More problems covering asymmetric synthesis. This time with examples of substrate control, chiral reagents, and chiral catalysis. Also another example of a synthesis.

1. Asymmetric
synthesis:
problems 2
reagent controlled
crotylation
substrate control carbonyl
addition
substrate controlled aldol
reaction
catalyst controlled Diels-
Alder reaction
an enantioselective
synthesis of kainic acid
1
B
O
O
iPrO2C
iPrO2C O
OBn 70%
88:12 dr
OH
OBn
Question 1
An easy start. Rationalise the observed diastereoselectivity.
For bonus marks is this an example of matched or mismatched substrate/reagent
stereoselectivity (sometimes called double asymmetric induction). Or in other words,
learn what matched/mismatched means!
2

2. B
O
O
iPrO2C
iPrO2C O
OBn 70%
88:12 dr
OH
OBn
Answer
This is taken from the synthesis of callipeltoside A and
is an example of the Roush asymmetric crotylation. The
Roush reagents are more user friendly than the Brown
crotylation reagents, being less moisture sensitive and
can be stored for longer periods of time. On the
downside they often give slightly lower
enantioselectivities than the Brown reagents.
Initially it looks like the selectivity may be harder to
rationalise than the Brown reagents …
Angew. Chem. Int. Ed. 2012, 51, 9366
O
OH3CO
H
O
OH
O O
NH
O
Cl
O
H
H3CO
3
H
O
B
H
H
O
O
O
OiPr
O
OiPr
BO O
O
O
OiPr
O
OiPr
H
H
OBn
H
BnO
disfavoured
lone pair
repulsion
vs.
anti-Felkin-Anh
disfavoured
OH
H
H
H
BnO
≡
BnO
H
H OH
H
attractive
interaction
Felkin-Anh
… with the
stereochemical
information being further
from the reactive centre.
It turns out that the key
interaction is an
electronic effect and not
a steric effect.
The lone pair of electrons
on the aldehyde have a
disfavoured interaction
with the lone pair of
electrons on the lower
ester group of the
tartrate derivative. This
means the aldehyde will
prefer to be further from
this lower ester.
There is also thought to
be an attractive
interaction … a formyl
hydrogen bond as shown
in the scheme.
4

3. OH
OBn
i. OsO4, NMO
ii. NaIO4
iii. Zn...............
72% (3 steps)
85:15 dr
Br OH
OBn
OH
Question 2
This time an example of substrate control (taken from the same synthesis as Q1). Is the
reaction under Felkin-Anh control or chelation control? Justify your answer with the
appropriate sketch.
5
OH
OBn
i. OsO4, NMO
ii. NaIO4
iii. Zn...............
72% (3 steps)
85:15 dr
Br OH
OBn
OH
Answer
The first step converts the alkene into an aldehyde. The OsO4 causes dihydroxylation of
the alkene. The presence of NMO (N-methylmorpholine N-oxide) allows the use of a
sub-stoichiometric quantity of osmium (which is expensive and very toxic).
The 1,2-diol is then oxidatively cleaved by treatment with NaIO4. Basically is a
practically simpler version of ozonolysis.
6

4. H
O
O
ZnLn
OH
OBn
O
OLnZn
H
H
OLnZn
H
H
R
H
H
R
H
H
disfavoured
O
LnZn
O
O
LnZn O H
R
H
H
ZnLn
H
H
R
or
≡
OH
R
OH
H
HO
H
OH
H
R
≡
The second step involves
formation of a
propargylic zinc reagent.
This is the nucleophile.
The zinc also coordinates
to both the alcohol and
the aldehyde. The two
possible half-chair
conformations are shown
on the right. The one with
the large substituent in
the psuedo-axial position
is disfavoured.
Once we have the
probably conformation of
the chelated substrate the
nucleophile can either
add from the top or
bottom (Si) face. The
approach that gives the
chair intermediate and
not the twist-boat is
favoured.
7
BzO
O
i. c-hex2BCl,
Et3N
ii. EtCHO
94%
> 20:1 dr
i. TBSOTf
ii. xs CH2=CHCH2MgBr
iii. NaIO4
83%
O OTBS
i. .......
BF3•OEt2
ii. HCl
82%
10:1 dr
OTMS
OEt
O
OH
O
A
C15H20O4
Question 3
Give the structure of
intermediate A. Pay
special attention to the
diastereoselectivity.
Rationalise the observed
high selectivity.
Lets face it, I give
you the
diastereoselectivity
(as long as you
understand the
second set of
reactions). So this
question is all about
determining why we
observe this
stereochemistry
because this example
does not obey the
normal steric
arguments!
8

