Question

An organic compound A with the molecular formula C$_6$H$_4$NO$_2$Br is treated with magnesium in dry ether gives an organometallic compound which on treatment with CO$_2$ (dry ice) gives compound B. B on hydrolysis gives C. The name of the products A and C are

• A. 2-bromoaniline and terephthalic acid
• B. 2-nitrobromobenzene and terephthalic acid
• C. 3-nitrobromobenzene and 3-nitrobenzoic acid
• D. 3-nitrobromobenzene and 3-bromobenzoic acid
• E. 3-nitrobromobenzene and 2-nitrobenzoic acid

Hint:

Chemistry Tip: Grignard reagent + CO$_2$ + hydrolysis converts halide position into $-COOH$.

Answer 1

The Correct Option is C

Concept:
This is a standard Grignard reagent + CO$_2$ carboxylation sequence. General steps: \[ R-Br \xrightarrow[\text{dry ether}]{Mg} R-MgBr \] \[ R-MgBr + CO_2 \rightarrow RCOOMgBr \] \[ \text{Hydrolysis} \rightarrow RCOOH \] Thus bromine is replaced ultimately by: $$-COOH$$
Step 1: Interpret molecular formula.
Given: $$C_6H_4NO_2Br$$ This is bromonitrobenzene. So compound A must be a nitrobromobenzene isomer.
Step 2: Effect of Grignard formation.
When aryl bromide reacts with Mg: $$ArBr \rightarrow ArMgBr$$ Then with CO$_2$ and hydrolysis: $$ArMgBr \rightarrow ArCOOH$$ So bromine position becomes carboxylic acid position. Nitro group remains unchanged.
Step 3: Check positional isomer.
If A is: $$3\text{-nitrobromobenzene}$$ Then replacing Br by COOH gives: $$3\text{-nitrobenzoic acid}$$ This matches option (C).
Step 4: Reject others.

• Aniline formula does not match.
• Terephthalic acid needs two substituents COOH.
• Bromobenzoic acid means Br remains, impossible here.
• Wrong positional isomer in option (E).

Step 5: Final answer.
Thus: $$A = 3\text{-nitrobromobenzene}$$ $$C = 3\text{-nitrobenzoic acid}$$ Hence correct option is: $$\boxed{\text{(C)}}$$ :contentReference[oaicite:1]{index=1}