When phenyl magnesium bormide reacts with `t-butanol` the product would be :

Options

A. Benzene
B. Phenol
C. t-Butyl benzene
D. t-Butyl pheyl ether

Explanation

To determine the product formed when phenyl magnesium bromide reacts with t-butanol, we can follow these steps:

### Step 1: Identify the Reactants
- The reactants in this case are phenyl magnesium bromide (C6H5MgBr), which is a Grignard reagent, and t-butanol (C4H10O).

### Step 2: Understand the Nature of Grignard Reagents
- Grignard reagents are highly reactive organometallic compounds. The phenyl group (C6H5-) in phenyl magnesium bromide carries a negative charge, while the magnesium bromide (MgBr) part carries a positive charge.

### Step 3: Analyze the Reaction with t-Butanol
- t-Butanol is a tertiary alcohol, which means it has a hydroxyl (-OH) group attached to a carbon that is bonded to three other carbon atoms. The hydroxyl group in t-butanol can release a proton (H+) due to its acidic nature.

### Step 4: Reaction Mechanism
- When phenyl magnesium bromide reacts with t-butanol, the negatively charged phenyl group (C6H5-) will react with the proton (H+) from the hydroxyl group of t-butanol. This results in the formation of benzene (C6H6) and a t-butoxide ion (C4H9O-).

### Step 5: Write the Balanced Equation
- The balanced reaction can be represented as:
\[ C6H5MgBr + (CH3)3COH \rightarrow C6H6 + (CH3)3CO^- \]
- Here, benzene (C6H6) is formed as the main product, and the remaining product is the t-butoxide ion.

### Conclusion
- The product formed when phenyl magnesium bromide reacts with t-butanol is **benzene (C6H6)**.

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