5. BzO
O
BzO
O
B
Cl
Cy Cy
H
H
:NEt3 BzO
O
BCy2
Answer
The first step is formation of the boron enolate. This can
give either the E- or Z-boron enolate. It is believed that
the bulky chlorodicyclohexyborane exclusively gives the
E-enolate as shown.
The boron coordinates the carbonyl trans to the bulky
stereocentre or cis to the ethyl group to minimise steric
interactions. The cyclohexyl substituents then force the
methyl group downwards as shown on the right.
Deprotonation then gives the E-enolate.
O
R
BCy2Cl
H C
H
vs.
O
R
BCy2Cl
H C
H
9
O
O
B
H
H
Et
L
L
H
BzO
HO
O
H
H
Et
H
BzO
≡ BzO
O
Et
OH
incorrect
minimisation
of A1,3 strain
Directed delivery of
the aldehyde …
Initially, we may think that the stereocentre on the
enolate substrate will control the diastereoselectivity as
shown above. Here we determine the conformation of
the enaolte by minimising A1,3 strain by having the
smalles substituent parallel to the enolate double bond.
The aldehyde then approaches the enolate anti to the
bulky benzoate ester as shown above. Unfortunately, this
gives the wrong diastereoisomer.
Instead it is thought that there is a formyl hydrogen
bond between the ester and aldehyde that directs the
approach as shown on the right.
C O BL2C
H H
O
O
PhO
Et H
10

6. O
O
B
H
H
Et
L
L
H O
O
Ph
O
O
B
H
H
Et
L
L
H
BzO
OH
O
H
H
Et
H OBz
≡BzO
O
Et
OH
If we assume hydrogen
bonding directs the
delivery of the aldehyde
then the Zimmerman-
Traxler transition state
must look like the
diagram on the right.
This predicts the correct
diastereomer.
11
BzO
O OTBS
MgBr
O OTBS
O
MgBr
Ph
O
MgBr
OH OTBS
HO
Ph
OH
I
OO
O O
O OTBS
O
I
HO
HO O
O
O OTBS
The next bit is not part of
the question but is how
the ketone required for
the question 4 is formed.
First the Grignard reagent
attacks the ketone and
removes the ester
‘protecting group’ on the
alcohol. This furnishes a
diol. The idol can be
cleaved with NaIO4 to gibe
the necessary ketone.
12

7. BzO
O
i. c-hex2BCl,
Et3N
ii. EtCHO
94%
> 20:1 dr
i. TBSOTf
ii. xs CH2=CHCH2MgBr
iii. NaIO4
83%
O OTBS
i. .......
BF3•OEt2
ii. HCl
82%
10:1 dr
OTMS
OEt
O
OH
O
A
C15H20O4
Question 4
Using the curly arrow
notation and appropriate
Newman projections,
explain how the ketone on
the bottom row is
transformed into the
lactone.
13
O OTBS
OTMS
OEt
BF3•OEt2
OTBSO
EtO
OH
R
H
O
≡
≡
R
H
O
OTMS
OEt
R
H
HO
OEt
O
Answer
This is simply an example of substrate control so we need to use the Felkin-Anh model
to predict/rationalise the diastereochemical outcome.
First draw the Newman projection. Then rotate the C–C bond so that the substituent of
the closest stereocentre is perpendicular to the carbonyl group. The silyl ketene acetal
then attacks along the Bürgi-Dunitz angle.
14

