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Chapter 23: Q34P (page 1232)

Aldohexoses A and B both undergo Ruff degradation to give aldopentose C. On treatment with warm nitric acid, aldopentose C gives an optically active aldaric acid. B also reacts with warm nitric acid to give an optically active aldaric acid, but A reacts to give an optically inactive aldaric acid. Aldopentose C is degraded to aldotetrose D, which gives optically active tartaric acid when it is treated with nitric acid. Aldotetrose D is degraded to (+)-glyceraldehyde. Deduce the structures of sugars A,B,C and D, and use Figure 23-3 to determine the correct names of these sugars.

Short Answer

Expert verified

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 catalyzes oxidation reaction with hydrogen peroxide and bond cleavage between carbon-1 and carbon-2 occurs forming an aldehyde. When aldohexoses A and B undergo Ruff degradation, then resulting aldopentose C will have one less carbon than parent chain. A gives optically inactive aldaric acid whereas B and C gives optically active aldaric acid on warming with nitric acid. Oxidation of aldehyde and alcoholic group takes place on nitric acid addition. From this statement, structures of A, B and C are deduced.

D-galactose D-talose D-lyxose

(C)

Optically active D-galactose D-talose Optically inactive

(A) (B)

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 catalyzes oxidation reaction with hydrogen peroxide and bond cleavage between carbon-1 and carbon-2 occurs forming an aldehyde. When aldohexoses A and B undergo Ruff degradation, then resulting aldopentose C will have one less carbon than parent chain. A gives optically inactive aldaric acid whereas B and C gives optically active aldaric acid on warming with nitric acid. Oxidation of aldehyde and alcoholic group takes place on nitric acid addition. From this statement, structures of A, B and C are deduced.

D-galactose D-talose D-lyxose

(C)

Optically active D-galactose D-talose Optically inactive

(A) (B)

02

Step-2

Aldopentose C is degraded to aldotetrose D, which when treated with nitric acid gives optically active tartaric acid which is not meso compound. Aldotetrose D is further degraded to (+)-glyceraldehyde. Structure of D can be deduced by carrying out Ruff degradation of C and on nitric acid addition, aldehyde and alcoholic group of D gets oxidised to carboxylic acid. (+)-glyceraldehyde gets formed on Ruff degradation of D as hydroxyl group is on right side of chiral carbon in glyceraldehyde.

D-lyxose D-threose

(C) (D)

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Most popular questions from this chapter

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 鈥渆pimer鈥� of D-xylose. Notice that this 鈥渆pimer鈥� is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

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.

Fructose is found in many fruits. From memory, draw fructose in

  1. the Fischer projection of the open chain.
  2. The most stable chair conformation of the most stable pyranose anomer.
  3. The Haworth projection of the most stable pyranose anomer

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.

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鈥� OH group of nucleosides.

The trityl ether derivative of just the 5鈥� 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鈥� oxygen.

(b) Explain why the trityl derivative is selective for the 5鈥� OH group. Why doesn鈥檛 it react at 2鈥� or 3鈥�? (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?

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