Chapter 28, Problem 28.47P

Interpretation Introduction

(a)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 1

Explanation of Solution

The given cyclic monosaccharide is,

Figure 2

The steps for the conversion of Haworth projection into Fischer projection are as follow:

Step 1 In the Haworth projection, if ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

Step-2 The substituents which are present below the ring in the Haworth projection are drawn on the right side in the Fischer projection. Similarly, the substituents which are present above the ring in the Haworth projection are drawn on the left side in the Fischer projection.

The acyclic form of given cyclic monosaccharide is,

Figure 3

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 4

Explanation of Solution

The given cyclic monosaccharide is,

Figure 5

The steps for the conversion of Haworth projection into Fischer projection are as follow:

Step 1 In the Haworth projection, if ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

Step-2 The substituents which are present below the ring in the Haworth projection are drawn on the right side in the Fischer projection. Similarly, the substituents which are present above the ring in the Haworth projection are drawn on the left side in the Fischer projection.

The acyclic form of given cyclic monosaccharide is,

Figure 6

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 7

Explanation of Solution

The given cyclic monosaccharide is,

Figure 8

In the chair form, if the ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

The substituents on a carbon are present above the plane (i.e. either equatorial or axial position) in chair form are drawn to left side in the Fischer projection and vice versa.

The acyclic form of given cyclic monosaccharide is shown below.

Figure 9

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 7.

Interpretation Introduction

(d)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 10

Explanation of Solution

The given cyclic monosaccharide is,

Figure 11

The steps for the conversion of Haworth projection into Fischer projection are as follow:

Step 1 In the Haworth projection, if ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

Step-2 The substituents which are present below the ring in the Haworth projection are drawn on the right side in the Fischer projection. Similarly, the substituents which are present above the ring in the Haworth projection are drawn on the left side in the Fischer projection.

The acyclic form of given cyclic monosaccharide is,

Figure 12

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 10.

Interpretation Introduction

(e)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 13

Explanation of Solution

The given cyclic monosaccharide is,

Figure 14

The steps for the conversion of Haworth projection into Fischer projection are as follow:

Step 1 In the Haworth projection, if ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

Step-2 The substituents which are present below the ring in the Haworth projection are drawn on the right side in the Fischer projection. Similarly, the substituents which are present above the ring in the Haworth projection are drawn on the left side in the Fischer projection.

The acyclic form of given cyclic monosaccharide is,

Figure 15

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 13.

Interpretation Introduction

(f)

Interpretation: The given cyclic monosaccharide is to be converted into its acyclic form.

Concept introduction: The structural representation of sugar molecule in cyclic form is known as Haworth projection. Sugar molecule that has six-membered-ring is known as pyranose and sugar molecule that has five-membered-ring is called furanose. In Fischer projection formula, the horizontal and vertical line represents the bonds that are present above and below the plane, respectively. The verticals bonds are represented as dashed wedge and horizontal bonds as dark wedge.

Answer to Problem 28.47P

The acyclic form of given cyclic monosaccharide is,

Figure 16

Explanation of Solution

The given cyclic monosaccharide is,

Figure 17

The steps for the conversion of Haworth projection into Fischer projection are as follow:

Step 1 In the Haworth projection, if ${\text{CH}}_{\text{2}}\text{OH}$ group is present above the ring, indicates that the $\text{OH}$ group is present on the right side of the $\text{C5}$ in Fischer projection. Thus, the $\text{OH}$ group is drawn on the right side of the $\text{C5}$ and vice versa.

Step-2 The substituents which are present below the ring in the Haworth projection are drawn on the right side in the Fischer projection. Similarly, the substituents which are present above the ring in the Haworth projection are drawn on the left side in the Fischer projection.

The acyclic form of given cyclic monosaccharide is,

Figure 18

Conclusion

The acyclic form of given cyclic monosaccharide is shown in Figure 16.