Chapter 28, Problem 28.66P

Interpretation Introduction

(a)

Interpretation: The glycosidic linkages of the stachyose are to be labeled.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Answer to Problem 28.66P

The red colored oxygen atoms are a part of glycosic linkages as hown below.

Figure 1

Explanation of Solution

The structure of stachyose,

Figure 1

The acetals containing alkoxy groups attached to anomeric carbons in the stachyose represent the glycosidic linkages. The red colored oxygen atoms are a part of glycosic linkages.

Conclusion

The glycosidic linkages of the stachyose are shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: Each glycosidic linkages of the stachyose are to be classified as $\text{α}$ or $\text{β}$ and the numbers are to be used to designate the locations between two rings.

Concept introduction: Monosaccharides are the small units of simple sugars. Polysaccharides are made up of small units of monosaccharides. These small units are joined together by glycoside linkage.

Answer to Problem 28.66P

The glycosidic linkage $\text{I}$ is $\text{α }1\to 6$, $\text{II}$ is $\text{α }1\to 6$ and $\text{III}$ is $\text{α}\mathrm{ 1}\to 2\text{β}$ as shown below.

Figure 1

Explanation of Solution

The structure of stachyose,

Figure 1

The glycosidic linkages which are composed of alkoxy group above the plane are $\text{β}$-glycosidic linkages and which are composed of alkoxy groups below the plane are $\text{α}$-glycosidic linkages. The glycosidic linkage $\text{I}$ is $\text{α }1\to 6$, $\text{II}$ is $\text{α }1\to 6$ and $\text{III}$ is $\text{α}\mathrm{1 }\to 2\text{β}$.

Conclusion

The glycosidic linkage $\text{I}$ is $\text{α }1\to 6$, $\text{II}$ is $\text{α }1\to 6$ and $\text{III}$ is $\text{α}\mathrm{ 1}\to 2\text{β}$ shown in Figure 1.

Interpretation Introduction

(c)

Interpretation: The products formed on hydrolysis of stachyose are to be predicted.

Concept introduction: Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol. Acetals in the presence of acid undergo hydrolysis to form cyclic hemiacetals.

Answer to Problem 28.66P

The products formed on hydrolysis of stachyose are,

Figure 2

Explanation of Solution

On acidic hydrolysis of stachyose, the glycosidic linkages are cleaved to form the four products. The corresponding chemical reaction is shown below.

Figure 3

The products formed are $\text{α}$-D-glucose, $\text{β}$-D-fructose and two units of $\text{α}$-D-galactose.

Interpretation Introduction

(d)

Interpretation: The validation to the corresponding fact that whether stachyose is reducing sugar or not is to be stated.

Concept introduction: The reducing sugars contain hemiacetal and they undergo mutarotation. These sugars are in equilibrium with cyclic monosaccharide forms.

Answer to Problem 28.66P

Stachyose is not a reducing sugar.

Explanation of Solution

The given tetrasaccharide stachyose does not contain any hemiacetal and does not undergo mutarotation. Thus, the given tetrasaccharide is not a reducing sugar.

Conclusion

The product formed on treatment of stachyose with excess ${\text{CH}}_{\text{3}}\text{I}$, ${\text{Ag}}_{\text{2}}\text{O}$ is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation: The product formed on treatment of stachyose with excess ${\text{CH}}_{\text{3}}\text{I}$, ${\text{Ag}}_{\text{2}}\text{O}$ is to be determined.

Concept introduction: The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide. Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol.

Answer to Problem 28.66P

The product formed on treatment of stachyose with excess ${\text{CH}}_{\text{3}}\text{I}$, ${\text{Ag}}_{\text{2}}\text{O}$ is,

Figure 4

Explanation of Solution

On treatment of stachyose with excess ${\text{CH}}_{\text{3}}\text{I}$, ${\text{Ag}}_{\text{2}}\text{O}$, the hydroxyl groups of stachyose are converted into the methoxy groups. The corresponding chemical reaction is shown below.

Figure 5

Conclusion

The product formed on treatment of stachyose with excess ${\text{CH}}_{\text{3}}\text{I}$, ${\text{Ag}}_{\text{2}}\text{O}$ is shown in Figure 4.

Interpretation Introduction

(f)

Interpretation: The product formed on treatment of product in (e) with ${\text{H}}_3{\text{O}}^{+}$ is to be determined.

Concept introduction: Glycosidic linkages are hydrolyzed in presence of acid to form cyclic hemiacetal and corresponding alcohol. Acetals in the presence of acid undergo hydrolysis to form cyclic hemiacetals.

Answer to Problem 28.66P

The products formed on treatment of product in (e) with ${\text{H}}_3{\text{O}}^{+}$ are,

Figure 6

Explanation of Solution

On acidic hydrolysis of the given compound, the glycosidic linkages are cleaved to form the four products. The corresponding chemical reaction is shown below.

Figure 7

Conclusion

The products formed on treatment of product in (e) with ${\text{H}}_3{\text{O}}^{+}$ are shown in Figure 6.