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CEM 850 FS20 final exam
2
CEM 850 Final Exam Review
December 15, 2020 This exam consists of 6 pages. Please make certain that your exam has all of the necessary pages. Total points possible for this exam are 150. In answering your questions, please write legibly and draw all structures clearly.
You are not to use any notes or outside resources when taking this exam.
You may send me a PDF of your answers that has been scanned or created from ChemDraw. You may also send me the original ChemDraw. I will not except any other image formats.
Good luck.
I. Pd-catalyzed cross-coupling reactions often involve a catalytic cycle that includes oxidative addition, trans-metallation, and reductive elimination steps. Illustrate a catalytic cycle involving those three steps for the generic cross-coupling reaction shown below. (9 pts)
R–X + R'–Mcatalytic Pd(0)
R–R' + MX
CEM 850 FS20 final exam
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II. Throughout the semester, we saw examples of amines reacting with carbonyl compounds. Provide all products for each of the four examples shown below and provide complete arrow (electron) pushing mechanisms that explain your answer. (24 pts)
1.
2.
3.
4.
O excess NH2OH
catalytic H+
NOH+ H2O
excess
catalytic H+O
NH
N + H2O
1 equiv
ClO
NH
ON
+Cl
O + NH • HCl
(1:1:1)
catalytic
MeOO
N N
PhCHO
MeOO
PhOH
MeOO
N N
MeOO
N
N
MeOO
PhOH
PhO
MeOO
N
N OPh
HPT
MeOO
N
N OHPh
+ N N
CEM 850 FS20 final exam
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III. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (10 pts)
OBnOMe
O
HCHOaq, Me2NH•HCl,
Ac2O, 95 °C
MeI, DMF;
then KCN
OBnOMe
O
NC
88% (two steps)
a mixture ofβ-aminoketone
and enone
OBnOMe
O
OBnOMe
O
NC
HCHO+
aq, Me2NH•HCl,N
Me Me+ H2O
Cl
OBnOMe
OH
NMe Me OBn
OMe
O
N
H
H2OCl
OBnOMe
O
N
H3O+H
OBnOMe
O
NH
H
ClH2O
OBnOMe
O
NH
OBnOMe
O
NH
I–Me
OBnOMe
O
NH
OBnOMe
O
+KCN
I–
—CN
—CN
H+
CEM 850 FS20 final exam
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IV. The reaction of 2-methyl cyclohexanone with lithium diisopropylamide (LDA) at –78 °C can generate the less substituted (kinetic) lithium enolate preferentially over the more substituted (thermodynamic) lithium enolate. To assure the highest levels of selectivity it is best to add the ketone to an LDA solution as opposed to adding the LDA to a solution of ketone. Provide a detailed arrow (electron) pushing mechanism that explains this operational observation. (6 pts)
THF, –78 °CO
NMe Li
OLi
Me
+ OLi
Me
major minor
O
Me
OLi
Me
Ketone added to LDA
N Li
N Li
N Li
N LiH
As ketone is added to the reaction it "sees" an excess of LDA. Kinetic deprotonation occurs and is irreversible. Thus the kinetic enolate predominates.
:B–Li
+ excess LDA
rxn mixture
ketone continues to be added OLi
Me
major
O–Li
Me
OLi
Me
LDA added to ketone
N Li
As LDA is added to the reaction it "sees" an excess of ketone. Kinetic deprotonation occurs, however the kinetic enolate can behave as a base and deprotonate some of the excess ketone. This sets up an equlibrium affording more of the thermodynamic enolate in the final mixture.
rxn mixture
OLi
Me
O
Me
O
Me O
MeO
Me
+excessketone
+
O
MeH
O
Me
+excessketone
LDAcontinues
to be added
OLi
Me
CEM 850 FS20 final exam
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V. Our last homework assignment involved the reaction sequence shown below.
The starting dithiane undergoes deprotonation with n-BuLi in step one, the product of which then reacts with the epoxide in step two. Were one to start with the corresponding silyl-dithiolane the desired ring opening of the epoxide does not occur. Provide a detailed arrow (electron) pushing mechanism that explains why. (6 pts)
VI. In class, we saw an example where an organocerium reagent was prepared from a Grignard reagent and used as a nucleophilic addition to an ester.
What advantage can organocerium reagents hold over organolithium and/or Grignard reagents? (3 pts) They are less basic and therefor less likey to deprotonate carbonly substrates.
