123.713A/B. Description of the Jacobsen synthesis of muconin. This is an example of total synthesis, retrosynthesis and asymmetric synthesis and shows the kind of information required in the assigment for this course.

1. synthesis of
muconin
©Reinaldo Aguilar@flickr

2. O
HO
O
OH
C12H25
O
O
OH
This molecule is called muconin.
This document will act as an example of what is expected in the 123.713A/B
assessment and it will discuss the synthesis of muconin.
A glance at the mark scheme reveals that you should describe the retrosynthesis or the
overall plan before outlining a possible synthesis. The emphasis is on the creation of
the stereocentres within the molecule so this synthesis will not involve any chiral pool
materials.
I will go into more detail than you need to as I will use this as an opportunity to
reinforce some of the principles we have covered.

3. O
HO
O
OH
C12H25
O
O
OH
The Retrosynthesis
Ideally we want a convergent synthesis. This requires the target to be split in half.
The alcohol functionality is the obvious point to split the molecule. C–C bond formation
at this point is simple, it involves nucleophilic addition to a carbonyl group.

4. O
HO
O
OH
C12H25
O
O
OH
C–C
addition to
aldehyde
O
O
O
OH
C12H25 O
O
OH
fragment A fragment B
M
We could disconnect the
molecule on either side of
the alcohol.
Two factors favour the
disconnection shown:
1. Ease of generation of
the nucleophile.
An alkyl organometallic
reagent should be simpler
to prepare than an
anomeric nucleophile
(although there are a
number of methods for
achieving this).
2. Chelation control.
A heteroatom 𝛂-to the
carbonyl might allow
stereocontrol in the
nucleophilic addition step.
This gives two fragments.

5. O
O
O
OH
C12H25
O
O
O
OH
C12H25
O
O
O
OH
C12H25
hetero-Diels-Alder
challenging
Fragment A is by far the more challenging half of the molecule. It contains 5
contiguous stereocentres with 4 of them build into tetrahydrofuran and
tetrahydropyran (THP) rings.
The formation of a 6-ring is often possible by a variant of the Diels-Alder reaction. In
this case a hetero-Diels-Alder (HDA) reaction. It would still leave some stereochemistry
to be installed but would be a good starting point …

6. O
O
O
OH
C12H25
O
O
O
OH
C12H25
O
O
O
OH
C12H25
hetero-Diels-Alder
challenging
… formation of the 5-ring is more challenging. One might argue that formation of C–O
bonds should be easy as we have a host of nucleophilic addition reactions available to
us but this would be a very linear process (making each by addition to a carbonyl).
Alternatively, we could consider installing some by Sharpless Asymmetric Epoxidation
or Dihydroxylation. This might be productive train of thought (consider the work of Tim
Donohoe). But we’re going to stick to the HDA as we have already mentioned Diels-
Alder reactions.

7. The idea will be to install at least one stereocentre of the THP ring by a catalytic
asymmetric HDA reaction.
This means we need to disconnect the tetrahydrofuran ring from the THP ring. It might
be possible to create the requisite C–C bond with control of two adjacent stereocentres
by using a [3,3]-sigmatropic rearrangement such as a Claisen or Cope rearrangement.
Such rearrangements involve 6 atoms in a cyclic transition state, 5 of the atoms have
been highlighted above. The 6th must be on the tetrahydrofuran side …
O
O
O
OH
C12H25

8. O
O
O
OH
C12H25
FGI
FGI
protecting
groups
alkene
O
PG2O
O
OPG1
C12H25
To prepare for such a rearrangement we require more functionality in the molecule.
Technically the two alkenes above would allow a Cope rearrangement but that doesn’t
simplify the problem for us; we still have two rings that would need to be joined by a C–
C bond.
Instead, only one of these alkenes is required for the rearrangement. The other has
been incorporated to aid construction of the tetrahydrofuran ring.

