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Chapter 23: Question 23.10 (page 1218)

(a) Figure 23-2 shows that the degradation of D-glucose gives D-arabinose, an aldopentose. Arabinose is most stable in its furanose form. Draw D-arabinofuranose.

(b) Ribose, the C2 epimer of arabinose, is most stable in its furanose form. Draw D-ribofuranose.

Short Answer

Expert verified

Answer

(a.)

D-arabinofuranose

(b.)

D-ribofuranose

Step by step solution

01

Haworth projections

Haworth projections are used to depict cyclic sugars. It indicates the stereochemical arrangement of groups on the ring system via up and down orientation of groups or atoms. The H and OH attachment orientation can be drawn in sugar molecules.

02

Furanose form of arabinose

  1. D-arabinofuranose is the furanose form of arabinose sugar. It is formed via hemiacetal formation, and arabinose is most stable in its furanose form, which is a five-membered cyclic form.

D-arabinofuranose

03

Furanose form of ribose


  1. Ribose, an aldopentose, commonly exists in furanose structure, a five-membered cyclic form. The cyclisation of ribose occurs via hemiacetal formation, which is due to an attack on the aldehyde by the carbon-4 hydroxyl to produce furanose form.

D-ribofuranose

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Most popular questions from this chapter

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine at left. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and electron-rich pair of nonbonding electrons.

(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?

Does lactose mutarotate? Is it a reducing sugar? Explain. Draw the two anomeric forms of lactose.

In 1891, Emil Fischer determined the structures of glucose and seven other D-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry and symmetry. He received the Nobel Prize for this work in 1902. Fischer has determined that D-glucose is an aldohexose, and he used Ruff degradation to degrade it to (+)-glyceraldehyde. Therefore, the eight D-aldohexose structures shown in Figure 23-3 are the possible structures for glucose.

Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and sue the following results to prove which of these structures represent glucose, mannose, arabinose, and erythrose.

(a)Upon Ruff degradation, glucose and mannose gives the same aldopentose: arabinose.Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are the two possible structures of arabinose?

(b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive aldaric acid, meso-tartaric acid. What is the structure of erythrose?

(c) Which of the two possible structures of arabinose is correct? What are the possible structures of glucose and mannose?

(d) Fischer鈥檚 genius was needed to distinguish between glucose and mannose. He developed a series of reactions to convert the aldehyde group of an aldose to an alcohol while converting the terminal alcohol to an aldehyde. In effect, he swapped the functional groups on the ends. When he interchanged the functional groups on D-mannose, he was astonished to find that the product was still D-mannose. Show how this information completes the proof of the mannose structure, and show how it implies the correct glucose structure.

(e) When Fischer interchanged the functional groups on D-glucose, the product was an unnatural L sugar. Show which unnatural sugar he must have formed, and show how it completes the proof of the glucose structure.

Some protecting groups can block two OH groups of a carbohydrate at the same time. One such group is shown here, protecting the 4-OH and 6-OH groups of 尾 -D-glucose.

(a) What type of functional group is involved in this blocking group?

(b) What did glucose react with to form this protected compound?

(c) When this blocking group is added to glucose, a new chiral center is formed. Where is it? Draw the stereoisomer that has the other configuration at this chiral center. What is the relationship between these two stereoisomers of the protected compound?

(d) Which of the two stereoisomers in part (c) do you expect to be the major product? Why?

(e) A similar protecting group, called an acetonide, can block reaction at the 2鈥� and 3鈥� oxygens of a ribonucleoside. This protected derivative is formed by the reaction of the nucleoside with acetone under acid catalysis. From this information, draw the protected product formed by the reaction.
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