1
CH402 MW component in 20011-2012 academic year ANSWERS
For workshop 26/10/2011.
Examples of questions that may be asked:
1) Azadirachtin.
16 – Give a mechanism for the hydroxymethylalkyne formation.
(either Br can be removed)
Br
Br
Br
:
R
R
R
R
R
H
H
O
iPr
Me
R
PRODUCT
MsO Ms O
17 – Explain the stereochemistry of the Diels-Alder step.
OMe
OMe
O
S
CO2Me
PhMe2Si
O
OMe
OMe
O
Diels-Alder
reaction
S
PhMe2Si
S
H
MeO2C
The selectivity is reversed if the
silane is not present.
H
OH
S
2.4:1 preference for this isomer.
Explanation requires a diagram of the transition state, to get the product illustrated, the alkene has
to be above the diene, The transition state involves an ENDO interaction of the CO2Me with the
diene as is typical for Diels-Ander reactions, however the bulk of the silyl group is also important:
O
O
H
PhMe2Si
CO2Me
S
R
S
O H
H
PhMe2Si
O
CO2Me
S
R
S
O H
H
PhMe2Si
CO2Me
S
R
S
O H
18 – Give a mechanism for the first step and explain why O-alkylation is observed.
The 15-C-5 chelates the Na+ and hence the resulting enolate is unhindered and
has no metal co-ordinating to the alkoxide, hence it reacts at O:
R
OBn
O
TESO
NaH,
MsO
R
[15-crowno
H
H
H
5], 0 C
TESO
O
O
O
H
O H
MeO C
MeO2C
2
18 – Why is one PMB ether used in the substrate for step 3, and no more?
Because it can be selectively removed using DDQ (dichlorodicyanoquinone
- oxidative).
18 – What reagent is used to remove the silyl group?
Fluoride of some sort, either HF (very strong) or a salt such as TBAF
(tertiarybutylammonium fluoride).
2
18 – Provide a mechanism for the rearrangement step.
It is not essential to draw the entire molecule, however
enough should be drawn to indicate the correct reagent:
R
OPMB
OBn .
O
H
O
H
H
O
O H
18 – Provide a mechanism for the radical cyclisation:
N
AIBN=
N
CN
-N2
(Bu3n)Sn
.
(x 2)
O
.
R
H
O
CN
(Bu3n)Sn
. Sn(nBu3)
.
S
OBn OMe
R
.
H
O
H
.
OBn OMe
OBn OMe
R
O
O
O
. Sn(nBu
(tin radical goes on to
initiate a further cycle
.
CN
NC
MeS
H
O
OBn OMe
3)
R
O
O
Problem question example; ‘why is the formation of a direct bond between the two halves of
the target molecule difficult to achieve?
Answer would be – the two halves are very complex and highly functionalised, and leave
little room for addition of functional groups to assist their connection ina reaction.
2) Strychnine.
25 – Give a mechanism for the key step.
This is essentially from the notes:
N
HN
usual
condensation
H
HO
NR2
OtBu
A
H
HO
NR2
OtBu
N
N
HO
NR2
H
OtBu
O
NR2
H
OtBu
B
26 - What is the mechanism of the step?
3
EtS
EtS
N
EtS
N
EtS
H
O
NH2
N
H
H
OSEM OPMB
B
H
C
OSEM OPMB
H
H
N
O
H
H
N
OH
H
N
27- Give a mechanism for the key step and explain how the stereochemistry is controlled.
What other protecting groups could you use?
be sure to show how one part of the molecule is positionally related to the other in the t.s.
On then, you could use a carbamate such as a Boc group.
N
N
KOtBu
THF, 0.02M
80oC
N
H
N
H H
O
O
N
N
N
H
H
H
H
H
N
H
N
H H
O
O
O tBu
H
OtBu
N
H H
O
28 – key step A-> C mechanism.
O
Ag
Br
O
NBn
NBn
AgOTf
N
H
A
N
H
CO2Me
O
NBn
H
B
CO2Me
NBn
DBU
(base)
N
CO2Me Show
deprotonation
next to ester then
O
protonation at O
NBn gamma
position as above.
N
H H
N
H H
CO2Me
C
CO2Me
Problem question example ‘A-> C why is this an efficient way to make the tetracyclic
system?’
4
An answer here would be that it makes two rings, one after the other, in an efficient process.
Compare this with the previous route – indicate which is shorter, how convenient are the
reagents etc. advantages, diadvantages.
3) Epothilones.
26 – provide a mechanism for the key Grubbs metathesis step.
