The sequence \(\text{R-X} \xrightarrow{\text{Mg}} \text{R-MgX} \xrightarrow{\text{CO}_2, \text{H}^+} \text{R-COOH} \xrightarrow{\text{Soda lime}} \text{R-H}\) is a classic method for converting an alkyl/aryl halide back to the parent hydrocarbon.
Step 1: Formation of Grignard Reagent (A)
The reactant is p-bromonitrobenzene. When treated with Magnesium (Mg) in the presence of dry ether, it forms a Grignard reagent. The Mg inserts into the carbon-bromine bond.
\[ \text{p-NO}_2\text{-C}_6\text{H}_4\text{-Br} + \text{Mg} \xrightarrow{\text{ether}} \text{p-NO}_2\text{-C}_6\text{H}_4\text{-MgBr} \ (A) \]
(Note: While nitro groups can interfere with Grignard reagents in practice, in the context of this idealized reaction sequence, we proceed with the functional group transformation of the halide). Step 2: Carbonation (Formation of Carboxylic Acid B)
The Grignard reagent reacts with Carbon Dioxide (\(\text{CO}_2\)) followed by acid hydrolysis (\(\text{H}_3\text{O}^+\)) to form a carboxylic acid.
\[ \text{R-MgBr} + \text{CO}_2 \rightarrow \text{R-COO-MgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{R-COOH} \]
Here, \(\text{R} = \text{p-nitrophenyl}\).
So, \(B\) is p-nitrobenzoic acid (\(\text{p-NO}_2\text{-C}_6\text{H}_4\text{-COOH}\)).
Step 3: Decarboxylation (Formation of C)
Compound B reacts with Na to form the sodium salt (Sodium p-nitrobenzoate). This salt is then heated with soda lime (\(\text{NaOH} + \text{CaO}\)). Soda lime causes decarboxylation, removing the carboxyl group (\(-\text{COO}^-\)) and replacing it with a hydrogen atom.
\[ \text{R-COOH} \xrightarrow{\text{Na}} \text{R-COONa} \xrightarrow{\text{NaOH + CaO}, \Delta} \text{R-H} + \text{Na}_2\text{CO}_3 \]
Substituting \(\text{R} = \text{p-NO}_2\text{-C}_6\text{H}_4\), the product is \(\text{p-NO}_2\text{-C}_6\text{H}_5\), which is Nitrobenzene.
Conclusion:
The sequence converts the Bromine group into a Hydrogen atom. The product C is Nitrobenzene.
Final Answer: Option (A).
