Question

The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is :

• A. $HNO_3/H_2SO_4$, $Br_2/AlBr_3$, $Mg/ether$, $CO_2$, $H_3O^+$
• B. $Br_2/AlBr_3$, $NaCN$, $H_3O^+$, $HNO_3/H_2SO_4$
• C. $Br_2/AlBr_3$, $HNO_3/H_2SO_4$, $NaCN$, $H_3O^+$
• D. $Br_2/AlBr_3$, $HNO_3/H_2SO_4$, $Mg/ether$, $CO_2$, $H_3O^+$

Hint:

Always check the directing effect. If you need meta products, perform the meta-directing substitution (like nitration) first.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To prepare 3-nitrobenzoic acid, we need a carboxyl group and a nitro group in a meta relationship. This requires utilizing the directing effects of existing groups on the benzene ring.
Step 2: Detailed Explanation:
1. Nitration: Reaction with $HNO_3/H_2SO_4$ converts benzene to nitrobenzene.
2. Bromination: The $-NO_2$ group is a strong deactivating and meta-directing group. Reaction with $Br_2/AlBr_3$ yields 1-bromo-3-nitrobenzene.
3. Grignard formation: Reaction with $Mg/\text{ether}$ converts the aryl bromide to a Grignard reagent, 3-nitrophenylmagnesium bromide.
4. Carboxylation: Reaction with $CO_2$ followed by acid hydrolysis ($H_3O^+$) converts the Grignard reagent into the carboxylic acid.
Result: 3-nitrobenzoic acid.
Other options fail because bromobenzene followed by nitration gives ortho/para products.
Step 3: Final Answer:
The correct sequence is (A).