What are the products of the following reactions `?`
`C_(6)H_(5)OCH_(2)CH_(2)OH underset("Heat")overset("Excess HBr")(rarr)`

Options

A. `C_(6)H_(5)OH+BrCH_(2)CH_(2)OH`
B. `C_(6)H_(5)OH+HOCH_(2)CH_(2)OH`
C. `C_(6)H_(5)OH+BrCH_(2)CH_(2)Br`
D. `C_(6)H_(5)Br+HOCH_(2)CH_(2)H`

Answer

To determine the products of the reaction of C₆H₅OCH₂CH₂OH with excess HBr upon heating, we can follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- C₆H₅OCH₂CH₂OH (which is an ether with a hydroxyl group)
- Excess HBr (hydrobromic acid)

### Step 2: Understand the Mechanism
In the presence of HBr, the ether and alcohol can undergo protonation. The oxygen atom in the ether and the hydroxyl group can both act as nucleophiles. The HBr will protonate the ether oxygen, leading to the formation of an oxonium ion.

### Step 3: Formation of Oxonium Ion
When HBr is added, the ether oxygen (C₆H₅O) gets protonated:
\[ C₆H₅OCH₂CH₂OH + HBr \rightarrow C₆H₅OCH₂CH₂OH₂⁺ + Br⁻ \]
This results in the formation of an oxonium ion (C₆H₅OCH₂CH₂OH₂⁺).

### Step 4: Carbocation Formation
The oxonium ion can lose a water molecule (H₂O), generating a carbocation:
\[ C₆H₅OCH₂CH₂OH₂⁺ \rightarrow C₆H₅OCH₂CH₂⁺ + H₂O \]

### Step 5: Nucleophilic Attack by Bromide Ion
The bromide ion (Br⁻) can now act as a nucleophile and attack the carbocation:
\[ C₆H₅OCH₂CH₂⁺ + Br⁻ \rightarrow C₆H₅OCH₂CH₂Br \]
This results in the formation of a brominated ether.

### Step 6: Further Reaction with Excess HBr
Since we have excess HBr, the hydroxyl group (–OH) can also be protonated and undergo a similar reaction:
\[ C₆H₅OCH₂CH₂OH + HBr \rightarrow C₆H₅OCH₂CH₂Br + H₂O \]
This leads to the formation of phenol (C₆H₅OH) and a vicinal dibromide.

### Final Products
The final products of the reaction are:
1. Phenol (C₆H₅OH)
2. Bromoethanol (C₆H₅OCH₂CH₂Br)

Thus, the products of the reaction are:
- **Phenol (C₆H₅OH)**
- **Bromoethanol (C₆H₅OCH₂CH₂Br)**