Galactose is epimeric with glucose at `C_(4)`. Which of the following structures represents `beta-D` galactorpyronose?

Options

A.
B.
C.
D.

Explanation

To determine the structure of β-D-galactopyranose, we need to follow these steps:

### Step 1: Understand the Structure of D-Galactose
D-Galactose is an aldohexose sugar that is an epimer of D-glucose at carbon 4. The structure of D-galactose can be derived from D-glucose by changing the configuration at C4.

### Step 2: Draw the Fischer Projection of D-Galactose
1. Draw the Fischer projection of D-galactose:
- The carbon chain has six carbons.
- The carbonyl group (C=O) is at C1.
- The hydroxyl groups (OH) and hydrogen atoms (H) are arranged as follows:
- C1: CHO
- C2: OH on the right, H on the left
- C3: OH on the left, H on the right
- C4: H on the right, OH on the left (this is the epimeric carbon compared to glucose)
- C5: CH2OH

### Step 3: Convert Fischer Projection to Haworth Structure
1. To convert the Fischer projection to a cyclic form (Haworth structure), we need to form a ring:
- The hydroxyl group on C4 will react with the carbonyl carbon (C1) to form a hemiacetal.
- This will create a six-membered ring (pyranose form).

### Step 4: Determine the Anomeric Configuration
1. In the cyclic form, the anomeric carbon is C1:
- For β-D-galactopyranose, the hydroxyl group (OH) at C1 must be on the same side as the CH2OH group at C5 when the ring is drawn.
- This means that the OH group at C1 will be above the plane of the ring.

### Step 5: Draw the β-D-Galactopyranose Structure
1. Draw the six-membered ring:
- C1 (anomeric carbon) has the OH group above the ring.
- C2 has OH on the right.
- C3 has OH on the left.
- C4 has H on the right and OH on the left.
- C5 has the CH2OH group.

### Final Structure
The final structure of β-D-galactopyranose will have the following characteristics:
- The OH group on C1 is above the ring.
- The configurations of the other carbons follow the D-galactose structure.