EBK ORGANIC CHEMISTRY
8th Edition
ISBN: 8220102744127
Author: Bruice
Publisher: PEARSON
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Chapter 20, Problem 49P
The 1H NMR spectrum of D-glucose in D2O exhibits two high-frequency doublets. What is responsible for these doublets?
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The 1H NMR spectrum of d-glucose in D2O exhibits two high-frequency doublets. What is responsible for these doublets?
(d) Draw the structure of the expected product when monosaccharide B undergo
mutarotation upon dissolving in water in the presence of Tollens reagent (AGNO3,
NHẠOH).
он
OH
O.
OH
OH
OH
monosaccharide B
Two sugars differing in configuration at a single
asymmetric carbon atom are known as
epimers.
D-mannose is a C-2 epimer of glucose
(Structure I), while D-galactose is a C-4 epimer
of glucose. Structure Il and IIl are
H-
-OH
но-
-H
H OH
HO H
но-
H
HO H
H
-OH
H-
-OH
HO OH
H -OH
H-
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H OH
HO.
HO,
OH
II
II
Which of the following represents an
aldopentose?
OH
H O
HO O
HO H
но н
HO H
но-
HO H
HO H
HO-
H OH
H OH
H OH
H OH
OH
HO
Он
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HO O
OH
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III
IV
These are chemical messengers that are
secreted by endocrine glands and carried
through the bloodstream to target tissues.
prostaglandin
deoxysugar
glycoside
hormones
Which of the following is NOT correctly paired?
cellulose: beta-1,4-glycosidic linkage
amylose: alpha-1,4-glycosidic linkage
chitosan: alpha-1,4-glycosidic linkage
cellubiose: beta-1,4-glycosidic linkage
Chapter 20 Solutions
EBK ORGANIC CHEMISTRY
Ch. 20.1 - Prob. 1PCh. 20.2 - Prob. 2PCh. 20.2 - Prob. 3PCh. 20.3 - Prob. 4PCh. 20.3 - Prob. 5PCh. 20.3 - Prob. 6PCh. 20.4 - Prob. 7PCh. 20.4 - Prob. 8PCh. 20.5 - Prob. 9PCh. 20.5 - Prob. 10P
Ch. 20.5 - Prob. 11PCh. 20.6 - Prob. 12PCh. 20.6 - Prob. 13PCh. 20.6 - Prob. 14PCh. 20.7 - Prob. 15PCh. 20.8 - Prob. 16PCh. 20.9 - Prob. 18PCh. 20.10 - Prob. 20PCh. 20.10 - Prob. 21PCh. 20.10 - Prob. 22PCh. 20.11 - Prob. 23PCh. 20.11 - Prob. 24PCh. 20.12 - Prob. 25PCh. 20.12 - Prob. 26PCh. 20.14 - Prob. 28PCh. 20.15 - Prob. 29PCh. 20.15 - Prob. 30PCh. 20.16 - Prob. 31PCh. 20.17 - Prob. 32PCh. 20.18 - Refer to Figure 20.5 to answer the following...Ch. 20 - Prob. 34PCh. 20 - Prob. 35PCh. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - Prob. 39PCh. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - Prob. 44PCh. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - The 1H NMR spectrum of D-glucose in D2O exhibits...Ch. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - A hexose is obtained when the residue of a shrub...Ch. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - Prob. 67PCh. 20 - Prob. 68PCh. 20 - Prob. 69PCh. 20 - Prob. 70PCh. 20 - Prob. 71PCh. 20 - Prob. 72PCh. 20 - Prob. 73P
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- (a) Label compounds A, B, and C as D- or L-sugars. (b) How are compounds A and B related? A and C? B and C? Choose from enantiomers, diastereomers, or constitutional isomers.arrow_forwardMonosaccharides can be drawn in a variety of ways, and in truth, often a mixture of cyclic compounds is present in a solution. Identify each monosaccharide, including its proper D,L designation, drawn in a less-than-typical fashion.arrow_forwardDraw the isomers for each aldotetrose and ketopentose in the figure below and designate each isomer as a D or L sugar and designate also the R and S in every chiral centers. Label the enantiomers and diastereomers respectively. CH,OH C=0 Н—С—ОН Н-С—ОН Н—С—ОН H-C-OH ČH,OH ČH,OHarrow_forward