9. O
PG2O
O
OPG1
C12H25
O
PG2O
O
OPG1
C12H25C=C
RCM
FGI
O
PG2O
O
OPG1
C12H25
OMeO
O
PG2O
O
OPG1
C12H25
OO
1
2
3
1
2
3
[3,3]-sigmatropic
rearrangement
Ireland-Claisen
rearrangement
ring-closing
metathesis
C=C bond formation is
relatively simple. There
are many different
possibilities, Witting
reaction, Peterson
olefination, Julia
olefination, McMurry
coupling and metathesis.
The latter is both powerful
but mild.
Things are now looking
much better. We have two
alkenes with terminals 6-
atoms apart. This is
perfect for a
rearrangement reaction.
But if we did this we
would still have a
challenging C–C bond to
make.
The alternative …

10. O
PG2O
O
OPG1
C12H25
O
PG2O
O
OPG1
C12H25C=C
RCM
FGI
O
PG2O
O
OPG1
C12H25
OMeO
O
PG2O
O
OPG1
C12H25
OO
1
2
3
1
2
3
[3,3]-sigmatropic
rearrangement
Ireland-Claisen
rearrangement
ring-closing
metathesis
… is shown in the left
hand corner; the Ireland-
Claisen rearrangement.
This converts a C–O bond
into a C–C bond (during
the rearrangement we
break a bond between 1 &
1 and form one between 3
& 3).
Therefore we need to
carry out a series of
functional group
interconversions (FGI) to
instal an ester or in the
forward sense we must be
able to convert an ester
into an alkene. This is
relatively simple (as I
said, formation of C=C
bonds is not hard).

11. And now all the hard work pays off with a massive simplification of our problem.
The Ireland-Claisen rearrangement has the power to convert an ester (easily formed)
into a C–C bond while creating two contiguous stereocentres. More importantly the new
target (the right hand side) looks simple to disconnect. We have two C–O bonds that
can easily be removed (ester and acyclic ether). This leaves …
O
PG2O
O
OPG1
C12H25
OMeO
C–C
rearrangement
O
OPG2
O
O
O
C12H25 OPG1
12
3
1
2
3
Ireland-Claisen

12. O
OPG2
O
O
O
C12H25 OPG1
hetero-Diels-Alder
O
OPG2
O
O
O
C12H25 OPG1
substrate control
… a dihydropyran that can be prepared by asymmetric HDA, a functionalised acetate
and a chiral diol. The latter could be prepared by a number of routes; reagent or
catalyst controlled crotylation of a greasy aldehyde or substrate controlled alkenylation
of a chiral aldehyde.
This shows there are many correct answers. All would get you the marks (as long as you
could rationalise the stereochemical outcome).

13. O
OPG2
O
O
O
C12H25 OPG1
esterification
alkylation
2x C–O
C12H25
OPG1
OH
Br
Br
O
O
OPG2
OH
So this slide shows what units we require to prepare fragment A. Doesn’t look to
challenging at all anymore?

14. C12H25
OPG1
OH
C–C
Cram-chelation
control
C12H25
OPG1
O
M
C12H25
OH
OH
FGI
C12H25
O
C–O
Jacobsen's hydrolytic
kinetic resolution
The alcohol stereocentre
could be formed by Cram
Chelation control as long
as we can set up the
initial stereocentre.
Here it comes from a
chiral diol that will be
formed using Jacobsen’s
powerful hydrolytic kinetic
resolution but there are
many other routes to the
same compound.
Other possibilities include
asymmetric epoxidation
of the alkene by either
Jacobsen or Shi
chemistry. Sharpless
Asymmetric
Dihydroxylation might
also give the diol.

15. O
OPG2
OH
FGI
O
OPG2
O
C–O
C–C
HDA O
OPG2
OTMS
H3CO
hetero-
Diels-Alder
The dihydropyran comes from substrate controlled reduction of a enone, which in turn
is derived from catalytic asymmetric HDA. There are a number of catalysts that can
mediate this transformation. We have seen an example of a chromium ‘half-salen’
version already. There are a number of diol-based organocatalysts that can also control
this reaction (Rawal kick started this area).

