When phenol is treated with `CHCl_(3)` and `NaOH`, the product fromed is

Options

A. Benzaldehyde
B. Salicylaldehyde
C. Salicylic acid
D. Benzoic acid

Answer

To solve the question of what product is formed when phenol is treated with CHCl3 (chloroform) and NaOH (sodium hydroxide), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Phenol (C6H5OH)
- Chloroform (CHCl3)
- Sodium Hydroxide (NaOH)

### Step 2: Understand the Reaction Conditions
The reaction occurs under basic conditions, where sodium hydroxide acts as a base. The temperature is typically around 343 K (approximately 70°C).

### Step 3: Recognize the Type of Reaction
This reaction is known as the Reimer-Thiemann reaction. It involves the ortho-formylation of phenol, where an aldehyde group is introduced to the aromatic ring.

### Step 4: Determine the Product Structure
In the Reimer-Thiemann reaction, the product formed is 2-hydroxybenzaldehyde, which is also known as salicylaldehyde. The hydroxyl group (-OH) from phenol remains on the aromatic ring, and a formyl group (-CHO) is introduced at the ortho position relative to the hydroxyl group.

### Step 5: Write the Chemical Equation
The overall reaction can be summarized as:
\[ \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{NaOH} \rightarrow \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + \text{by-products} \]
Where the product is salicylaldehyde (C7H6O2).

### Conclusion
The product formed when phenol is treated with CHCl3 and NaOH is **salicylaldehyde (2-hydroxybenzaldehyde)**.

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