An optically active compound, having molecular formula `C_6 H_12 O_6` is found in two isomeric forms. When isomers dissolved in water, they show the following equilibria
`underset(alpha=111^(n))("[A]") hArr underset(52.5^@)("equilibrium mixture") hArr underset(19.2)("[B]")`
Such isomers are called

Options

A. Anomers
B. Enantiomers
C. Positional isomers
D. Geometrical isomers

Answer

To solve the question, we need to identify the type of isomers represented by the compound with the molecular formula C₆H₁₂O₆, which is known to exist in two isomeric forms. The question provides information about the optical activity of these isomers and their equilibrium in water.

### Step-by-Step Solution:

1. **Identify the Compound**:
- The molecular formula C₆H₁₂O₆ corresponds to glucose, which is a common sugar.

2. **Understand Optical Activity**:
- The question mentions that the compound is optically active, which means it can rotate plane-polarized light. This property is typically associated with chiral molecules.

3. **Analyze the Given Equilibrium**:
- The equilibrium provided shows two forms of the compound: one with an angle of rotation of 111° (denoted as [A]) and another with an angle of rotation of 19.2° (denoted as [B]).
- This suggests that [A] and [B] are two different isomers of glucose.

4. **Determine the Isomeric Forms**:
- The two isomers in question are α-D-glucose and β-D-glucose. These forms differ in the configuration around the anomeric carbon (the carbon that is part of the carbonyl group in the open-chain form).

5. **Draw the Structures**:
- The structure of α-D-glucose has the hydroxyl group (OH) on the anomeric carbon (C1) pointing down, while in β-D-glucose, the hydroxyl group on the anomeric carbon points up.

6. **Identify the Type of Isomers**:
- Since α-D-glucose and β-D-glucose differ only at the anomeric carbon, they are classified as **anomers**. Anomers are a specific type of stereoisomer that differ in configuration at the anomeric carbon in cyclic sugars.

7. **Conclusion**:
- Based on the analysis, the isomers are called **anomers**. Therefore, the correct answer is option 1.