Alkylation of acetone:

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This reaction is done with a strong base (LDA in this case). Is this because the protons aren't particularly acidic? Secondly it should be done at low temperatures; why is this so? -78 degrees Celsius is typical.


Why should alkylation of enols/carbonyl compounds be done at low temperature with a strong base?

If you start with an unsymmetrical ketone, then there are two different sides of the molecule you could remove a proton from. Let's use methyl ethyl ketone as an example. Removing a proton from the methyl group leads to what is termed the kinetically favored anion (e.g. this hydrogen is the most acidic and the corresponding enolate anion is the fastest formed anion). Removing a proton from the methylene group on the ethyl side leads to the thermodynamically favored anion (e.g the enolate anion with the most highly substituted [most stable] double bond).

Alkylation of carbonyls/enols is not always done using strong base and low temperature. Strong base and low temperature generally leads to the kinetically favored enolate anion, while weaker bases and higher temperatures generally lead to the thermodynamically favored enolate anion.

In your case, acetone is symmetric so it doesn't really matter which side you alkylate. But, once you form your desired product, methyl ethyl ketone, it can react further to give polyalkylated products. These secondary products are more stable and thermodynamically preferred. To reduce the likelihood of thermodynamic control leading to these undesired products, the reaction is run under kinetic control - strong base and low temperature - to maximize formation of the monoalkylated product. Here is an earlier answer that further highlights kinetic vs. thermodynamic control in alkylation reactions.


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