8. OTBSO
EtO
OH
HCl
OHO
EtO
OH
HCl
O
OH
O
Finally, if we treat this molecule with acid we perform two reactions. The first removes
the silyl TBS protecting the group. The second is the acid promoted transesterification
or lactonisation.
15
PhS
OEt
O N
Cu
N
OO
Ph Ph
2
2SbF6 cat. 10%,
92%
15:1 endo:exo
> 95% ee
SPh
OEtO SPh
O
OEt
endo exo
Question 5
Welcome to the world of asymmetric catalysis. Explain both the enantio- and
diastereoselectivity of the Diels-Alder reaction shown above.
16

9. SPh
O
EtO
SPh
O
EtO
HOMO
LUMO
EtO
O
SPh
HH
SPh
OEtO
Answer
This example comes from the synthesis
of norbornenone.
The first task in this question is to
explain the diastereoselectivity. In terms
of the Diels-Alder reaction this is called
endo/exo selectivity. In endo selectivity
the dienophile approaches the diene in
such a manner that the electron
withdrawing group is under the diene.
This allows the electron rich diene to
interact with the electron poor carbonyl
group. This interaction is called
secondary orbital overlap. It is shown in
three different representations on the
right.
The endo product is the kinetically
favoured product from the Diels-Alder
reaction (but it is sterically disfavoured).
Chem Commun. 1998, 1985
17
PhS
PhS
HH
SPh
O
OEt
OEt
O
O
OEt
The alternative transition state is known
as the exo transition state as the
dienophile is orientated away from the
diene (there is no overlap between the
carbonyl group and the diene).
This transition state is sterically less
demanding but is only observed under
very forcing conditions which allow the
reversibility of the Diels-Alder reaction to
be exploited.
Normally Diels-Alder reactions give the
previous product, the endo compound.
The Diels-Alder reaction is very powerful
and I recommend that you learn more
about it. A very brief introduction is given
in one of my courses:
http://www.massey.ac.nz/~gjrowlan/
stereo2.html
Lecture 10
18

10. N
Cu
N
OO
Ph PhO
EtO
S
Ph
H
H
N
Cu
N
OO
Ph PhO
EtO
S
Ph
H
H
vs.
disfavoured favoured
O
EtO
SPh
H
H
SPh
OEtO
≡
Once we have determined
that the reaction proceeds
through the endo
transition state we have to
consider the influence of
the chiral Lewis acid as
this will control the
enantioselectivity.
A copper(II) catalyst is
most likely to be square
planar. As a result of the
sulfide substituent the
substrate (the dienophile)
is bidentate and forms a
rigid chelate. The
orientation of
coordination is
unimportant as the box
ligand is C2 symmetric.
When the substrate
coordinates to the
copper ...
19
N
Cu
N
OO
Ph PhO
EtO
S
Ph
H
H
N
Cu
N
OO
Ph PhO
EtO
S
Ph
H
H
vs.
disfavoured favoured
O
EtO
SPh
H
H
SPh
OEtO
≡
... the phenyl substituent
of the sulfide is forced
away from the ligand as
shown in the diagram.
This minimises non-
bonding interactions
between the phenyl group
and the oxazoline ring
and/or the phenyl on the
ligand.
The diene (in this case
cyclopentadiene, a very
common test substrate)
then approaches anti to
the phenyl of the sulfide.
This avoids the interaction
between the diene and
the phenyl. In the
drawings on the right this
means it approaches from
the bottom face and gives
the molecule shown.
20

