Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 25.SE, Problem 63AP
Interpretation Introduction
Interpretation:
Isotrehalose and neotrehalose, which differ from trehalose by their linkage.
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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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- a) 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_forwardTrehalose, 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_forwardd-(-)-Erythrose has the formula HOCH2¬CH(OH)¬CH(OH)¬CHO, and the d in its name implies that it can be degraded to d-(+)-glyceraldehyde. The (-) in its name implies that d-(-)-erythrose is optically active (levorotatory). When d-(-)-erythrose is reduced (using H2 and a nickel catalyst), it gives an optically inactive product of formula HOCH2¬CH(OH)¬CH(OH)¬CH2OH. Knowing the absolute configuration of d-(+)-glyceraldehyde (Section 5-14), determine the absolute configuration of d-(-)-erythrose.arrow_forward
- 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. To determine which ring is formed, ribose is treated with methanol in the presence of an acid catalyst. The products are then isolated and treated with NaIO, then with H₂O*. HO OH OH OH Ribose, C5H10O5 H 9.81 ** MeOH H* A & B isomeric cyclic acetals with formula C6H12O5 Assuming that ribose formed a five-membered ring cyclic hemiacetal, draw the structure of the sodium periodate digestion products. ▼ • Use the wedge/hash bond tools to indicate stereochemistry where it exists. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. 1. NalO4, 2. H₂O* MeOH products 4 SIF Previous Nextarrow_forwardRibose, 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. To determine which ring is formed, ribose is treated with methanol in the presence of an acid catalyst. The products are then isolated and treated with NaIO4 then with H3O+. OH HO OH OH Ribose, C5H10O5 H MeOH H* A & B isomeric cyclic acetals with formula C6H₁2O5 1. NalO4 2. H₂O* MeOH products Assuming that ribose formed a six-membered ring cyclic hemiacetal, draw the structure of the sodium periodate digestion products. • Use the wedge/hash bond tools to indicate stereochemistry where it exists. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner.arrow_forwardIdentify the organic functional group and reaction type for the following reaction. The reactant is a(n) - carboxylic acid hexose - Aldohexose - aldotetrose -deoxyhexose -carboxylic acid tetrose - ketohexose The product is a(n) - carboxylic acid tetrose - aldotetrose -alcohol hexose -aldohexose -carboxylic acid hexose - alcohol tetrose The reaction type is - hemiacetal formation -hydrolysis -oxidation( Benedict’s) -acetal formation -reduction( hydrogenation) - mutarotationarrow_forward
- Shown here is a/anglycosidic linkage. CH₂OH CH₂OH I OH Н OH Н O Alpha Beta OL- O D- О Н I ОН HO OH -О Н Н H OH OHarrow_forwardWhich of the following is the correct pair of monosaccharides needed 2 to form the given glycoside? * CH2OH CH-OH CH2OH OH он он OH OH a-D-fructofuranose & B-D-glucopyranose a-D-sorbofuranose & B-D-allopyranose a-D-mannofuranose & B-D-glucopyranose O a-D-sorbofuranose & B-D-galactopyranosearrow_forwardTrehalose and maltose are both dimers of glucose. However, they have considereably different reactivities. Concisely explain why these differences are observed. но но HO HO но "HO он но он OH O HO OHOH но trehalose maltose 1. Malthose is a reducing sugar while trehalose is not. 2. Trehalose is very resistant to acid hydrolysis while maltose can be acid-hydrolyzed with ease.arrow_forward
- Please don't provide handwritten solutionarrow_forward(g) Using appropriate prefixes/infixes/suffixes (ketohentese, aldahentase, etc.), classify each of the monosaccharides shown below. СНО CH2OH O: HOI но ОН OH HO H. CH2OH CH2OH I II (h) Identify B-D-altrose and oa-D-altrose from the monosaccharides shown below. CH2OH CH2OH CH2OH CH2OH O. ОН OH OH OH OH OH ОН OH OH OH ОН OH OH OH OH OH I II III IV | (i) Identify B-D-altrose and oa-D-altrose from the monosaccharides shown below. VI I CH2OH CH2OH II V OH III ОН IV H OH Harrow_forward5. Provide suitable responses for questions (a) – (). 6 CH2OH 4 OH OH 3 OH (a) What is the relative configuration of the above monosaccharide? (b) Which labeled carbon is the anomeric carbon? (c) Trace and identify the acetal in the above monosaccharide. (d) Draw the hemiacetal that results from above acetal. (e) Draw the open chain equivalent of the sugar in part (d). (f) Classify the monosaccharide below as a D-sugar or an L-sugar. H. OH O. OH CH,OH OH OHarrow_forward
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