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

We want to reach 3-nitrobenzoic acid starting from simple benzene. That target molecule possesses two substituents on the ring :

$$COOH \;( \text{meta-director})$$ and $$NO_2 \;( \text{meta-director}).$$

Because both groups are meta directing, the relative orientation we want (NO2 meta to COOH) will automatically arise if the nitro group is put on the ring before the carboxyl group is introduced. Hence we first introduce $$NO_2$$ by the classical nitration of benzene.

We have

$$C_6H_6 \xrightarrow{HNO_3/H_2SO_4 C_6H_5NO_2}$$

Now the ring already possesses a strong meta-directing nitro group. When we next brominate, electrophilic substitution will occur meta to the nitro group (because the o- and p-positions are deactivated). Using the usual Lewis-acid-promoted bromination we write

$$C_6H_5NO_2 \xrightarrow{Br_2/AlBr_3 3\!-\!BrC_6H_4NO_2}$$

The product is 3-bromonitrobenzene.

To transform the bromine atom into a carboxyl group we employ the sequence “Grignard formation → carbonation → acidic work-up.” Stating the general formula first :

$$ArBr + Mg \;\xrightarrow{\ ether\ ArMgBr} \qquad ArMgBr + CO_2 \to ArCO_2MgBr \qquad ArCO_2MgBr + H_3O^+ \to ArCO_2H$$

Applying it stepwise to our 3-bromonitrobenzene :

1. Formation of the Grignard reagent :

$$3\!-\!BrC_6H_4NO_2 + Mg \xrightarrow{\ dry\ ether\ 3\!-\!NO_2C_6H_4MgBr}$$

2. Carboxylation with dry carbon dioxide :

$$3\!-\!NO_2C_6H_4MgBr + CO_2 \to 3\!-\!NO_2C_6H_4CO_2MgBr$$

3. Acidic hydrolysis gives the desired acid :

$$3\!-\!NO_2C_6H_4CO_2MgBr + H_3O^+ \to 3\!-\!NO_2C_6H_4COOH$$

Combining all of the above, the reagent order must be

$$HNO_3/H_2SO_4 \;\longrightarrow\; Br_2/AlBr_3 \;\longrightarrow\; Mg/ether \;\longrightarrow\; CO_2 \;\longrightarrow\; H_3O^+$$

This precise sequence is listed in Option D of the question set.

Hence, the correct answer is Option 4.