Give the product of the reaction of `Ph_(2)CHCH_(2)OH` with `HBr` and explain its formation.

Answer

To determine the product of the reaction of diphenylcarbinol (Ph₂C(OH)CH₂) with HBr, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is diphenylcarbinol, which can be represented as Ph₂C(OH)CH₂. Here, Ph represents the phenyl group (C₆H₅).

2. **Protonation of the Alcohol**:
When HBr is added to the reaction, the hydroxyl group (–OH) of the alcohol gets protonated by the H⁺ from HBr. This forms a better leaving group (water, H₂O) and generates a carbocation intermediate.
\[ \text{Ph}_2C(OH)CH_2 + H^+ \rightarrow \text{Ph}_2C(OH_2)^+CH_2 \]

3. **Formation of Carbocation**:
The protonated alcohol (Ph₂C(OH₂)⁺CH₂) can lose a water molecule (H₂O) to form a carbocation. The loss of water leads to the formation of a highly stabilized carbocation due to resonance from the adjacent phenyl groups.
\[ \text{Ph}_2C(OH_2)^+CH_2 \rightarrow \text{Ph}_2C^+CH_2 + H_2O \]

4. **Phenyl Shift**:
The carbocation formed is stabilized by the adjacent phenyl groups. A phenyl group can shift to the carbocation center, resulting in a more stable carbocation. The shift occurs from one of the phenyl groups to the carbon bearing the positive charge.
\[ \text{Ph}_2C^+CH_2 \rightarrow \text{Ph}^+CH(Ph)CH_2 \]

5. **Nucleophilic Attack by Bromide Ion**:
The bromide ion (Br⁻) from HBr can now attack the carbocation, leading to the formation of the final product.
\[ \text{Ph}^+CH(Ph)CH_2 + Br^- \rightarrow \text{Ph}CH(Br)CH_2Ph \]

6. **Final Product**:
The final product of this reaction is 1-bromo-1,2-diphenylethane (PhCH(Br)CH₂Ph).

### Final Product:
The product of the reaction of diphenylcarbinol with HBr is:
\[ \text{Ph}CH(Br)CH_2Ph \]