CHEM 633: Advanced Organic Chem: Physical
Problem Set 3
Please do not look up references until after you turn in the problem set unless otherwise noted.
For the following problems, please use Excel (or another graphing program), when necessary. Please submit your graphs
with your problem set.
1. Edwards, Montoya-Peleaz and Crudden have studied the mechanism of epoxidation of benzaldehyde (B) with
sulfonium SM (Org. Lett. 2007, 9, 5481). The general mechanism for this transformation is believed to proceed via sulur
ylide 1 and then intermediate 2.
N
H3C
S
Ph
CH3
BF4
SM
N
(DBU)
step 1
H3C
O
PhCHO (B)
S
Ph
CH3
1
O
Ph
Ph
step 2
S
2
step 3
Ph
Ph
P
(a) Under pseudo-first-order conditions (large excess of benzaldehyde B and DBU), the authors monitored the
disappearance of sulfonium tetrafluoroborate SM. What is the kinetic order of [SM] in this reaction?
(b) Under pseudo-first-order conditions, the following data was obtained. In each of these reactions, [B] = 0.0174 M
and temperature = 298 K. What is the kinetic order of [DBU]?
Entry
[DBU] (M)
Trendline from plot of ln[SM] vs. time (seconds/100)
1
0.0291
y = –0.0068x – 1.8296
2
0.0751
y = –0.0199x – 1.6405
3
0.113
y = –0.0329x – 1.6278
4
0.156
y = –0.0466x – 6.3816
(c) Under pseudo-first-order conditions, the following data was obtained. In each of these reactions, [DBU] = 0.156 M
and temperature = 298 K. What is the kinetic order of [B]?
Entry
[B] (M)
Trendline from plot of ln[SM] vs. time (seconds/100)
1
0.0174
y = –0.0466x – 6.3816
2
0.0433
y = –0.1103x – 1.8947
3
0.0866
y = –0.1986x – 1.7842
4
0.130
y = –0.2474x – 1.7569
5
0.173
y = –0.3022x – 1.7994
6
0.260
y = –0.3192x – 1.7368
1
(d) Based on the proposed mechanism and this data, derive a kinetic rate law for this transformation.
(e) Assuming that step 3 is fast, what is the ratio of k–1/k2?
2. Consider two possible mechanistic sequences for an acid-catalyzed aldol reaction, (i) and (ii).
(i)
(ii)
H+
O
+
Me
H
K1
H
Me
A
H
Me
H
C
H
Me
H
+
Me
O
k3
O
OH
H
OH
H
H
B
+
OH
H+
Me
C
Me
H
B
C
O
k3
O
+
H
P
H+
+
H
k–2
B
C
H
B
k2
O
H+
+
O
Me
A
H
B
OH
Me
H
H
K1
O
+
H
OH
k–2
H
H+
B
k2
O
O
OH
H
Me
P
(a) Derive a rate expression for sequence (i), using the steady state approximation where appropriate. Your final equation
+
should only contain terms that are experimentally quantifiable (i.e., A, H ). You may assume that the initial carbonyl
protonation is a fast equilibrium and that the overall reaction is irreversible.
(b) Derive a rate expression for sequence (ii), using the steady state approximation where appropriate. Your final equation
should only contain terms that are experimentally quantifiable. You may assume that the initial carbonyl protonation is
a fast equilibrium and that the overall reaction is irreversible.
(c) Show that the two sequences are kinetically distinguishable when the final step (i.e., C–C bond formation) is ratelimiting.
(d) Show that the two sequences are kinetically indistinguishable when the last step is rapid compared with the initial
steps.
3. Consider the following theoretical reaction coordinate diagrams, describing the transformation of starting material A into
product C via intermediate B. For each reaction coordinate diagram, please answer the following questions:
(a) Is there a possibility that intermediate B may be observable using standard spectroscopic methods, such as NMR
spectroscopy?
