Heating D-altrose with dilute acid produces a nonreducing anhydro sugar (C6H1005). Methylation of the anhydro sugar followed by acid hydrolysis yields 2,3,4-tri-O-methyl-D-altrose. The formation of the anhydro sugar takes place through a chair conformation of ẞ- D-altropyranose in which the -CH2OH group is axial. What is the structure of the anhydro sugar? CH3 OH HO OH I H₂C CH3 0. 0 H H H H HO OH OH I CH3 H₂C H H H H CH₂ -OH OH HO eTextbook and Media Explain how is it formed. The anhydro sugar is formed when the axial -CH2OH group reacts with C to form a cyclic SUPF Because the anhydro sugar is (i.e., an internal glycoside), it is a sugar. Methylation followed by acid hydrolysis converts the anhydro sugar to 2,3,4-tri-O-methyl-D-altrose.

Heating D-altrose with dilute acid produces a nonreducing anhydro sugar (C6H1005). Methylation of the anhydro sugar followed by acid hydrolysis yields 2,3,4-tri-O-methyl-D-altrose. The formation of the anhydro sugar takes place through a chair conformation of ẞ- D-altropyranose in which the -CH2OH group is axial. What is the structure of the anhydro sugar? CH3 OH HO OH I H₂C CH3 0. 0 H H H H HO OH OH I CH3 H₂C H H H H CH₂ -OH OH HO eTextbook and Media Explain how is it formed. The anhydro sugar is formed when the axial -CH2OH group reacts with C to form a cyclic SUPF Because the anhydro sugar is (i.e., an internal glycoside), it is a sugar. Methylation followed by acid hydrolysis converts the anhydro sugar to 2,3,4-tri-O-methyl-D-altrose.

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter16: Aldehydes And Ketones

Section: Chapter Questions

Problem 16.67P: Treatment of -D-glucose with methanol in the presence of an acid catalyst converts it into a mixture...

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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 trehalose
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.
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.
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 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.
1. 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 protein
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 Next
When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react to form an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of D-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 °C, but only about 0.1% of D-glucose exists in the anhydro form under the same conditions.
The formation of Br2 from NBS first involves the reaction of NBS with HBr to form an iminol intermediate and molecular bromine. The intermediate then undergoes acid-catalyzed tautomerism to form succinimide, the byproduct of the reaction. Propose a curved-arrow mechanism for the conversion of NBS into succinimide that also accounts for the formation of Br2.
The first step in the metabolism of glycerol, formed by digestion of fats, is phosphorylation of the pro-Râ€”CH2OH group by reaction with adenosine triphosphate (ATP) to give the corresponding glycerol phosphate plus adenosine diphosphate (ADP). Show the stereochemistry of the product.
How many chirality centers are contained in (a) the aldotetrose and (b) the ketopentose just given below? How many stereoisomers would you expect from each general structure? CH,OH C=0 Н-С—ОН Н-С—ОН Н—С—ОН Н—С—ОН ČH,OH CH,OH A
tion 6 of 17 > The structures of four isomers of a ketopentose are shown. OH کر کو 3 E D C 80 OH H-C-H 0=C HOLC-H HOC CH H-C-H OH A Dc $ 4 D B D D A Which of the structures are diastereomers of A? R F A DOD F4 H-C-H V OH C=0 HICIOH Но..C.....H HOLC-H HOCH H-C-H OH Which of the structures are stereoisomers of C? % 5 B T G H-C-H Es C=0 B H-C-H OH 6 C FG Y H & 7 OH H-C-H N 0=C HOLC…H HICIOH H-C-H 99 U OH D J * 8 M Which of the structures are enantiomers of C? - Ов OD A DII FB ( - 0 9 K DD FO H O < ) -O L command J F10 P A. : c F11 { [ option + ? 11 I 1 Attempt 5 F12 } 1 delete
When an optically active D-aldopentose was subjected to Kiliani-Fischer synthesis, followed by NaBH4/H2O, it produced a mixture of an optically active and an optically inactive alditol. When the same D-aldopentose was subjected to Wohl degradation followed by HNO3 it produced an optically inactive aldaric acid. Provide the structure of this D-aldopentose. Н- НО- н- H -ОН НО -H Н- -Н CH₂OH 1 A) 1 B) II C) III осн H D) IV НО осн -H Н- -OH H- H HỌ CH₂OH П -OH H -OH H Н -H CH₂OH Ш _H -OH HO -OH H- -OH H- осн -H -OH CH₂OH IV -OH CH₂OH V