16. O
O
OH
fragment B
FGI
O
O
OPG3M
Fragment B
This fragment is considerably easier to prepare; there are only two stereocentres and
plenty of functionality to play with.
First we have to find a method of creating the nucleophilic fragment. This could be
achieved with a halide and conversion to the Grignard but such functionality is hard to
carry through an entire synthesis so we have masked it here as an alkyne.

17. The key to this fragment is recognising that one (and in fact both) of the stereocentres
could be formed from a chiral epoxide (and hence an alkene if we want to go back to
cheap achiral starting materials).
An alcohol four carbon atoms from an carbonyl as shown above is readily formed by
ring opening an epoxide with an enolate. So all we have to do is find reliable chemistry
to instal the double bond of the lactone as that then frees up the enolate …
O
OPG3
O
( )4
1
2
3
4 1,4-diO
enolate
and
epoxide

18. O
OPG3
O
( )4
FGI
O
OPG3
O
( )4
PhS
C–C
OPG3
( )4 O
O
≡
PhS
O
( )4
enolate
alkylation
Sulfoxides readily undergo syn-elimination so can be used to install the double bond.
Such functionality would also increase the acidity of the 𝛂-protons making enolate
formation easier. I haven’t used the sulfoxide for the other steps as it adds unnecessary
stereochemistry to the system (giving us a problem with diastereomers) but it can be
easily added at the last minute (oxidation). Alternatively, we could use palladium
chemistry to add the double bond.

19. O
OPG3
O
( )4
FGI
O
OPG3
O
( )4
PhS
C–C
OPG3
( )4 O
O
≡
PhS
O
( )4
enolate
alkylation
This allows the epoxide-enolate disconnection.
How to we make this epoxide?

20. O
( )4
C–C ( )3
O
≡
≡
( )3
O
MgBr Cl
organometallic
It is readily prepared from epichlorohydrin (the chemistry of which we covered in the
third year).
This can either be bought (if we are cheating) or can be prepared. In this synthesis it
will be prepared by Jacobsen’s hydrolytic kinetic resolution again (for two reasons;
firstly I think this one of the methods used commercially and the second will soon
become clear …)

21. O
O
PhS
OH
PhS
H3CO O
C–C
H3CO
O
SPh
OH
≡
≡O
H3CO
O
SPh
C–O
lactone
formation
epoxide
opening
The lactone fragment is
again prepared by the
simple ring opening of an
epoxide with an enolate;
standard undergraduate
chemistry.
The only difficulty is
sourcing the chiral
epoxide. Once again, this
is so simple it is
undoubtedly
commercially available.
But in the spirit of the
assignment we should
make it so we could use
Shi or Jacobsen epoxide
or, of course …
Jacobsen’s hydrolytic
kinetic resolution.
And now for the synthesis.

22. C12H25
i. mCPBA
ii. hydrolytic kinetic
resolution
C12H25
OH
OH
C12H25
O
i. CH(OCH3)3, H+
ii. DIBAL-H
C12H25
OMOM
OH
i. TEMPO, NaOCl
ii. CH2=CHMgBr,
MgBr2C12H25
OMOM
OH
Synthesis
Starting with Fragment A.
The first important step in
this reaction is the
Jacobsen hydrolytic
kinetic resolution. This
reaction takes a racemic
mixture of epoxide and
hydrolyses one
enantiomer far quicker
than the other. This
results in a mixture of
diol and epoxide being
formed. As these are
different molecules they
are readily separated. The
only major draw back of
this methodology is that
the maximum yield is
50%.

