Synthesis of cyclohexane -1,2 diol from cyclohexene may be accomplished in two ways :
I `MnO_(4)` dilute `OH, O^(@)C` dihydroxylation
II . Peracid epoxidation followed by NaOH opening of the epoxide ring .
Which of the following statement about the products from these reactions is correct ?

Options

A. Methods I and II give same product
B. Method I gives resolvable racemic mixture while method II will give non- resolvable achiral product
C. Method I gives resolvable optically inactive compound while method II gives resolvable racemic mixture
D. Products obtained through method I and II will have diasteriomeric relationship

Explanation

To synthesize cyclohexane-1,2-diol from cyclohexene, we can use two methods:

### Method I: Dihydroxylation using KMnO4
1. **Reagents**: Use cold and dilute potassium permanganate (KMnO4) in a basic medium.
2. **Reaction Mechanism**:
- Cyclohexene reacts with KMnO4, leading to the formation of a cyclic intermediate.
- This intermediate undergoes syn-dihydroxylation, where both hydroxyl (OH) groups are added to the same side of the double bond.
3. **Product Formation**: The product is a syn addition of hydroxyl groups, resulting in a compound that has a plane of symmetry, making it optically inactive.

### Method II: Epoxidation followed by Hydrolysis
1. **Reagents**: Use a peracid (like m-chloroperbenzoic acid) to form an epoxide from cyclohexene.
2. **Reaction Mechanism**:
- Cyclohexene reacts with the peracid to form an epoxide.
- The epoxide ring is then opened by sodium hydroxide (NaOH) in a nucleophilic attack, leading to the formation of two hydroxyl groups.
- This process results in anti-dihydroxylation, where the hydroxyl groups are added to opposite sides of the double bond.
3. **Product Formation**: The product is an anti addition of hydroxyl groups, resulting in a compound that does not have a plane of symmetry, making it chiral.

### Comparison of Products
- **Method I** yields a syn-diol which is optically inactive due to the presence of a plane of symmetry.
- **Method II** yields an anti-diol which is chiral and can exist as a racemic mixture.

### Conclusion
The correct statement regarding the products from these reactions is:
- Method I gives a resolvable optically inactive compound, while Method II gives a resolvable racemic mixture.

### Final Answer
The correct statement is: "Method 1 gives a resolvable optically inactive compound while method 2 gives a resolvable racemic mixture."

---