Name_______Key__________________
215 F010-Exam No. 2
Page 2
I. (15 points) For each of the following pairs of compounds, predict which compound is more acidic.
Compare the two underlined Hs for each pair and circle the compound that is more acidic for each pair.
The more acidic compound for each pair will be (circle one):
F
H
(1) a. F C CH2-OH
b. H C CH2-OH
F
H
H
H
N
H
(2) a.
Cl
(3) a.
SH
H
O
(5) a.
H
H
CH3O
H
b.
H
2
H
H
N
H
b.
H
(4) a.
H
b. H
OCH3
H
O
N
H
2
O
(6) a.
SH
b.
H
H
2
N
b. H
O
H
H
3
CH3
CH3
3
O
H
CH3
3
II. (a) (9 points) Show the structures of the three possible conjugate acids of benzotriazole and explain, in
several words, as to which of the three conjugate acids is most stable and would be expected to form
preferentially. Use drawings of pertinent resonance form(s) and several words to explain your answer.
explanation:
N
N
N
#1 is the most stable conjugate acid.
Unlike the other two, #1 has two idential
resoance structures where the positive
charge is delocalized over the two N
atoms.
benzotriazole
H
three possible conjugate acid structures:
1.
N
N
N
N
H
H
N
N
N
N H
N
H
H
3.
2.
H
N
N
N
H H
N
N
N
H
H
3
6
(b) (6 points) The pKa values of indole and benzotriazole are estimated to be 20.1 and 11.9, respectively.
Explain this difference in pKa values using resonance structures of the conjugate bases of the two
compounds.
explanation:
indole
N
H pKa ~20.1
The conjugate base of benzotriazole has two identical resonance
structures where the negative charge is delocalized over the
two N atoms
N
benzotriazole
benzotriazole
N
N
N
N
N
N
N
N
N
N
H
pKa ~11.9
indole
6
Name_____Key__________________
215 F10-Exam No. 2
Page 3
III. (28 points) Glycoside 1 is a new steroidal saponins isolated from the rhizomes of Polygonatum
kinigianum, one of the original plants commonly known as Huang-jing in traditional Chinese medicine
[Helv. Chhem. Acta 2010, 93, 1086]. Treatment of the glycoside, 1, with the enzyme β−glycosidase
provides trisaccharide 2. Answer the following questions about this glycoside 1 and trisaccharide 2.
O
H
O
HO
HO
HO
O
O
O
α
HO
OH
1
O
β
O
H
O
HO
H
HO
HO
O
H
O
HO
O
C
OH
O
enzyme
HO
α
A
OH
H
O
HO
B
HO
2
HO
O
OH
OH
(1) (3 points) Label each of the glycosidic bonds in glycoside 1 using a small arrow ( → ).
(2) (3 points) Classify each glycosidic linkage in 1 as α or β right next to each of the arrows you have shown.
(3) (2 points) How many D-sugar units does trisaccharide 2 have? (circle one) 0 1 2 3
(4) (2 points) Is the hemi-acetal OH in trisaccharide 2 α or β?
(circle one)
(5) (2 points) Is trisaccharide 2 a reducing sugar? (circle one)
Yes
α
β
No
(6) (4 points) Describe the linkages between the following sugar units [e.g., α(1 -> 6)]:
For A and C: α(1−>3)
α(1−>2)
For B and C:
(7) (12 points) Draw Fischer projections for the open-chain forms of the three carbohydrates produced when
glycoside 1 is hydrolyzed with a warm dilute aqueous acidic solution.
Carbohydrate A
H
H
Carbohydrate B
H
O
OH
Carbohydrate C
H
O
H
OH
H
OH
HO
O
OH
HO
H
H
HO
H
HO
H
H
OH
HO
H
H
OH
CH2OH
CH2OH
CH3
4
H
4
4
Name______Key__________________
215 F10-Exam No. 2
Page 4
IV. (14 points) Treatment of glycosyl trichloroacetimidate 1 with isopropyl alcohol (2) in the presence of a
catalytic amount of a strong acid results in the formation of an anomeric mixture of glycosides 3 [Angew.
Chem. Int. Ed. 2010, 49, 8724]. Propose in the box below a step-by-step, curved-arrow reaction mechanism
for this transformation from 1 to 3. You may use H-A and A- for the catalytic acid and its conjugate base,
respectively. You do not need to balance each step.
