1. The document describes several organic reactions and asks questions about determining product structures and rationalizing stereochemical outcomes. 2. Key concepts discussed include: conformational analysis to determine reactivity; Cram chelation control to set stereochemistry; Ireland-Claisen rearrangements maintaining configuration; and using chiral auxiliaries to induce diastereoselectivity through chelation. 3. Rationalizations of stereochemical outcomes involve analyzing transition states, identifying favored conformations, and determining approach selectivity based on steric interactions.

1. Asymmetric
synthesis:
problems 1
Ph
OH
Br
NaOH
Ph
OH
Br
NaOH
A
C12H14O
(chiral)
B
C12H14O
(achiral; ir = 1715 cm–1)
Question 1
Products A & B are structural isomers. Examine the conformations of the two
diastereomeric starting materials and use these to determine the structures of A & B.

2. Ph
H
O
H
Br
H
H OH
Ph
H
O
H
Br
H
O
Ph
H
Answer
The key to this question is the conformation of the cyclohexane and the necessity for
functionality to be trans-diaxial or antiperiplanar for reactions to occur. This structural
requirement is the result of orbital overlap and the need for electrons to flow into the
C–Br σ* antibonding orbital.
The product is chiral as it has no plane of symmetry (it has no improper rotation axis).
Ph
H
O
H
Br
H
H OH
Ph
H
O
H
Br
H
O
Ph
H
The first reaction involves the formation of an epoxide (oxirane). Deprotonation of the
acidic hydroxyl group forms an alkoxide that participates in a substitution reaction with
backside attack.
It is fortunate (?) that the phenyl substituent encourages the ring to sit in a
conformation that places the hydroxyl group and the bromide axial but it is not
essential. Ring-flipping occurs in most systems and as it is the only reactive
conformation the energy required to adopt this conformation will comprise some of the
activation energy.

3. Ph
H
H
O
Br
H Ph
H
H
OH
Br
H Ph
H
OH Ph
H
OOH
In the second example it is impossible for the hydroxyl group to be antiperiplanar to the
bromide. So epoxide formation is impossible.
In one conformation of the molecule a hydrogen atom is antiperiplanar to the bromide.
Thus we can get an E2 elimination (whether elimination occurs with the alcohol or the
alkoxide is debatable and, ultimately, unimportant). Elimination gives an enol (or
enolate) the tautomeric form of a carbonyl (ir stretch) (or its resonance form).
The molecule is achiral as it has a plane of symmetry running through the middle.
Question 2
With the aid of a Newman projection determine the structure of the product. It is
formed as a single diastereoisomer. Note: under the acidic conditions of the reaction
the PMB protecting group is lost.
TBDPSO H
O
OPMB
TMS
Br TiCl4
85%
C24H33BrO3Si

4. TBDPSO H
O
OPMB
TMS
Br TiCl4
85%
TBDPSO
OH
OHH
Br
Answer
This reaction is taken from a synthesis of amphidinol 3.
It is basically a chance for you to practice manipulating
molecules between skeletal and Newman projections.
An extension of this question would have involved the
addition of a substituent to the alkene so that you had
to deal with the formation of two new stereocentres.
How would you approach that problem? (and for once a
6-membered ring would not help you as this reaction
proceeds through an open transition state).
TBDPS = is tert-
butyldiphenylsiyl (a bulky
protecting group) and
PMB = para-methoxybenzyl
(a substituted benzyl
group that is far easier
to remove than the
standard benzyl group;
either acid or an
oxidising agent such as
DDQ
(dichlorodicyanobenzoqui
none)).
Org. Biomol. Chem. 2012, 10, 9418
R
H
PMBO
R
H
O
OPMB
≡
O
C
H
PMBO
R
H
O
C
H
Cln
Ti
TMS
Br
TMS
Br
R
H
PMBO
C
H
OH
Br
≡
R
HPMBO
C
HHO
Br
≡ R
H OPMB
H OH
Br
First, convert the skeletal
representation into a
Newman projection.
This is an example of
Cram Chelation control
so the two Lewis basic
atoms are tethered
together and this fixes the
conformation of the
substrate.
The nucleophile then
approaches along the
Bürgi-Dunitz angle. One
approach is hindered by
the R group. The other
isn’t.
Finally, convert back to
skeletal representation
remembering that you
haven’t changed one
stereocentre.

