Grignard's Reagent will generate a strong base and nucleophile, $\ce{R-}$. Can this not show nucleophilic substitution at the C atom attached to the O atom of the ether, causing $\ce{RO-}$ to depart as a leaving group? Since $\ce{RO-}$ is a better leaving group than $\ce{R-}$ this reaction should be feasible. Additonally, $\ce{R-}$ is a very good nucleophile.
Answer
The free nucleophile $\ce{R-}$ never really exists in a Grignard reagent. That's why we write it $\ce{RMgX}$. Ethers are used as a solvent as the oxygen coordinates to the $\ce{Mg}$, improving solubility and stability.
Regarding your mentioned substitution reaction, it probably would happen if a Grignard reagent was left long enough. There's a reason you make Grignard reagents immediately before use. They don't last if left on the shelf!
But Grignard reagents are usually reacted with carbonyl compounds. These are much more electrophilic than ethers, because the oxygen atom is pulling electrons away from one carbón atom instead of spreading its pull between 2, and because a double $\ce{C=O}$ bond is inherently weaker than two $\ce{C-O}$ bonds. Hence the desired reaction happens much faster than the undesired one.
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