Which of the following reagent(s) is required for the conversion of Benzene to methyl benzoate?
(A) \( Br_2/FeBr_3 \)
(B) Mg, dry Ether
(C) \( CO_2, H_3O^+ \)
(D) Methanol, Conc. \( H_2SO_4 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B), (C) and (D)
Benzene has to gain a –\( \ce{COOCH3} \) group. The route goes through a Grignard reagent: the ring is first given a bromine atom, the C–Br bond is turned into a C–Mg bond, the Grignard reagent adds to carbon dioxide to give benzoic acid, and the acid is finally esterified. Each of the four reagents does one step.
First, benzene is brominated by (A) \( \ce{Br2/FeBr3} \). \( \ce{FeBr3} \) is a Lewis acid that polarises \( \ce{Br2} \) and generates the electrophile \( \ce{Br+} \) for electrophilic substitution:
\( \displaystyle \ce{C6H6 + Br2} \xrightarrow{\ce{FeBr3}} \ce{C6H5Br + HBr} \)
Bromobenzene then reacts with (B) Mg in dry ether to give phenylmagnesium bromide. The ether must be dry because Grignard reagents react with water:
\( \displaystyle \ce{C6H5Br + Mg} \xrightarrow{\text{dry ether}} \ce{C6H5MgBr} \)
With (C) \( \ce{CO2} \), then \( \ce{H3O+} \), the phenyl carbanion of the Grignard reagent adds to the carbon of \( \ce{CO2} \), and acid hydrolysis of the salt gives benzoic acid:
\( \displaystyle \ce{C6H5MgBr} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{CO2}} \ce{C6H5COOH + Mg(OH)Br} \)
Finally, (D) methanol with conc. \( \ce{H2SO4} \) esterifies benzoic acid; the acid is the catalyst:
\( \displaystyle \ce{C6H5COOH + CH3OH} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{C6H5COOCH3 + H2O} \)