Statement I: Glycosides mutarotate.
Statement II: The anomeric `OH` is etherified and the equilibrium with the free carbonyl form is destroyed.

Options

A. Statement `I` is true, Statement `II` is true, Statements `II` is the correct explanation of Statement `I`.
B. Statement `I` is true, Statement `II` is true, Statement `II` is not the correct explanation of Statement `I`.
C. Statement `I` is true, Statement `II` is false.
D. Statement `I` is false, Statement `II` is true.

Explanation

To solve the question regarding the statements about glycosides and mutarotation, we can break it down step by step.

### Step-by-Step Solution:

1. **Understanding Mutarotation**:
- Mutarotation is the change in the optical rotation that occurs when an anomeric carbon (the carbon that was part of the carbonyl group in the open-chain form) is converted between its α and β forms. This process occurs in sugars that can form hemiacetals or hemiketals.

**Hint**: Recall that mutarotation is a phenomenon observed in sugars that can interconvert between cyclic forms.

2. **Analyzing Statement I**:
- Statement I claims that "glycosides mutarotate." Glycosides are formed when the anomeric hydroxyl group of a sugar reacts with an alcohol, resulting in the formation of a glycosidic bond. In this process, the anomeric carbon's hydroxyl group is replaced by an ether group, which means it cannot participate in mutarotation.

**Hint**: Consider the structure of glycosides and how the anomeric carbon is involved in the formation of glycosidic bonds.

3. **Conclusion for Statement I**:
- Since glycosides do not have a free anomeric hydroxyl group (it is etherified), they cannot undergo mutarotation. Therefore, Statement I is **false**.

4. **Analyzing Statement II**:
- Statement II states that "the anomeric OH is etherified and the equilibrium with the free carbonyl form is destroyed." This is true because when the anomeric hydroxyl group is replaced by an ether group in a glycoside, the equilibrium between the open-chain form and the cyclic form is disrupted. The free carbonyl form cannot exist since the anomeric carbon is now part of a glycosidic bond.

**Hint**: Think about the implications of etherification on the structure and reactivity of the sugar.

5. **Conclusion for Statement II**:
- Since the anomeric hydroxyl group is indeed etherified in glycosides, and this prevents the equilibrium with the free carbonyl form, Statement II is **true**.

### Final Conclusion:
- **Statement I**: False
- **Statement II**: True

Thus, the correct assessment of the statements is that Statement I is false and Statement II is true.