(b) If there is the possibility that B may be observable, what is the maximum possible energy difference between A and B
to allow observation of B by NMR spectroscopy? Please explain your reasoning.
(c) What is the rate law for the reaction?
(a)
(b)
B
ΔG
B
ΔG
A
A
C
C
reaction coordinate
reaction coordinate
4. The temperature-dependent rates of rearrangement of allyl vinyl ether (eq 1) was studied in the gas phase.
O
O
(1)
2
–1
Temp (K)
469.1
469.4
473.7
427.7
456.7
451.6
440.2
k (s )
–3
2.875 x 10
–3
3.021 x 10
–3
3.838 x 10
–3
0.120 x 10
–3
1.166 x 10
–3
0.788 x 10
–3
0.341 x 10
a. Provide a depiction of the transition structure for this transformation.
b. Draw a reaction coordinate diagram for this transformation, clearly labeling all intermediates and transition states.
‡
‡
c. Determine ΔH and ΔS from the data (in kcal/mol and eu, respectively).
‡
d. Explain whether your proposed transition structure is consistent with the experimentally measured value of ΔS .
5. One proposed mechanistic sequence for the Baylis-Hillman reaction is shown below (e.g., J. Org. Chem. 2003, 68,
692). This reaction has recently been studied using the initial rates method (Org. Lett, 2005, 7, 147). You may assume
that no observable intermediates accumulate during the course of the reaction.
O
O
O
OMe
+
M
k1
N
N
OMe
k–1
O
+
N
D
(DABCO)
I
R
H
R
N
k2
O
–DABCO,
proton transfer
N
N
A
OH O
fast
OMe
R
OMe
(a) Derive a rate expression for this sequence, using the steady-state approximation where appropriate.
(b) Show that the steady-state approximation and the pre-equilibrium approximation (for saturation kinetics) are
equivalent when k–1 is large compared with k2.
(c) Initial rate experiments showed that the initial rate is first order in [D], first order in [M], and second order in [A].
Evaluate the plausibility of the proposed mechanistic sequence on the basis of this initial rate data.
(d) The authors also evaluated this reaction by analyzing isotope effects. For the illustrated mechanism below (the
generally accepted mechanism), please predict the isotope effects that should be observed at the indicated
positions.
N
O
MeO
O
MeO
N
H(D)
O
H(D)
N
Ar
O
O
H(D)
Ar
MeO
H(D)
N
rate-determining
step
N
O
elimination
H(D)
MeO
& protonation
OH
H(D)
Ar
N
(e) In a recent study, the following isotope effects were measured. What do these isotope effects suggest about the
rate-determining step?
N
O
O
H(D)
MeO
kH/kD = 5.2 ± 0.6
+
O
N
(D)H
OH
MeO
DMSO
NO2
kH/kD = 0.75 ± 0.05
NO2
(f) Propose an alternative mechanistic sequence that accounts for the kinetic and isotope effect data.
6. Is there a linear free energy relationship between enantioselectivity and the size of the substituent (R) in the following
aziridination reaction (ACIE 1998, 37, 3392)? Please propose an explanation for why or why not.
Ph
catalyst
pyridine, Ts2O
pyridine N-oxide
R
CH2Cl2, 3 h
TsN
Ph
R
H
R
H
Me
n-Pr
i-Pr
ee (%)
41
85
90
94
H
N N N
Mn
O
O
catalyst
H
3
7. The Rh(I)-catalyzed hydrogenation of methyl-(Z)-acetamidocinnamate (A) has been studied. The catalytic cycle shown
below has been proposed.
