Step 1: Read the sequence:
Phenol is treated with CCl4 and aqueous NaOH to give A. A is heated with NaOH to give B, and B is acidified with H3O+ to give C. This is a Reimer-Tiemann type reaction using CCl4.
Step 2: Formation of A:
In NaOH, phenol becomes the phenoxide ion, which is rich in electrons at the ortho position. It attacks the carbon of CCl4 (through dichlorocarbene-like species) and puts a \(-\text{CCl}_3\) group at the ortho position. A is o-trichloromethyl phenoxide.
Step 3: Formation of B:
On treatment with NaOH, the \(-\text{CCl}_3\) group is hydrolysed. Each C-Cl bond is replaced and the group becomes \(-\text{COO}^-\text{Na}^+\). B is sodium salicylate.
Step 4: Formation of C:
Acidification with H3O+ turns the carboxylate into \(-\text{COOH}\) and the phenoxide into \(-\text{OH}\). C is salicylic acid (2-hydroxybenzoic acid), with OH and COOH on adjacent ring carbons. This is option (B).
Step 5: Why the other options are wrong:
Option (A) has CHO instead of COOH. That is the product of CHCl3/NaOH (the normal Reimer-Tiemann reaction), not CCl4. Option (C) has two OH groups and option (D) has a methyl group, and neither can form from a CCl3 group.
Final Answer:
The final product C is salicylic acid, option (B).
\[ \boxed{\text{Salicylic acid (option B)}} \]