S S
SiR3
n-BuLi (1.3 equiv)
Et2O, 0 °C, 115 min1.2 equiv
OSiR3
O
(1.35 equiv)
Et2O, –30 °C, 2 hS
S Li
R3SiS
S
R3Si LiO OSiR3
S S
SiR3
n-BuLi (1.3 equiv)
Et2O, 0 °C, 115 min1.2 equiv
OSiR3
O
(1.35 equiv)
Et2O, –30 °C, 2 hS
S Li
R3Si
unreacted epoxiderecoverd
SS Li
R3Si
CH2CH2
+ TMS SLi
S
TBDPSO
O
OEt
O
THF, –78 °C, 2 h; rt, 12 h
Ph
TBDPSO
O81%
PhSiMe3MgClMe3Si
2 M aq HCl (21 equiv)
THF, 0 °C
CeCl3 (5.0 equiv)
THF, –78 °C
CEM 850 FS20 final exam
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VII. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (10 pts)
O
O
OTIPS
Meexcess MeMgBr
excess CeCl3
THF
Swern [O]
OTIPS
OHO
O
O
H3CMgCl
Cl2Ce ClH3C
CeCl2 + MgCl2
H3CCeCl2
CeCl2
OTIPS
OHHO
OTPS
O
OTPS
—O CH3O—
OTPS
O CH3 H3CCeCl2
OTIPS
O——O
CeCl2 CeCl2CeCl2
aqueous
workup R R'
OH
SO
SO Cl
ClO
OSO
OCl
Cl O
SO
OCl
O
Cl
SCl O
OCl
O
CO2 + CO + Cl
R OH SCl
R OS
H
R OSCH2
H
Et3N
R OSCH2H
+ Me2S
Cl
Et3NH+ Cl
SO S
OEt3N
HClCl
2SO
+
R' R'
R'
R'
OTIPS
OHO
Swern
CEM 850 FS20 final exam
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VIII. In class, we saw how NaBH4 of of enones can proceed in a 1,2- or 1,4-fashion. However, in the presence of CeCl3, NaBH4 reductions of enone tend to be highly 1,2-selective.
The presence of CeCl3 can also alter the selectivity of ketone vs. aldehyde reductions. Typically, aldehydes are more reactive than ketones. Thus, while the selectivity is often low, reductions of compunds bearing both a ketone and an aldehyde will proceed with the aldehyde being preferncially reduced.
In contrast, if CeCl3 is added to the process, the major product after acidic workup is that in which the ketone has been reduced. Provide an explanation for this seeming preferential reduction of the ketone in the presence of the aldehyde. Note: I am not asking for an arrow pushing mechanism, but you may want to think about intermediates that may form in the presence of CeCl3. (6 pts)
O
NaBH4(1 equiv)
MeOH OH OH O
+ +
O
+
NaBH4CeCl3
MeOHOH
O
NaBH4(1 equiv)MeOH;
then H3O+
CHO
O
CH2OH
+
OH
CHO
major minor
CeCl3MeOH;
then NaBH4(1 equiv)
then H3O+O
CH2OH
+
OH
CHO
minor major
O
CHOCeCl3MeOH
OH
CHO
major
CeCl3 is a Lewis acid
O
OMe
OMe
in situ protection of the aldehyde as an acetal
NaBH4(1 equiv)
OB(OR)Na
OMe
OMe H3O+Acidic workup hydrolyzes the borate salt and deprotects the aldehyde
CEM 850 FS20 final exam
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IX. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (6 pts)
OTIPS
OMe
HOS CH3
NN
N NPh
OOKN(SiMe3)2 (2.2 equiv)
MeOCH2CH2OMe, –55 °C, 80 min
1.0 equiv
1.0 equiv
MeOCH2CH2OMe, –55 °C, 1 h; rt, 12 h
H2O
OTIPS
Me
HO
SNN
N NPh
OOH
HH
K N(SiMe3)2 SN
N NPh
OO
N
K
HH2N(SiMe3)2+
R R'
O
NN
N N
S
O
NNN N R'
OO
PhPh
K
NNN N
Ph
K
O
SO
O
NNN N
Ph
OK
+
+ SO2
S
O
OO
R
K
OTIPS
OMe
HO
2 equiv
SNN
N NPh
OOH
HH
K N(SiMe3)2 SN
N NPh
OO
N
K
2 equiv
KN(SiMe3)2
+
OTIPS
OMe
KO
KN(SiMe3)2
HN(SiMe3)2+
R R'
O
R'
R
RR'
OTIPS
Me
KO
H2O
OTIPS
Me
KONN
N NPh
OH
CEM 850 FS20 final exam
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X. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (8 pts)
OO
OMeO
cat. TsOH
HB
1.
