Question

A disaccharide, \(\mathrm{G}, \mathrm{C}_{11} \mathrm{H}_{20} \mathrm{O}_{10}\), is hydrolyzable by \(\alpha\) glucosidase, yielding D-glucose and a D-pentose. The disaccharide does not reduce Fehling's solution. Methylation of \(\mathrm{G}\) with dimethyl sulfate in NaOH yields a hepta-methyl ether, \(\mathrm{H}\), which upon acid hydrolysis yields \(2,3,4,6\) -tetra-O methyl-D- glucose and a pentose tri-O- methyl ether, (I). Oxidation of I by bromine water yields \(2,3,4\) -tri-O-methyl-D ribonic acid. Assign structures to compounds G through I compatible with these data.

Short answer

In summary, the structures of compounds G, H, and I are: - G: α-ᴅ-glucose-(1→5)-ᴅ-ribose - H: 2,3,4,6-tetra-O-methyl-ᴅ-glucose-(1→5)-2,3,4-tri-O-methyl-ᴅ-ribose - I: 2,3,4-tri-O-methyl-ᴅ-ribose

Step-by-step solution

Identify the non-reducing sugar in disaccharide G

Since G does not reduce Fehling's solution, it must contain a non-reducing sugar. There are only two non-reducing sugars: sucrose and trehalose. But since G is hydrolyzed by α-glucosidase, we can eliminate sucrose (which is hydrolyzed by sucrase). Therefore, the non-reducing sugar in G is trehalose, which contains two glucose molecules connected by an α-1,1-glycosidic linkage.

Identify the D-pentose and the glycosidic linkage

We know that the disaccharide G yields D-glucose and a D-pentose after hydrolysis. Therefore, G must be composed of D-pentose and D-glucose connected by a glycosidic linkage. Based on the oxidation reaction of I, we can conclude that the D-pentose is D-ribose because the oxidation product is 2,3,4-tri-O-methyl-D-ribonic acid. Now we need to determine the glycosidic linkage between D-glucose and D-ribose. As G is hydrolyzable by α-glucosidase, the linkage must be an α-linkage, which means that the anomeric carbon (C1) of glucose is involved in the linkage. This means that the glucose molecule is connected with ribose via an α-glycosidic linkage. Since both glucose and ribose are pentoses, they have 5 available hydroxyl groups. By considering that G is methylated with dimethyl sulfate in NaOH and forms H, which is a heptamethyl ether, we can conclude that the glycosidic linkage is formed between the O1 of glucose and the O5 of ribose. Thus, we have α-ᴅ-glucose-(1→5)-ᴅ-ribose as the structure of the disaccharide G.

Determine the structure of compound H

Methylation of G with dimethyl sulfate in NaOH yields a hepta-methyl ether, H. Since G contains two glucose molecules and one ribose molecule, and each sugar has five hydroxyl groups, 7 out of the 10 hydroxyl groups in G are converted to methyl ethers in H. The remaining 3 hydroxyl groups must be at the glycosidic linkage (O1 of glucose and O5 of ribose) and C6 of glucose. We can write the structure of compound H as 2,3,4,6-tetra-O-methyl-ᴅ-glucose-(1→5)-2,3,4-tri-O-methyl-ᴅ-ribose.

Determine the structure of compound I

Acid hydrolysis of H yields 2,3,4,6-tetra-O-methyl-ᴅ-glucose and a pentose tri-O-methyl ether, I. Based on the information given, compound I, after being oxidized, gives 2,3,4-tri-O-methyl-D-ribonic acid. Since oxidation doesn't affect the methylated hydroxyl groups, we can conclude that compound I is 2,3,4-tri-O-methyl-ᴅ-ribose. In conclusion, the structures of compounds G, H, and I are: - G: α-ᴅ-glucose-(1→5)-ᴅ-ribose - H: 2,3,4,6-tetra-O-methyl-ᴅ-glucose-(1→5)-2,3,4-tri-O-methyl-ᴅ-ribose - I: 2,3,4-tri-O-methyl-ᴅ-ribose

Key concepts

Understanding Non-Reducing Sugars

When dealing with the category of carbohydrates known as non-reducing sugars, one must appreciate their inability to participate in certain chemical reactions. Unlike reducing sugars, non-reducing sugars do not have a free anomeric carbon available to act as a reducing agent. This characteristic makes them unreactive towards tests such as Fehling's solution, which is commonly used to identify reducing sugars.

A classic example pertinent to our exercise is trehalose, a disaccharide composed of two glucose units. Due to its unique bonding, it does not possess a free aldehyde or ketone group crucial for the reduction of Fehling's solution. Understanding the behavior of non-reducing sugars is essential, particularly in deciphering the structure of complex carbohydrates and their behavior in biological systems.

The Role of Glycosidic Linkage

The glycosidic linkage is a fundamental bond that connects two sugar molecules in carbohydrates. It forms between an anomeric carbon of one sugar molecule and a hydroxyl group of another. This bond is pivotal in the structure and function of various disaccharides and polysaccharides. In our exercise scenario, we determine the type of glycosidic linkage through enzymatic activity; for instance, alpha-glucosidase specifically hydrolyzes an α-linkage.

Depending on the orientation of the linked anomeric carbon (alpha or beta) and which carbons are connected, the properties of the sugar can dramatically change. This is vital when predicting the reactivity of sugars, their digestion, and their role in living organisms. The discovery of an α-(1→5)-glycosidic linkage between glucose and a pentose sugar in our disaccharide deciphered through careful analysis of enzymatic hydrolysis and product identification is key to understanding the structure and function of the carbohydrate in question.

Methylation of Sugars Analysis

The process of methylation of sugars is a critical step in structure elucidation. In the context of our exercise, methylation protects the hydroxyl groups in the sugar molecules by converting them to methyl ethers. Using reagents like dimethyl sulfate and base (NaOH), this treatment results in derivatized sugars that can resist hydrolysis. After this, acid hydrolysis of the methylated product can reveal which hydroxyl groups were originally part of the glycosidic linkage and which were free.

This method gives us a map of protected sites and can lead us to the exact position of glycosidic bonds within complex carbohydrates. It is particularly useful when it is used in conjunction with other techniques such as spectroscopy or chromatography for a comprehensive structural analysis. In the given exercise, methylation helped to identify the hepta-methyl ether configuration of compound H and to deduce the structure of compounds G and I effectively.