Grubbs
metathesis
catalyst
S
HO
S
HO
N
N
O
O TBSO
O
O (1.2:1 E:Z)
Key Step
(3:2 with other
trans isomer)
O TBSO
O
37 – provide a mechanism for the DCC/DMAP step, and the metathesis step.
DCC/DMAP:
(show deprotonation
with DMAP first)
S
HO
HO
O
O
.
O TBSO
O
TBSO
O
NMe2
NCy
Get H+ from
NCy protonated
DMAP
O
NCy
N
NCy
TBSO
O TBSO
(complete
by
deprotonation)
ROH
N
NCy
O
O
O
NMe2
N
NHCy
TBSO
O
Metathesis:
You can abbreviate the Grubbs catalyst as Ru=CH2. You can also abbreviate the structure
as well as long as you define the abbreviation.
CH2
Ru
S
S
HO
HO
N
N
O
O
O TBSO
(1.2:1 E:Z)
Key Step
O TBSO
O
O
CH2
Ru
Ru
Ru
S
HO
N
S
HO
N
O
O TBSO
O
O
O TBSO
O
Problem question example ‘why is formation of a triple bond by metathesis sometimes
advantageous over formation of a double bond?’.
Because the double bond can form as an E/Z mixture. The triple bond can be reduced to a Z
bond selectively.
5
38 – Provide a mechanism for the aldol step.
S
TBSO
TBSO
N
N
O
O
H
TPSO
O
TMS
O
N
TPSO
O
O
S
TMS
TBSO
N
O
Product shown after
protonation.
TPSO
OH
O
Problem question example ‘why is the protecting group changed in the sequence?’.
39 – Yamaguchi lactonisation mechanism.
O
S
S
HO
HO
N
O TBSO
O
First the acid is
deprotonated by a
mild base such as Et3N
Cl
O
O
Cl
N
O
OTBS
O
Cl
Following loss of TBS
by adddition of Cl-
O
TBSO
O
O
Cl
TBS
Cl
S
S
N
HO
HO
N
OTBS
OTBS Cl
O
O TBSO
O
O
O TBSO
Cl
O
Cl
O
Cl
Cl
O
Cl
For workshop 18/11/2011.
4) Amphotericin B (6 slides, 4 key steps).
44/45 – give mechanisms of Wadsworth-Horner-Emmons reactions in one of these two steps.
6
This is a mechanism you should already know. The first step is a deprotonation which you should illustrte (even though I
have not). The anion is delocalised, although I have shown the C-based form. Partial structures are acceptable as long
as you make it clear which compounds they refer to.
OMe
OMe
OTBS
O
O
O
OTBS O
O
O
MeO
MeO
OTBS O
O
CO2Me
CO2Me
O
P
MeO
MeO O
OMe
OTBS
P
O
OMe
OTBS O
O
OTBS
O
MeO
MeO
CO2Me
O
OTBS O
O
O
OTBS
O
O
OMe
OTBS
O
CO2Me
OTBS O
O
O
MeO
MeO
P
O
CO2Me
P
O
45 - Problem question example ‘What could go wrong in the macrocyclisation step’.
It could do an intermolecular reaction, or polymerise, or not cyclise at all.
46 – cycloaddition mechanism.
(But)O
H
Cl
Cl
TBSO
N
OH
TBSO
N
O
H
BnO
O
O
O
O
O
O
But
O
TBSO
TBSO
XN
N
N
O
O
O
O
O
O
HO
MeO
O
O
HO
MeO
XN
O
O
47 – provide a mechanism for the key step – complete what’s shown below – this is basically
from the slide and has been completed now.
7
Cl R enantiomer
Ph2
P
10 mol%
O
of catalyst.
>99% ee with a 30:1
preference for this
isomer over the meso.
OMe
OMe
Ir
O
Ph2P
OH
OH
Cl
OH
O2 N
OH
Cl
OAc Cs2CO3, 110oC
H
H
Ir
Ir
- OAc
(abbreviated
catalyst)
Ir
OAc
OAc
H
H
H
Ir
O
OH
O
Ir
H
H
OH
OH
Ir
O
OH
+
H
Ir
H
OAc
48 – Krische strategy – why take this approach, i.e. what are the advantages?
It provides asymmetric and regiocontrol, you can build a chain in both directions in same
step.
5) Taxol (5 slides, 6 key steps).
52 – Give a mechanism for the cycloaddition. Explain how the stereochemistry is controlled.
O
OH
+
O
CO2Et
O
PhB(OH)2 90oC
O
EtO2C
(Diels-Alder)
O
OH
BPh
O
O
O
O
O
O CO2Et
B
Ph
(Endo TS)
O
O
O CO2Et
B
Ph
53 – Shapiro mechanism? ‘Why are lithiated alkenes difficult to form?’
