Chapter 18, Problem 18.83EP

(a)

Interpretation Introduction

Interpretation: The open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$ has to be drawn.

Concept introduction: The Fischer projection formula can be drawn by using the following rules.

• Ø For $\alpha$ and $\beta$ configurations, if in the cyclic form of Haworth projection, $\text{OH}$ group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
• Ø If in the cyclic form of Haworth projection, $\text{OH}$ group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
• Ø For $\beta$ configuration, the position of $\text{OH}$ group at the first carbon with respect to ${\text{CH}}_{\text{2}}\text{OH}$ group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

Answer to Problem 18.83EP

The open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$ is,

Explanation of Solution

The given Haworth projection formula is shown as,

The open-chain form of the given Haworth projection formula of an $\alpha -\text{D}-\text{monosaccharide}$ is shown as,

Thus, in this open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$, ${\text{CH}}_{\text{2}}\text{OH}$ group is placed at the last carbon. The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide. The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(b)

Interpretation Introduction

Interpretation: The open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$ has to be drawn.

Concept introduction: The Fischer projection formula can be drawn by using the following rules.

• Ø For $\alpha$ and $\beta$ configurations, if in the cyclic form of Haworth projection, $\text{OH}$ group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
• Ø If in the cyclic form of Haworth projection, $\text{OH}$ group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
• Ø For $\beta$ configuration, the position of $\text{OH}$ group at the first carbon with respect to ${\text{CH}}_{\text{2}}\text{OH}$ group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

Answer to Problem 18.83EP

The open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$ is,

Explanation of Solution

The given Haworth projection formula is shown as,

The open-chain form of the given Haworth projection formula of an $\alpha -\text{D}-\text{monosaccharide}$ is shown as,

Thus, in this open-chain form of the given $\alpha -\text{D}-\text{monosaccharide}$, ${\text{CH}}_{\text{2}}\text{OH}$ group is placed at the last carbon. The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide. The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(c)

Interpretation Introduction

Interpretation: The open-chain form of the given $\beta -\text{D}-\text{monosaccharide}$ has to be drawn.

Concept introduction: The Fischer projection formula can be drawn by using the following rules.

• Ø For $\alpha$ and $\beta$ configurations, if in the cyclic form of Haworth projection, $\text{OH}$ group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
• Ø If in the cyclic form of Haworth projection, $\text{OH}$ group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
• Ø For $\beta$ configuration, the position of $\text{OH}$ group at the first carbon with respect to ${\text{CH}}_{\text{2}}\text{OH}$ group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

Answer to Problem 18.83EP

The open-chain form of the given $\beta -\text{D}-\text{monosaccharide}$ is,

Explanation of Solution

The given Haworth projection formula is shown as,

The open-chain form of the given Haworth projection formula of a $\beta -\text{D}-\text{monosaccharide}$ is shown as,

Thus, in this open-chain form of the given monosaccharide, ${\text{CH}}_{\text{2}}\text{OH}$ group is placed at the last carbon. The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide. But as it is a $\beta -\text{D}-\text{monosaccharide}$, thus $\text{OH}$ group at the first carbon with respect to ${\text{CH}}_{\text{2}}\text{OH}$ group is drawn at the left side of the open-chain structure. The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.

(d)

Interpretation Introduction

Interpretation: The open-chain form of the given $\beta -\text{D}-\text{monosaccharide}$ has to be drawn.

Concept introduction: The Fischer projection formula can be drawn by using the following rules.

• Ø For $\alpha$ and $\beta$ configurations, if in the cyclic form of Haworth projection, $\text{OH}$ group is placed below the ring then the same hydroxyl group is drawn at the right position in Fischer projection formula.
• Ø If in the cyclic form of Haworth projection, $\text{OH}$ group is placed above the ring then the same hydroxyl group is drawn at the left position in Fischer projection formula.
• Ø For $\beta$ configuration, the position of $\text{OH}$ group at the first carbon with respect to ${\text{CH}}_{\text{2}}\text{OH}$ group always remains in the same direction and in the Fischer projection formula it is placed at the left side.

Answer to Problem 18.83EP

The open-chain form of the given $\beta -\text{D}-\text{monosaccharide}$ is,

Explanation of Solution

The given Haworth projection formula is shown as,

The open-chain form of the given Haworth projection formula of a $\beta -\text{D}-\text{monosaccharide}$ is shown as,

Thus, in this open-chain form of the given monosaccharide, ${\text{CH}}_{\text{2}}\text{OH}$ group is placed at the last carbon. The hydroxyl group that is placed below the cyclic ring is then drawn at the right-side position of the open-chain form of monosaccharide. The hydroxyl group that is placed above the cyclic ring is then drawn at the left-side position of the open-chain form of monosaccharide.