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Draw the products you would obtain by reaction of \(\beta\) -D-ribofuranose with: (a) \(\mathrm{CH}_{3} \mathrm{I}, \mathrm{Ag}_{2} \mathrm{O}\) (b) \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\), pyridine

Short Answer

Expert verified
(a) Tetra-O-methyl ether of \(\beta\)-D-ribofuranose. (b) Tetra-O-acetate of \(\beta\)-D-ribofuranose.

Step by step solution

01

Understanding the Problem

The problem involves reacting \( \beta \)-D-ribofuranose with two different reagents and predicting the products of each reaction. \( \beta \)-D-ribofuranose is a cyclic form of ribose, a five-carbon sugar, with specific stereochemistry.
02

Reaction with \( \mathrm{CH}_{3} \mathrm{I} \) and \( \mathrm{Ag}_{2} \mathrm{O} \)

This reaction involves methylation. \( \mathrm{Ag}_{2} \mathrm{O} \) acts as a base and facilitates the substitution of hydrogen atoms on the hydroxyl groups of the sugar with \( \mathrm{CH}_{3} \) groups from \( \mathrm{CH}_{3} \mathrm{I} \). All hydroxyl groups on \( \beta \)-D-ribofuranose will be converted into methoxy groups, resulting in the formation of \( \beta \)-D-ribofuranose tetra-O-methyl ether.
03

Reaction with \( \left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \) and pyridine

This reaction involves acetylation of the hydroxyl groups in the presence of pyridine, which acts as a base and catalyst. The acetyl groups from \( \left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \) will replace the hydrogen atoms of the hydroxyl groups to form esters. Therefore, all the hydroxyl groups of \( \beta \)-D-ribofuranose will be converted into acetoxy groups, yielding \( \beta \)-D-ribofuranose tetra-O-acetate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carbohydrate Chemistry
Carbohydrates are essential biomolecules made of carbon, hydrogen, and oxygen. They are commonly known as sugars and come in different forms, including simple sugars like glucose and complex forms like starch. Carbohydrates play a crucial role in energy storage and supply in living organisms.
They exist in different isomeric forms depending on the arrangement of atoms. One fascinating aspect of carbohydrates is their ability to form cyclic structures.
These structures are prevalent in nature and add to the complexity and functionality of sugars.
Methylation Reaction
Methylation is a chemical reaction in which a methyl group (\( \mathrm{CH}_{3} \)) is added to a molecule.
This process is essential in modifying the properties of various organic compounds, including sugars. In the context of carbohydrate chemistry, methylation involves substituting the hydroxyl (\( \mathrm{OH} \)) groups on a sugar with methoxy (\( \mathrm{OCH}_{3} \)) groups.
Using reagents like \( \mathrm{CH}_{3} \mathrm{I} \) and \( \mathrm{Ag}_{2} \mathrm{O} \), the hydroxyl groups on the sugar are replaced. The outcome is the transformation of \( \beta \)-D-ribofuranose into \( \beta \)-D-ribofuranose tetra-O-methyl ether.
This modification can protect the sugar from further reaction and alter its solubility and reactivity.
Acetylation Reaction
Acetylation is the process of introducing an acetyl group (\( \mathrm{CH}_{3} \mathrm{CO} \)) into a compound, often to modify its chemical properties.
This reaction is common in the modification of sugars, and in our exercise, it involves the use of \( \left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \) and pyridine.
During acetylation, the hydroxyl groups on \( \beta \)-D-ribofuranose are replaced with acetoxy groups, forming \( \beta \)-D-ribofuranose tetra-O-acetate.
Acetylation changes the polarity and reactivity of the sugar, making it more hydrophobic.
Cyclic Sugars
Cyclic sugars are sugars that form ring structures instead of remaining linear.
This occurs due to the reaction between a carbonyl group and a hydroxyl group within the same molecule.
In the case of \( \beta \)-D-ribofuranose, it forms a five-membered ring, known as a furanose ring.
These cyclic forms are more stable and prevalent in biological systems.
  • The cyclic structure influences the sugar's reactivity and interaction with other molecules.
  • The stability of these rings plays a crucial role in energy metabolism and the structural integrity of nucleic acids like RNA.
Understanding these cyclic forms helps in predicting the behavior of sugars in chemical reactions and biological processes.

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Most popular questions from this chapter

(+)-Arabinose, an aldopentose that is widely distributed in plants, is systematically named \((2 R, 3 S, 4 S)-2,3,4,5\) -tetrahydroxypentanal. Draw a Fischer projection of ( \(+\) )-arabinose, and identify it as a o sugar or an L sugar.

Draw the structure of L-galactose, and then answer the following questions: (a) Which other aldohexose gives the same aldaric acid as L-galactose on oxidation with warm \(\mathrm{HNO}_{3}\) ? (b) Is this other aldohexose a D sugar or an L sugar? (c) Draw this other aldohexose in its most stable pyranose conformation.

The Kiliani-Fischer chain extension is a method for lengthening an aldose chain by one carbon, giving two new aldoses that differ in stereochemistry at C2. D-Erythrose, for instance, yields a mixture of D-ribose and D-arabinose on Kiliani-Fischer chain extension. \(\begin{array}{lll}\text { D-Erythrose } & \text { D-Pibose } & \text { D-Arabinose }\end{array}\) (a) What product(s) would you expect from Kiliani-Fischer reaction of D-ribose? (b) What aldopentose would give a mixture of L-gulose and \(\mathrm{L}\) -idose on Kiliani-Fischer chain extension?

ane of the steps in the biological pathway for carbohydrate metabolism is the conversion of fructose 1,6 -bisphosphate into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. Propose a mechanism for the transformation. \(\begin{array}{ccc}\text { Fructose } & \text { Dihydroxyacetone } & \text { Glyceraldehyde } \\ \text { 1,6-bisphosphate } & \text { phosphate } & \text { 3-phosphate }\end{array}\)

Reduction of D-glucose leads to an optically active alditol (D-glucitol), whereas reduction of D-galactose leads to an optically inactive alditol. Explain.

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