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D-Raffinose is a trisaccharide that does not react with Fehling's solution. Upon hydrolysis it produces \(\mathrm{D}\) - glucose, D-galactose, and D-fructose. Completely methylated raffinose is hydrolyzed to \(2,3,4-\) tri- O -methylglucose, \(1,3,4,6-\) tetra-O-methylfrueto se, and \(2,3,4,6-\) tetra- \(\mathrm{O}-\) methylgalactose. What structures for raffinose are consistent with these data? What further information, if any, is required to identify completely the structure of raffinose?

Short Answer

Expert verified
D-Raffinose is a trisaccharide consisting of D-glucose, D-fructose, and D-galactose monosaccharides linked together. Based on the methylation pattern of the hydrolyzed products, D-raffinose has D-glucose connected at its first carbon atom to the second carbon atom of D-fructose, which in turn is connected at its first carbon atom to the first carbon atom of D-galactose. Thus, the structure can be represented as D-glucose (1→2) D-fructose (1→1) D-galactose. However, to completely identify the structure, additional information is needed about the type of glycosidic bonds (α or β) between the monosaccharides.

Step by step solution

01

Analyze the given information for D-raffinose hydrolysis

D-raffinose is a trisaccharide that hydrolyzes to D-glucose, D-galactose, and D-fructose. This means that it consists of these three monosaccharides linked together in some configuration.
02

Analyze the given information for completely methylated D-raffinose hydrolysis

When completely methylated D-raffinose is hydrolyzed, it produces 2,3,4-tri-O-methylglucose, 1,3,4,6-tetra-O-methylfructose, and 2,3,4,6-tetra-O-methylgalactose. This information will help us in finding the possible linkage points in D-raffinose.
03

Determine monosaccharide connections

Based on the methylation pattern of the hydrolyzed products, the connection points on the monosaccharides can be inferred: - In D-glucose, the 2,3,4 positions are methylated, meaning the original connection point must be at the 1 position. - In D-fructose, the 1,3,4,6 positions are methylated, meaning the original connection point must be at the 2 position. - In D-galactose, the 2,3,4,6 positions are methylated, meaning the original connection point must be at the 1 position.
04

Propose a structure for D-raffinose

Given the connection points determined in Step 3, D-raffinose has D-glucose connected at its first carbon atom to the second carbon atom of D-fructose, which in turn is connected at its first carbon atom to the first carbon atom of D-galactose. Therefore, the structure can be represented as: D-glucose (1→2) D-fructose (1→1) D-galactose.
05

Determine if further information is needed

To completely identify the structure of D-raffinose, we need information about the type of glycosidic bonds (α or β) between the monosaccharides. Since this information is not given, additional information is required to fully identify the structure.