23. H
R
H
O
t-Bu
t-Bu
N
Co
O
N
t-Bu t-Bu
L
O
H
R
O
t-Bu
t-Bu
Co
O
t-But-BuOH2
N
N
H
H
H
H
H
O
t-Bu
t-Bu
N
Co
O
N
t-Bu t-Bu
L
O
O
t-Bu
t-Bu
Co
O
t-But-BuOH2
N
N
H
H
disfavoured
The hydrolytic kinetic resolution involves the cobalt-salen-mediated addition of water to
an epoxide. Extensive kinetic studies have revealed that the reaction is second order
with respect to the salen catalyst and the model above has been proposed to explain
the selectivity. It is thought that both the nucleophile and the electrophile are both
activated by the cobalt and it is the fact that both are now chiral that leads to such high
selectivity in the reaction.
Science 1997, 936 & J. Am. Chem. Soc. 2004, 126, 1360

24. The salen ligand basically acts as a sloped floor (or
ceiling) that blocks approach of a substrate to three
sides (The tert-butyl groups block approach from the
front. The aromatic ring blocks approach from the side
that is sloped downwards while the axial hydrogen of the
diamine hinders approach from behind). Only if the two
‘open faces’ of the catalyst meet can the nucleophile
attack. Further more the orientation of the epoxide is
controlled by minimising its interactions with the various
substituents.
H
R
H
O
t-Bu
t-Bu
N
Co
O
N
t-Bu t-Bu
L
O
H
R
O
t-Bu
t-Bu
Co
O
t-But-BuOH2
N
N
H
H
H
H
H
O
t-Bu
t-Bu
N
Co
O
N
t-Bu t-Bu
L
O
O
t-Bu
t-Bu
Co
O
t-But-BuOH2
N
N
H
H
disfavoured
H
OH
OH
R
≡
R
OH
OH

25. C12H25
i. mCPBA
ii. hydrolytic kinetic
resolution
C12H25
OH
OH
C12H25
O
i. CH(OCH3)3, H+
ii. DIBAL-H
C12H25
OMOM
OH
i. TEMPO, NaOCl
ii. CH2=CHMgBr,
MgBr2C12H25
OMOM
OH
The next step involves a
selective protection of a
secondary alcohol in the
presence of a primary
alcohol. This is a
challenging reaction;
primary alcohols are
frequently more reactive
than secondary alcohols
due to reduced steric
hindrance.
In this case the selectivity
is achieved by first forming
an orthoformate and then
selectively cleaving one
bond (see next slide).
In all honesty in your
assignments I will not be too
picky about protecting groups
as long as there are no
screaming errors. If you have
selected them it normally
means you are following a
paper (someone else picked
them for you).

26. C12H25
O
O
OCH3
C12H25
O
O
OCH3
AlHR2
C12H25
O
O
OCH3
AlHR2C12H25
OMOM
OH
This the probable mechanism for the selective protection. Orthoformate formation is
followed by reduction with DIBAL. The Lewis acidic aluminium coordinates to the most
accessible oxygen atom. This activates this oxygen atom as a leaving group. The lone
pair of electrons on a second oxygen atom kicks it out giving an oxonium ion that is
reduced by the DIBAL.
Remember in your assignments to add the mechanisms to interesting reactions. This will show you are
thinking about the chemistry and not copying it from a paper. You will get higher marks.

27. C12H25
i. mCPBA
ii. hydrolytic kinetic
resolution
C12H25
OH
OH
C12H25
O
i. CH(OCH3)3, H+
ii. DIBAL-H
C12H25
OMOM
OH
i. TEMPO, NaOCl
ii. CH2=CHMgBr,
MgBr2C12H25
OMOM
OH
Mild oxidation of the
primary alcohol to an
aldehyde is followed by
Cram Chelation
Controlled addition of an
alkenyl Grignard reagent.
Remember what half
this module was
about? Asymmetric
synthesis so you
must go through
the selectivity …

28. C12H25
OMOM
O
O
H
C12H25
O
H
H
C12H25 OMOM
OH
H
Mg
Br Br
OCH3
BrMg
C12H25
OMOM
OH
≡
This is an example of
substrate control and
Cram Chelation control.
There are two ways of
representing this. One
uses the Newman
projection (shown right).
First translate the skeletal
representation into a
Newman projection. Then
tie the two Lewis basic
oxygen atoms together.
The nucleophile
approaches along the
Bürgi-Dunitz angle.
Finally, take the new
Newman projection and
convert it back to a
skeletal drawing. The
second method uses the
skeletal representation
and redraws the chelate
as a 5-membered ring
and attacks from the least
hindered face.

