In the reaction
`CF_(3)-CHCl_(2) underset(Delta)overset(LDA)rarr CF_(2) = CCl_(2)`
Intermediate formed is :

Options

A. carbocation
B. carbanion
C. free radical
D. No intermediate formed

Explanation

To solve the question regarding the intermediate formed in the reaction of `CF3-CHCl2` with LDA, we will follow these steps:

### Step 1: Identify the Reactants and Conditions
The reactant is `CF3-CHCl2`, and it is treated with LDA (Lithium Diethyl Amine), which is a strong base.

**Hint:** Recognize the role of LDA as a strong base that can facilitate elimination reactions.

### Step 2: Analyze the Type of Reaction
The formation of `CF2=CCl2` indicates that an elimination reaction is taking place. In elimination reactions, we typically remove a leaving group and a hydrogen atom to form a double bond.

**Hint:** Look for the products to determine the type of reaction occurring (elimination in this case).

### Step 3: Determine the Elimination Mechanism
There are three types of elimination mechanisms: E1, E2, and E1CB. Since LDA is a strong base, it is unlikely to follow the E1 mechanism, which typically involves carbocation formation.

**Hint:** Recall that E2 mechanisms do not form intermediates, while E1CB mechanisms do.

### Step 4: Eliminate E2 Mechanism
In E2 elimination, no intermediate is formed; only a transition state exists. Since we are looking for an intermediate, we can rule out the E2 mechanism.

**Hint:** Remember that E2 reactions proceed in a single step without intermediates.

### Step 5: Confirm E1CB Mechanism
Given that E2 is ruled out, the remaining possibility is the E1CB mechanism, which involves the formation of a carbanion intermediate. The presence of an electron-withdrawing group (CF3) supports the formation of a carbanion.

**Hint:** Identify the role of electron-withdrawing groups in facilitating the E1CB mechanism.

### Step 6: Formation of the Carbanion
In the E1CB mechanism, the strong base (LDA) abstracts a proton (H) from the carbon adjacent to the leaving group (Cl), resulting in the formation of a carbanion (`CF3-CCl2-`).

**Hint:** Visualize the process of proton abstraction leading to carbanion formation.

### Step 7: Finalize the Product
After the carbanion is formed, the negative charge can shift, leading to the formation of the double bond and the final product `CF2=CCl2`.

**Hint:** Understand how charge rearrangement leads to the final product in elimination reactions.

### Conclusion
The intermediate formed in the reaction is a **carbanion**.

**Final Answer:** The intermediate formed is a carbanion.