What would be the correct structure of `beta-D-`Glucose?

Options

A.
B.
C.
D.

Answer

To determine the correct structure of beta-D-glucose, we will follow these steps:

### Step 1: Understand the Fischer Projection of D-Glucose
- D-glucose is an aldohexose, meaning it has six carbons and an aldehyde group. The Fischer projection of D-glucose shows the arrangement of its functional groups.
- The structure can be represented as follows:
- C1: C=O (aldehyde)
- C2: -OH (on the right for D-glucose)
- C3: -OH (on the left)
- C4: -OH (on the right)
- C5: -OH (on the right)
- C6: -CH2OH

### Step 2: Cyclization of D-Glucose
- D-glucose can cyclize to form a cyclic structure. This occurs when the hydroxyl group (-OH) on the fifth carbon (C5) reacts with the aldehyde group (C1).
- This reaction forms a hemiacetal, resulting in a six-membered ring structure known as a pyranose.

### Step 3: Identify the Anomeric Carbon
- In the cyclic form, the carbon that was originally the carbonyl carbon (C1) becomes the anomeric carbon.
- The orientation of the -OH group on the anomeric carbon determines whether the glucose is in the alpha or beta form.

### Step 4: Determine the Structure of Beta-D-Glucose
- In beta-D-glucose, the -OH group on the anomeric carbon (C1) is positioned above the plane of the ring.
- If the -OH group on C1 is on the same side as the -CH2OH group on C5, it is beta-D-glucose.

### Step 5: Drawing the Structure
- Draw the six-membered ring with the following configuration:
- C1: -OH (above the plane)
- C2: -OH (below the plane)
- C3: -OH (above the plane)
- C4: -OH (below the plane)
- C5: -CH2OH (above the plane)

### Step 6: Verify the Structure
- Ensure that the structure has six carbons and that the -OH group on C1 is indeed above the plane, confirming it is beta-D-glucose.

### Final Structure
The correct structure of beta-D-glucose can be represented as follows:

```
O
/ \
H-C1 C2-H
| |
OH C3-H
|
H-C4 C5-CH2OH
```