29. Simple alkylation then gives one half of the Ireland-Clausen rearrangement precursor.
C12H25
OMOM
OH
I
ONa
O
C12H25
OMOM
O CO2H

30. The dihydropyran moiety is going to be prepared by a
hetero-Diels-Alder reaction of a reactive, electron rich,
diene (known as Danishefsky’s diene) and an aldehyde
(the protecting group PBB is para-bromobenzyl). The
reaction is catalysed by the chromium-salen complex on
the right.
The rgiochemistry is most readily explained by
considering the polarity of the bonds. For the aldehyde
the carbonyl carbon is δ+. For the diene, the electrons
are being pushed towards the top (as drawn above;
consider the resonance forms). Matching the δ+ and δ–
together gives the regiochemistry shown.
TMSO
OCH3
O
OPBB catalyst
O
TMSO
OPBB
OCH3
H
O
O
OPBB
N
Ot-Bu
t-Bu
N
t-Bu
t-Bu
Cr
O
BF4

31. H
Cr
O
L O
N
N
O
H
t-Bu
t-Bu
t-Bu
t-Bu
H
OPBB
TMSO
OCH3
H
H
H
The stereochemistry can be explained by considering the sloped shape of salen
catalysts. The aldehyde coordinates to the Lewis acidic chromium. It is orientated so
that the metal is trans to the aldehyde substituent. It is suspected that aldehyde points
away from the axial hydrogen atom to minimise non-bonding interactions as shown.
The diene cannot approach from the front due to the tert-butyl groups. It can’t
approach from the back due to the axial hydrogen and it can’t approach from the left
(re) face of aldehyde due to the slope. So it must approach the si face as shown.

32. O
H
OPBBTMSO
OCH3
H
H
H
O
OCH3
H
TMSO
H
H
H
OPBB
Approach from the si face is shown above and the resulting product has the correct
stereochemistry at the CH2OPBB group.
I have draw the endo transition state above but this is ultimately unimportant. The
endo/exo selectivity only influences the methoxy stereocentre and as we shall see on
the next slide this is removed …

33. O
O
OPBB
O
O
O
OPBB
CH3H
(H3C)3Si
Treatment of the cycloaddition product with acid results in β-elimination of the
methoxy group to give the enone as shown above. The precursor for the Ireland-Claisen
rearrangement requires selective 1,2-reduction of the enone …

34. O
O
OPBB
NaBH4,
CeCl3
O
HO
OPBB
O
H
OPBB
O
H
vs.
O
H
O
H
OPBB
Classic Luche reaction
conditions (NaBH4, CeCl3)
result in exclusive
reduction of the carbonyl
with no 1,4-reduction.
The origin of this
chemoselectivity is
described in the the notes
for this course. Arguably
the CeCl3 is a selective
Lewis acid for the
methanolysis of NaBH4.
The resulting reagents are
harder reducing agents.
The diastereoselectivity is
rationalised by axial
attack (minimises
torsional strain) on the
more stable half-chair
conformation. The
hydride approaches from
the least hindered face.

35. O
HO
OPBB C12H25
OMOM
O CO2H
O
OPBB
O
O
O
C12H25 O
MOM
The precursor is formed by simple esterification.

36. O
OPBB
O
O
O
C12H25 O
MOM
LDA,
TMSCl,
HMPA
O
OPBB
O
OTMS
O
C12H25 O
MOM
Substrate control (the existing stereocentre of the dihydropyran) only controls one of
the two new stereocentres. The other is controlled by the geometry of the enolate as we
shall see on the next slide).
Normally, enolate (silyl ketene acetal to be accurate) formation with an ester favours
formation of the E(OSi)-enolate (as shown by the Ireland model). This is due to the fact
the ester (the alkoxy moiety) counts as a small substituent as the R group (the
dihydropyran) can rotate out the way).

