Concept explainers
Interpretation:
The structure of the anhydrous sugar (
Concept introduction:
舧 Chair conformations: It is the most stable conformation, which accurately shows the spatial arrangement of atoms.
舧 Equatorial bonds are parallel to the average plane of the ring, while axial bonds are perpendicular to the average plane of the ring.
舧 The conformation having bonds at the equatorial positions are more stable than those with bonds at the axial position.
舧 On flipping the cyclohexane ring, axial bonds become equatorial bonds and equatorial bonds becomes axial bond.
舧 Bulkier group acquires equatorial positions to form stable conformer due to steric factors.
舧 Sugars that have moisture content below 0.5% are said to be anhydrous sugars. They are usually high purity crystalline sugars obtained from
舧 Altrose is an aldohexose sugar with the chemical formula
舧 The most stable configuration of aldopyranoses is when the
舧 Stereochemistry: The equatorial orientation refers to the spatial arrangement of
舧 The anomeric effect is lowest for sugars with equatorial orientation, which results in lower energetic state, and consequently this type of orientation confers higher stability.
舧 The anomeric effect is highest for sugars with axial orientation, which results in higher energetic state, and consequently this type of orientation confers lower stability.
舧 The anhydrous sugar is produced when the axial
舧 Anhydrous sugar is an acetal, it is a nonreducing sugar, and therefore it is called anhydrous non-reducing sugar.
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Organic Chemistry
- 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_forwardAll 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_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
- 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_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_forwardNonearrow_forward
- 3a. 3b. 3c 3d. 3e. CO₂ clavulanic acid CH₂-OH H Answer the following questions about clavulanic acid. Does clavulanic acid inhibit D-alanyl-D-alanine transpeptidase? Does clavulanic acid contain a ß-lactam? Does clavulanic acid contain a thiazolium ring? What is the result of the treatment of penicillinase with clavulanic acid? Does clavulanic acid form a covalent acyl-enzyme intermediate with penicillinase?arrow_forwardPropose a mechanism for this conversion and account for the fact that only the -OH on carbon 1 is transformed into an -OCH3 group.arrow_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 H30*. OH Меон 1. NalO4 2. H30* A & B MEOH + products HO H. H* isomeric cyclic acetals with formula CgH1205 ÕH ÕH Ribose, C5H1005 Assuming that ribose formed a six-membered ring cyclic hemiacetal, draw the structure of compounds A and B.arrow_forward
- One of the later steps in glucose biosynthesis is the isomerization of fructose 6-phosphate to glucose 6-phosphate. Propose a mechanism, using acid or base catalysis as needed.arrow_forwardThe 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.arrow_forwardTreatment 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_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning