Aswer all parts of the question.A freshly prepared solution of αα ‑D‑glucose shows a specific rotation of +112°.+112°. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [?]25 °CD=+52.5°.[α]D25 °C=+52.5°. In contrast, a freshly prepared solution of ?β‑D‑glucose has a specific rotation of +19°.+19°. The rotation of this solution increases over time to the same equilibrium value as that shown by the ?α anomer. A solution of one enantiomer of a given monosaccharide rotates plane‑polarized light to the left (counterclockwise) and is the levorotatory isomer, designated (−). The other enantiomer rotates plane‑polarized light to the same extent but to the right (clockwise) and is the dextrorotatory isomer, designated (+). An equimolar mixture of the (+) and (−) forms does not rotate plane‑polarized light.The optical rotation, the number of degrees by which plane‑polarized light rotates on passage through a given path length of a solution of the compound at a given concentration, quantitatively describes the optical activity of a stereoisomer. The specific rotation ([?]??)([α]λt) of an optically active compound is specific for a particular temperature (?)(t) and wavelength of light (?).(λ). Usually, as here, ?λ is the D line of sodium, 589 nm. **Question:** Which Haworth perspective formula represents β-D-glucose?**Description of Structures:**1. **Structure 1:** - Hexagonal ring structure with one oxygen atom included. - Hydroxyl (OH) groups attached to the carbon atoms. - The terminal carbon with an OH group is positioned above the plane of the ring.2. **Structure 2:** - Similar hexagonal ring structure with one oxygen atom. - Hydroxyl groups attached to carbon atoms. - Both the anomeric carbon's OH group and the hydroxyl group on the last carbon appear above the ring plane.3. **Structure 3:** - Hexagonal ring including one oxygen atom. - Hydroxyl groups are attached to the carbon atoms, indicating an opposite orientation compared to Structure 1. - The terminal carbon bonded OH group is positioned below the plane of the ring.4. **Structure 4:** - Hexagonal ring with one oxygen atom. - Alternating arrangement of hydroxyl groups around the ring. - Anomeric carbon’s hydroxyl group positioned above ring plane, terminal carbon’s OH group below the plane.**Answer Choices:**- 1- 3- 4- 2**Explanation for Educational Purposes:** The Haworth projection of a carbohydrate depicts the molecule as a cyclic hemiacetal. In β-D-glucose, the hydroxyl group on the anomeric carbon (the first carbon) is oriented above the plane of the ring. Therefore, the β form of D-glucose has the OH group on the anomeric carbon in the same direction as the CH2OH group on the fifth carbon, which is above the plane in common Haworth representations of D-glucose.The correct answer is **2**. **Question:**Why does the specific rotation of a freshly prepared solution of the α form gradually decrease with time? Why do solutions of the α and β forms reach the same specific rotation at equilibrium?**Options:**- ○ Glucose interconverts between cyclic and linear structures. The linear structure is most favorable. When the system reaches equilibrium, the net amounts of each structure no longer change.- ○ Glucose interconverts between pyranose and furanose forms. The furanose form is most favorable. When the system reaches equilibrium, the net amounts of each form no longer change.- ○ Glucose interconverts between α and β configurations. The β configuration is the most favorable. When the system reaches equilibrium, the net amounts of each configuration no longer change.---**Task:**Calculate the percentage of each of the two forms of D-glucose present at equilibrium.- **α-D-glucose:** [Input Box] %- **β-D-glucose:** [Input Box] %

Given that, the specific rotation of the α ‑D‑glucose and β‑D‑glucose are +112° and +19°,…

Answered: Why does the specific rotation of a…

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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A freshly prepared solution of αα ‑D‑glucose shows a specific rotation of +112°.+112°. Over time, the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to [?]25 °CD=+52.5°.[α]D25 °C=+52.5°. In contrast, a freshly prepared solution of ?β‑D‑glucose has a specific rotation of +19°.+19°. The rotation of this solution increases over time to the same equilibrium value as that shown by the ?α anomer. A solution of one enantiomer of a given monosaccharide rotates plane‑polarized light to the left (counterclockwise) and is the levorotatory isomer, designated (−). The other enantiomer rotates plane‑polarized light to the same extent but to the right (clockwise) and is the dextrorotatory isomer, designated (+). An equimolar mixture of the (+) and (−) forms does not rotate plane‑polarized light.

The optical rotation, the number of degrees by which plane‑polarized light rotates on passage through a given path length of a solution of the compound at a given concentration, quantitatively describes the optical activity of a stereoisomer. The specific rotation ([?]??)([α]λt) of an optically active compound is specific for a particular temperature (?)(t) and wavelength of light (?).(λ). Usually, as here, ?λ is the D line of sodium, 589 nm.

**Question:** Which Haworth perspective formula represents β-D-glucose?

**Description of Structures:**

1. **Structure 1:**
- Hexagonal ring structure with one oxygen atom included.
- Hydroxyl (OH) groups attached to the carbon atoms.
- The terminal carbon with an OH group is positioned above the plane of the ring.

2. **Structure 2:**
- Similar hexagonal ring structure with one oxygen atom.
- Hydroxyl groups attached to carbon atoms.
- Both the anomeric carbon's OH group and the hydroxyl group on the last carbon appear above the ring plane.

3. **Structure 3:**
- Hexagonal ring including one oxygen atom.
- Hydroxyl groups are attached to the carbon atoms, indicating an opposite orientation compared to Structure 1.
- The terminal carbon bonded OH group is positioned below the plane of the ring.

4. **Structure 4:**
- Hexagonal ring with one oxygen atom.
- Alternating arrangement of hydroxyl groups around the ring.
- Anomeric carbon’s hydroxyl group positioned above ring plane, terminal carbon’s OH group below the plane.

**Answer Choices:**
- 1
- 3
- 4
- 2

**Explanation for Educational Purposes:**

The Haworth projection of a carbohydrate depicts the molecule as a cyclic hemiacetal. In β-D-glucose, the hydroxyl group on the anomeric carbon (the first carbon) is oriented above the plane of the ring. Therefore, the β form of D-glucose has the OH group on the anomeric carbon in the same direction as the CH2OH group on the fifth carbon, which is above the plane in common Haworth representations of D-glucose.

The correct answer is **2**.

**Question:**

Why does the specific rotation of a freshly prepared solution of the α form gradually decrease with time? Why do solutions of the α and β forms reach the same specific rotation at equilibrium?

**Options:**

- ○ Glucose interconverts between cyclic and linear structures. The linear structure is most favorable. When the system reaches equilibrium, the net amounts of each structure no longer change.

- ○ Glucose interconverts between pyranose and furanose forms. The furanose form is most favorable. When the system reaches equilibrium, the net amounts of each form no longer change.

- ○ Glucose interconverts between α and β configurations. The β configuration is the most favorable. When the system reaches equilibrium, the net amounts of each configuration no longer change.

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**Task:**

Calculate the percentage of each of the two forms of D-glucose present at equilibrium.

- **α-D-glucose:** [Input Box] %

- **β-D-glucose:** [Input Box] %

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