Name__Key__________________________
215H W06-Exam No. 3
Page 2
I. (12 points) A disaccharide is cleaved by a β-glycosidase, an enzyme that specifically hydrolyzes a βglycosidic linkage. When the disaccharide is treated with excess dimethyl sulfate [(CH3)2SO4]/NaOH/H2O
and then hydrolyzed under aqueous acidic conditions at around 90 °C, two monosaccharide products are
formed: 2,3,4-tri-O-methyl-D-mannopyranose (1) and 2,3,4,6-tetra-O-methyl-D-galactopyranose (2).
CH2OH
H
H
HO
CH2OH
O
OH
HO
1
HO
HO
H
HO
H
H
O
OH
H
1
H
H
H
H
!-D-mannopyranose
OH
!-D-galactopyranose
(1) Draw in the box below the structure of the disaccharide. Each of the monosaccharide units should be
drawn in the Haworth projection formula and the stereochemistry of the anomeric hydroxyl group in the βform.
CH2OH
HO
O
HO
H
O
H
CH2
1
H
H
HO
H
H
H
OH
O
OH
HO
1
H
HO
H
H
6
(2) Draw in the boxes given below the structures of the two methylated monosaccharide products 1 and 2 in
their most stable chair conformers. Assume the stereochemistry at the anomeric hydroxyl group to be in the
β-form.
For 1:
For 2:
OH
OCH3
O
H
H3CO
H3CO
H3CO
H
H3CO
OH
H
H
1
H
OCH3
H
O
OH
H
H CO H
H 3
H
3
1
3
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215H W06-Exam No. 3
Page 3
II. (12 points) Draw a Fischer projection formula for each of the following two compounds and assign its
absolute stereochemistry using the D/L-designation.
H
(1)
(2)
OH OCH3
H
HOCH2
H
H
O
HO
O
N
H3C
CH3
C
H
CH3 OH
OCH3
stereochemical
designation:
H
N(CH3)2
D ; L
HO
H
HO
H
2
Circle the one that
applies.
O
H
HO
C
H
H
OH
H
OH
O
HO
stereochemical
designation:
D ; L
2
Circle the one that
applies.
H
H
OH
H
OH
HO
H
CH3
CH2OH
4
Draw in an open-chain
aldehyde form.
4
Draw in an open-chain
aldehyde form.
III. (10 points) Provide in the boxes below the structure of the expected reaction product from each of the
following carbohydrate molecules (J. Org. Chem. 2002, 67, 5654).
(1)
O
H3C
H3C
H3C
O
H3C
O
O
O
O
CHCl3
H3C
O
H-I (1 mol equiv)
(anhydrous)
O
O O
CH3
H3C
H3C
O
O
O
O
O
O
O
O
+
H3C
OH
O
H3C
O
I
O
5
!-anomer
(2)
O
H3C
O
O
O
CH3
O
H3C
CH3
O
O
L-saccharide!
O
O
CH3
BF3•O(CH2CH3)2
(catalytic)
PhSH (1.1 mol equiv)
CHCl3
H3C
O
OO
CH3
O
H3C
O
O
O
SPh
+
CH3
5
!-anomer
H3C
OH
O
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215H W06-Exam No. 3
Page 4
IV. (19 points) For each of the following pairs of compounds, predict which compound is the more acidic.
Compare the two underlined protons for each pair.
The more acidic compound for each pair will be (circle one):
(1)
O
BrCH2
(4)
O
OH
H3C
(2)
H3C
3
NH2
H
H
O
O
3
NH3+
3
(6)
O
O
O
H
O
N+
-O
NH3+
H3CO
O
H
O
3
(5)
NH2
O
OH
OH
OH
O
(3)
O
H3C
H3C
OCH2CH3
N(CH3)2
3
4
V. (7 points) The pKa values of the two acidic hydrogens of the enediol part of L-ascorbic acid (vitamin C)
are 4.10 and 11.79. Which one of the two hydroxyl hydrogens at C-2 and C-3 is expected to be more acidic?
