid
NH
O
C
H
H
stronger base
weaker base
H
C
O
H
C
H
H
C
H
C
H
equilibrium lies on the left, reaction right to left is faster, the weaker base has lower energy less
reactive electrons, both bases are resonance stabilized but the weaker base can put the negative
charge on the more electronegative oxygen compared to nitrogen in the case of the stronger base
‡
Energy
NH
H
O
H
C
C
H
H
H
C
H
C
H
Question 12) Rank the following in order of increasing acidity. Give a brief explanation, illustrated
using resonance contributors (ignore keto-isomers and also intramolecular hydrogen-bonding)
OH
OH
O
OH
O
A
B
A
C
O
C
B
most
least
<
<
relative acidity determined by stabilities of anions, these increase in stability C < A < B
because of 2, 3 and 4 resonance contributors, respectively (resonance contributors not
shown to conserve space)
*
O
O
O
O*
*
O*
O
C
A
B
the stars represent the locations of the negative charges in the other resonance contributors
*
*
CHM 233 Final Exam : Practice Exam #3
- 10 -
NAME
Question 14 Provided are spectra for a compound with molecular formula C5H11Br
degrees of unsaturation
a) Give the degrees of unsaturation 0________________
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.
sp3
H
C
c) draw the structure and clearly indicate which hydrogens correspond to which signals in the
proton nmr spectrum ONLY
A 6H
doublet
B/C
2H
quartet
and
C
CH2
Br
CH2
D
1H
CH3 A
CH B
CH3
A
D
2H
triplet
multiplet
(overlapping)
CHM 233 Final Exam : START
OF Practice Exam #4
-2-
NAME
Question 1. Give the IUPAC name for the following. Specify stereochemistry as appropriate.
a)
Br
Br
trans-1,4-dibromocyclohexane
(no asymmetric centers here)
Cl
b)
(6R)-chloro-(3,4S)-methyl-(2E)-heptene
Question 2 For the following reactions
a) Given the provided curved arrows, give the major products, include all important resonance
conrtributors as appropriate
b) Indicate the Lewis acid/base and whether they are also Brønsted acids/bases
c) Indicate whether the reaction is endothermic or exothermic and justify your answer with a
BRIEF 1 sentence explanation
H
H
C
a)
+
C
H Br
H
H
LB/BB
b)
H
H
endothermic, breaks 2 bonds only makes one bond
LA/BA
H
d) H C
H
LB/BB
H
H
C C
+
H C
C H
H H H H
LA/BA
Na
H
LB/BB
C
H C
H H
H
H
C
H C
H H
C
C H
Na
H
C
C H
H
+ H
exothermic, makes resonance stabilized anion
H
+
Br
+
C H
C
H
H
H
H
H
H
C C
H C H
H
H
LA/BA
H
H
+
H
C C
H
exothermic,puts electron pair is an sp2 hybrid orbital instead
of an sp3 orbital, lowers energy of electrons
H
-3-
CHM 233 Final Exam : Practice Exam #4
Question 3) For the following compound, rank the pairs of electrons indicated as A, B, C
and D, in order of INCREASING energy. Give a BRIEF explanation.
H
A (nonbonding electrons)
N
B (electrons in C=C π bond)
(electrons in C–C σ bond) C
(electrons in C–C σ bond) D
D
<
C
<
B
A
<
highest
lowest
lowest energy in sigma bond made from sp3-sp2 A.O.s, next is sigma bond made from
sp3-sp3 A.O.s, next are electrons in p-M.O., highest are non-bonding electrons
Question 4). For each pair of electrons indicated, draw the wavefunction on TOP OF THE
STRUCTURE, and INDICATE THE ATOMIC ORBITALS THAT ARE USED TO BUILD THE
MOLECULAR ORBITALS AS APPROPRIATE
H
C
N
H
N = sp A.O.
H H = 1s A.O.
C
N
H
H
N = p A.O.
C = p A.O.
Ψ for C-N π-antibonding orbital
Ψ for N-H σ-bonding orbital
Question 5) Rank in order of increasing rate of E1 elimination in hot methanol. Give a BRIEF
explanation.
