In the following reaction,
`C_(6)H_(5)CH_(2)Br underset((ii)H_(3)O^(+))overset((i)" Mg, ether")toX`,
the product 'X' is

Options

A. `C_(6)H_(5)CH_(2)OCH_(2)C_(6)H_(5)`
B. `C_(6)H_(5)CH_(2)OH`
C. `C_(6)H_(5)CH_(3)`
D. `C_(6)H_(5)CH_(2)CH_(2)C_(6)H_(5)`

Explanation

To solve the given reaction, we need to analyze the steps involved in the transformation of C₆H₅CH₂Br in the presence of magnesium (Mg) in ether and then with hydronium ion (H₃O⁺).

### Step-by-Step Solution:

1. **Identify the Reactant:**
The reactant is C₆H₅CH₂Br, which is benzyl bromide. It consists of a benzene ring (C₆H₅) attached to a CH₂ group, which is further attached to a bromine atom (Br).

2. **Reaction with Magnesium (Mg):**
When benzyl bromide (C₆H₅CH₂Br) reacts with magnesium in an ether solvent, it forms a Grignard reagent. The bromine atom is replaced by magnesium, resulting in the formation of benzyl magnesium bromide (C₆H₅CH₂MgBr).
\[ \text{C₆H₅CH₂Br} + \text{Mg} \rightarrow \text{C₆H₅CH₂MgBr} \]

3. **Reaction with Hydronium Ion (H₃O⁺):**
The next step involves the addition of H₃O⁺ to the Grignard reagent (C₆H₅CH₂MgBr). The Grignard reagent is highly reactive and will react with the hydronium ion. The magnesium bromide part will dissociate, and the CH₂ group will be protonated by H₃O⁺, leading to the formation of toluene (C₆H₅CH₃).
\[ \text{C₆H₅CH₂MgBr} + \text{H₃O⁺} \rightarrow \text{C₆H₅CH₃} + \text{MgBrOH} \]

4. **Final Product:**
The final product 'X' is toluene (C₆H₅CH₃), and the by-product is magnesium bromide hydroxide (MgBrOH).

### Conclusion:
The product 'X' formed in the reaction is **C₆H₅CH₃ (toluene)**.

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