11. PMBO
O
OBn
N
Ru
H2N
PhPh
Ts
Cl HCO2H:Et3N
83%
95% ee
PMBO
OBn
OHH
Question 6
Here we are going to go through the synthesis of (–)-α-
kainic acid, a neuroexcitatory amio acid used in research
into the CNS.
First, investigate the Noyori transfer hydrogenation and
see if you can rationalise the stereoselectivity.
N
H
CO2H
CO2H
21
N
Ts
N
Ph
H
HPh
Ru H
N
Ts
N
Ph
H
HPh
Ru
O
H
H
O
H
N
Ts
N
Ph
H
HPh
Ru
O
H
H
O
H
N
Ts
N
Ph
H
HPh
Ru
H
H
H
Answer
The following questions
concern the synthesis of
kainic acid.
This first step is an
example of a Noyori
transfer hydrogenation.
This reaction employs
formic acid as a source of
hydrogen as this is more
practical than the use of
hydrogen gas. The
mechanism on the right
shows how this occurs.
Hydrogen bonding
between acid and amine
positions the formic acid
next to the reactive
centre. Reductive
elimination gives a metal-
hydride and carbon
dioxide.
J. Org. Chem. 2013, 78, 3355
22

12. N
Ts
N
Ph
H
HPh
Ru
H
H
H
O
R
R
N
Ts
N
Ph
H
HPh
Ru
H
H
H
OR
R
H
H
OR
R
≡
OHH
OBn
OPMB
vs.
One of the advantages of
the Noyori diamine
catalysts is that they do
not require additional
functionality within the
substrate to coordinate
the catalyst and substrate
together. This can be
achieved by a simple
hydrogen bond between
the carbonyl and the
ammonium group.
There are two possible
orientations for the
substrate. Calculations
(and the selectivity of the
reaction) suggest that
substrates with
unsaturated functionality
(alkynes, alkenes &
aromatics) form an
attractive interaction ...
23
N
Ts
N
Ph
H
HPh
Ru
H
H
H
O
R
R
N
Ts
N
Ph
H
HPh
Ru
H
H
H
OR
R
H
H
OR
R
≡
OHH
OBn
OPMB
vs.
... between the π-system
and a hydrogen atom of
the aromatic ring of the
catalyst.
This is a simplification of
the mechanism, which
undoubtedly involves the
solvent molecules as well
as the substrate and
catalyst but it is a useful
predictive model.
24

13. PMBO
OBn
OHH
Red-Al®
86%
PMBO
OHH
OBn
Red-Al® = sodium bis(2-
methoxyethoxy)aluminiu
m hydride
Question 7
This to is a stereoselective reaction (although many
people forget that alkenes are stereocentres as soon as
they move onto asymmetric synthesis). How does
reduction with Red-Al® result in formation of the E
(trans) alkene exclusively?
As a clue I will just say that the alcohol is important and
that the structure of Red-Al® is given on the right.
O
Al
H
O H
O
O
Na
25
PMBO
R2
OHH
O
Al
H
O H
O
O
Na
PMBO
R2
OH Al(OR)2
O
Al
H
O H
O
O
PMBO
AlO
R2
OR
OR
H
PMBO
OHH
OBn
H
H
H
The Red-Al serves two functions in this reaction; it activates the alkyne and acts as the
hydride source. It is thought that the alcohol reacts with Red-Al first, the aluminium
Lewis acid is then tethered to the substrate and this is essential for stereoselectivity.
The electron deficient aluminium activates the electron rich alkyne. A second equivalent
of Red-Al attacks the activated alkyne, delivering a hydride and forcing the alkyne to
attack the aluminium to give a five-membered ring. The formation of the ring forces the
hydride and the aluminium onto opposite faces of the alkene. The C–Al bond can be
protonated (or halogenated) with retention of stereochemistry.
26

