[jerm174]

Hi! I'm writing a proposal for a lab and need to provide the volumes/amounts of reactants I will be using to create isopentyl propionate. If I understand correctly, the acid I would be using is propanoic, while the alcohol is isopentanol.

Now.. my question is in regards of the amounts of each to use. I need to synthesize 10 g (theoretical yield) of ester. My lab manual says this: "Normally the alcohol is the excess reagent, but there are sometimes advantages to using an excess of the acid, particularly if the acid is soluble in water and the ester is not. The cost of chemicals is also a factor - to save money." In my case (please correct me if I'm wrong) the acid IS soluble in water and the ester is not. So if I just answered my own question correctly, I'd say I use an excess of propanoic acid. 

In addition, the manual states "...use a 2.5 fold excess, or specify otherwise and justify". Does this mean I would use 2.5 times more of the excess compound? From what I remember, a 1:10 molar ratio produces a 97% yield, while 1:2 ~ 1:3 produces 85% and 90% respectively (assuming K=4). Do I not want to produce as much ester as possible? 

A point in the right direction would be greatly appreciated!

[discodermolide]

Yes use an excess of acid as its water solubility is very good. So use a 2.5 molar excess of the acid. This is just a Fischer esterification. The propionic acid should catalyse the reaction, so you should get a good yield.