Write a balanced chemical equation using condensed structural formulas for

b. the saponification (base hydrolysis) of methyl benzoate.

Video transcript

Welcome back, everyone. Write the balanced chemical equation for the base catalyzed saponification reaction of prop benzoate wet sodium hydroxide. As the name suggests, we have prophyl benzoate as the starting material which is in ether. And we know that ester have a general formula of R coo R have prop benzoate and benzoate tells us that the first alkyl group are essentially consists of the benzene ring. So what we want to do is first, we'll draw a six member ring. So let's introduce six carbons and let's add the double bonds, one of them, the second one and the third one. Now we want to make sure that each carbon has four bonds. So let's add the missing hydrogens and let's bond the last carbon to the carbon from the Cabano. Now we will have coo R. So let's add two oxygens. And then according to the name, it's prop benzoate and that propi is essentially bonded to oxygen of coo right, based on the general formula and the propi group would be CH two C two C three is a three numbered alal group. According to the problem, this ester reacts with sodium hydroxide. And what do we know about the base catalyzed hydrolysis. Well, essentially similarly to hydrolysis of esters in water, we're going to release two products. One of them is going to be an alcohol that alcohol is going to be obtained from the Al Coxie group given to us. Right? So we're going to take the propel group and we are going to bond it to, oh, this will give us our alcohol which is propane 10, we get ch three ch two another siege to and oh, what is our next product? Well, we know that in water acids hydroly to produce acids in addition to alcohols. But if we're using a basic medium, we're producing or boxy instead of a carboxylic acid. So we're going to take our ester and we're going to turn it into a corresponding carboxylate. First of all, we will remove the alkyl group that we used for the formation of the alcohol. And we simply want to add a negative charge to our carboxylate. So we get coo negative. What about sodium? Well, essentially it is an iron spectator, right? If it is an iron spectator and we know that sodium hydroxide, we can just remove sodium from both sides because it's the sodium cion and simply add a negative charge to hydroxide. And now we have our complete equation. Let's label it as our final answer. Thank you for watching.