5. TBDPSO H
O
OPMB
TMS
Br TiCl4
85%
TBDPSO
OH
OHH
Br
Obviously I’m focusing on the stereochemical outcome of this reaction. I do not have
time to discuss the simple chemistry/reactivity as that should be covered in the
undergraduate courses.
If you do not know about the chemistry of allylsilanes then I recommend you read up
on these valuable nucleophiles. To make life easier I can tell you this is sometimes
called the Hosomi-Sakurai reaction.
I have some brief notes on these reagents in “Strategy in Synthesis” lecture 3, a second
year paper:
http://www.massey.ac.nz/~gjrowlan/strat.html
OH
C6H13
O
H Ph
O SmI2 (15 mol%)
94%
> 99:1 dr
O
C6H13
H OH
O
Ph
Question 3
With the aid of the appropriate drawings, rationalise the stereochemistry of this
transformation.

6. OH
C6H13
OH Ph
O
I2Sm
O
C6H13
OPh
HOSmIn
O
C6H13
H OH
O
Ph
In this reaction samarium(II)
iodide is behaving as a Lewis
acid but it should be
remembered that it is a very
valuable reagent for single
electron transfer (SET).
It is also fun to note that
the student on this paper is
now a very successful
academic in his own right and
we might look at some of his
work later (and if we don’t
you should read it anyways).
Answer
This is an example of what is now known as the Evans-
Tishcenko reaction (but the reference below is by Evans
and only refers to the reaction as the Tishcenko
reaction).
Here is the basic mechanism (badly drawn). The
samarium acts as a Lewis acid and activates the
aldehyde allowing the alcohol to add. The samarium
then activates the ketone (not shown) and mediates the
internal hydride transfer.
J. Am. Chem. Soc. 1990, 112, 6447
O
H
H
OO Sm
II
C6H13 O
H
H
OOH
C6H13
Ph
Ph
≡
C6H13
HHO H O
O
Ph
To understand the diastereoselectivity of the reaction we need to look at the probable
conformation. The drawing above shows the samarium tethering the ketone to the
hemi-acetal-like oxygen. As you can see the hydride is 5-atoms away from the carbonyl
(like a 1,5-hydride shift) so we can use a 6-membered transition state and so we can
model this with chair conformation. The two oxygen atoms adopt the axial position to
allow the substituents to be in the pseudo-equatorial position.

7. O
H
H
OO Sm
II
C6H13 O
H
H
OOH
C6H13
Ph
Ph
≡
C6H13
HHO H O
O
Ph
The facial selectivity (Si in this case) of the hydride approach is controlled by the
existing stereocentre. Approach from the top (Re) face is only possible if the isopropyl
group adopts the pseudo-axial position. This is, of course, disfavoured. So approach is
from the bottom face (away from us or Si) as this places the isopropyl group pseudo-
equatorial.
The tetrahedral intermediate collapses to reform the carbonyl group and reduce the
ketone in an analogous fashion to the Cannizzaro reaction or the Meerwein-Ponndorf-
Verley oxidation.
O
i. LDA
ii. TBSCl
iii. heat
49%
> 90% ee
OH
O
O
O
i. LDA
ii. TBSCl
iii. heat OH
O
O
➎➊ ➋ ➌ ➍
➎
➊ ➋ ➌ ➍
Question 4
With the aid of the appropriate drawings explain why the two diastereomers of the
starting material both undergo rearrangement to give the same enantiomer of product.

8. O
i. LDA
ii. TBSCl
iii. heat
49%
> 90% ee
OH
O
O
O
i. LDA
ii. TBSCl
iii. heat OH
O
O
➎➊ ➋ ➌ ➍
➎
➊ ➋ ➌ ➍
Answer
Hopefully these reactions are not too hard. They are an example of the Ireland-Claisen
rearrangement. With a bit of luck my numbering might have given you a clue as to the
nature of the reaction.
In steps i & ii a silyl ketene acetal is formed (silyl enol ether). Warming the reaction
then promotes the rearrangement (often warming means returning the reaction to rt).
J. Org. Chem. 1993, 58, 4589
O
OTBS
O
H
H
OTBS
O
H
H
OTBS
OH
O
The Ireland-Claisen
rearrangement is an
example of pericyclic
(sigmatropic) reaction. It
involves 6 atoms and the
movement of 6 electrons
(3 curly arrows). As such
you should instantly be
thinking about using a
chair transition state to
model the reaction.
The key control element is
the methyl substituent.
This can either be
pseudo-equatorial or
pseudo-axial with each
different conformation
resulting in a different
enantiomer.
If you ever spot two double
bonds whose ends are 6
atoms apart think about a
rearrangement.

