Name__Key__________________________
215H W06-Exam No. 2
Page 2
I. (24 points) Pheromones are those chemicals that play a role in communication between individual
animals or insects. Structure 1 is the pheromone of the Japanese peach moth and acts as a sexual attractant
for the male. Devise the synthesis of racemic 1 starting from readily available 4-pentynol (2). You may use
any monofunctional reagents of eight or fewer carbons. Show reagents/chemicals needed for each step, the
product(s) after each step, and specify a solvent if this is important for reagent stability. Select methods that
are likely to give the desired compound as the major product at each step.
H
CH3(CH2)7 C (CH2)2
OH
(CH2)5CH3
C H
C
H 1 (racemate)
HO-CH2CH2CH2 C C H
2
O
HO-CH2CH2CH2 C C H
p-TsOH
(catalytic)
2
O
O-CH2CH2CH2 C C H
1. Na / NH3 (liq)
2. I CH (CH ) CH
2
24
3
HO-CH2CH2CH2 C C (CH2)5CH3
H3O+
!
O
O-CH2CH2CH2 C C (CH2)5CH3
PCC
H-C-CH2CH2 C C (CH2)5CH3
O
1. CH3(CH2)6-CH2MgBr
2. aq NH4Cl
H
CH3(CH2)7-C-CH2CH2 C C (CH2)5CH3
OH
H2, Pd, BaSO4/quinoline
1 (racemate)
II. (6 points) Treatment of the multi-functional group containing compound, 3, with NaBH4 in THF
followed by the protection of the selectively produced alkoxide with TBSCl results in the formation of TBS
ether 4 in 81% overall yield from 3 (Org. Lett. 2005, 7, 5593). Draw in the box below the structure of 4.
Ph
O
Ph
O
O
H
CH3O
O
1. NaBH4 in THF
(room temperature)
CH3O
2. TBSCl, imidazole, DMF
CH3O
O
H
3
CH3O
note: Ph = phenyl; TBSCl = (tert-butyl)dimethylsilyl chloride
imidazole:
NH
N
O TBS
O
DMF: dimethylformamide
(solvent)
O
4
H
6
Name___Key________________________
215H W06-Exam No. 2
Page 3
III (10 points) When trans-diol 5 was treated with one mol equivalent of lead tetraacetate,
Pb[OC(=O)CH3]4, in the presence of a catalytic amount of trichloroacetic acid (pka~1.0), diketone 6 was
produced together with lead diacetate and acetic acid (2 mol equivalents). In the box provided, draw the
structure of the neutral intermediate and show the mechanism of its formation from trans-diol 5 and
transformation of the intermediate X to diketone 6 through the use of the curved-arrow convention. You
don’t need to balance each step.
OH
OH
O
O-C-CCl3
H
CH3-C-O
C CH3
O
O
Pb
2
O
O
CH3-C-O
OH
C CH3
O
H3C
5
O
O
Pb
2
H
O
C
OH
O
OH
Pb
O
O
O
O
+ 2 HO-C-CH3
O
6
CH3-C O
O
O
O
Pb
2
O
H O-C-CCl3
+
Pb(O-C-CH3)2
O
+
CCl3C-OH
O
(catalytic)
5
CH3-C O
O
O
CH3-C-O
O
Pb(O-C-CH3)4
O
H3C
O
6
O
CH3-C-O
C
O
H
H3C
2
O
C
O
OH
4
neutral intermediate X
IV. (12 points) Complete the following reaction scheme by providing in the boxes the structures of the
corresponding products (J. Org. Chem. 2006, 71, 1390). Also, indicate for each of the compounds how
many 13C NMR peaks are expected to be observed when run by the proton-decoupled method.
H
OCH3
OH
HO
HO
OH
H
H
O
(1 mol equiv)
p-TsOH
(catalytic)
NaH
(2.2 mol equiv)
OCH3
O
HO
Br
O
4
10
O
O
O
O
C C H
(2.8 mol equiv)
DMF
0 °C to room temp.
HO
Number of 13C peaks expected:
OCH3
2
Number of 13C peaks expected:
4
16
2
Name____Key________________________
215H W06-Exam No. 2
Page 4
V. (8 points) Treatment of keto-diol 7 with a catalytic amount of p-toluenesulfonic acid (p-TsOH) produced
a mixture of two stereoisomeric tricyclic ketals. Draw in the boxes provided below the structures of these
products. Make sure to clearly indicate their stereostructures.
H
H
OH
O
OH
+
p-TsOH
(catlytic)
7
H
O
O
O
H
+ H2O
O
H
4
4
VI. (12 points) Provide the structures of the expected major products for each of the following reactions.
Make sure to clearly indicate their stereochemistry and, where applicable, label isotopes.