14. PMBO
OHH
OBn
H
H
HO
O
DCC, DMAP
80%
PMBO
OBn
O
O
i. LiHMDS, TMSCl,
–78°C to rt
ii. CH2N2
79%
O
O
OBn
PMBO
DCC =
dicyclohexylcarbodiimide
DMAP =
dimethylaminopyridine
LiHMDS = lithium
hexamethyldisilazide
[lithium
bis(trimethylsilyl)amide]
Question 8
If you completed last weeks problems then this
sequence should offer no challenge (if you haven’t
looked at last weeks questions then you can either cheat
and look at the answers (thus spoiling the enjoyment) or
have a go now).
Reagents are given on the right.
N
C
N
N
N
27
PMBO
OBn
O
O
PMBO
OBn
O OTMS
Answer
The first step is esterification and I will not go through the mechanism of this reaction
(if you don’t know it or can’t work it out go through your undergraduate notes, it will in
be in there somewhere).
The second step involves stereoselective enolate/silyl ketene acetal formation. It is
essential that we control the geometry of the ‘enolate’ as this effects the relative
stereochemistry of the two new stereocentres formed in the rearrangement.
Under standard conditions such as this esters invariably give the (OSi)-E-enolate. One
simplistic interpretation of the Ireland model says this is due to the fact the ester
substituent can rotate out of the way so is smaller than the substituents of the base
and hence the enolate substituent will be better accommodated on the same face as
the alkoxy substituent.
28

15. O
H
HPMBO
OTMS
H
OBn
O
H
HPMBO
OTMS
H
OBn
≡
H H
O
O
OBn
PMBO
H
H
Once the silyl ketene
acetal has formed the
molecule is set-up for the
Ireland-Claisen
rearrangement. This
proceeds through the
standard Zimmerman-
Traxler-like 6-membered
transition state. The
existing stereocentre
controls the conformation
of the transition state
with the larger substituent
adopting the pseudo-
equatorial position. All
other substituents
positions are fixed as they
are on alkenes.
The representations on
the right show the
reaction and the
stereochemical outcome.
29
O
O
OBn
PMBO
i. DIBAL
ii. Ac2O,
DMAP, Et3N
iii. DDQ
73% (3 steps)
O
OBn
HO
O
DDQ = 2,3-dichloro-5,6-
dicyano-1,4-
benzoquinone
DDQ is a mild oxidant
Simple functional group transformations. I have only
included it so that you know how we get to each of the
key intermediates (and so that you can practise your
synthesis since that is the point of the course!).
O
O
NC
NC Cl
Cl
30

16. O
O
OBn
PMBO
i. DIBAL
ii. Ac2O,
DMAP, Et3N
iii. DDQ
73% (3 steps)
O
OBn
HO
O
DDQ = 2,3-dichloro-5,6-
dicyano-1,4-
benzoquinone
DDQ is a mild oxidant
DIBAL reduces the ester to an alcohol.
Acetic anhydride with dimethylaminopyridine as a
catalyst and triethylamine as a base forms a new ester
(an acetate).
The DDQ deprotects the PMB ether. It acts as an
oxidant, accepting a hydride from the PMB group and
effectively, oxidising the PMB ether to an aldehyde and
releasing the free alcohol. This only works for the
electron rich PMB group and not normal benzyl ethers.
O
O
NC
NC Cl
Cl
31
AcO
OBn
HO
Ti(iOPr)4, (–)-DET,
t-BuOOH
82%
100% de
AcO
OBn
HO
O
Question 9
This is another example of the Sharpless Asymmetric
Epoxidation. See if you can determine the product,
obviously paying special attention to the
chemoselectivity (or is it regioselectivity?) (and not
looking at the next slide).
32

17. AcO
OBn
HO
Ti(iOPr)4, (–)-DET,
t-BuOOH
82%
100% de
AcO
OBn
HO
O
Answer
The chemoselectivity is easy to understand. For the Sharpless Asymmetric Epoxidation
to proceed it must coordinate to the substrate. It does this through an alcohol group
and as a result the SAE invariably occurs at allylic alcohols (although I realise there is
an example in my notes that is not allylic). So the allylic alcohol reacts considerably
faster than the isolated alkene.
33
OH
"O" D-(–)-DET unnatural isomer
R
You are not expected to draw the
postulated transition state. All you need
to say is that the Sharpless Asymmetric
Epoxidation is predictable and that the
(–)-enantiomer of the tartrate ligand
would be expected to cause epoxidation
from the top face.
The mnemonic that shows the selectivity
of each ligand is given on the right.
Remember, the alcohol goes in the
bottom right corner and the alkene goes
backwards into the plane of the page.
34