H
CH2OH
OH
O
H
L-ascorbic acid
3
O
H
O
Explain briefly in the box below why the hydroxy hydrogen you selected
is more acdic than the other by using resonance structures of their conjugate
bases.
CH2OH
H
OH
O
2
O
H
H
2-OH or
3-OH
3
Circle one that applies.
3
O
H
H
O
O
2
H 3
O
H
O
O
2
O
CH2OH
OH
O
H
O
O
O
H
O
O
H
O
H
The conjugate base from the 2-OH has the resonance form
that places the negative charge at C-3, whereas the negative
charge in the conjugate base from the 3-OH can be
delocalized over the larger area.
O
O
H
O
O
O
H
4
VI. (15 points) Complete each of the following Aldol condensation reaction sequences as necessary.
(1) Tetrahedron Lett. 2006, 47, 1637.
CH3O
NaOH
40 °C
CH3OH
O
O
CH3
+
H
Br
CH3O
OCH3
88%
O
CH3O
+ H2O
Br
CH3O
OCH3
5
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215H W06-Exam No. 3
Page 5
VI. (continued)
(2) Org. Lett. 2006, 8, 1387.
N
O
O
OCH3
H3C
+
NaOCH2CH3
60 °C
CH3CH2OH
OH
O
N
O
OH
O
OCH3
~60%
H
+ H2O
5
(3) Tetrahedron Lett. 2006, 47, 2083.
O
H
O
H
H
H H
(PhCH2)2NH2+ -OC(=O)CF3
OMEM
O
64%
O
THPO
+ H2O
OMEM
benzene, 50 °C
O
THPO
MEM = CH2OCH2CH2OCH3
5
VII. (16 points) Treatment of enone 3 with NaOH/H2O in ethanol resulted in the formation methyl ketone 4
and sodium acetate (J. Med. Chem. 2006, 49, 1433). In the box below draw a stepwise mechanism by the
use of the curved arrow convention. You may use –OH as the base. You don’t need to balance each step.
O
O
O
OCH3
O
NaOH/H2O
CH3CH2OH
CH3
O-Na+
O
CH3
3
O
+
H3C
O-Na+
OCH3
4
O
O
3
O
OCH3
CH3
H OR
CH3
O
OH
O
O
O
O
O
CH3
OCH3 O
O
OCH3 O
OH
CH3
CH3
OH
O
OCH3
CH3
O
O
O
O
H OR
O
O CH3
OCH3 OH
O
(R = H or
CH2CH3)
O
O
OCH3
O
O
OH
CH3
H
O
CH3 +
O
O
O CH3
OCH3 O H
CH3
O
O-Na+
OCH3 4
O
OH
H3C
O-Na+
16
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215H W06-Exam No. 3
Page 6
VIII. (29 points) Complete each of the following reaction sequences as necessary.
(1)!(Org. Lett. 2006, 8, 1495).
1.
O
+
O
(2 mol equiv)
Li-enolate
O
O
O
N
H
2. H3O+workup
2
O
O
N
H
O
THF
O
OCH3
O-Li+
LDA (2 mol equiv)
(2 mol equiv)
C18H23O4N
5
p-TsOH
toluene (solvent)
90 °C
O
CH3
N
H
O
C13H15O2N
5
(2)
O
O
HO
H
O
heating
OCH3
NH
HO
H
1. LiAlH4
N
O
2. aqueous workup
O
O
5
+ HOCH3
Note: Boc =
O
N
O
OH
OH
O
2
Provide the reagents.
(3) J. Med. Chem. 2006, 49, 567.
BocNH
O
HO
H
O
O
H2N
DCC =
OCH2Ph
N C N
DCC
OH
BocNH
H
N
OH
O
OCH2Ph
CF3COOH
H3N+
H
N
O
OCH2Ph
O
O
5
CF3COO- salt
5