General, Organic, and Biological Chemistry (3rd Edition)
3rd Edition
ISBN: 9780134042428
Author: Laura D. Frost, S. Todd Deal
Publisher: PEARSON
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Question
Chapter 6, Problem 2IA.6Q
Summary Introduction
To draw:
The anomer of D-Glucose.
Introduction:
Epimers are the stereoisomers in which the configuration differs only at one stereogenic center. Anomers are the epimers in which the configuration of the hydroxyl group at
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7. Draw the Haworth structure for the following monosaccharides
a.
H
НО
H
c=0
H
+
но - н
3
4
Н
5
- ОН
- OH
애
CH₂OH
6
D-Mannose
b.
H
0
=
с
H - C - OH
HO-C - H
H - C - OH
H - C - ОН
CH₂OH
Glucose
2. Draw a Haworth structure for the B-pyranose form of the monosaccharide shown (Fischer
projection)
HC
HO-C-H
H-C-OH
HO-C-H
H-C-OH
H2C-OH
3. Draw a Haworth structure for the a-furanose form of the monosaccharide shown (Fischer
projection)
H C-OH
c=0
HO-C-H
HO-C-H
H-C-OH
H₂C-OH
This monosaccharide is in its
form.
CHO
H-OH
HO -H
HO
H
CH₂OH
Select one:
O a. It is "L" since the C4 OH group is on the left.
Ob. It is "D" since the sugar is not symmetric.
O c. It is "D" since the C2 OH group is on the right.
Od. It is "L" since the C3 OH group is on the left.
Chapter 6 Solutions
General, Organic, and Biological Chemistry (3rd Edition)
Ch. 6 - Prob. 6.1PPCh. 6 - Prob. 6.2PPCh. 6 - Prob. 6.3PPCh. 6 - Prob. 6.4PPCh. 6 - Prob. 6.5PPCh. 6 - Classify each of the following alcohols as a...Ch. 6 - Prob. 6.7PPCh. 6 - Prob. 6.8PPCh. 6 - Prob. 6.9PPCh. 6 - Prob. 6.10PP
Ch. 6 - Prob. 6.11PPCh. 6 - Prob. 6.12PPCh. 6 - Prob. 6.13PPCh. 6 - Prob. 6.14PPCh. 6 - Prob. 6.15PPCh. 6 - Prob. 6.16PPCh. 6 - Prob. 6.17PPCh. 6 - Prob. 6.18PPCh. 6 - Prob. 6.19PPCh. 6 - Prob. 6.20PPCh. 6 - Prob. 6.21PPCh. 6 - Prob. 6.22PPCh. 6 - When an aldehyde undergoes oxidation, the...Ch. 6 - Prob. 6.24PPCh. 6 - Prob. 6.25PPCh. 6 - Prob. 6.26PPCh. 6 - Prob. 6.27PPCh. 6 - Prob. 6.28PPCh. 6 - Prob. 6.29PPCh. 6 - Prob. 6.30PPCh. 6 - Prob. 6.31PPCh. 6 - Prob. 6.32PPCh. 6 - Prob. 6.33PPCh. 6 - Prob. 6.34PPCh. 6 - Prob. 6.35PPCh. 6 - Prob. 6.36PPCh. 6 - Prob. 6.37PPCh. 6 - Prob. 6.38PPCh. 6 - Prob. 6.39PPCh. 6 - Prob. 6.40PPCh. 6 - Prob. 6.41PPCh. 6 - Prob. 6.42PPCh. 6 - Prob. 6.43PPCh. 6 - Prob. 6.44PPCh. 6 - Prob. 6.45APCh. 6 - Prob. 6.46APCh. 6 - Prob. 6.47APCh. 6 - Prob. 6.48APCh. 6 - Prob. 6.49APCh. 6 - Prob. 6.50APCh. 6 - Prob. 6.51APCh. 6 - Prob. 6.52APCh. 6 - Prob. 6.53APCh. 6 - Classify each of the following as primary,...Ch. 6 - Prob. 6.55APCh. 6 - Prob. 6.56APCh. 6 - Prob. 6.57APCh. 6 - Prob. 6.58APCh. 6 - Prob. 6.59APCh. 6 - Prob. 6.60APCh. 6 - Prob. 6.61APCh. 6 - Prob. 6.62APCh. 6 - Prob. 6.63APCh. 6 - Prob. 6.64APCh. 6 - Prob. 6.65APCh. 6 - Prob. 6.66APCh. 6 - Prob. 6.67APCh. 6 - Prob. 6.68APCh. 6 - Prob. 6.69APCh. 6 - Draw the product of the following 1 4...Ch. 6 - Prob. 6.71APCh. 6 - Prob. 6.72APCh. 6 - Prob. 6.73APCh. 6 - Prob. 6.74APCh. 6 - Prob. 6.75APCh. 6 - Prob. 6.76APCh. 6 - Prob. 6.77CPCh. 6 - Prob. 6.78CPCh. 6 - Prob. 6.79CPCh. 6 - Prob. 6.80CPCh. 6 - How much energy is produced if a person eats 50 g...Ch. 6 - Prob. 6.82CPCh. 6 - Prob. 1IA.1QCh. 6 - Prob. 1IA.2QCh. 6 - Prob. 1IA.3QCh. 6 - Prob. 1IA.4QCh. 6 - Prob. 1IA.5QCh. 6 - Prob. 1IA.6QCh. 6 - Prob. 1IA.7QCh. 6 - Prob. 1IA.8QCh. 6 - Prob. 1IA.9QCh. 6 - Prob. 2IA.1QCh. 6 - Which oxygen n the hemiacetal product in Figure 1...Ch. 6 - Prob. 2IA.3QCh. 6 - Prob. 2IA.4QCh. 6 - Where did you place the OH for C1 (top or bottom)?Ch. 6 - Prob. 2IA.6QCh. 6 - Prob. 2IA.7QCh. 6 - Prob. 1ICCh. 6 - Prob. 2ICCh. 6 - Prob. 3ICCh. 6 - Prob. 4ICCh. 6 - Prob. 5IC
