1.
Answer ALL parts. MW CH3B) OLD COURSE 2010 exam questions.
(a) Illustrate the frontier orbitals, in each molecule, which are involved in the
cycloaddition reactions shown below.
(i)
A [2+2] cycloaddition.
Ph
Ph
Ph
irradiation (hυ)
+
Ph
Ph
LUMO
(ground state)
Ph
Ph
Ph
Ph
HOMO
Ph
(excited state)
Ph
Ph 1 mark per molecule
[10%]
(ii)
A [4+2] cycloaddition.
O
O
heat
+
LUMO
HOMO
O
(or alternative HOMO/LUMO
combination)
1 mark per molecule
[10%]
(b) The conversion of 1 into 2 may be achieved in a two step process. The use of
photochemical irradiation results in formation of a reactive intermediate 3, which
subsequently rearranges to 2 under non-photochemical conditions.
OH
Me
H
OH
nonphotochemical
Photochemical
irradiation (hυ)
3
HO
HO
1
(i)
2
Identify the structure of intermediate 3.
Me
OH
HO
3
2 marks
[10%]
(ii)
Write a mechanism for each step of the reaction above, and describe the
type of pericyclic reaction that is taking place.
OH
Me
H
OH
nonphotochemical
Photochemical
irradiation (hυ)
3
HO
HO
1
2
OH
Me
Photochemical
irradiation (hυ)
H
H
antarafacial
6 e process
HO
1
OH
HO
3
nonphotochemical
antarafacial,
8 e process
OH
HO
2
1 mark for each mechanism and 1 for each process description.
[20%]
(c) Provide a mechanism for the following reaction and identify the product 4.
H
Cl
i) Et3N (a base)
Cl
Cl
4
ii)
O
Et3N
H
Cl
Cl
Cl
H
Cl
Cl
O
C
O
Cl
Cl
H
O
Cl
Cl
C
O
[4+2] cyclisation favoured because
two pi systems on the ketene are involved.
(covered in detail in lectures)
1 mark for step 1, 2 marks for step 2, 2 marks for explanation.
[25%]
(d) Account for the stereochemistry of the product 5 formed in the sigmatropic
rearrangement shown below. Your answer should include an illustration of the
transition state of the reaction.
Ph
O
Ph
Ph
heat
O
Ph
5
Ph
Ph
O
Ph
Ph
O
Ph
Ph
5
Ph
Ph
O
via
Ph
O
O
Ph
[25%]
2.
Answer ALL parts.
(a) The synthesis of the alkaloid Pulimiotoxin C may be achieved through the reaction
sequence illustrated below, which proceeds via the formation of intermediate 6.
O
O
+
H
heat
O
steps
Pumiliotoxin C.
NH
N
H
(i)
OBn
6
BnO
O
Account for the control of the regiochemistry of 6 in the cycloaddition.
δ+
δ+
O
δ−
O
δ−
N
H
OBn
The electron donating N group and electron
withdrawing C=O create complimentary areas
of positive and negative electron density.
2 marks for diagram , 1 mark for explanation.
[15%]
(ii)
Draw a transition state for the cycloaddition reaction, and thus predict the
relatively configurations of the three chiral centres in 6.
O
NH O
LUMO of dienophile aligns
with HOMO of diene.
Endo transition state due
NHCO2Bn
to secondary orbital
overlap.
CHO
BnO
3 marks for diagram, 2 for explanation.
[25%]
(iii)
The use of a Lewis acid increase the rate of the cycloaddition reaction.
With aid of appropriate energy level diagrams, explain why?
LUMO+LA
LA
diene HOMO
O
LUMO no LA
Lewis acid co-ordinates
to C=O, lowering LUMO
energy and giving better
overlap with diene HOMO
2 marks for energy level diagram, 2 marks for explanation.
[20%]
(b) The reaction of hydroxylamine 7 with aldehyde 8 results in the formation of a
reactive intermediate 9. Intermediate 9 subsequently undergoes a reaction with
styrene 10 to give the product 11.
Ph
Ph
N
H
OH
+
Ph
7
(i)
O
Ph
9
Ph
8
10
N
O
11 Ph
Propose a structure for 9, and provide a mechanism for its formation.
O
Ph
N
N attacks C=O and then elimination of water.
Ph
1 for product and 2 for mechanism.
[15%]
(ii)
Illustrate the mechanism of the reaction of 9 with 10 to form 11.
Ph
Ph
(iii)
N
O
Ph
Ph
Ph
N
O
11 Ph
2 mark for mechanism.
[10%]
(c) Explain why the reaction below fails under basic conditions, but gives a product in
100% yield under acidic conditions. Provide a mechanism for the successful
cyclisation.
O
O
HO
Using NaOH; 0% yield.