Br
Br
Br
B
A
C
slowest
B
<
C
<
A
fastest
The rate determining step formation of a carbocation intermediate. That from B is tertiary,
which is not as stable as that from A which is tertiary and resonance stabilized. That from C is
secondary and not resonance stabilized, thu least stable, thus slowest reaction.
CHM 233 Final Exam : Practice Exam #4
-4-
NAME
Question 3) Give a curved arrow mechanism and a reaction energy diagram for the following
reaction. Indicate LB, LA, BB, BA as usual and state the number of sets of intermediates and
transition states. Indicate the positions of the transition states on the diagram but do not draw them.
Indicate the rate determining steo on the mechanism and the activation energy for the rate
determining step on the diagram. Briefly justify your choice of rate determining step.
H
LA/BA
H O
H Cl OR CH3
CH3
H3C
HCl
C CH2
H3C C O CH3
CH3OH
H3C
CH3
LB/BB
rate
determining
2 sets of
step
intermediates and 3
H3C
transition states
CH
H3C
3
C CH2
LA
O C CH3
H3C
H
H
LA/BA
CH3
CH3 O
LB
CH3 O
H
H
‡
energy
LB/BB
‡
‡
EaRDS
reaction coordinate
Question 4) Explain which of the following two reactions will be faster. Include in your explanation an
indication of the reaction mechanism and also a discussion of exactly how the presence of Br or F
influences the reaction rate (i.e. not just "Reaction X is faster because it is Y as a substituent...")
Br
Ph
DMF
Ph
F
Ph
Ph
Na+ –OMe
A
Ph
Ph
Ph
Na+ –OMe
DMF
B
Ph
these are E2 reactions, reaction A is faster, bromide is a better leaving group (it is a weaker
base) since it has a weaker bond to carbon than fluorine, fluoride is a poorer leaving group
CHM 233 Final Exam : Practice Exam #4
-5-
NAME
Question 5. For the following reactions:
a) Give the missing major ORGANIC PRODUCT or reagents/conditions
b) Properly describe all stereoisomeric products if any and identify any MESO compounds
c) Explain whether a solution of the product(s) would be optically active and why or why not
d) Assign each reaction to addition, elimination, rearrangement or substitution
1. BH3 . THF
a)
not optically
active, racemic
mixture
(±)
2. –OH, H2O2
OH
Br
Br
Excess HBr
b)
(±)
ROOR
HO
H2O
heat
(±)
(substitution product only, no elimination)
c)
Br
Br
d)
DMF
no,
racemic
substitution
N3
no,
achiral
substitution
(good nuclephile)
N N N
Na
not optically
active, racemic
mixture
good nuclephile, E2 not possible - SN2
Br
1. NBS / hν
e)
no,
achiral
2. Na+ –O-t-Bu/DMF
substitution
3° halide + bulky base gives ANTISayetzeff ELIMINATION product
1) BH3 . THF
f)
addition
(±)
2) H2O2, –OH
OH
elimination
K+ –O-t-Bu
g)
Br
H D
no,
racemic
yes, single enantiomer formedE2
acetone
D
elimination
CHM 233 Final Exam : Practice Exam #4
-6-
NAME
Question 8 For each of the following reactions
a) give the major organic product
b) state whether the mechanism is SN1, SN2, E1 or E2 and give a brief justification for your choice
c) state whether a solution of the product would be optically active and give a brief explanation
Me
Br
a)
Me
H
t-Bu
E2, 2° halide, but strong
(bulky) base, poor
nuclophile, aprotic solvent
t-BuO– +K
Me
H
Me
DMF
Br
t-Bu
Me
t-Bu
optically active, single enantiomer formed
OCH3
OTs
b)
CH3OH
(±)
heat
(substitution product only)
SN1, 2° tosylate, but
weak nucleophile/base,
protic solvent and heat
not optically active, racemic mixture formed
N
C
Br
SN2, alylic bromide,
strong nucleophile,
aprotic solvent E2 not
possible
Na+ –:C N:
c)
acetonitrile
optically active, single enantiomer formed
Question 9 For (1S,2S)-1-bromo-1,2-diphenylpropane, draw