- 4. Identify the component monosaccharides of each of the following compounds and describe the type of glycosidic linkage in each. Но он Но OH HO он Но- Но- (a) OH (c) CH,OHO. (b) CH2OHO Lon OH HO H ČHOH H OH ÓH ОНarrow_forward(d) Use the diagram below to complete the cyclic alpha form of structure V (e) Circle the hemiacetal in cyclic alpha form of structure V. (f) Redraw the cyclic alpha form of structure V but replace the OH group on the anomericcarbon with a methoxy group. Is this modified monosaccharide a reducing sugar or anonreducing sugar?arrow_forwardClassify the structures as being either an enantiomer, diastereomer or diastereomer/epimer of D-glucose. Structure A: CH H- -OH Structure B: но- H- H- OH Structure C: H- он H- -H D-Glucose: OH || CH CH CH он H- он но -H H- он но он но но- H- но H- OH OH но -- H- -H H -Ç-H Structure A: OH Structure B: OH Structure C: OHarrow_forward
- m) Which pyranose ring (A, B, C, or D) in the tetrasaccharide below is derived from D-altrose? The Fischer projections for the four aldohexoses that make up this tetrasaccharide are shown below. B C D HOH CHO CHO CHO CHO HO+H HTOH Fонно н HOH HOH H-OH H-OH H-+-OH H-TOHHOH HOH H+OH HOHнтон нон CHOH CHOH CHOH CHOH De D-glue Dalose Datrone i NH Į HOH STEP 1 OH, HO- answer 1. Just add arrows and charges for steps 1, 2, and 3 in formation of this enamine з no arrows needed here OH н Н ЮН -OH STEP 3 :ÖH н H STEP 2 на это про за почarrow_forwardThe most stable conformation of the pyranose ring of most D-aldohexoses places the largest group, CH2OH, in the equatorial position. An exception to this is the aldohexose D-idose. Draw the two possible chair conformations of either the a or ß anomer of D-idose. Explain why the more stable conformation has the CH2OH group in the axial position.arrow_forwardBarrow_forward
- The following is the structure of salicin, a bitter-tasting compound found in the bark of willow trees: The aromatic ring portion of this structure is quite insoluble in water. How would forming a glycosidic bond between the aromatic ring and β - D -glucose alter the solubility?arrow_forwarda) The D-aldopentose A, C5H1005, reacts with HNO3 to yield an optically active aldaric acid B. Kiliani-Fischer chain extension of A produces a pair of D-aldohexoses C and D. C is converted by HNO3 to an optically active aldaric acid, but D is converted by HNO3 to an optically inactive aldaric acid. Write acyclic Fischer projections for A, B, C, D.arrow_forwarda) The D-aldopentose A, C5H1005, reacts with HNO3 to yield an optically active aldaric acid B. Kiliani-Fischer chain extension of A produces a pair of D-aldohexoses C and D. C is converted by HNO3 to an optically active aldaric acid, but D is converted by HNO3 to an optically inactive aldaric acid. Write acyclic Fischer projections for A, B, C, D. b) Disaccharide E is a reducing sugar. It is hydrolyzed by an α-glycosidase enzyme, which means it contains an α- glycoside link. Treatment of E with Ag2O and excess Mel gives an octamethyl derivative F. Hydrolysis of F in dilute aqueous acid gives the pair of molecules shown below. Write the structures of E and F. (If the stereochemistry at a particular carbon is not determined by the above data, indicate this with a wavy line as shown below.) HO OMe OMe MeO MeO MOH OMe mOH OMe OMearrow_forward
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