O
Using HCl; 100% yield.
OMe
OMe
The base catalysed process requires a disfavoured 5-endo-trig reaction (2 marks)
whilst the acid catalysed reaction involves protonation of the C=O and
rearrangement to structure below, which cyclises via a favoured 5-exo-trig
cyclisation (3 marks), the mechanism of which should be illustrated.
OH
HO
OMe
[25%]
3.
Answer ALL parts.
(a) Propose a synthesis of trisaccharide 12 from the three reagents 13, 14 and 15,
paying particular attention to the stereocontrol of formation of each anomeric centre.
Reagents should be given for each step, together with a mechanism for each step.
[80%]
Ph
O
O
HO
O
O
OH
HO
OH
O
TBDPS=Si(tBu)Ph2
TBDMS=Si(tBu)Me2
Bn=CH2Ph
OMe
O
12
O
OH
OH
O
O
Ph
Br
Ph
Bn
O
O
O
13
HO
O
Br
O
BnO
OBn
O
TBDMSO
O
OMe
OBn
O
O
O
O
TBDPS
14
15
Ph
Neighbouring group participation.
Ph
Bn
Ag
O
O
O
O
Br
O
Ph
Ag
Bn
O
O
O
O
Ph
Bn
Ph
Bn
O
O
O
O
O
O
1 mark
O
2 marks
13
Ph
O
Bn
O
O
O
HO
O
BnO
O
O
OBn
O
TBDMSO 14
O
2 marks (mechanism)
2 marks for explanation of anomeric control.
O
remove TBDMS Ph
O
O
O
with TBAF (1 mark)
O
O
O
OBn
Bn
OBn
OAc
OAc
O
O
BnO
OMe
BnO
OMe
Ag
O
TBDMSO
2 marks
Br
2 marks (mechanism)
Anomeric control
OBn
Ag
O
OBn
O
O
O
O
O
O
O
Ph
OBn
O
O
Ph
OAc
O
TBDPS
15
BnO
OMe
TBDPS
2 marks for
explanation of
anomeric control
O
O
O
TBAF to remove TBDPS
then hydrogenation to remove Bn groups
2 marks
12
OBn
O
O
Ph
TBDPS
(b) Predict the structure of the product X from the cycloaddition reaction shown below,
and give a mechanism for its formation.
[20%]
+ EtO2C
N
heat
CO2Et
N
X
N
H
EtO2C
OMe
CO2Et
EtO2C
2 marks for product and 2 for mechanism.
H
CO2Et
X
( -yfrtya^
O
M.
>q
/
ow/i-7
OS tHJ
5-((.'Al '
r~\
A
<-\ /v
^1
c
7$
\
fr\
9
0
DN)
u)
r
r
»w
T
3
o
«
M
J
3
\
rS
£
a
j s
W
r\
I
1
c.
VU
c t
*
11
-J
X
4
i
4
<C
VI
v*
-^
x
i ^ -H
o j? ^
\j
-8
^o
^Q
&Y~-
O
H
-o
'H
VI
(
. J
£>/• 14.
O
-8
£"'-73
\
\
n
\
I/I
<r-"
1.1
I U
#
f
6-0
^^
6~?XI/
2 S^T-
;
C?
"A*
/^-t-
CPT^
1^
o
^>«e
^
r
CH3B0
2006-2007
Course taught by M. Shipman.
CO
ft?
\J
£. L ti£
V
*—*./>£
O
ir
f-7
V)
I?
a
1}
\
s
M
ri
I
T
\M
V)
Vj
I
1
3V
c.
CT
°^,
S>;
^
r
0rts
5-
(A3
-^)
i-K-
0
9*
ov
A
-S
\
~~\s
0
v ^sf
f
I
NJ
v
o
O-
>-
Ox,
C)
c
(0
r\
C
/
^
- ;
h
* /O
^"^\*
Kvf
0
rw
^
0
k^
-^>
<5
/v«.
a
~A\
f
H
o
A
I ne(~tt
J
^I
'C
/
(f)
ri
\
?
I *;)
x A
Wl
^ 0V
?<?,
V ^
H-f-
0*
1
J^."
~-nVJ'
^~
H
O
1.5
&
c\ ±
,^f~^T
n &
-—^
-9 H
tf/1 t~^r-f^ -
•- ^jv-t'fei{
-"-'"
far\ ,—o CIA. 1'iun
Wj^-,.
£
«^H
^
* ->oW
o
/-V-IA ^ >«_i
1
£
31
—C,
4s
\i
-ja
^
£
,u
5
H
c
U!
-»-
'i X
4 -i,
jj
^
t
-i.
LC I
o
c[
+
X
^
• •>
£
'"O
XJ
' ^
V*
i*
K
\ Cj.
-p, t, v (
y^'
-i
•