O
PhCH2O
O
CCl3
N
PhCH2O
HO
H
PhCH2O
1
O
PhCH2O
O
PhCH2O
1
CCl3
N
PhCH2O
O
PhCH2O
O
O
O
PhCH2O
PhCH2O
3
O
PhCH2O
mech. arrows for each step: 2 pts
each intermediate: 2 pts
CCl3
H
PhCH2O
H
PhCH2O
3
N
H
PhCH2O
A
anomeric mixture
PhCH2O
PhCH2O
H
H
PhCH2O
(2)
CH2Cl2 (solvent)
Mechansim:
O
O
PhCH2O
acid catalyst
O
H
PhCH2O
A
O
PhCH2O
O
PhCH2O
PhCH2O
14
V. (22 points) Treatment of hydroxy ester 4 with a catalytic amount of p-toluenesulfonic acid (TsOH; pKa
-0.51) results in the formation of its lactone derivative, 5, shown below [Angew. Chem. Int. Ed. 2010, 49,
5887]. Provide in the box below a step-by-step, curved-arrow reaction mechanism for this transformation
from the hydroxy ester. You may use H-A and A- for the acid TsOH and its conjugate base, respectively.
H
O
TsOH
(catalytic)
CH3O
HO
H
Mechanism:
H
O
CH3O
HO
H
4
H
H
O
+
O
4
CH3OH
5
H
H
A
H
O
H
HO
CH3O
CH3O
HO
H
H
H 3C
H
A
H
H
O
H
H
H
A
H
H
O
O
O
CH3O
A
HO
H
H
HO
mech. arrows: 2 pts/set
interm. structure: 2 pts each
H
O
O
O
H
H
O
H
5
H
22
Name_______Key_________________
215 F10-Exam No. 2
Page 5
VI. (18 points) Complete the following reaction schemes by providing in the boxes the structures of the
corresponding products. Make sure to indicate the stereochemistry where applicable.
(1) [Synthesis 2010, 2512]
O
O
O
NH
O
O
K2CO3 (2 equiv)
HO
OH
NH
O
CH3O
CH3I (2 equiv)
O
+ 2 KI
OCH3
O
3
(2) [Synthesis 2010, 2512]
O
NH2
O
HN
CH3O
OCH2CH3
+ HOCH2CH3
CH3O
O
O
4
(3) [J. Med. Chem. 2010, 53, 7377]
O
O
O
CH3O
N3
O
O
O
O
CH3O
O
NH2
OH
+
O
N
H
N3
3
(4) [J. Med. Chem. 2010, 53, 7377]
HO
O
NaOCH3
(1.1 equiv)
O
HOCH3
O
HO
O
O
CH3O
O
O
OH
C13H22O6
4
(5) [Chem. Eur. J. 2010, 16, 12788]
or
O
O
H
N
O
NaOCH3
(1.1 equiv)
O
O
H
HOCH3
O
O
N
O
H
OCH3
C13H21NO5
HN
O
H
O
C12H17NO4
O
O
4
OCH3
Name____________________________
215 F10-Exam No. 2
Page 6
VII. (18 points) Complete the following reaction schemes by providing in the boxes the structures of the
corresponding products. Make sure to indicate the stereochemistry where applicable.
(1) [J. Med. Chem. 2010, 53, 7296]
Br
O
O
N
O
NaH
O (1 equiv)
O
Na
or the enolate form
O
NH2
O
O
H
O
N
2
N
H
O
w/o stereo.
designation
acceptable
+ H2
N
NH2
O
+ enatiomer
4
+ NaBr
(2) [J. Org. Chem. 2010, 75, 7146]
O
O
5% KOH
CH3CH2OH
O
OK
+
O
O
OH
H
+
O
C11H14O2
H2O
4
(3) [Chem. Asian J. 2010, 5, 2192]
1.
MgBr
BrMg
O
OH
O
2. aq NH4Cl
meso!
HO
C8H16O2
4
(4) [Org. Lett. 2007, 9, 4041]
OCH2Ph
1. LDA
(1.1 equiv)
-78 °C
2.
O
OCH3
OCH2Ph
+ LiI
I
O
OCH3
4