MeO2C
NHCOMe
P
MeO2C
Ph Ph
P
Rh
P
Ph Ph
1
Ph
k4
Ph
k–1
Ph Ph
Ph
P
Rh
O
P
Ph
CH3
k–3
k3
A
k1
Ph Ph Ph
P
H
CO2Me
Rh
P O
NH
Ph
Ph
4
NHCOMe
Ph Ph
P Ph
H
Rh
CO2Me
P H O
NH
Ph
Ph
H3C
3
Ph
CO2Me
NH
H 3C
2
k–2
k2
H2
(a) Assuming that the second step (2 –> 3) is rate-limiting and all subsequent steps to regenerate 1 are rapid and
irreversible, write down a “one plus” catalytic rate expression. Use the rate constants depicted in the catalytic
cycle above, and assume that any steps prior to the rate-limiting step are rapid and reversible compared with the
rate-limiting step. (ie, use the pre-equilibrium approximation). Express your rate law in terms of A, H2, and [Rh]total.
(b) Assuming that the second step (2 –> 3) is rate-limiting and all subsequent steps to regenerate 1 are rapid and
irreversible, apply the steady-state approximation to 2 and derive a rate law for the reaction in terms of A, H2, and
[Rh]total.
(c) Show that the equations derived in (a) and (b) are equivalent when k2 is small compared with k–1.
(d) The asymmetric hydrogenation of A using a chiral diphosphine ligand has an analogous mechanism. Draw a pair
of catalytic cycles that accounts for the formation of both enantiomers. You may depict the chiral diphosphine
generically.
(e) With chiral diphosphine ligand: Assuming that the second step (2 –> 3) is rate-limiting and all subsequent steps to
regenerate 1 are rapid an irreversible, write down a rate law for the total reaction rate using the steady-state
approximation. (Hint: Use your “1+’ rate law for the non-asymmetric reaction as a basis; it is not necessary to
provide a new derivation for this rate law.)
For R-enantiomer pathway:
(f) With chiral diphosphine ligand: Show that the relative rates of
4
–1 –1
k1 = 1.06 x 10 M s
formation of the two enantiomers depends only on [H2].
–1
k–1 = 3.2 s
(g) With chiral diphosphine ligand: Use the boxed experimental data to
–1 –1
k2 = 630 M s
determine the enantiomeric ratio expected from a reaction run at
infinitely low [H2].
For S-enantiomer pathway:
(h) With chiral diphosphine ligand: Use the experimental data to determine
3
–1 –1
k1 = 5.3 x 10 M s
the enantiomeric ratio expected from a reaction run at infinitely high
–1
k–1 = 0.15 s
[H2].
–1 –1
k2 = 1.1 M s
(i) With chiral diphosphine ligand: Explain why it is necessary to use the
steady-state approximation (rather than the pre-equilibrium approximation) to account for the kinetic profile of this
reaction.
8. Please read this article: Resek & Beak. J. Am. Chem. Soc. 1994, 116, 405.
(a) What is an intramolecular kinetic isotope effect experiment?
(b) Why are the values of the intramolecular and intermolecular kinetic isotope effects different?
From The Art of Writing Reasonable Organic Reaction Mechanisms, Ch. 5: Please draw reasonable arrow-pushing
mechanisms for the following transformations:
5. 9.
Me
O
Me
O
S
O
O
Me
O
MeS
CN
O
Me
Me
O
Me
O
O
O
Me
Bu3SnH
cat. AIBN
O
Me
NC
4
CH3
hv
6.10.
H3C
O
i-Pr
i-Pr
7.
11.
H3C CH3
cat. Bu3SnH
I
I
H3C
H3C
From The Art of Writing Reasonable Organic Reaction Mechanisms, Ch. 6: Please draw reasonable arrow-pushing
mechanisms for the following transformations. If the metal reagent is used catalytically, be sure to regenerate the catalyst
for the next catalytic cycle.
S
SiMe3
12.
H
Ph
cat. Pd(PPh3)4
cat. CuI
I
S
SiMe3
Et3N
Ph
OH
I
13.
cat. Pd(OAc)2
CHO
Et3N
Bn
14.
Fcm
O
N
Cl
CPh2
Cy3P Ru
Cy3P
Cl (5 mol %)
Bn
Fcm
N
O
5