`CH_3CH=CHCH_3+CHCl_3+t-BuOK rarr A. ` A is

Options

A.
B.
C. `CH_3underset(OBu)underset(|)"CH"-underset(Cl)underset(|)CHCH_3`
D.

Explanation

To solve the given reaction step by step, let's break it down:

### Step 1: Identify the Reactants
The reactants provided are:
- 2-butene (CH₃CH=CHCH₃)
- Chloroform (CHCl₃)
- Tert-butoxide (t-BuOK, which provides the base t-BuO⁻)

### Step 2: Understand the Role of the Base
Tert-butoxide (t-BuO⁻) is a strong base. It will abstract a proton (H⁺) from the alkene (2-butene) during the reaction. This results in the formation of a carbanion.

### Step 3: Proton Abstraction
The base (t-BuO⁻) abstracts a proton from the double bond of 2-butene:
- The double bond between the second and third carbon (C2=C3) is broken.
- This leads to the formation of a carbanion at C2.

### Step 4: Formation of Dichlorocarbene
When the proton is abstracted, chloroform (CHCl₃) reacts with the carbanion. The t-BuO⁻ will facilitate the elimination of a chloride ion (Cl⁻) from chloroform, resulting in the formation of dichlorocarbene (CCl₂).

### Step 5: Electrophilic Attack
The dichlorocarbene (CCl₂) is an electrophile and will attack the double bond of the alkene. This results in the formation of a cyclic compound:
- The CCl₂ adds across the double bond of 2-butene, forming a cyclopropane derivative.

### Step 6: Final Product
The final product of the reaction is a compound where the two carbons of the former double bond are now bonded to the dichlorocarbene, resulting in:
- The structure CH₃CH(CHCl₂)CH₂.

### Conclusion
Thus, the product A is:
**A = CH₃CH(CHCl₂)CH₂**