Because alkene protons are not very acidic. Other positions tend to deprotonate first. The
mechanism is shown below.
8
TBSO
O
TBSO
H
i) nBuLi
(shapiro reaction)
OTBS
OBn
OTBS
ii)
OBn
O
O
H
O
HO
H
N NSO2Ar
nBu
O
H
nBu
TBSO
After protonation,
which you should
show in the answer
OTBSOBn
TBSO
TBSO
O
H
O
N N
SO2Ar
O
N N
HO
O
OH
O
OBn
O
H
O
O
O
OBn
TiCl3.DME
Zn-Cu
McMurry
Coupling
O
H
O
O
O
O
O
TiCl3 + Zn/Cu -> Ti(II) (or Ti(0)
O
Ti(II)
O
O
Upon work up and
hydrolysis.
If you use Ti(0) it will be
oxidised to Ti(IV)
Ti(IV)
O
OBn
OBn
53 – Mechanism of McMurry coupling, What are strengths of this approach?
The mechanism is shown above. There is some debate as to whether it is Ti(0) or Ti(II). I will
accept either. The important thing is the chelation control. It forms a C-C bond in a
stereoselective manner.
55 – Why does the epoxide open, i.e. what drives the reaction?
Release of a lot of ring strain drives it.
56 – mechanisms of both key steps should be known.
A->B is a standard dihydroxylation which you should know. C->D is an SN2-type
cyclisation..
56 – why does the cyclisation of C-D proceed as shown. – Because it goes with inversion of
configuration.
6) FR182877/abyssomicin C (5 slides, 5 key steps).
9
64 – Cyclisation step – why is a ‘transannular’ intramolecular cyclisation employed. Provide
a mechanism for the reaction.
Because the two reacting components are held close together by the ring. Mechanism is
below, make sure you show the overlaps if you are asked to explain why the outcome is as
shown. The mechanism is also shown on slide 65, for the other enantiomer – careful!
TESO
TESO
H
OTES
O
H
OTES
O
HH
OtBu
OtBu
H
H
OTMS
O
OTMS
O
TESO
H
OTES
O
H
H
OtBu
H
H
H
OTMS
O
65 – Mechanism of the addition step.
I
PPh3
I
I
I
TBSO
PPh3
TBSO
OTBS
O
OTBS
I2, PPh3
CO32H
OEt
O
then CsCO3
OEt
Br
OH
O
I
PPh3
O
OEt
OTBS
An acceptable alternative
would be to form the allylic
iodide then displace it with
the anion.
O
Ph3P
O
66 – Provide a mechanism for the reaction.
O
OTBS
10
O
O
Pri
N
H
Pri
O
OMe
O
product formed after addition
of acid in workup, which should
be illustrated.
OMe
OTBS O
O
67 – Provide a mechanism for the reaction.
O
O
O
O
O
O
OMe
O
O
O
OMe
O
O
O
O
O
O
OMe
OMe
O
68 – Give a mechanism for the reaction. Why do the addition before the metathesis, and not
the other way round?
The mechanism is shown below. If you tried to do the metathesis first, you would not have
the advantage of an intramolecular cyclisation and you’d get homodimers as well as the
required product.
i) tBuLi then
O
HO
O
O
OTES
PMBO
OH
O
O
O
First the tBuLi deprotonates to form the anion shown.
7) Statins (6 slides, 5 key steps).
O
OTES
11
77 – provide a mechanism for one or both reactions, explain how the stereochemistry is
controlled in the cycloaddition.
I have already shown a mechanism for the Wadsworth-Emmons reaction, which is the first
step. The second step is a cycloaddition. See below – use appropriate overlaps to indicate
stereochemical control.
OBn
OBn
O
OBn
O
O
O
+
O
O
O
P(O)(OMe)2
O
NaH
O
THF
H
H
OTBS
O
OTBS
H
OTBS
78 – Provide a mechanism and explain how the stereochemistry is controlled.
TMSO
TMSO
O
O
H
H
H
TMSO
H
Endo Diels-Alder
reprotoantes
O
TMSO
TMSO
O
deprtonates
H
O
H
H
H
H
H
79 – Give a mechanism for the key step. What advantage does the reaction have over enolate
formation?
Mechanism is shown below. Note that each Sm(II) provides 1 electron in a reductive process
so you need 2 eq. Of SmI2.
12
O
O
Br
O
O
O
O
Sm(III)
O
O
H
O
O
2 x SmI2
H
(reductive coupling)
HO
H
O
H
O
O
O
H
Product after protonationnote that protonation should
be shown.
H
80 – Give a mechanism for the reaction.
It is the same mechanism as for slide 18 above, but a different substrate.