A solution of (-l)- chloro -1 phenyletane in toluene recemises slowly in the presence of a small amunt of `SbCI_(5)` due to the formation of .

Options

A. carbanion
B. carbene
C. carbocation
D. free radical

Explanation

To solve the question regarding the reaction of (-l)-chloro-1-phenyletane in the presence of SbCl5, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is (-l)-chloro-1-phenyletane (C6H5CHClCH3) and the reagent is antimony pentachloride (SbCl5) in toluene.

**Hint**: Recognize the structure of the reactant and the role of SbCl5 as a Lewis acid.

2. **Mechanism of Reaction**:
When (-l)-chloro-1-phenyletane reacts with SbCl5, the SbCl5 acts as a Lewis acid and facilitates the departure of the chloride ion (Cl-). This results in the formation of a carbocation intermediate.

**Hint**: Remember that Lewis acids can stabilize the formation of carbocations by accepting electron pairs.

3. **Formation of Carbocation**:
The chloride ion (Cl-) leaves, forming a carbocation (C6H5C^+HCH3). This carbocation is planar, which means that it can be attacked from either side.

**Hint**: Visualize the planar structure of the carbocation to understand how it can be attacked from multiple directions.

4. **Nucleophilic Attack**:
The chloride ion (Cl-) can now attack the carbocation from either the front or the back. This leads to the formation of two different stereoisomers (enantiomers).

**Hint**: Consider the stereochemistry of the attack; the two possible attack directions will yield different configurations.

5. **Formation of Racemic Mixture**:
Since the carbocation can be attacked from both sides, the final product will be a racemic mixture of the two enantiomers (D and L forms of the product).

**Hint**: Recall that a racemic mixture contains equal amounts of both enantiomers, leading to optical inactivity.

6. **Conclusion**:
The presence of the carbocation and its planar nature is what leads to the formation of a racemic mixture when the chloride ion attacks from either side.

**Hint**: Summarize the key points: carbocation formation, planar structure, and racemic mixture outcome.

### Final Answer:
The reaction of (-l)-chloro-1-phenyletane with SbCl5 in toluene leads to the formation of a racemic mixture due to the generation of a planar carbocation that allows for nucleophilic attack from both sides.