Question: 52: \mathrm{RMgX}+\mathrm{CO}_{2} \xrightarrow[\text { ether }]{\mathrm{dry}} \mathrm{Y} \xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{RCOOH}
What is Y in the above reaction?
(1) \mathrm{RCOO}^{-} \mathrm{X}^{+}
(2) (\mathrm{RCOO})_{2} \mathrm{Mg}
(3) \mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}
(4) \mathrm{R}_{3} \mathrm{CO}^{-} \mathrm{Mg}^{+} \mathrm{X}
Answer: Option (3)
Explanation:
\mathrm{RMgX} is a Grignard reagent, which behaves as a nucleophile due to the polar carbon–magnesium bond.
When a Grignard reagent reacts with carbon dioxide in dry ether, the carbon atom of \mathrm{CO}_{2} is attacked by the nucleophilic carbon of \mathrm{RMgX}.
This leads to the formation of a magnesium carboxylate intermediate.
The product formed after the reaction with \mathrm{CO}_{2} is \mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}.
On subsequent acidic hydrolysis with \mathrm{H}_{3}\mathrm{O}^{+},
this magnesium carboxylate is converted into the corresponding carboxylic acid \mathrm{RCOOH}.
Therefore, Y is \mathrm{RCOO}^{-} \mathrm{Mg}^{+} \mathrm{X}.