Write the products of the following reactions:
b. 2, 2-Dimethylpentanoic acid + KOH → ?

Question

Write the products of the following reactions:

b. 2, 2-Dimethylpentanoic acid + KOH → ?

Accepted Answer

Welcome everyone. Our next problem says, write the compounds produced when Cyclopedic acid is reacted with sodium hydroxide. And we have three different sets of possible products. A B and C along with choice D no reaction. So we'll look at our possible products. But first, let's think about what kind of reaction we could expect. So, Cyclopedic acid is a carboxylic acid. We have the cyclops ring attached to a carbon carbon. So C double V tono and then that has an oh group attached to that carbon. I want to draw out the bond between the O and the H there. Now, this is reacting with sodium hydroxide. When we look at sodium hydroxide, what should we think of right away? Well, it's a strong base that oxygen has two double bonds and it will exist in a pionic form. So we'll go ahead and write as a sodium ion positively charged and a negatively charged hydroxide ion with three lone pairs and a negative charge on that oxygen. And what do we know about our psy carboxylic acid? Our cyclops acid. Well, carboxylic acids are weak bases so it can lose that hydrogen. So we'll see when it reacts with this strong base. Those lone pairs on the hydroxide ion will bond with our hydrogen on the oh group of the carboxylic acid. The electrons from that bond to that hydrogen go on to the oxygen giving it a negative charge on three lone pairs. So we'll have a product involving our cyclopedic acid becoming cyclopentolate ion. So we have our cyclops ring bond to our carbonyl carbon that's still there. But now that carbon, carbon, carbon has a negatively charged oxygen and it's a negatively charged ion. So we can form an ionic bond with our positively charged sodium ion. So we'll put that there next to our negatively charged oxygen. And of course, the other product is water. So an acid base reaction going on here. So let's look at our answer choices. Well, we can eliminate choice D no reaction because we do have our acid base reaction. We know that water is one of our products. And Choice B shows um the, the product being hydrogen peroxide rather than water. So choice B is incorrect. Since our product. Second, one of our products should be water. Choices A and C correctly have water. But we need to look at the on it produces in choice A, we see our negatively charged oxygen bound to the carbon carbon, but we see sodium Britain as a covalent bond next to the carbonyl oxygen. So the oxygen is double bonded there. So this is not going to be correct that oxygen has a partial negative charge. But our sodium ion is going to be attracted to the complete negative charge on the other oxygen. So choice A will be incorrect. Choice C then shows us our correct product. It has water, our cyclopentolate ion with the negative charge on that oxygen that has a single bond to the carbon carbon and then this positively charged sodium ion next to it. So choice C gives us the correct structure of our products. When cyclopedic acid reacts with sodium hydroxide. See you in the next video.

Write the products of the following reactions:
b. 2, 2-Dimethylpentanoic acid + KOH → ?

Verified video answer for a similar problem:

Key Concepts

Acid-Base Reaction

Carboxylic Acids

Potassium Hydroxide (KOH)

Write the products of the following reactions:b. 2, 2-Dimethylpentanoic acid + KOH → ?

Verified video answer for a similar problem:

Key Concepts

Acid-Base Reaction

Carboxylic Acids

Potassium Hydroxide (KOH)

Write the products of the following reactions:
b. 2, 2-Dimethylpentanoic acid + KOH → ?