Reaction of phenol with chloroform`//`sodium hydroxide to give `o-`hydroxy benzaldehyde involves the formation of

Options

A. Dichloro carbene
B. Trichloro carbene
C. Chlorine atoms
D. Chlorine molecules

Answer

To solve the question regarding the reaction of phenol with chloroform and sodium hydroxide to yield ortho-hydroxybenzaldehyde, we will break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are phenol (C6H5OH), chloroform (CHCl3), and sodium hydroxide (NaOH).

2. **Understand the Reaction Type**:
- This reaction is known as the Reimer-Tiemann reaction, which involves the introduction of a formyl group (CHO) into the ortho position of phenol.

3. **Formation of Dichlorocarbene**:
- When chloroform (CHCl3) reacts with sodium hydroxide (NaOH), the hydroxide ion (OH-) abstracts a proton from chloroform. This leads to the formation of a dichlorocarbene intermediate (CCl2).
- The reaction can be summarized as:
\[ CHCl_3 + NaOH \rightarrow CCl_2 + NaCl + H_2O \]
- Here, CCl2 is known as dichlorocarbene.

4. **Nucleophilic Attack by Phenol**:
- The phenol molecule (C6H5OH) can lose a proton to form the phenoxide ion (C6H5O-). This ion is more nucleophilic and can attack the dichlorocarbene.
- The reaction can be represented as:
\[ C_6H_5O^- + CCl_2 \rightarrow C_6H_5O-CCl_2 \]

5. **Formation of Ortho-Hydroxybenzaldehyde**:
- The intermediate formed will undergo further reactions, including the loss of a chlorine atom and rearrangement, leading to the formation of ortho-hydroxybenzaldehyde (C6H4(OH)(CHO)).
- The final product can be represented as:
\[ C_6H_4(OH)(CHO) \]

6. **Conclusion**:
- The key intermediate formed during this reaction is dichlorocarbene (CCl2). Therefore, the answer to the question is that the reaction involves the formation of dichlorocarbene.

### Final Answer:
The reaction of phenol with chloroform and sodium hydroxide to give ortho-hydroxybenzaldehyde involves the formation of **dichlorocarbene**.

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