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Chapter 23: Q10P. (page 1218)

Question. (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.

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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

(a) 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.

03

Furanose form of ribose

(b) 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.

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Most popular questions from this chapter

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

Draw the following monosaccharides, using chair conformations for the pyranoses and Haworth projections for the furanoses.

(a)α‐D‐mannopyranose(C2epimerofglucose)(b)β‐D‐galactopyranose(C4epimerofglucose)(c)β‐D‐allopyranose(C3epimerofglucose)(d)α‐D‐arabinofuranose(e)β‐D‐ribofuranose(C2epimerofarabinose)

(a) Show the product that results when fructose is treated with an excess of methyl iodide and silver oxide.

(b) Show what happens when the product of part (a) is hydrolyzed using dilute acid.

(c) What do the results of parts (a) and (b) imply about the hemiacetal structure of fructose?

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.

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyl-disiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.

(a) The TIPDS group is somewhat hindered around the Siatoms by the isopropyl groups. Which OHis more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen.

(b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected.

(c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

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