Phenol reacts with chloroform in the presence of aq. KOH at 340 K followed by hydrolysis of the resulting product giving salicyladehdye
The electrophile in this electrophilic substitution reaction is:

Options

A. `""^(-)C Cl_(3)`
B. `: C Cl_(2)`
C. `CHCl_(2)^(+)`
D. `Cl^(-)`

Answer

To determine the electrophile in the reaction between phenol and chloroform in the presence of aqueous KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Type**: The reaction described is a Reimer-Tiemann reaction, which involves the electrophilic substitution of phenol.

2. **Write the Reactants**: The reactants are phenol (C6H5OH) and chloroform (CHCl3) in the presence of aqueous KOH.

3. **Understand the Role of KOH**: Aqueous KOH acts as a base in this reaction. It will deprotonate phenol, forming phenoxide ion (C6H5O⁻), which is more nucleophilic than phenol itself.

4. **Formation of the Electrophile**: When chloroform reacts with KOH, it undergoes a reaction where KOH abstracts a hydrogen atom from chloroform. This leads to the formation of a trichloromethyl anion (CCl3⁻).

5. **Electrophile Generation**: The trichloromethyl anion (CCl3⁻) can then undergo further reaction. The negative charge on the carbon of CCl3 shifts, resulting in the formation of dichlorocarbene (CCl2), which is the actual electrophile in this reaction.

6. **Electrophilic Attack**: The dichlorocarbene (CCl2) then attacks the phenoxide ion, leading to the substitution reaction that ultimately gives salicylaldehyde as the product after hydrolysis.

7. **Conclusion**: The electrophile in this electrophilic substitution reaction is dichlorocarbene (CCl2).

### Final Answer:
The electrophile in this electrophilic substitution reaction is **CCl2** (dichlorocarbene).