37. O
OPBB
O
O
O
C12H25 O
MOM
LDA,
TMSCl,
HMPA
O
OPBB
O
OTMS
O
C12H25 O
MOM
In this example we need the Z(OSi)-enolate. Fortunately this is readily achieved by the
addition of HMPA (hexamethylphosphoramide) to the reaction. This highly Lewis basic
molecule coordinates the lithium and breaks up any aggregates or coordination. The
reaction no longer can be modelled by a Zimmerman-Traxler-like 6-membered
transition state and it gives the desired Z(OSi)-enolate.
On warming (–78°C to rt) the rearrangement proceeds. It does not go through a chair-
like transition state.

38. O
O
H
O
OTMS
OPBB
H
R O
O
TMSO
O
H
H
OPBB
R
H
H
vs.
O
O
TMSO
O
H
H
OPBB
R
H
≡O
H
O
PBBO
OTMSO
R
As you can see, the chair-
like transition state
suffers from considerable
non-bonding interactions.
Effectively the OTMS and
ether groups suffer
unfavourable 1,3-diaxial
interactions with the
dihydropyran ring.
If the reaction proceeds
through a boat-like
transition state instead
then we minimise these
unfavoured interactions
as shown on the right.
The reaction then
proceeds to communicate
one stereocentre into two
new stereocentres (with
destruction of the
original) as shown.

39. O
H
O
PBBO
OTMSO
R
O
PBBO
O
OMOM
C12H25
OMeO
CH2N2
DIBAL
O
PBBO
O
OMOM
C12H25
O
i. CH2=PPh3
ii. TMSBr
iii. TBSCl
O
PBBO
O
OTBS
C12H25
Standard functional group
transformations give the
diene required for ring
closing metathesis and
formation of the
tetrahydrofuran ring.
Mild esterification with
diazomethane gives the
methyl ester.
Low temperature
reduction with DIBAL
converts the ester into an
aldehyde.
Wittig reaction to give the
alkene is followed by a
change in protecting
group.

40. Ring-closing metathesis forms the tetrahydrofuran ring.
Hydrogenation removes the two alkenes and deprotects
the primary alcohol. Selective oxidation with Dess-
Martin’s periodinane gives Fragment A.
O
PBBO
O
OTBS
C12H25
i. RCM
ii. H2, Pd/C
iii. Dess-Martin
periodinane
O
O
O
OTBS
C12H25
O
I
O
OAc
OAc
AcO

41. Cl
O
N
Ot-Bu
t-Bu
N
t-Bu
t-Bu
Co
O
OAc
H2O
Cl
O
Cl
OH
HO
MgBr
TMS
( )4
Cl
OH
TMS
( )4
i. NaOH
ii. TBAF
( )4
O
( )4
OTBS
I
i. LiI
ii. TBSCl
Fragment B
The synthesis of the fragment B is more straight forward. Epichlorohydrin is
commercially available as a racemic mixture of enantiomers. Another example of
Jacobsen’s hydrolytic kinetic resolution permits isolation of the pure enantiomer we
require. The selectivity was described earlier. After this it is standard functional group
transformations. Opening of the more electrophilic epoxide (oxygen is more
electronegative than chlorine and ring strain) gives the alcohol.

42. Cl
O
N
Ot-Bu
t-Bu
N
t-Bu
t-Bu
Co
O
OAc
H2O
Cl
O
Cl
OH
HO
MgBr
TMS
( )4
Cl
OH
TMS
( )4
i. NaOH
ii. TBAF
( )4
O
( )4
OTBS
I
i. LiI
ii. TBSCl
Treatment with base forms a second epoxide.
Ring-opening with iodide and protection of the alcohol to prevent reformation of the
epoxide gives the iodide above.

43. O N
Ot-Bu
t-Bu
N
t-Bu
t-Bu
Co
O
OAc
H2O
O
OH
HO
CO2HPhS
LDA
CO2HPhS
OH
H
O
PhS
O
( )4
OTBS
IO
OTBS
O
( )4
PhS
The lactone is readily formed from an enantiomerically pure epoxide. Yet again, this
epoxide can be accessed via non-selective epoxidation of an alkene follow by kinetic
resolution. The enolate formation and epoxide ring opening gives the chiral secondary
alcohol, which is readily cyclised by treatment with acid.
Formation of a second enolate permits alkylation with the iodide formed on the
previous slide.

