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Question:

a) Draw D-allose, the C3 epimer of glucose.

b) Draw D-talose, the C2 epimer of D-galactose.

c) Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.

d) Draw the C4 “epimer” of D-xylose. Notice that this “epimer” is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

Short Answer

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Answer

(a)

(b)

(c)

(d)

Step by step solution

01

Epimers

Epimers are the compounds which are stereoisomers and differ at only one stereogenic center like the placing the OH group is different from the other. For example, D-glucose and D-galactose are epimers having only one chiral center with different configuration.

02

The structure of D-allose, D-talose and 2-idose and their comparison

(a) Epimers are stereoisomers which differs in configuration at any one stereogenic or chiral center. C3 epimer of glucose means configuration differs at carbon-3 of glucose which gives rise to different stereoisomer that is, D-allose. Position of hydroxyl group in glucose at carbon-3 is at left whereas position of hydroxyl group at carbon-3 of D-allose is at right.

The structure of D-allose


(b) Similarly, D-talose is C2 epimer of D-galactose and differs in configuration at carbon-2. Position of hydroxyl group in D-galactose is at right at carbon-2 whereas in D-talose, it is at left. D-idose is C3 epimer of D-talose and differs in configuration at carbon-3.

The structure of D-talose


(c) Position of hydroxyl group is at left at carbon-3 in D-talose whereas it is at right in D-idose.

The structure of D-idose

03

The C4 epimer of D-xylose

(d) C4 epimer of D-xylose is L-arabinose. Change in configuration at carbon-4 of D-xylose with respect to hydroxyl group changes the sugar series from “D” to “L” as the change occurred at bottom chiral center which as per convention, leads to formation of enantiomer. The D/L system names molecules by relating them to glyceraldehyde molecule.

The C4 epimer of D-xylose

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Most popular questions from this chapter

Except for the Tollens test, basic aqueous conditions are generally avoided with sugars because they lead to fast isomerizations.

  1. Under basic conditions, the proton alpha to the aldehyde (or ketone) carbonyl group is reversibly removed, and the resulting enolate ion is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either the original structure or its epimer. Because a mixture of epimer results, this process is called epimerization. Propose a mechanism for the base-catalyzed equilibration of glucose to a mixture of glucose and its C2 epimer, mannose.
  2. Propose a mechanism for the isomerization of a ketose to an aldose, via the enediol intermediate, shown immediately above. Note that the enediol has twoprotons and removing one or the other gives two different enolate ions.

When the gum of the shrub Sterculia setigera is subjected to acidic hydrolysis, one of the water-soluble components of the hydrolysate is found to be tagatose. The following information is known about tagatose:

(1) Molecular formula C6H12O6

(2) Undergoes mutarotation.

(3) Does not react with bromine water.

(4) Reduces Tollens reagent to give d-galactonic acid and d-talonic acid.

(5) Methylation of tagatose (using excess CH3 I and Ag2O) followed by acidic hydrolysis gives 1,3,4,5-tetra-O-methyltagatose.

(a) Draw a Fischer projection structure for the open-chain form of tagatose.

(b) Draw the most stable conformation of the most stable cyclic hemiacetal form of tagatose.

(a) Which of the D-aldopentoses will give optically active aldaric acids on oxidation with HNO3 ?

(b) Which of the D-aldotetroses will give optically active aldaric acids on oxidation withHNO3 ?

(c) Sugar X is known to be a D-aldohexose. On oxidation with HNO3 , X gives an optically inactive aldaric acid. When X is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X.

(d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically active products?

(e) Show what products results if the aldopentose formed from degradation of X is further degraded to an aldotetrose. DoesHNO3 oxidize this aldotetrose to an optically active aldaric acid?

Draw the following sugar derivatives.

  1. ethyl-β-D-glucopyranoside
  2. 2,3,4,6-tetra-O-ethyl-D-mannopyranose
  3. 1,3,6-tri-O-ethyl-D-fructofuranose
  4. Ethyl-2,3,4,6-tetra-O-methyl-β-D-galactopyranoside

Question. Cellulose is converted to cellulose acetateby treatment with acetic anhydride and pyridine. Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into fibres. Show the structure of cellulose acetate.

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