hour exam iii wednesday, apr. 23 7:00 – 8:15 pm foellinger auditorium
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Hour Exam III Wednesday, Apr. 23 7:00 – 8:15 pm Foellinger Auditorium. Conflict Exam 5:30-6:45 161 NL. Help session: Mon. 7:00 – 9:00 100 MSEB. Hour Exam I Wednesday, Apr. 23 7:00 – 8:15 pm. 141 Wohlers Ford (CQF, G, J ,L) Livingston (CQP). 100 Noyes Lab Gorski (CQC, H) - PowerPoint PPT PresentationTRANSCRIPT
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Hour Exam III Wednesday, Apr. 23
7:00 – 8:15 pmFoellinger Auditorium
Conflict Exam5:30-6:45 161 NL
Help session:Mon. 7:00 – 9:00 100 MSEB
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Hour Exam I Wednesday, Apr. 23
7:00 – 8:15 pm141 WohlersFord (CQF, G, J ,L)Livingston (CQP)
100 Noyes LabGorski (CQC, H)Carberry (CQI, K)Tumuluru (CQD, E)
Conflict Exam5:30-6:45 161 NL
Help session:Mon. 7:00 – 9:00 100 MSEB
217 Noyes LabMohan (CQA, B)
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Carbohydrates
I
I optically active? yes
*
*
how many possible stereoisomers? 4how many actual stereoisomers? 4Is this D- or L-? Is this (+) or (-)? ?
II optically active? how many possible stereoisomers?how many actual stereoisomers?Is this D- or L-? Is this (+) or (-)?
yes
*
*
*
23 = 88
?
II
CnH2nOn
1 degree of unsaturation
C
OH
C O
C OHH
C OHH
CH2OH
C OH
COH
COH
CH2OH
H
H
H
CH2OH
C (H2O)
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I IICarbohydrates can be:
simple monosaccharidescan’t be hydrolyzed
complex disaccharideshydrolyzed to 2 monosaccharidespolysaccharideshydrolyzed to 3 - 1000’s
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I IIClassifying monosaccharides
1. carbonyl group “aldo” or “keto”
2. number of carbon atoms“tri”, “tetr”, “pent”, “hex”
3. suffix “ose”
I aldotetrose II ketohexose22 isomers 23 isomers
C=O
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Physical properties
2-deoxyriboseDNA
Classify: aldopentose
What are IMF ?
H-bond donors and acceptorshigh b.p.solids at room Tsoluble in H2O
chiral 2-deoxyribose has ______ isomers 4
Is this L- or D- isomer
*
*
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CHO
aldo
CH2OH
hexose
OH
OHOH
HOH
HHH
____stereocenters
****
4
______stereoisomers24=16
glucoseD-
CHO
CH2OH
H
HH
HHO
OHHOHO
L-glucose
natural monosaccharides = D
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monosaccharides to know:
D-glucose D-galactose
aldohexose diastereomers
D-fructose
ketohexose
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+ R’-OHH+
....
H+
..
..+ H+
aldehyde alcohol
hemi-acetalcyclic
R CO
H
R CO
H
RO
HOR’
H
OH
hemi-acetal formation
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O=C_HOH
C
OH
C
OH
CHO C
CH2OH
6
OHHO
OHOH
O=C
_H12345
CH2OH
6HOH2C
OH
5
right = down
aldehyde +left = up
alcohol
O=C_H
OH
C
OHC
CH2OH
OH
C
OH
C
H+
+
+
..
..
Fischer Haworth
H+
2 1
3
4
5
6
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OHHO
OHOH
CH2OH
O=C
_H12345
6
1
23
4
5
6
*
**
*
**
*
*
*
created a new C* at C1
25 isomers
differ only at C1
OH up =
OH down = racemic mixture
-D-glucose
-D-glucose
anomers
C1 = anomeric C
CC
C
C OCH2OH
OH
OHH
OH
C
HO
CC
C
C OCH2OH
OH
OH
H
OH
C
HO
hemi-acetal unstable
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3OH
1
2
4
5
6
OH
OH
OCH2OH
OH
H
3OH
1
2
4
5
6
OH
OH
OCH2OH
OH
H
-D-glucose
-D-glucose
anomers diastereomersdifferent physical properties
b.p. = 150o
b.p. = 146o
OH
HO
OH
CH2OH
OH
HO
OH
CH2OH
OH
OH
more stable
123
4 56
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OHHO
OHOH
CH2OH
O=C
_H
OH
HO
OH
CH2OH
OH64% 36%
0.01% hemi-acetals
alcohol + aldehyde hemi-acetal
aldehyde carboxylic acid
Fehling’s reagent - reducing sugarshemi-acetals
OH
HO
OH
OH
CH2OH
-D-glucose -D-glucose
unstable
[O]+ Cu2+ + Cu+Ag+ Ag
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cyclic hemi-ketals1
2
3
4
5
6OHHOH2C
23
4
5
OOH
HO
CH2OH
C2 is anomeric C
OH
CH2OH
D-fructose
OH
CH2OH
5 sided ring
--D-fructose
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oxidation of ketoses
ketone no reaction[O]
ketose carboxylic acid[O]
ketose “enol” aldose
all monosaccharides are reducing sugars