a) a 3D structure showing sterochemistry in terms of wedged and dashed bonds
b) a Newman projection of BOTH the lowest energy conformer for rotation around the C1-C2 bond,
AND a Newman projection of the conformer that would undergo an E2 elimination
c) the alkene product of an E2 elimination
Me
Br
C
H
Ph
Ph
C
Br
Ph
H
H
3D structure
Br
H
Ph
Me
lowest
energy conformer
Me
Ph
H
H
Me
Ph
Ph
Ph
H
E2 conformer
alkene product
CHM 233 Final Exam : Practice Exam #4
NAME
-7-
Question 10) For the following two reactions
Na+ –OCH3
Br
A
DMF
FASTER
Na+ –OCH3
Br
B
SLOWER
DMF
a) draw the transition states for both reactions, clearly indicate which is which
Me
H
O
Me –
–
Br
H
Br
O
B
A
b) draw a reaction energy diagram for BOTH reactions on the SAME diagram (same axes)
clearly indicate BOTH activation energies, and BOTH exothermicities on the diagram
‡
A
Energy
‡
EaB
B
EaA
Br
Na+ –OCH3
exo B
exo A
Reaction Coordinate
c) Indicate which reaction would be faster, A or B, and give an explanation
Reaction A is faster because a more substituted, and thus more stable product is formed.
The Hammond postulate says that the more exothermic reaction will have a lower energy
transition state. The more stable alkene will have a stronger partial π bond in the
transition state.
-8-
CHM 233 Final Exam : Practice Exam #4
NAME
Question 10) For each reaction give a full curved-arrow pushing mechanism, indicate the Lewis
and Bronsted acids/bases for each bimolecular step and give the number of transition states and
sets of intermediates
O
OH
HCl
a)
H
Cl
LB/BB
LA/BA
Cl
LB/BB
H
L Base
LA/BA
O
O H
3 transition states
2 sets of intermediates
L Acid
CH3OH
b)
Br
OCH3
boil
H3C
O
LB
H
O
H
LA
3 transition states
2 sets of intermediates
LA/BA
LB/BB
CH3
CH3
O
H
-9-
NAME
CHM 233 Final Exam : Practice Exam #3
Question 12)
a) Draw a structure (using wedged and dashed bonds) of (2S)-chloro-(3S)-methylpentane.
Cl
b) Draw a structure (using wedged and dashed bonds) of the product of the substitution
reaction of (2S)-chloro-(3S)-methylpentane shown below. Assign the absolute configurations
of all asymmetric centers.
SH
H–S Na
(R)
(S)
(2S)-chloro-(3S)-methylpentane
DMF
c) Draw a 3-D structure of a diastereomer of the product of the reaction you gave above.
Assign the absolute configurations of all asymmetric centers.
SH
(S)
SH
(S)
(R)
or
(R)
Question 13) Provide detailed (arrow pushing) mechanism for the following reaction, provide a
reaction energy diagram, indicate the positions of any intermediates,where appropriate label the
Lewis/Bronsted Acids/bases.
OH
LB
Cl
Cl
LA
Cl2
O
CCl4
Cl
(±)
••OH LB
LA/BA
O+
H
Cl
(±)
Cl+
energy
‡
LB/BB
‡
note, you would
not have to put
the (±) on each
intermediate as I
have done here
the reaction above is basically the
same as
except that the alcohol is attached
to the molecule in this case
••
Cl
••
••
‡
••
(±)
LA
reaction coordinate
Ph
Br2
MeOH
Ph
OMe Br
- 10 -
NAME
Question 14 Provided are spectra for a compound with molecular formula C8H10O2
degrees of unsaturation
a) Give the degrees of unsaturation 4________________
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.
O
H
C
C
H
H
sp3
sp2
c) draw the structure and clearly indicate which hydrogens correspond to which signals in the
proton nmr spectrum ONLY 2 possible answers in this case (assignment on proton nmr
given for one answer to save space)
3H
A
C
CH3O
E
D
A
OH
B
OR
HO
OCH3
2H
C
2H
H3CO
OH
this position less deshielded, thus
less downfield, thus protons E
have a smaller chemical shift (δ)
2H
D
1H
E
B