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- For each monosaccharide: [1] determine all acetal and hemiacetal carbons; [2] select the anomeric carbon; [3] designate the compound as an a or ß anomer, 6 CH2OH 5. H H CH2OH O 1CH2OH OH OH H a. 4 b. H H 12 OH ÓH Но H ÓH ÓH a. Carbon 1 is the hemiacetal carbon. Carbon 2 is the v acetal carbon. Carbon 4 is the v anomeric carbon. The compound is B b. Carbon 2 is the hemiacetal carbon. There is not an v acetal carbon. Carbon 4 is the v anomeric carbon. The compound is a v I-arrow_forward5. Use B-D-glucose (below) to draw the disaccharides that are connected by 1-3 and 2-4 linkages. JOH НО HO OH B-D-glucose OH OH OH Ex OH НО CHarrow_forwardPlease draw a tetrasaccharide. The cyclic monosaccharide needed is pictured below. Step 1: Connect 3 glucose units through a (1-4) glycoside bonds Step 2: Connect the 4th glucose unit to the second glucose unit through a a (1-3) glycoside bond glucose CH₂OH OH OH OH OHarrow_forward
- Identify the type of glycosidic bond in the following disaccharide. CH,OH H H ОН Н CH2 H H H ОН Н HO ОН H. ОН HO H H OH O B-1,4-glycosidic bond O a-1,4-glycosidic bond O B-1,6-glycosidic bond O a-1,6-glycosidic bondarrow_forwardN-NHPH CHO CH2OH H- C=N-NHPH Но- -H- 3 equiv Но- -H- 3 equiv Но- PHNHNH2 PHNHNH, H- H- OH H- -O- H- H- -HO- H- -HO- ČH2OH ČH2OH ČH2OH D-glucose osazone D-fructose + NH3 + PHNH2 + 2 H20 The final step in the formation of the osazone from glucose is the reaction of the keto imine with 2 equivalents of phenylhydrazine to yield the osazone plus ammonia. This reaction involves the following intermediate steps: 1. Addition of phenylhydrazine to the imine and proton transfer to yield intermediate 1; 2. Elimination of ammonia to yield phenylhydrazone 2; 3. Addition of phenylhydrazine to the ketone to yield tetrahedral intermediate 3; 4. Proton transfer yields carbinolamine 4; 5. Elimination of water yields the final product osazone. Write out the mechanism on a separate sh of paper and then draw the structure of tetrahedral intermediate 3. NH НО- -H H- -ОН H- -ОН ОН Glucose keto imine Previous Nearrow_forward5. The Haworth projection of a simple sugar is given below. Note that the-OH group on the anomeric carbon can be either up or down. Using only this sugar, draw the Haworth projections of the following: a) an a-disaccharide with a B (1 3) glycosidic bond b) an a-disaccharide with an a (1→ 3) glycosidic bond c) a B-disaccharide with an a (1→4) glycosidic bond d) a B-disaccharide with an B (1 4) glycosidic bond CH2OH HO H H OH group can be either up or down OH H. OH Harrow_forward
- What type of linkage is between these monosaccharide derivatives? CH₂OH சிப் H NH H Он H OH H H 10 H NH CH₂OH C=0 CH3 B(1–6) a(1→6) a(1-4) ẞ (1-4)arrow_forwardFollowing is the Haworth structure of a disaccharide. Identify the type of glycosidic bond in this disaccharide. CH₂OH H HO H OH H H OH H H a-1,6-glycosidic bond O a,B-1,2-glycosidic bond O B-1,4-glycosidic bond O a-1,4-glycosidic bond CH₂OH H OH H H OH H OH B a hparrow_forwardIdentifying a given carbon in a cyclic monosaccharide Make the following changes on the molecule in the drawing area below: 1. First, make sure this is the ß anomer. Edit the molecule if necessary. 2. Then replace the hydroxyl group on carbon #2 with a chlorine atom. OH H CH₂OH H OH H H OH H OHarrow_forward
- Check the box under each ketose. If there are none, check the None of the above box under the table. H H- НО НО c=0 ОН H Н CH2OH Н. Н- HO- Н- None of the above. c=0 ОН -Н ОН CH2OH H- НО- CH₂OH 1 c=0 ОН -Н CH2OH НО НО- c=0 H H— ОН CH₂OH ЗЕР Н c=0 Н- -ОН I I H Н- - ОН ОН CH₂OH H— xarrow_forwardDraw the Haworth projection of the disaccharide made by joining two D-glucose molecules with a B(1-4) glycosidic bond. If the disaccharide has more than one anomer, you can draw any of them. Click and drag to start drawing a structure. X Ś èarrow_forwardOHC CH₂OH H OH 1. Draw the Fischer Projection of the following monosaccharide. Determine whether it has an R S configuration orarrow_forward
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