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Key Concepts

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

Trisaccharides
Trisaccharides are a type of carbohydrate made of three monosaccharide units. They are essentially three simple sugars linked together. Unlike simpler carbohydrates, like monosaccharides and disaccharides, trisaccharides have more complex structures.
This complexity comes from the arrangement and the type of glycosidic linkages between the monosaccharide units.
  • For example, raffinose is a trisaccharide consisting of D-glucose, D-fructose, and D-galactose.
  • These sugars are connected via specific glycosidic bonds, which determine the overall structure and properties.
Trisaccharides serve various functions in organisms, from energy storage to serving as protective agents against extreme environmental conditions.
Monosaccharides
Monosaccharides are the simplest form of carbohydrates and are often called single sugars. They are the building blocks for more complex carbohydrates, including disaccharides, oligosaccharides, and polysaccharides.
In the context of D-raffinose, the monosaccharides involved are D-glucose, D-galactose, and D-fructose.
  • D-glucose is a primary source of energy for living cells.
  • D-galactose is similar to glucose but differs in orientation of one hydroxyl group.
  • D-fructose is a sugar found in many plants and is sweeter than glucose.
Monosaccharides have unique properties, such as forming rings in solution, which are crucial for forming glycosidic bonds.
Methylation
Methylation involves the addition of a methyl group ( ext{-CH}_3) to a molecule. In the context of carbohydrates, methylation of sugars can provide insights into the points where glycosidic bonds form.
For example, the methylation pattern observed in the hydrolysis products of completely methylated D-raffinose reveals the connection points of the monosaccharides.
  • In D-glucose, methyl groups at positions 2, 3, and 4 indicate an original link at position 1.
  • For D-fructose, methyl groups at positions 1, 3, 4, and 6 point to a link at position 2.
  • D-galactose, with methylation at 2, 3, 4, and 6, shows a link at position 1.
Methylation is a powerful technique in structural biochemistry for elucidating sugar linkages.
Glycosidic bonds
Glycosidic bonds are chemical linkages that connect sugar molecules through an oxygen atom. They are crucial in forming oligosaccharides and polysaccharides from simpler monosaccharides.
In D-raffinose, the glycosidic bonds hold together the D-glucose, D-fructose, and D-galactose molecules, forming the trisaccharide structure.
  • The type of bond (either α or β) affects the sugar's properties and how it interacts with other substances.
  • The position of these bonds, as inferred through methylation data, helps in predicting the core structure of trisaccharides.
Determining whether a glycosidic bond is α or β helps in fully understanding the structure and reactivity of the carbohydrate.
Fehling's Solution
Fehling's solution is a chemical test used to differentiate between water-soluble carbohydrate and ketone functional groups, and it helps to identify reducing sugars.
This solution contains copper(II) ions in alkaline solution, which make it blue.
  • Reducing sugars will reduce the copper(II) ions to copper(I) oxide, which makes a brick-red precipitate.
  • D-raffinose does not react with Fehling's solution, indicating it is a non-reducing sugar.
This specific behavior helps in testing sugars and understanding their reactivity based on their structure and composition.

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Most popular questions from this chapter

(a) A certain sugar and its methyl glycoside can be isolated in both \(\alpha-\) and \(\beta\) -anomeric forms. Also, the ring structure of the glycoside is known to be present in the pyranose form. How could you ascertain from rotation data whether or not the parent sugar also exists in the pyranose form? (b) A D-glucose derivative with a 1,2 -epoxide ring gives a methy, \(\beta\) -D-glucoside on solvolysis with methanol. Explain how this reaction helps to establish the configuration of the \(\alpha-\) and \(\beta\) -anomers of D-glucose.

(a) \((+)\) - Trehalose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), a non-reducing sugar found in young mushrooms, gives only D-glucose when hydrolyzed by aqueous acid or by maltase. Methylation gives an octa-DMethyl derivative that, upon hydrolysis, yields only \(2,3,4,6-\) tetra-O-methyl-D- glucose. What is the structure and systematic name for \((+)\) -trehalose? (b) (-)-Isotrehalose and \((+)-\) neotrehalose resemble trehalose in most respects. However, isotrehalose is hydrolyzed by either emulsin or maltase, and neotrehalose is hydrolyzed only by emulsin. What are the structures and systematic names for these two carbohydrates?

(a) How many chiral centers are there in (+)-fructose? (b) How many stereo isomeric 2-ketohexoses should there be? (c) Draw a "cross" formula of one enantiomer of each pair, placing \(\mathrm{C}=\mathrm{O}\) near the top, and \(-\mathrm{OH}\) on the right on the lowest chiral center \((\mathrm{C}-5)\).

Identify each of the following glucose derivatives: (a) \(\mathrm{A}+4 \mathrm{HIO}_{4} \rightarrow 3 \mathrm{HCOOH}+\mathrm{HCHO}+\mathrm{OHC}-\mathrm{COOH}\) (b) \(\mathrm{B}+5 \mathrm{HIO}_{4} \rightarrow 4 \mathrm{HCOOH}+2 \mathrm{HCHO}\) (c) \(\mathrm{C}+3 \mathrm{HIO}_{4} \rightarrow 2 \mathrm{HCOOH}+2 \mathrm{OHC}-\mathrm{COOH}\) (d) \(\mathrm{D}+4 \mathrm{HIO}_{4} \rightarrow 4 \mathrm{HCOOH}+\mathrm{OHC}-\mathrm{COOH}\)

How can the D-glucoside units of cellulose produce a polymer with a stronger, more compact physical structure than the D-glucose units of starch?

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