RMgX + CO_2 dry ether Y H_3O^+ RCOOHWhat is Y in the above reaction ?
1. \((RCOO)_2Mg\)
2. \(RCOO^-Mg^+X\)
3. \(R_3CO^-Mg^+X\)
4. \(RCOO^-X^+\)
1. \((RCOO)_2Mg\)
2. \(RCOO^-Mg^+X\)
3. \(R_3CO^-Mg^+X\)
4. \(RCOO^-X^+\)
Best answer
Answer: 2. \(RCOO^-Mg^+X\)
Key idea: a Grignard reagent adds to \(\mathrm{CO_2}\) to give a magnesium carboxylate salt, which acid hydrolysis converts to the acid.
\(\mathrm{RMgX + O{=}C{=}O \to RCOO^-Mg^+X}\) (Y), then \(\mathrm{H_3O^+}\) gives \(\mathrm{RCOOH}\).
Y is the halomagnesium carboxylate, option 2. Option 1 would need two carboxylate groups on one Mg, but one \(\mathrm{RMgX}\) adds to one \(\mathrm{CO_2}\), and option 3 is what a ketone would give after further addition.
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