The smallest aldose which is able to form cyclic hemiacetal `is//are`:

Options

A. D-glyceraldehyde
B. D-Erythrose
C. D-Threose
D. D-Ribose

Answer

To determine the smallest aldose that can form a cyclic hemiacetal, we need to analyze the structures of the given aldoses and understand the requirements for hemiacetal formation.

### Step-by-Step Solution:

1. **Identify the Aldoses**: The aldoses given in the question are:
- D-glyceraldehyde
- D-erythrose
- D-threose
- D-ribose

2. **Draw the Structures**:
- **D-glyceraldehyde**: This is a 3-carbon aldose (C1: aldehyde, C2: hydroxyl, C3: hydroxyl).
- **D-erythrose**: This is a 4-carbon aldose (C1: aldehyde, C2: hydroxyl, C3: hydroxyl, C4: hydroxyl).
- **D-threose**: This is also a 4-carbon aldose (C1: aldehyde, C2: hydroxyl, C3: hydroxyl, C4: hydroxyl).
- **D-ribose**: This is a 5-carbon aldose (C1: aldehyde, C2: hydroxyl, C3: hydroxyl, C4: hydroxyl, C5: hydroxyl).

3. **Determine the Minimum Carbon Requirement**:
- To form a cyclic hemiacetal, a minimum of 4 carbons is required. This is because the aldehyde group on the first carbon must react with a hydroxyl group on one of the other carbons to form a ring structure.

4. **Analyze Each Aldose**:
- **D-glyceraldehyde**: With only 3 carbons, it cannot form a cyclic hemiacetal.
- **D-erythrose**: With 4 carbons, it can form a cyclic hemiacetal.
- **D-threose**: With 4 carbons, it can also form a cyclic hemiacetal.
- **D-ribose**: Although it has 5 carbons, it is not the smallest aldose that can form a cyclic hemiacetal.

5. **Conclusion**: The smallest aldoses that can form cyclic hemiacetals are D-erythrose and D-threose.

### Final Answer:
The smallest aldoses which can form cyclic hemiacetals are **D-erythrose and D-threose**.