Question

What is the product formed in the following reaction \[({C_6}{H_5}OH + CC{l_4} \xrightarrow[(ii)H^{+}]{(i)NaOH}) \]
A. p-hydroxybenzoic acid
B. o-hydroxybenzoic acid
C. Benzaldehyde
D. Salicylaldehyde

Answer 1

Hint: The reaction between phenol and carbon tetrachloride in presence of sodium hydroxide is a type of nucleophilic substitution reaction. Here, carbon tetrachloride is used as a solvent.

Complete Step by Step Solution:
The reaction between phenol and carbon tetrachloride in presence of sodium hydroxide takes place in four steps.
Step 1: In the first step the lone pair of hydroxide ions of sodium hydroxide abstract the proton of phenol to form phenoxide ion and water.

Image: Formation of sodium phenolate

Step 2: In the second step, the phenoxide ion attack the carbon of carbon tetrachloride is form a complex.

Image: Reaction of phenoxide ion and carbon tetrachloride

Step 3: In the third step the complex reacts with sodium hydroxide. Here, a nucleophilic substitution reaction takes place. First, the halide group (leaving group) is detached from the carbon and then the hydroxide ion (nucleophile) attacks the carbon to form an unstable complex.

Image: Formation of unstable complex

Step 4: In the fourth step, the water molecule is removed to form a stable compound with a carboxylic group forming a benzoic acid and hydroxy group at the ortho position. The compound name of o-hydroxy benzoic acid.

Image: Formation of o-hydroxy benzoic acid

Therefore, the correct option is B.

Note: In the given reaction, NaOH should be taken in excess amount because phenoxide ion is formed first. The nucleophilic substitution involved is SN₁ which takes place in two-step and the reaction rate depends on one reactant.