Consider the folliwng reactions:
I. `Me_(3)C - Br underset(SN^(1))overset(H_(2)O + NaBr)(rarr) "Product"`
II. `Ph_(3)C - Br underset(SN^(1))overset(H_(2)O + NaBr)(rarr) "Products"`
Which of the folliwng statements are correct about the above reactions?

Options

A. The products in reactions `(I)` and `(II)` are mixture of `(Me_(3) - OH + Me_(3)Br)` and `(Ph_(3)C-OH + Ph_(3)C - Br)`, respectively.
B. The product is `(I)` is `(Me_(3)C - OH)` and in `(II)` is `(Ph_(3)C - OH + Ph_(3)C - Br)`.
C. The product in `(I)` is `(Me_(3)C - OH + Me_(3)C - Br0` and in `(II)` is `(Ph_(3)C - OH)`.
D. `Ph_(3) C^(o+)` is more stable than `MeC^(o+)`.

Explanation

To analyze the given reactions and determine the correct statements, we will break down the reactions step by step.

### Step 1: Analyze Reaction I
The first reaction involves a tertiary alkyl bromide, `Me3C-Br`, undergoing an SN1 reaction with water (`H2O`) and sodium bromide (`NaBr`).

1. **Formation of Carbocation**: The tertiary alkyl bromide will first undergo ionization to form a tertiary carbocation (`Me3C+`). This step is the rate-determining step in an SN1 mechanism.

**Hint**: Tertiary carbocations are more stable due to hyperconjugation and inductive effects.

2. **Nucleophilic Attack**: The water molecule acts as a nucleophile and attacks the carbocation, leading to the formation of a tertiary alcohol (`Me3C-OH`).

**Hint**: In SN1 reactions, the nucleophile attacks the carbocation after its formation.

### Step 2: Analyze Reaction II
The second reaction involves a triphenylmethyl bromide, `Ph3C-Br`, also undergoing an SN1 reaction with water and sodium bromide.

1. **Formation of Carbocation**: The triphenylmethyl bromide will ionize to form a benzylic carbocation (`Ph3C+`). This carbocation is highly stabilized by resonance due to the three phenyl groups attached to it.

**Hint**: Resonance stabilization significantly increases the stability of carbocations.

2. **Nucleophilic Attack**: Similar to the first reaction, water will attack the benzylic carbocation, forming triphenylmethanol (`Ph3C-OH`). However, due to the partial double bond character of the carbon-bromine bond in aryl halides, the reaction is less favorable, and the product may be formed in minor amounts.

**Hint**: Aryl halides are less reactive in SN1 reactions due to the strength of the C-Br bond.

### Step 3: Evaluate the Statements
Now we will evaluate the statements based on our analysis of the reactions:

1. **Statement A**: Product one is a mixture of products.
- **Evaluation**: This statement is incorrect. Product one is a single tertiary alcohol (`Me3C-OH`).

2. **Statement B**: Product one is a tertiary alcohol while product two is a mixture.
- **Evaluation**: This statement is correct. Product one is indeed a tertiary alcohol, and product two (triphenylmethanol) is formed in minor amounts.

3. **Statement C**: Benzyl carbocation is more stable than tertiary carbocation.
- **Evaluation**: This statement is correct. The benzylic carbocation is resonance-stabilized by three phenyl groups, making it more stable than the tertiary carbocation.

4. **Statement D**: The reaction of triphenylmethyl bromide does not proceed via SN1 due to the partial double bond character.
- **Evaluation**: This statement is partially correct. While the reaction does proceed via SN1, it is less favorable due to the strong C-Br bond in aryl halides.

### Final Conclusion
Based on the evaluations:
- **Correct Statements**: B and C.
- **Incorrect Statements**: A and D.

### Summary of the Solution
1. **Reaction I**: Forms a tertiary alcohol (`Me3C-OH`).
2. **Reaction II**: Forms triphenylmethanol (`Ph3C-OH`) but is less favorable due to the strength of the C-Br bond.
3. **Correct Statements**: B and C.