Salicylic acid is produced when phenol in alcoholic KOH is treated with

Options

A. `CH_(3)Cl`
B. `CHCl_(3)`
C. `CH_(2)Cl_(2)`
D. `"CCl"_(4)`

Answer

To produce salicylic acid from phenol using alcoholic KOH, we need to follow the Riemann-Tiemann carboxylation reaction. Here’s a step-by-step breakdown of the process:

### Step 1: Understanding the Reactants
- **Phenol**: This is our starting compound, which contains a hydroxyl group (-OH) attached to a benzene ring.
- **Alcoholic KOH**: This serves as a base and helps in deprotonating phenol to form phenoxide ion.

### Step 2: Deprotonation of Phenol
- When phenol is treated with alcoholic KOH, the hydroxyl group (-OH) of phenol is deprotonated.
- This results in the formation of the phenoxide ion (C6H5O⁻) and potassium ion (K⁺):
\[ \text{C}_6\text{H}_5\text{OH} + \text{KOH} \rightarrow \text{C}_6\text{H}_5\text{O}^- \text{K}^+ + \text{H}_2\text{O} \]

### Step 3: Reaction with CCl4
- The phenoxide ion is then reacted with carbon tetrachloride (CCl4).
- The oxygen of the phenoxide ion attacks the carbon atom of CCl4, leading to the formation of an intermediate compound where chlorine atoms are replaced.

### Step 4: Formation of the Intermediate
- The reaction results in the formation of a chlorinated intermediate:
\[ \text{C}_6\text{H}_5\text{O}^- + \text{CCl}_4 \rightarrow \text{Intermediate} \]
- In this intermediate, the chlorine atoms can be replaced by hydroxyl groups.

### Step 5: Hydrolysis and Formation of Salicylic Acid
- When the intermediate undergoes hydrolysis, the chlorine atoms are replaced by hydroxyl groups, leading to the formation of salicylic acid:
\[ \text{Intermediate} \rightarrow \text{Salicylic Acid (C}_7\text{H}_6\text{O}_3\text{)} \]

### Final Reaction
- The final product, salicylic acid, has both a hydroxyl group and a carboxylic acid group attached to the benzene ring.

### Conclusion
- The correct reagent to produce salicylic acid from phenol in alcoholic KOH is **CCl4**.

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