Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 25.6, Problem 20P
Interpretation Introduction
Interpretation:
To determine the active aldaric acid and inactive aldaric acid.
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Which D-aldopentose is oxidized to an optically active aldaric acid and undergoes the Wohl degradation to yield a Daldotetrose that is oxidized to an optically active aldaric acid?
WHICH OF THE FOLLOWING IS
an alpha anomer
a ketohexose
has an L- configuration
an epimer of DEF
Which D-aldopentoses are reduced to optically inactive alditols using NaBH4, CH3OH?
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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- does structure E represent fructofuranose? explainarrow_forwardSugar 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 Xarrow_forwardWhich of the following are reducing sugars? Comment on the common name sucrosefor table sugar.(a) methyl a-d-galactopyranoside (b) b-l-idopyranose (an aldohexose)(c) a-d-allopyranose (d) ethyl b-d-ribofuranosidearrow_forward
- expanded structure of the D-tagatose Fischer projection of D-tagatose complete steps of Haworth projection of D-tagatosearrow_forwardShow the steps in converting Fischer to Haworth to: Convert D-Talose to its aIpha pyranose structure. Convert D-Tagatose to its beta furanose structure.arrow_forwardIn 1891, Emil Fischer determined the structures of glucose and the seven other d-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry andsymmetry. He received the Nobel Prize for this work in 1902. Fischer had determined thatd-glucose is an aldohexose, and he used Ruff degradations to degrade it to (+)-glyceraldehyde. Therefore, the eight d-aldohexose structures shown in Figure 23-3 are the possiblestructures for glucose.Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and usethe following results to prove which of these structures represent glucose, mannose,arabinose, and erythrose.(a) Upon Ruff degradation, glucose and mannose give the same aldopentose: arabinose.Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are thetwo possible structures of arabinose?(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive…arrow_forward
- In 1891, Emil Fischer determined the structures of glucose and the seven other d-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry andsymmetry. He received the Nobel Prize for this work in 1902. Fischer had determined thatd-glucose is an aldohexose, and he used Ruff degradations to degrade it to (+)-glyceraldehyde. Therefore, the eight d-aldohexose structures shown in Figure 23-3 are the possiblestructures for glucose.Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and usethe following results to prove which of these structures represent glucose, mannose,arabinose, and erythrose.(a) Upon Ruff degradation, glucose and mannose give the same aldopentose: arabinose.Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are thetwo possible structures of arabinose?(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive…arrow_forwardUsing Fischer projection structure where no cyclization took place, what product/s formed when D-galactose reacts with the following: CN- , H+ ; H2O (hydrolysis); NaBH4 NaBH4 CH3OH, HCl Acetic anhydridearrow_forwardTo synthesize D-galactose, a student went to the stockroom to get some D-lyxose to use as a starting material. She found that the labels had fallen off the bottles containing D-lyxose and D-xylose. How can she determine which bottle contains D-lyxose?arrow_forward
- Glucose is an aldose sugar but it does not react with sodium hydrogensulphite. Give reason.arrow_forwardAn oligosaccharide isolated from an organism is found tocontain two glucose residues and one galactose residue.Exhaustive methylation followed by hydrolysis producedtwo glucoses with methoxy groups at positions 2, 3,and 6 and galactose with methoxy groups at positions2, 3, 4, and 6. What is the structure of the originaloligosaccharide?arrow_forward
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