44. O
OTBS
O
( )4
PhS
O
O
O
OTBS
C12H25
i. Cy2BH
ii. ZnEt2
O
OTBS
O
( )4
PhS
EtZn
O
O
OTBS
O
HO
O
OTBS
C12H25
PhS
Now we can join the two fragments together. Hydroboration of the alkyne gives an
alkenylborane. Transmetallation converts the borane into an organozinc species.
Addition of this reagent to the Fragment A aldehyde should give the desired alcohol.
The diastereoselectivity can be rationalised by chelation control as shown on the next
slide.

45. HO
O
H
H
O O
H
Zn
HO
nucH
OH
O
HO
O
OTBS
C12H25
R
nuc
As before there are two methods to predict the stereochemistry. I prefer the use of the
Newman projection but it can be achieved with the skeletal representation.
First draw the Newman projection then chelate the two Lewis basic oxygen atoms
together. The nucleophile then attacks along the Bürgi-Dunitz angle. Then we have to
unravel this representation.
In theory we get the desired diastereomer (in practice the selectivity is quite poor).

46. The final steps are shown here.
First, hydrogenation removes the
unwanted alkene. Next we have to install
the desired alkene in the lactone ring.
This is achieved by oxidation of the
sulfide with mCPBA to give the sulfoxide.
Heating promotes syn-elimination and
formation of the alkene. Potentially we
could form an endo or exo alkene; both
are in conjugation with the lactone. Both
are trisubstituted. Arguably, reduced
conformational freedom in the ring
encourages elimination in the ring and
possibly increases the degree of
hyperconjugation (holds the C–H bond in
place). So I think (and experiment
agrees) that the desired product will be
favoured.
Acid then removes the protecting groups
and finishes the synthesis of muconin.
O
O
OTBS
O
HO
O
OTBS
C12H25
PhS
i. H2, Pd/C
ii. mCPBA
iii. heat
O
O
OTBS
O
HO
O
OTBS
C12H25
H
O
HO
O
OH
C12H25
O
O
OH

47. Hopefully, this chemistry gives you more
insight into planning total syntheses and
asymmetric synthesis.
Hopefully it also acts as an example of
what is expected in your assignment.
I confess I would not give myself full
marks. This synthesis is more or less
directly taken from a single paper (J. Org.
Chem. 1998, 63, 4876 … by Jacobsen,
hence the use of his methodology in
many steps). Higher marks will be
awarded for originality (designing your
own) or from at least assimilating
information from two or more syntheses.
My excuse? I use this synthesis to cover
other material and not just to act as an
example of the assignment.
O
O
OTBS
O
HO
O
OTBS
C12H25
PhS
i. H2, Pd/C
ii. mCPBA
iii. heat
O
O
OTBS
O
HO
O
OTBS
C12H25
H
O
HO
O
OH
C12H25
O
O
OH

48. Notes based on the 1999 postgraduate course on
Retrosynthesis delivered by gjr at the University of
Sussex, UK.
Overhauled in 2015 while listening to Neu! & Neu! 75 by
Neu!, Maniac OST by Rob., The Leftovers OST & The
Congress OST by Max Richter and Strategies Against
Architecture IV by Einstürzende Neubauten.
Hope it is of use to somebody.
gareth

49. O
O OH
OH
O
H
H
AcO
O
O
O
AcO
O
O
O
OH
O
H H
OCH3
HO
O
O
O
Ac
O
OHHN
OO
O
O O
HO
HO
OH
O
HO
HO
O
O
O
O
O
OH
OH
H
H

50. O
O OH
OH
O
H
H
6
AcO
O
O
O
2
AcO
O
O
O
OH
8
O
H H
OCH3
HO
O
O
O
Ac
5
O
OHHN
OO
O
1
O O
HO
HO
OH
7
O
HO
HO
O
O
O
3
O
O
OH
OH
H
H
4