# Acetophenone reacted with LDA and diethyl carbonate. Why do I need HCl?

I'm working on a practice problem that asks for the product of acetophenone and LDA reacted with diethyl carbonate. After this, it says step 2 is HCl. After I react the first step, I get the correct answer to the problem - ethyl benzoylacetate. I'm asking why the problem says that HCl is added even when everything is neutral in the product I got and it's correct.

## 1 Answer

If you draw what you have carefully, without omitting any charges or adding magically appearing protons, notice that the actual product you initially get is either the ethanolate addition product of ethyl benzoylacetate, a mixture of ethyl benzoylacetate and lithium ethanolate or the lithium enolate anion of ethyl benzoylacetate and ethanol. In any case, you still have an anionic product.

For isolation and purification, you want to get rid of that negative charge (initially introduced by the strongest base you added: LDA). This is called quenching and typically performed with weak acids. I personally would use ammonium chloride for this type of reaction, but carefully adding dilute $\ce{HCl}$ will do the trick just as well. You should use $\ce{HCl}$’s proton to remove any remaining negative charges and eliminate $\ce{LiCl}$.

• So is it a matter of quenching the negative charges on the minor products that aren't ethyl benzoylacetate? Nov 19 '16 at 20:52
• @5antoro I wouldn’t call them minor. 1) They are stoichiometric; 2) not too safe on the $\mathrm{p}K_\mathrm{a}$ values, but I could well assume ethyl benzoylacetate to be equally or more acidic than ethanol.
– Jan
Nov 19 '16 at 20:57