Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 25.SE, Problem 54AP
Interpretation Introduction
Interpretation:
Gentiobiose is a disaccharide composed of two units of D-glucose joined with a β(1→6) linkage.
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Ribose, a carbohydrate with the formula shown, forms a cyclic hemiacetal, which, in
principle, could contain either a four-membered, five-membered, or six-membered ring.
When D-ribose is treated with methanol in the presence of an acid catalyst, two cyclic
acetals, A and B, are formed, both with molecular formula C,H,0, These are separated,
and each is treated with sodium periodate (Section 10.8C) followed by dilute aqueous
acid. Both A and B yield the same three products in the same ratios.
он о
CHO
СНО
H+ CH,OH A +B
ÕH
1. NalO,
2. H,0*
НО
CHO + CHOH + CH,OH
ÕH
CH,OH
Isomeric cyclic
acetals with molecular
formula CH12O,
D-Ribose
(C;H1605)
From this information, deduce whether the cyclic hemiacetal formed by D-ribose is four-
membered, five-membered, or six-membered.
Treatment with NaBH 4 converts aldose U into an optically inactive (meso) alditol V. Ruff degradation ofU gives W, whose alditol is optically inactive. Ruff degradation of W forms D-glyceraldehyde, thesimplest aldose. Upon Kiliani-Fischer synthesis, U is converted to two aldoses, X and Y. X is oxidized toan optically active aldaric acid Z. Y is oxidized to an optically inactive aldaric acid. Draw the structuresof D-glyceraldehyde, V, W, X, Y, and Z. Structure of compound U is shown below.
Aldohexoses A and B both undergo Ruff degradation to give aldopentose C. On treatment with warm nitric acid, aldopentose C gives an optically active aldaric acid. B alsoreacts with warm nitric acid to give an optically active aldaric acid, but A reacts to givean optically inactive aldaric acid. Aldopentose C is degraded to aldotetrose D, whichgives optically active tartaric acid when it is treated with nitric acid. Aldotetrose D isdegraded to (+)@glyceraldehyde. Deduce the structures of sugars A, B, C, and D, and useFigure 23-3 to determine the correct names of these sugars.
Chapter 25 Solutions
Organic Chemistry
Ch. 25.1 - Prob. 1PCh. 25.2 - Prob. 2PCh. 25.2 - Prob. 3PCh. 25.2 - Prob. 4PCh. 25.2 - Prob. 5PCh. 25.3 - Prob. 6PCh. 25.3 - Prob. 7PCh. 25.4 - Prob. 8PCh. 25.4 - Prob. 9PCh. 25.4 - Prob. 10P
Ch. 25.5 - Prob. 11PCh. 25.5 - Prob. 12PCh. 25.5 - Prob. 13PCh. 25.5 - Prob. 14PCh. 25.5 - Prob. 15PCh. 25.6 - Prob. 16PCh. 25.6 - Prob. 17PCh. 25.6 - Prob. 18PCh. 25.6 - Prob. 19PCh. 25.6 - Prob. 20PCh. 25.6 - Prob. 21PCh. 25.6 - Prob. 22PCh. 25.6 - Prob. 23PCh. 25.7 - Prob. 24PCh. 25.8 - Show the product you would obtain from the...Ch. 25.SE - Prob. 26VCCh. 25.SE - Prob. 27VCCh. 25.SE - Prob. 28VCCh. 25.SE - Prob. 29VCCh. 25.SE - Prob. 30MPCh. 25.SE - Prob. 31MPCh. 25.SE - Glucosamine, one of the eight essential...Ch. 25.SE - D-Glicose reacts with acetone in the presence of...Ch. 25.SE - Prob. 34MPCh. 25.SE - Prob. 35MPCh. 25.SE - Prob. 36APCh. 25.SE - Prob. 37APCh. 25.SE - Prob. 38APCh. 25.SE - Prob. 39APCh. 25.SE - Prob. 40APCh. 25.SE - Assign R or S configuration to each chirality...Ch. 25.SE - Prob. 42APCh. 25.SE - Prob. 43APCh. 25.SE - Prob. 44APCh. 25.SE - Prob. 45APCh. 25.SE - Prob. 46APCh. 25.SE - Prob. 47APCh. 25.SE - Prob. 48APCh. 25.SE - Prob. 49APCh. 25.SE - Prob. 50APCh. 25.SE - Prob. 51APCh. 25.SE - Prob. 52APCh. 25.SE - Prob. 53APCh. 25.SE - Prob. 54APCh. 25.SE - Prob. 55APCh. 25.SE - Prob. 56APCh. 25.SE - Prob. 57APCh. 25.SE - Prob. 58APCh. 25.SE - Prob. 59APCh. 25.SE - Prob. 60APCh. 25.SE - Prob. 61APCh. 25.SE - Prob. 62APCh. 25.SE - Prob. 63APCh. 25.SE - D-Mannose reacts with acetone to give a...Ch. 25.SE - Prob. 65APCh. 25.SE - Prob. 66APCh. 25.SE - Prob. 67APCh. 25.SE - Prob. 68APCh. 25.SE - Prob. 69APCh. 25.SE - Prob. 70APCh. 25.SE - Prob. 71AP
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- Treatment of -D-glucose with methanol in the presence of an acid catalyst converts it into a mixture of two compounds called methyl glucosides (Section 25.3A). In these representations, the six-membered rings are drawn as planar hexagons. (a) Propose a mechanism for this conversion and account for the fact that only the OH on carbon 1 is transformed into an OCH3 group. (b) Draw the more stable chair conformation for each product. (c) Which of the two products has the chair conformation of greater stability? Explain.arrow_forwardCompound A is a D-aldopentose that can be oxidized to an optically inactive aldaric acid B. On Kiliani-Fischer chain extension, A is converted into C and D; C can be oxidized to an optically active aldaric acid E, but D is oxidized to an optically inactive aldaric acid F. What is the structure of compound F? • Use the wedge/hash bond tools to indicate stereochemistry where it exists. You do not have to explicitly draw H atoms. If a group is achiral, do not use wedged or hashed bonds on it. • Show stereochemistry in a meso compound. • Do not include lone pairs in your answer. They will not be considered in the grading.arrow_forward1. Trehalose is a disaccharide that can be obtained from fungi, sea urchins and insects. Acid hydrolysis of trehalose yields only D-glucose. Trehalose is hydrolysed by α-glucosidase and not by β-glucosidase enzymes. Methylation of trehalose followed by hydrolysis yields two molar equivalents of 2,3,4,6-tetra-O-methyl-D-glucopyranose.From the following experimental data, deduce the structure of trehalose.What will be the effect of trehalose on Fehling’s solution? 2.Suggest a test you will use to show that a given food substance contains proteinarrow_forward
