Ester Reactions: Saponification - Video Tutorials & Practice Problems

Ester Reactions: Saponification Concept 1

In this video, we're gonna take a look at a particular ester reaction called saponification. Now, under this reaction, the hydroxide ion so oh minus dissolved in water reacts with an ester. This reaction cuts or cleaves Fiester bond to create a carboxylate anion and an alcohol. Now, here, what is a carboxylate anion? Well, that's just the conjugate base form of a carboxylic acid. And here we consider this to be the opposite process of auster education. Now, so pod application here, we have our Esther and this here represents our ester linkage when we use oh minus dissolved in water. So aqueous oh minus, it's going to cut this bond here, cut this connection by cutting this connection. This oxygen gains an age to reform our alcohol and this carbonyl gains an O minus. It becomes our carboxylate an ion. So remember the carboxylate, an ion is the conjugate base of a carbolic acid. Conjugate base just means we lose an H plus losing an H plus is why that oxygen now is negatively charged, right? So, keep this in mind when we encounter any type of saponification reaction

Ester Reactions: Saponification Example 1

Determine the products when propo butana weight is treated with an aqueous sodium hydroxide solution. So remember this is our ether bond. And when we use our aqueous hydroxide solution, it's going to basically cut or cleave this bond. And what occurs is that the carbon, carbon gains a negative oxygen to form a carboxylate an ion. And then this other oxygen gains an H to form an alcohol. So in this reaction, we form a carboxylate an ion and an alcohol as our two products.

Name alcohol formed when the following ester undergoes a saponification reaction.

• A. 4-methyl-4-butanol
• B. 2-methyl-4-butanol
• C. 2-methyl-1-butanol
• D. 2-methylhexanoate