Concept explainers
Interpretation:
The Fischer projection formulas for the given compounds, using given data are to be written.
Concept Introduction:
▸ Oxidation of aldoses by nitric acid produces optically inactive aldaric acids. This is due to the formation of meso compounds. Compounds that do not form meso structures are optically active.
▸ The Fischer projection is a way of representing
▸ The Ruff degradation reaction is used for decreasing the number of carbon atoms of the respective aldoses, thereby, shortening the carbon chain of the compound. It works in two steps: 1) Using bromine water to oxidize an aldose to the respective aldonic acid and, 2) using
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Organic Chemistry
- Compound 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_forwardLactose is a disaccharide in which a glycosidic linkage connects the monosaccharides galactose and glucose. OH НО OH (a) Identify the glycosidic linkage and the acetal carbon in lactose. (b) What type of glycosidic linkage does lactose have (i.e., is it 1,1'-, 1,2'-, etc., and is it a or B)? (c) People who are lactose intolerant are deficient in the enzyme lactase, and therefore cannot efficiently break down the disaccharide into its monosaccharides. When lactose is treated with aqueous acid, however, this hydrolysis can take place, though relatively slowly. Draw the complete, detailed mechanism and the products of the acid-catalyzed hydrolysis of lactose. Но ОН НО ОН ОН Lactosearrow_forwardDraw the structure of: (a) a polysaccharide formed by joining D-mannose units in 1->4-ß-glycosidic linkages; (b) a polysaccharide formed by joining D-glucose units in 1->6-a-glycosidic linkages. The polysaccharide in (b) is dextran, a component of dental plaque.arrow_forward
- A reddish color is obtained when compound A (a disaccharide) is reacted with Benedict solution. Is this compound more likely maltose or sucrosearrow_forwardWhich of the following statements is not true about glucose? (i) It is an aldohexose. (ii) On heating with HI it forms n-hexanearrow_forwardRegarding 4-O- (α-D-psychofuranosyl) -β-D-allopyranose. Please indicate the RIGHT alternative: (a) The disaccharide reacts with CH3OH in an acid medium to form a glycoside that cannot be oxidized with HNO3. (b) It is a reducing disaccharide only in basic medium. (c) In the structure there is only one glycosidic bond that is of the type β 1-O-4 ' (d) The hydrolysis products of this disaccharide do not show mutarrotation. (e) The disaccharide structure contains two six-membered rings.arrow_forward
- Like glucose, galactose mutarotates when it dissolves in water. The specific rotation ofa-d-galactopyranose is +150.7°, and that of the b anomer is +52.8°. When either ofthe pure anomers dissolves in water, the specific rotation gradually changes to +80.2°.Determine the percentages of the two anomers present at equilibrium.arrow_forwardCarbohydrates a. Compound Z is an aldopentose. If Z is oxidized with nitric acid the product is optically active. If Z undergoes one Ruff degradation and the product of that degradation is reduced to the alditol using H₂/Ni the resulting product is optically inactive. If compound Z undergoes two Ruff degradations, D-glyceraldehyde (shown below) is obtained. Draw the usual Fischer projection of Z (aldehyde at the top). (Hint: work backwards from D- glyceraldehyde.) H HIC OH CH₂OH D-Glyceraldehydearrow_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
- Which statement about saccharides is correct? (A). Alpha anomers are more common than beta anoomers. (B). Two monosaccharides can form a disaccharide and oxygen. (C). Polysaccharides with alpha or beta glycosidic linkages have the different properties. (D). Disaccharides can be cyclic or acyclic.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_forwardb) Compounds A and B have the following properties: Compound A (C5H1005) is an optically active D-pentose that exists exclusively as an acyclic molecule. (Hint: that sentence alone completely determines the structure of A! But we generously give you more information.) • Compound A does not react with Br₂/H₂O. • Reduction of A with NaBH4 gives only one product, B (C5H12O5), which is optically active. Write acyclic Fischer projection structures of A and B.arrow_forward
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