(1) My work from the last century!
O
CH3ONa
CH3OH
O
HO
H
CH3
OCH3
4
O
O
O
H
insect juvenile hormone
CH3
H218O
HClO4
(catalytic)
0 °C
O
O
18OH
CH3
H
OH
4
(2) J. Org. Chem. 2006, 71, 1139.
Si
Si
Si
HO
OCH2CH3
O
O
p-TsOH
(catalytic)
CH2Cl2
Si
O
OCH2CH3
HO
C21H34O4Si2
O
C21H34O4Si2
4
Name____Key________________________
215H W06-Exam No. 2
Page 5
VII. (28 points) Treatment of triol 8 with excess trimethyl orthoacetate [H3CC(OCH3)3] in the presence of a
catalytic amount of pyridinium p-toluenesulfonate (PPTS), a milder version of p-TsOH, cleanly produced a
diastereomeric mixture of cyclic orthoester 9 (98% yield) and 2 mol equiv of CH3OH. Subsequent treatment
of this unstable product 9 resulted in the formation of a new 5-membered cyclic ether 10 (91% yield)
together with one mol equiv of CH3OH (J. Org. Chem. 2006, 71, 1416).
H3CC(OCH3)3
PPTS
(catalytic)
H3C
CH2Cl2
room temp., 5 min
H
HO
H3C
OH
OH
8
OCH3
O
H3C
O
H
BF3•O(CH2CH3)2
(catalytic)
CH2Cl2
10 min
OH
9
+ 2 CH3OH
10 (C8H14O3)
IR: no peaks above 3100 cm-1
1739 cm-1 (strong)
13C NMR: ! 171.1 ppm and 7
sp3 13C peaks.
+ CH3OH
(1) Provide in the box below a stepwise mechanism for the formation of cyclic orthoester 9 from triol 8 by
the use of the curve-arrow convention. You may use H-B to represent PPTS. Make sure to indicate the
stereochemistry for each of the intermediates when applicable. Please also note that the entire reaction is
reversible. You don’t need to balance each step.
H
OCH3
CH3 C OCH3
OCH3
H B
OCH3
CH3 C OCH3
H B
OCH3
CH3 C OCH3
O
OCH3
CH3 C OCH3
OCH3
H
OCH3
H3C
OH
OH
H
O
OH
H3C
:B
O H
H3C
OH
OH
H3C
HO
H3C
C OCH3
H3C
OH
OCH3
CH3 C OCH3
CH3 C OCH3
CH3
O
H
OH
:B
OH
H3C
OH
8 OH
O
OCH3
O H
H
OH
9
16
(2) Draw in the box indicated the structure of compound 10 and show a stepwise mechanism for its
formation from 9 by the use of the curved-arrow convention. Use BF3 to represent the Lewis acid and
clearly indicate the stereochemical outcome for each step. You don’t need to balance each step.
H3C
H
OCH3
O
H3C
O
H
F3B OCH3
BF3
H3C
H
OH
9
H
H3C
H
O
8 points for the H C
3
mechanism
BF3
OCH3
O
O
H3C
O
OH
H
OH
O
O
O
H3C
H
H3C
H3C
H
H
F3B OCH3
O
H
H
O
H3C
10 (C8H14O3)
H
O
4
Name____Key________________________
215H W06-Exam No. 2
Page 6
VIII. (20 points) Complete each of the following reaction sequences as necessary.
(1) [Org. Lett. 2006, 8, 633].
O
O
H
N
H
CH3SO2Cl
(1 mol equiv)
(CH3CH2)3N
O
OH
OTBS
98%
N
O
O
KOC(CH3)3
(1 mol equiv)
O
SO2
CH3
OTBS
N
O
99%
OTBS
C22H39NO4Si
contains N-H
C22H37NO3Si
4
4
+ KOSO2CH3
(2) [Org. Lett. 2005, 7, 5163]
H
HCl (gas)
(catalytic)
N
O
OH
N
CH2Cl2
83%
O
C13H23NO3
O
H
O
O
trans-isomer
acceptable
+ enantiomer
4
C13H23NO3
(3) [J. Org. Chem. 2006, 71, 1696]
PhCH2
CH3
O
O
KOC(CH3)3
(1 mol equiv)
PhCH2
O
O
+ KOSO2CH3
CH2Cl2
O
SO2 OH
O
95%
4
(4) [Eur. J. Org. Chem. 2005, 4929]
1
BrMg
O
O
O
HO
10% aq H2SO4
O
+
2. aq NH4Cl
54%
+ enantiomer
4
cm-1
C6H12O2; IR: 3400
(strong and broad);
no 13C peaks in the ! 110-220 ppm region.
OH
OH