9. O
OTBS
O
H
H
OTBS
O
H
H
OTBS
OH
O
Once we have the chair
conformation we draw the
reaction, rearranging the
two double bonds.
This gives us the product
drawn in a 3D manner. Yet
again, the easiest place to
make an error is taking
this representation and
converting it to the
normal skeletal depiction.
As before, if you are
ever in doubt as to
whether you have
drawn the right
stereochemistry or
have inverted the
stereocentre simply
assign the
stereochemical
descriptor to both
drawings (in this case
the product is S).
O
OTBS
O
H
OTBS
H
O
H
OTBS
H
OH
O
If we apply the same
principles to the second
diastereomer we find we
form the same
enantiomer.
Again we place the methyl
substituent of the the
stereocentre in the
pseudo-equatorial
position. The methyl
substituent of the alkene
must be axial - we have
no choice as the alkene
cannot be rotated.
The rearrangement
generates the acid and we
have to unwrap the
drawing.

10. O
N O
O
Ph
F
i. LiHMDS
ii. BrCH2CN
80%
O
N O
O
Ph
F
NC
Question 5
Rationalise the diastereoselectivity of the alkylation reaction shown above.
O
N O
O
Ph
F
H N
LiO
N
O
Ph O
Si
Si
H
F
vs.
H N
LiO
N
O
Ph O
Si
SiH
FO
N O
O
Ph
F Li
H
H H
Answer
This is your introduction to the use of
chiral auxiliaries (unless you have read
my notes in which case it is revision) and
this example is taken from a synthesis of
PNP405 a drug used to prevent
transplant rejection.
To rationalise the diastereochemical
outcome of this reaction we first need to
determine the geometry of the enolate
formed (E or Z).
The Ireland model suggest that this
occurs through a 6-membered ring. The
two competing factors are repulsion
between the aryl substituent and the
oxazolidinone auxiliary or 1,3-diaxial
repulsion between the base and the aryl
substituent. With imides such as this the
auxiliary is considered larger than the
trimethylsilyl group and so the top
conformation is disfavoured and we get
the Z-enolate.
J. Org. Chem. 2002, 67, 6612

11. O
N O
O
Ph
F Li
NC
Br
H
O
N O
O
Ph
F
HNC
Once we know the geometry of the enolate the rest is easy.
The lithium cation coordinates to the oxygen of the enolate and the carbonyl of the
oxazolidinone auxiliary. The chelate prevents rotation of the C–N bond and so fixes the
conformation of the auxiliary.
The phenyl group blocks the bottom face and so the electrophile must approach from
the top face giving us the product shown.
ON
OO i. Bu2BOTf,
EtNiPr2
ii. aldehyde
O
i. KOH
ii. CH2N2
iii. (EtCO)2CO, Et3N
OCH3
OO
O
LDA,
TMSCl
OCH3
OO
OTMS
i. heat
ii. HCl
C
C13H21NO4
D
C11H18O4
Question 6
Using the appropriate drawings determine the structure of C & D. It goes without
saying that you should pay close attention to the stereochemistry.

12. ON
OO
B
Bu
OTf
Bu
ON
OO
H
B
Bu Bu
iPr2EtN
ON
OO
B
Bu Bu
Answer
This was taken from a 3rd year exam paper I set when I was teaching in the UK. I can’t
find the reference to the original paper (naughty me).
The first step is formation of the boron enolate. Remember that a tertiary amine is not
a strong enough base to directly deprotonate an imide; coordination of boron and
carbonyl activates the α-position. Once again the Z-enolate is favoured due to the bulk
of the auxiliary/amide.
O
N
O
B
O
H Bu Bu
vs.
ON
OO
B
O
H Bu Bu
O
The reaction follows a different pathway to the alkylation in the previous question. The
aldehyde must coordinate with boron. Without this Lewis acid activation the boron
enolate is insufficiently nucleophilic to attack the weakly electrophilic aldehyde.
When the aldehyde coordinates to the boron it prevents the boron from interacting with
the auxiliary (boron is already coordinatively saturated) yet we still get a favoured
conformation for the auxiliary (as demonstrated by high diastereoselectivities).

