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Chapter 23: Q32P (page 1231)

Ruff degradation of D-arabinose gives D-erythrose. The Kiliani-Fischer synthesis converts D-erythrose to a mixture of D-arabinose and D-ribose. Draw out these reactions and give the structure of D-ribose.

Short Answer

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D-arabinose D-erythrose D-arabinose D-ribose

Step by step solution

01

Step-1.

Ruff degradation is employed when carbohydrate chain needs to be shortened or degraded by a single carbon. Aqueous solution of bromine is used in first step which is used for oxidation of aldehyde to carboxylic acid and then in second step, ferric ion catalyses oxidation reaction with hydrogen peroxide and bond cleavage between carbon-1 and carbon-2 occurs forming an aldehyde. D-arabinose on Ruff degradation gets converted to D-erythrose.

D-arabinose D-erythrose

02

Step-2.

Kiliani-Fischer synthesis is employed when carbohydrate chain needs to be lengthened by one carbon. It involves addition of cyanide ion to aldoses which is then partially reduced and hydrolysed to give a new aldehyde. Kiliani-Fischer synthesis converts D-erythrose into mixture of D-arabinose and D-ribose. A new chiral carbon is generated in this process and due to this, position of hydroxyl group can be on left or right. That’s why we got two products, that is, D-arabinose and D-ribose. Diastereomeric mixture is obtained or products which are diastereomers of each other.

D-erythrose D-arabinose D-ribose

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Most popular questions from this chapter

Predict the products formed when the following sugars react with excess acetic anhydride and pyridine.

  1. α-D-glucopyranose
  2. β-D-ribofuranose

The Wohl degradation, an alternative to the Ruff degradation, is nearly the reverse of the Kiliani-Fischer synthesis. The aldose carbonyl group is converted to the oxime, which is dehydrated by acetic anhydride to the nitrile (a cyanohydrin). Cyanohydrin formation is reversible, and a basic hydrolysis allows the cyanohydrin to lose HCN. Using the following sequence of reagents, give equations for the individual reaction in the Wohl degradation of D-arabinose to D-erythrose. Mechanisms are not required.

(1) hydroxylamine hydrochloride

(2) acetic anhydride

(3) OH-,H2O

H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5′ OH group of nucleosides.

The trityl ether derivative of just the 5′ OH group is obtained by reaction of the nucleoside with trityl chloride, MMT chloride, or DMT chloride and a base like Et3N. The trityl ether derivative can be removed in dilute aqueous acid. DMT derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest.

(a) Draw the product with the trityl derivative on the 5′ oxygen.

(b) Explain why the trityl derivative is selective for the 5′ OH group. Why doesn’t it react at 2′ or 3′? (c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange when acid is added to a DMT derivative?

Emil Fischer synthesized L-gulose, an unusual aldohexose that reduces to give D-glucitol. Suggest a structure for this L sugar, and show how L-gulose gives the same alditol as D-glucose. (Hint: D-Glucitol has -CH2OHgroups at both ends. Either of these primary alcohol groups might have come from reduction of an aldehyde.)

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.

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