When chloroform reacts with NaOH an important reactive intermediate is formed . Type of reaction involved and formed intermediates are respectively .

Options

A. `E_(2) " and " CI -overset(Theta)underset(CI) underset(|)(C ) - CI`
B. `beta-" Elimination and "CI - underset(CI)underset(|)(C ":")`
C. `E_(2) " and " CI- underset(CI)underset(|)(C":")`
D. `alpha-" Elimination and " CI - underset(CI)underset(|)(C ":")`

Answer

To solve the question regarding the reaction of chloroform with NaOH and to identify the type of reaction and the reactive intermediate formed, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are chloroform (CHCl3) and sodium hydroxide (NaOH).

### Step 2: Understand the Role of NaOH
Sodium hydroxide (NaOH) is a strong base. In this reaction, the hydroxide ion (OH-) from NaOH will act as a nucleophile and will attack the hydrogen atom of chloroform.

### Step 3: Mechanism of Reaction
When chloroform reacts with NaOH, the hydroxide ion (OH-) will abstract a hydrogen atom from chloroform, leading to the departure of a chlorine atom (Cl-) as a leaving group. This process involves the elimination of a hydrogen and a chlorine atom from the alpha carbon (the carbon attached to the leaving group).

### Step 4: Identify the Intermediate Formed
The elimination of Cl- results in the formation of a reactive intermediate known as dichlorocarbene (CCl2). This intermediate has two chlorine atoms attached to the carbon atom that was originally part of chloroform.

### Step 5: Determine the Type of Reaction
The reaction that occurs is classified as an alpha elimination reaction because the chlorine atom is eliminated from the alpha carbon (the carbon that is directly bonded to the leaving group).

### Conclusion
Thus, the type of reaction involved is **alpha elimination**, and the reactive intermediate formed is **dichlorocarbene**.

### Final Answer
The correct answer is: **Alpha elimination and dichlorocarbene**.

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