Which one of the following will give racemised product
in `C_(2)H_(5)OH` ?

Options

A. `(1)` `CH_(3)-underset(CH_(3))underset(|)(CH)-Br`
B. `(2)` `Ph-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Cl`
C. `(3)` `Ph-underset(H)underset(|)overset(CH_(3))overset(|)C-Cl`
D.

Answer

To determine which of the following compounds will give a racemized product in ethanol (C₂H₅OH), we need to identify the conditions under which racemization occurs. Racemization typically occurs in reactions that follow the SN1 mechanism, where a carbocation intermediate is formed. Here’s a step-by-step solution:

### Step 1: Understand the SN1 Mechanism
The SN1 mechanism involves the formation of a carbocation intermediate. This mechanism is favored by:
- Good leaving groups
- Tertiary or highly substituted carbocations (due to steric hindrance)

### Step 2: Identify the Compounds
We need to analyze each compound provided in the options to see which one can undergo the SN1 mechanism.

### Step 3: Analyze Each Compound
1. **Compound 1**: Contains a bromine (Br) leaving group and is likely a secondary or tertiary carbon. Since Br is a good leaving group, this compound may undergo SN1.

2. **Compound 2**: Contains a chlorine (Cl) leaving group. Cl is a poorer leaving group compared to Br, which may hinder the SN1 reaction.

3. **Compound 3**: Also contains a chlorine (Cl) leaving group. Similar to compound 2, Cl is not a good leaving group, making SN1 less favorable.

4. **Compound 4**: Contains a bromine (Br) leaving group and is likely a tertiary carbon (due to steric hindrance). This compound has both a good leaving group (Br) and a favorable carbocation formation.

### Step 4: Conclusion
Among the options, **Compound 4** is the most likely to undergo the SN1 mechanism due to the presence of a good leaving group (Br) and the formation of a stable tertiary carbocation. Therefore, it will give a racemized product in ethanol.

### Final Answer
**Compound 4** will give a racemized product in C₂H₅OH.

---