Write a balanced chemical equation using condensed structural formulas for the saponification (base hydrolysis) of

a. methyl propionate

Video transcript

Welcome back, everyone. Write the balanced chemical equation of the saponification reaction of ethylbutyrate. We have to recall that whenever we want to write a saponification reaction of an ester. In this case, that's ethyl. Beter rate. We first of all have to draw the structure of that specific ether. And in this case, we're given ethyl berate. We have to recall that at the end of the name, we have our parent and berate. Essentially, it tells us that we have a four membered chain in our parent, right? And what does that mean? Well, essentially we have carbel and to have a total of four carbon atoms, we need three more. And that will be a prop group which is CH three CH two and another CH two, right? And we're going to add it on the left side of the chain. And because it's ethyl, right, we essentially want to add a carbon oxygen bond because ester have a general formula of R coo R and that ethyl group essentially belongs to the alcoy part of the ether. So we're going to draw oxygen bonded to an ethyl group, which is CH two CH three, basically a two membered LK group. So that's our ether. And in a saponification reaction, we're essentially using a, an aqueous solution of a strong base. So what we're going to do is essentially add a strong base that would be hydroxide. We don't need to choose any ion spectator. Of course, we can do that, but there is definitely no necessity because we are interested in the net ionic equation, right. So we're adding hydroxide and in order to get our products, we have to think about hydrolysis of esters. Now in a neutral solution, whenever we are using water, we are specifically producing and alcohol and an acid. But if it's a basic solution, we want to produce a carboxylate. So let's break our oxygen carbon bond and let's draw the result of carboxylate. First, we are going to introduce a negative charge on oxygen. Whenever we are performing a hydrolysis reaction, we are producing an acid. But in basic conditions, we get carboxylate. So our acid is deproteinate. And of course, the second product is an alcohol. The alcohol part is obtained from the alcoy group. So in this case, our alcohol would be ethanol because we have C three CH two and we simply want to add oh to get our alcohol. And that said we have a saponification reaction of ethyl beard rate. That's it. Thank you for watching.