`CH_(3)-overset(O)overset(||)(C)-O-H overset(NaH overset(14)(CO_(3)))rarr underset("(gas)")((A)) underset((ii) H_(3)O^(o+))overset((i) PhMgBr)rarr (B)`
`CH_(3)-underset(O)underset(||)overset(O)overset(||)(S)-O-H overset(NaHoverset(14)(CO_(3)))rarr underset("gas")((C )) underset((ii) H_(3)O^(o+))overset((i) PhMgBr)rarr (D)`
Product (B) and (D) in the above reaction are :

Options

A. `Ph-overset(O)overset(||)(C)-O-H, Ph-underset(O)underset(||)overset(O)overset(||)(S)-O-H`
B. `Ph-underset(14)overset(O)overset(||)(C)-O-H, Ph-overset(O)overset(||)(S)-O-H`
C. `Ph-underset(14)overset(O)overset(||)(C)-O-H, Ph-underset(14)overset(O)overset(||)(C)-O-H`
D. `Ph-overset(O)overset(||)(C)-OH, Ph-overset(O)overset(||)(S)-O-H`

Explanation

To solve the question, we need to analyze the reactions step by step. The reactions involve acetic acid and a sulfonic acid derivative, both of which react with sodium hydrogen carbonate (NaHCO₃) to produce carbon dioxide gas, which is then treated with phenyl magnesium bromide (PhMgBr) followed by hydrolysis.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The first reactant is acetic acid (CH₃C(=O)OH).
- The second reactant is a sulfonic acid derivative (CH₃S(=O)(=O)OH).

2. **Reaction with Sodium Hydrogen Carbonate (NaHCO₃):**
- Acetic acid reacts with NaHCO₃. The acidic proton from acetic acid is captured by NaHCO₃, resulting in the formation of sodium acetate (CH₃C(=O)O⁻Na⁺) and carbonic acid (H₂CO₃).
- H₂CO₃ decomposes into water (H₂O) and carbon dioxide (CO₂). Since the carbon in CO₂ is of the C-14 isotope, we denote it as C-14O₂.
- This produces a gas (A) which is C-14O₂.

3. **Formation of Product B:**
- The gas C-14O₂ then reacts with phenyl magnesium bromide (PhMgBr).
- In this reaction, the carbon atom in C-14O₂ (which has a partial positive charge) is attacked by the nucleophilic carbon from PhMgBr. This results in the formation of a magnesium alkoxide intermediate (PhC(=O)O⁻MgBr).
- Upon hydrolysis with H₃O⁺, this intermediate converts to a carboxylic acid, specifically PhC(=O)OH, which contains the C-14 isotope. This is product (B).

4. **Repeat for the Sulfonic Acid Derivative:**
- The sulfonic acid derivative (CH₃S(=O)(=O)OH) also reacts with NaHCO₃ in a similar manner.
- The acidic proton is captured, leading to the formation of a sulfonate salt (CH₃S(=O)(=O)O⁻Na⁺) and carbonic acid (H₂CO₃), which decomposes to produce C-14O₂ gas (C gas, denoted as C).

5. **Formation of Product D:**
- The C-14O₂ gas reacts with PhMgBr, similar to the previous reaction, forming the same magnesium alkoxide intermediate.
- Hydrolysis with H₃O⁺ yields the same product, PhC(=O)OH, which also contains the C-14 isotope. This is product (D).

6. **Conclusion:**
- Both products B and D are the same: PhC(=O)OH with the C-14 isotope.

### Final Products:
- Product (B): PhC(=O)C₁₄OH
- Product (D): PhC(=O)C₁₄OH