2. H2O2, NaOHH
OH
OO
O
OMeO
H+
O
OMeO
HHO
OMeO
PT
HOHO
O
OHO
MeH
H
PT
OO
OHO
MeH
HO
O
OHO
MeH
H
–H+ HO
HOHCO2Me +
OHOO
H+
OH
OHOH
HO
HPT
OH2O
HO
O
H2O
HOO
OH
OO
–H+
H BR2
OO
HBR2
H O ONa
OO
HBR2O O
H
OO
HO BR2
Na+–OHNa+
OO
HO B
R2
OHNa+
HONa
OO
+ R2BOH
H2O HOH
OO
NaOH
CEM 850 FS20 final exam
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XI. Provide a complete arrow (electron) pushing mechanism for the transformation shown below. (6 pts)
XII. Using your mechanistic knowledge provide the stuctures of intermediat products A–D. Note: I am not asking for arrow pushing mechanisms (12 pts)
OHONNHTs
excess MeLiTHF; then
H3O+ workupOHHO
R
N NSO
R
R
R
Me—H
Li+
LiO
OHHO
OHONNHTs
excess MeLiOLiO
NNLiTs RH
N N SO
OR
R
R2 equiv Me—HLi+
Me—Li
R'
R' OLiOLi OLiLiO
H3O+
workup
O
H
O OMe
PhMe, 160 °C16 h (92%)
A
48 HO
H
O
HH2O2, NaOH
MeOH
B
silicagel
O
H
ONaBH4
MeOH
C
NaH, CS2, MeI
THF, rt
AIBN, Bu3SnH
PhH, 80 °C
H
O
OH
H
O
DO
H
O
S
SMe
O
H
O
H
MeO
O
H
O
H
O
H
CEM 850 FS20 final exam
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XIII. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (10 pts)
HOO
O
MeO MeMeO NMe2
toluene, reflux82%
CF3CO2H,
THF/H2O, rt65%
OH
O
OH
NOMeMeO
N
OMe OMe
N
MeO
HOO
O
OO
O
N
MeO OO
O
N
HMeO–
OO
O
N
MeOH
OO
ON [3,3]
OO
CF3CO2H,
THF/H2O, rt65%
acetalhydroylsis
H+
OH
OH
NMe2
O
OH
OH
NMe2
O OH
NMe2 OH
OH
H
HO NMe2PT OH
OH
HO NMe2
H
OH
O
OH
H
HNMe2
–H+ OH
O
OH
CEM 850 FS20 final exam
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XIV. Provide a complete arrow (electron) pushing mechanism for the sequence shown below. (10 pts)
HO
OHSiMe3
SiMe3
NO2Se
CN
(3 equiv)n-Bu3P (3 equiv)
THF, rt, 4 h
30% aq H2O2 (89 equiv)
THF, rt, 4 hNaHCO3 (10 equiv) OH
SiMe3
SiMe3
NO2Se CN
:PBu3 NO2Se PBu3
R
O H
NO2Se PBu3
HCN
R
O
NO2Se
R
O PBu3
R
Se
O2N
OH2O
R
Se
O2N
O
HSe
O2N
HO
R
O PBu3
Se
O2NHO–OH
R
Se
O2N
OHO
H
H2O2
R
Se
O2N
OHO
HOH HOH
H
H2O+
CN
HO
OHSiMe3
SiMe3
1
2 2
1
OHSiMe3
SiMe3
+
CEM 850 FS20 final exam
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XV. Provide a complete arrow (electron) pushing mechanism for the transformation shown below. (6 pts)
2 M aq HCl (21 equiv)
THF, 0 °C
OHSiMe3
SiMe3
SiMe3
OHSiMe3
SiMe3
OSiMe3
SiMe3
HH Cl Cl
H2O+
SiMe3
SiMe3
H2O+
SiMe3
SiMe
Me
MeCl
SiMe3 ClSiMe3+
H O H
HCl
H2O+
CEM 850 FS20 final exam
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XVI. Provide a complete arrow (electron) pushing mechanism for the transformations shown below. (18 pts)
Bonus Question: Pfizer’s COVID vaccine is being manufactured at their Portage, MI facility. That facility was originally part of the Upjohn company. Over seventy years ago, Pfizer and Upjohn were aggressively looking to develop practical syntheses of cortisone and cortisol. (Merck’s cortisone synthesis was 36 steps.) Both companies ultimately discovered biocatalytic approaches that relied on 11-deoxycortisol. Upjohn’s original supply of 11-deoxycortisol came from the Glidden company, who in 1953 granted Pfizer exclusive license to their 11-deoxycortisol patents. Who was the Glidden chemist that invented their synthetic route to 11-deoxycortisol? (2 pts) (a) Carl Djerassi (b) Percy Julian (c) Herbert Murray (d) Durey Peterson (e) Arie Luyendyk
Me
HAcO
CF3CO3H
CHCl3
Me
HAcOO
160 °C
no solvent
TSOH
CHCl3
MeOAc
O H
H
MeH
O H
AcO
O
160 °C
no solvent
MeH
O H
AcO
Me
H
O
O MeOAc
O H
H
O
Me
H
O
OH+
O
Me
H
O
O
H
MeH
O H
OOH