13. O
N
O
B
O
H Bu Bu
vs.
ON
OO
B
O
H Bu Bu
O
favoured disfavoured
The conformation on the right is disfavoured due to dipole-dipole interactions. The
alignment of dipoles is unstable as the build-up of negative charges close to one-
another repel (like aligning to magnets in the same direction). Alternatively you could
argue that the lone pairs of electrons on each oxygen atom are in close proximity,
which is disfavoured. The conformation on the left is favoured as the dipole-dipole
interaction is minimised (as close to cancelling out as possible) or the lone pairs have
maximum separation.
B
Bu
N
O B
O
O
H
Bu
Bu
O
H
O O
H
N
H
Bu
O
O
vs.
OH O
H
N
H
O
O
≡N
OOH
O
O
Once again we have 6
atoms reacting with the
movement of 6 electrons.
The stereochemistry can
be rationalised with a
chair-like transition state.
This is known as the
Zimmerman-Traxler
transition state.
We have two choices; the
orientation of the
aldehyde and the facial
selectivity on the enolate.
The aldehyde will adopt
the position that places
its substituent in the
pseudo-equatorial
position.
The facial selectivity is the
result of the minimisation
of 1,3-diaxial-like
interactions.

14. B
Bu
N
O B
O
O
H
Bu
Bu
O
H
O O
H
N
H
Bu
O
O
vs.
OH O
H
N
H
O
O
≡N
OOH
O
O
If the aldehyde
approaches from the top
face of the enolate (Si
face) (top left picture)
then the isopropyl
substituent of the
auxiliary will clash with
the boron substituents.
Attack from the lower (Re)
face of the enolate avoids
such destabilising
interactions.
Once again, the hardest
part is probably
unravelling the 3D
representation and
drawing our skeletal
representation.
My only tip is to say it
should be obvious that
the two hydrogen atoms
are on the same face
(down).
ON
OO i. Bu2BOTf,
EtNiPr2
ii. aldehyde
O
i. KOH
ii. CH2N2
iii. (EtCO)2CO, Et3N
OCH3
OO
O
LDA,
TMSCl
OCH3
OO
OTMS
i. heat
ii. HCl
C
C13H21NO4
D
C11H18O4
The product of the aldol reaction is C.
The next three steps are simple functional group transformations. Potassium hydroxide
hydrolyses the imide to give a carboxylic acid. Mild esterification with diazomethane
(CH2N2) is followed by a second esterification, this time of the secondary alcohol.
Formation of the silyl ketene acetal proceeds to give the E-enolate.
Please note: the nomenclature for the enolates of esters is annoying. Simply swapping the silyl group
for a lithium changes the name to Z-enolate without changing the shape.

15. N
H
OLi
H
O
R
vs.
N
H
OLi
O
R
H
disfavoured
Why is the E-enolate favoured (the enolate with the methyl cis to the alkoxy
substituent)?
Again we can use the Ireland model of deprotonation to determine this. In this example
the repulsion between base and methyl group is key. The ester substituent can rotate
out of the way (unlike in the amide/imide example earlier). Of course, this is a
simplification, lithium compounds tend to form aggregates which complicate analysis.
More information can be found at an old version of my notes (lecture 4):
http://www.massey.ac.nz/~gjrowlan/stereo.html
OCH3
OO
OTMS
OH
TMSO
H O
OCH3
OH
TMSO
H O
OCH3≡HO
O
OCH3
OH
The final step is another
example of the Ireland-
Claisen rearrangement.
This is a useful reaction
as it permits the readily
formed C–O bond
(esterification) to be
converted to a C–C bond
with communication of
stereochemical
information.
The reaction proceeds
through a chair-like
transition state with the
largest substituent at the
stereocentre (not the
alkene) adopting the
pseudo-equatorial
position and controlling
the diastereoselectivity.
The hardest task is just
drawing this without
inverting any stereocentres.