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I am doing a malonic synthesis shown below, and I wonder why I should distill ethanol off before adding water in the first step of the reaction (First I add absolute ethanol and solid sodium).enter image description here

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  • $\begingroup$ I see no reason to distill off EtOH before adding water for the base hydrolysis step. The EtOH facilitates the reaction as the butyl malonate is readily soluble in it but not in water. $\endgroup$ – Waylander Sep 11 '18 at 15:25
  • $\begingroup$ I think it is because of this passmyexams.co.uk/GCSE/chemistry/… i.e alcohol will react with your product (carboxylic acid) to form an ester. $\endgroup$ – santimirandarp Sep 12 '18 at 0:06
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    $\begingroup$ Could you clear up the confusion and detail steps in more detail. Your reactions don't indicate when you boil off the ethanol... $\endgroup$ – MaxW Sep 12 '18 at 0:29
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I assume this is a real procedure for actual synthesis.

Distilling ethanol off does have an effect on the workup of this reaction. Usually one needs to separate the reaction product and educts or side products afterwards by separation between an aqueous and an organic phase. Having ethanol in the mixture prevents the two phases from separating. In order to isolate your product you will need to distill ethanol off before adding water, because water would also be distilled off, if you added it first.

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  • $\begingroup$ I disagree. You remove the ethanol after the base hydrolysis. $\endgroup$ – Waylander Sep 11 '18 at 18:36
  • $\begingroup$ Where does it say the ethanol is distilled off after base hydrolysis? $\endgroup$ – Chris Sep 11 '18 at 18:45
  • $\begingroup$ It doesn't. My point is that it makes little sense to remove the EtOH before the base hydrolysis, all you need do is add some water and an excess of KOH to conduct the hydrolysis, then concentrate as the first part of the workup. $\endgroup$ – Waylander Sep 11 '18 at 18:49
  • $\begingroup$ If you distill it off before, then do the hydrolysis, you can isolate your acid by acidifying your solution and extraction with apolar solvent. I think you just have a different procedure that you like to stick to, but OP describes the procedure like i explained. $\endgroup$ – Chris Sep 11 '18 at 18:58

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