18. AcO
OBn
HO
O
TBSCl,
imidazole
94%
AcO
OBn
TBSO
O
i. K2CO3, MeOH
ii. DIAD, Ph3P,
(PhO)2P(O)N3
45% (2 steps)
N3
OBn
TBSO
O
DIAD is the ispropyl
version of DEAD. Due
to its increased
steric bulk it is less
prone to side
reactions that result
in the formation of
hydrazides (attack at
the carbonyl group).
More functional group interconversions. See if you can
work out a mechanism for step ii.
As a clue this is an example of the Mitsunobu reaction
that was covered in the third year.
35
O N
O
N O
OPh3P
OiPr N
O
N OiPr
OPh3P
P
O
N3PhO
PhO
OiPr N
O
N OiPr
OPh3P
P(O)(OPh)2
R OH
R O
N3
N3
PPh3
R N3
Ph3P=O
The order of events is
protection of the primary
alcohol with the silyl
group.
The acetate is then
removed by base-
mediated hydrolysis.
The primary alcohol is
substituted with an azide
possibly by the
mechanism given on this
page.
First the
triphenylphosphine is
activated by reaction with
DIAD. This creates an
anion that can displace
the azide anion form
DPPA. The alcohol is then
activated by reaction with
the cationic phosphine.
Finally, SN2 displacement
gives the product.
36

19. N3
OBn
TBSO
O
i. TBAF
ii. PPh3, H2O
iii. Boc2O, Et3N
61% (3 steps)
N
H
Ot-BuO
OH
OBn
OH
Question 10
This is a straight forward, SN2-like, substitution reaction. Suggest a reason why the
reaction gives the pyrrolidine (5-membered) ring preferentially over formation of the
piperidine ring (6-membered).
Make sure you are happy that this is the only diastereomeric product.
37
R N
N
N
:PPh3
R N
N
N
Ph3P
R N
N N
PPh3
R N
PPh3
H
O
H
N2
R N
H
PPh3
O
H
R N
H
PPh3
O
H
R NH2
P3P=O
Answers
The use of triphenylphosphine and water to reduce an azide to an amine is a classic
reaction. For those of you that are interested the mechanism is given above. As is
frequently the case the driving force for this reaction is the formation of
triphenylphosphine oxide and, of course, nitrogen.
38

20. H
NH
BnO
H H
OH
H
O
H
NH
BnO
H H
OH
H
HO
The transition state of the cyclisation can be approximated as the diagram above. This
is based on a chair-like transition state with two of the three substituents in the
pseudo-equatorial position. It is impossible to get all of them equatorial; this is one of
the reasons the synthesis of kainic acid is so challenging.
In the cyclisation the amine has to approach the C–O σ* antibonding orbital of the
epoxide. This is 180° to the C–O bond. Hopefully the drawing above shows that this
should be possible for the formation of the pyrrolidine (5-membered ring) but would
be more challenging for the 6-ring.
This transition state also explains the diastereoselectivity.
39
N
H
Ot-BuO
OH
OBn
OH
i. NaIO4
ii. NaBH4
iii. Li•naphthalene
71%
N
H
Ot-BuO
OH
OH
i. Jones reagent
ii. TFA
62%
N
H
CO2H
H
CO2H
No more stereochemistry to worry about. These are just the ‘end-game’
transformations to get to the final product.
The diol is cleaved with the periodate and the resulting aldehyde is reduced to an
alcohol with sodium borohydride.
The lithium naphthalenide is a reductant (source of electrons) and is used to cleave the
benzyl ether.
The two primary alcohols are then oxidised to carboxylic acids with Jones reagent.
Surprisingly, this does not cleave the Boc group so deprotection is achieved with TFA
(trifluoroacetic acid).
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