- 3 Trehalose is a disaccharide that can be obtained from fungi, sea urchins and insects. Acid hydrolysis of trehalose yields only D-glucose. Trehalose is hydrolysed by α-glucosidase and not by β-glucosidase enzymes. Methylation of trehalose followed by hydrolysis yields two molar equivalents of2,3,4,6-tetra-O-methyl-D-glucopyranose.From the following experimental data, deduce the structure of trehalose.What will be the effect of trehalose on Fehling’s solution?arrow_forwardMonosaccharide A is a D-aldotetrose. Treatment of A with nitric acid gives an optically inactive aldaric acid (B). Killiani-Fischer chain extension of A gives two products (C & D). C on treatment with NABH4, gives an alditol (E) that is optically active. D gives an alditol (F) that is NOT optically active. Give Fischer projections for A, B, CD, E, and E.arrow_forwardOne step in the gluconeogenesis pathway for the biosynthesis of glucose is the partial reduction of 3-phosphoglycerate to give glyceraldehyde 3-phosphate. The process occurs by phosphorylation with ATP to give 1,3-bisphosphoglycerate, reaction with a thiol group on the enzyme to give an enzyme-bound thioester, and reduction with NADH. -OPO3²- Enz-SH H-C-OH ATP CH₂OPO3²- 3-phosphoglycerate O 0-0--0 O ADP CH₂CH3 substitute for 1,3-bisphosphoglycerate C H-C-OH CH₂OPO3²- 1,3-bisphosphoglycerate O=C CH3-SH substitute for Enz-SH H H-C-OH | CH₂OPO3²- PO4³- O. S-Enz H-C-OH glyceraldehyde 3-phosphate Propose a structure for the first intermediates in the reaction of 1,3-bisphosphoglycerate with a thiol group on the enzyme to form an enzyme-bound thioester. Assume a basic group on the enzyme catalyzes the formation of this intermediate. To simplify the drawing process, substitute the structures below for the 1,3-bisphosphoglycerate and Enz-SH. CH₂OPO3²- (Enzyme-bound thioester) NADH/H* NAD*,…arrow_forward
- Trehalose, C12H22O11, is a nonreducing sugar that is only 45% as sweet as sugar. When hydrolyzed by aqueous acid or the enzyme maltase, it forms only d-glucose. When it is treated with excess methyl iodide in the presence of Ag2O and then hydrolyzed with water under acidic conditions, only 2,3,4,6-tetra-O-methyl-d-glucose is formed. Draw the structure of trehalosearrow_forward(a) Which of the d-aldopentoses will give optically active aldaric acids on oxidation with HNO3 ?(b) Which of the d-aldotetroses will give optically active aldaric acids on oxidation with HNO3 ?(c) Sugar X is known to be a d-aldohexose. On oxidation with HNO3, X gives an optically inactive aldaric acid. WhenX is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine thestructure of X.(d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an opticallyactive aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form opticallyactive products?(e) Show what product results if the aldopentose formed from degradation of X is further degraded to an aldotetrose.Does HNO3 oxidize this aldotetrose to an optically active aldaric acid?arrow_forwardTrehalose, C12H22O11, is a nonreducing sugar that is only 45% as sweet as sugar. When hydrolyzed by aqueous acid or the enzyme maltase, it formsonly d-glucose. When it is treated with excess methyl iodide in the presence of Ag2O and then hydrolyzed with water under acidic conditions, only2,3,4,6-tetra-O-methyl-d-glucose is formed. Draw the structure of trehalose.arrow_forward
- All the glucose units in dextran have six-membered rings. When a sample of dextran is treated with methyl iodide and Ag2O and the product ishydrolyzed under acidic conditions, the final products are 2,3,4,6-tetra-O-methyl-d-glucose, 2,4,6-tri-O-methyl-d-glucose, 2,3,4-tri-O-methyl-d-glucose, and 2,4-di-O-methyl-d-glucose. Draw a short segment of dextran.arrow_forwardFarnesyl diphosphate is cyclized to sesquiterpene A, which is then converted to the bicyclic product epi-aristolochene. Write a stepwise mechanism for both reactions.arrow_forwardThe anticoagulant heparin is a polysaccharide that contains alternating residues of -D- glucuronic acid-6- sulfate and N-sulfo-D-glucosamine-6sulfate connected by (1 B 4)- glycosidic linkages. Draw a part of heparin that shows one each of the two residues.arrow_forward
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