Limonene is one of the compounds that give lemons their tangy ...

Question

Limonene is one of the compounds that give lemons their tangy odor. Show the structures of the products expected when limonene reacts with an excess of each of these reagents.

(c) ozone, then dimethyl sulfide

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the following problem. Murder scene is a pleasant smelling compound found in many fruits, including mangoes, predict the products obtained when mersin reacts with ozone and then with the metal sulfide. If we notice Mercy in has three double bonds and if you look at the reaction conditions based on the problem, it stays that murder scene reacts with ozone, meaning reaction number one is treating mercy in with ozone and number two because it says then with the metal sulfide, our second region is the metal sulfide. Looking at these two regions, we may recall that such a reaction is called the post analysis reaction. And that makes sense, right? Because we have double bonds here. Now, we will draw a general mechanism in which some hypothetical elkin undergoes oz analysis. It's not really a mechanism in the traditional sense, it's more like a scheme which will allow us to predict the products as easily as possible. So let's suppose our al king has one hydrant bonnet over here. Some al kill group are one and then two more alcohol groups on the other end, let's call them R two and R three. The general scheme in predicting the product of gnosis is just thinking about breaking this carbon carbon double bond. If you try to draw the products you get, not really products, but thinking about the two fragments that you get, you will have our two bonded to that carbon As well as hard three. And then on the other end you will still have that other part of that double bond with hydrogen and the alcohol group are one. Now you clearly see that if you join these fragments into one compound, you will get what you see on the left, right. But the thing is that you cannot form these two new carbon atoms. They're not seized two groups. You're only using ozone as your region, which is producing oxygen atoms. So this fall by the metal sulfide would essentially produce two oxygen's along each end of the double bonds. That's pretty easy to do. Right? So to repeat these steps, just break the double bond and add oxygen atom to each of your double bonds to get your products. Okay, so if we use this strategy, what if we think about all of these three double bonds? We may see that we have three possibilities. We can break this double bond, another one over here And then the last one which is over here. And if we think about breaking these three bonds, we can easily predict our products. What if we start with this position? If we break this double bond, notice that on the left side there are two implicit hydrogen atoms first of all. So we are breaking off A double bond with two hydrogen atoms. As we previously said, we have to add an oxygen animal over here to produce our products. We clearly see how we got our first product already. We just need to add an oxygen animal over here and we got our Aldo hide. Now on the other end, we will simply have our double bond ride. We have our double bond, there's an implicit Haijun over here. What else do we have? So that that's our carbon number one. Then we have carbon number two with another double bond. And we are also breaking at rice. We can just draw that double bond. Let's not worry about the top part yet. Let's just concentrate on this one. So how many more carbon atoms do we have? We have 12. We Right, so we are going to add one, two and three. And now if you think about it and that third carbon atom has another double bondage you're breaking. Right? So that essentially means that since we're breaking that bond, there will be another double bond in the casinos. This product in addition to that implicit hydrogen atom. So we immediately able the simplest hydrogen atom in green, Let's just add that hydrogen atom. And now because we were simultaneously analyzing three double bonds, notice how we're adding an auction atom over here. Right, let's add up another oxygen over here, let's add it and the final oxygen over here, let's add it. So that's our next product. Right? We have already analyzed this and of the double bond and all of this internal part. Right? So now let's think about the external parts. What about this double bond? It's very similar to this one isn't up. So we may just draw our double bond. There are two hydrogen atoms over here that are implicit. So let's add two hydrogen atoms and then we need to add an oxygen atom. So we are done with this part and the final external part is this one over here. Sophie through our double bond, we noticed that on this carbon we have two metal groups. So let's draw these two metal groups. And finally, let's add that oxygen atom. See it, That will be our fourth product. But now, if you think about all of these products, since we have seen all these combinations, right? We have analyzed the production of the internal carbonell compound. Right, adding an auction here here and here and then the three external parts. The one over here, here and here. You will notice that products one and three are the same. So let's label them as number one, because they're exactly the same. This will be our product # two and this will be our product # three. Meaning if we want to state our file products for this problem, we may just say that we are producing two moles of product number one because there are two molecules of that Allah high that we're seeing. Right? So we can just say There are two moles of the first one plus. Now we may draw structure # two. Right? So let's draw it. And finally we have our third product over here. So let's draw acetone, Right? So we got our three products. Let's label them as our final answer, because this will be our reaction scheme with three different products form, right? It's just that we have to most of the first product form. That's why we added a coefficient of to feel like the scheme once again, and we may simply conclude that the correct answer to this multiple choice question is C. However, if you have any questions, don't hesitate to leave their comment down below in the comment section, and thank you for watching.

Limonene is one of the compounds that give lemons their tangy odor. Show the structures of the products expected when limonene reacts with an excess of each of these reagents.

(c) ozone, then dimethyl sulfide

Key Concepts

Ozonolysis

Dimethyl Sulfide (DMS)

Product Structure Prediction

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