The product of the reaction of `Ph_(2)CHCh_(2)OH` with `HBr` is

Options

A. `Ph_(2)C = CH_(2)`
B. `Phunderset(Ph)underset(|)(CH)CH_(2)Br`
C. `PhCHBrCH_(2)Ph`
D. `PhCH = CHPh`

Explanation

To determine the product of the reaction of diphenylmethanol (Ph₂CHCH₂OH) with HBr, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is diphenylmethanol, which can be represented as Ph₂CHCH₂OH. Here, "Ph" stands for the phenyl group (C₆H₅).

2. **Protonation of the Alcohol**:
When HBr is added, the hydroxyl group (-OH) of the alcohol will be protonated by the H⁺ from HBr. This step converts the -OH group into a better leaving group (water, H₂O).
\[ \text{Ph}_2\text{CHCH}_2\text{OH} + \text{H}^+ \rightarrow \text{Ph}_2\text{CHCH}_2\text{OH}_2^+ \]

3. **Formation of Carbocation**:
The protonated alcohol (Ph₂CHCH₂OH₂⁺) will lose a water molecule (H₂O), leading to the formation of a carbocation. The carbocation formed is initially a primary carbocation, which is not very stable.
\[ \text{Ph}_2\text{CHCH}_2^+ + \text{H}_2\text{O} \]

4. **Carbocation Rearrangement**:
To stabilize the carbocation, a phenyl group can shift from the adjacent carbon to the carbocation center, forming a more stable tertiary carbocation. This is known as a phenyl shift.
\[ \text{Ph}_2\text{C}^+ \text{CH}_2\text{Ph} \]

5. **Nucleophilic Attack by Bromide Ion**:
The bromide ion (Br⁻) from HBr will then attack the carbocation, resulting in the formation of the final product.
\[ \text{Ph}_2\text{CBrCH}_2\text{Ph} \]

6. **Final Product**:
The final product of the reaction is diphenylbromomethane (Ph₂CBrCH₂).

### Conclusion:
The product of the reaction of diphenylmethanol (Ph₂CHCH₂OH) with HBr is diphenylbromomethane (Ph₂CBrCH₂).