When the following compound undergoes solvolysis in ethanol, t...

Question

When the following compound undergoes solvolysis in ethanol, three products are obtained. Propose a mechanism to account for the formation of these products.

Accepted Answer

Welcome back, everyone. We products are formed when the compound shown below undergoes Solis and ethanol proposing a reaction mechanism for each of the products formed. We were given our starting material. We noticed that it is a primary alkyl chloride, right? And it reacts with ethanol to produce repros, let's understand what's happening in this reaction. And first of all, because so many products are formed, it looks like we can possibly have a carbon canon involved in this reaction, right? Because we noticed that the position of the double bond changes, right. So first of all, even though it's a primary alcal chloride, we also noticed that they have that double bond present. What does that mean? Well, essentially, if we perform the loss of the lead group, we know that we are going to form a primary carbon Canion. And as we know, primary carbo canons are not really stable. However, a primary resonance stabilized carbo Canion is really stable. So we can essentially state that we are forming a primary resin stabilized carbo Canion, let's draw its resonance form when we shift the pi bond. And those two result on carbo canons will allow us to understand how we're getting so many different products. So we're essentially shifting the pi bond and getting a positive charge on a different carbon. OK. Well done. So we have managed to obtain the two possible carbo canons. Let's add the brackets to illustrate resonance. We're also forming the chloride anion. And now from here, we can start thinking about the formation of products one and two, they don't get any additional double bonds, right. So that means they are substitution products specifically, we will have an SN one and sn one substitution for products of one and two. So let's say that one, we are going to discuss the formation of products one and two. And now what we're going to do is just take the first carbo can I and perform the nucleic attack, right? Because we have a positive charge, you know, the Nile specifically, that would be ethanol well, at s the carbo can I am? OK. So when ethanol attacks it, we're going to get a protein at intermediate. And of course, we want to eventually deproteinate it to get the desired ether or so we're going to simply add oxygen that is bonded to an ethyl group, it's also bonded to hydrogen. So oxygen gets a formal positive charge. And another molecule of ethanol can essentially deproteinate the given intermediate because tin can also act as a weak base that is sufficient for the deep rotation stop. And this is how we get the first product, right, we are essentially going to and the ethyl group and basically this is it, right. We have got the first product, let's just make sure that we have all of the carbons present. Actually, we have to fix it slightly. Let's notice that oxygen is oxygen actually gets bonded to carbon and we are missing one carbon, right? So let's slightly modify the mechanism. So first of all, we have ch two bonded to oxygen and now oxygen has an ethyl group. OK? So let's fix the de protonation step. And now let's fix the product. OK. This is how we have got the first product. Now let's think about the formation of the second product. And for the second product, we're simply going to take the second carbo cion form. First of all, we are performing the nuclear file like a text step. We already know what the nuclear file is. In this case, that's ethanol. So we use the long pair on oxygen. We're performing the attack forming a proton at intermediate just as before. So let's add that carbon oxygen bond oxygen is still bonded to hydrogen. So we will need to deproteinate it. Once again, we can use another equivalent of ethanol for the dep protonation step. And this is how we get the second product. OK. So now we have got our second product. Indeed. That's correct. And now let's think about the formation of the third product. Now the third product will actually be an elimination product. How do we know that? Well, essentially we get that additional double bond within the rank and to get an elimination product, we need to think about each carbo Canion. Now, the first carbo Canion cannot undergo an elimination reaction. E one because we don't have any beta hydrogens available. But the second one can because we have a beta hydrogen. So we can only take one carbo cat I am. And specifically, we're going to take the second resins form because this is the form that can undergo elimination. Also, let's state that we are showing the formation of the third product. Now let's add the carbo Canion from the second resonance form. What we want to do from here is simply identify the beta hydrogen and now draw the elimination step. So we know that ethanol can act as a weak base, which is sufficient for an E one elimination reaction. We're taking hydrogen forming a pi bond. And this is how we end up with our product, which would be an elimination product. In this case, that would be a conjugated to die. In right, of course, in each case, our by products would be roten at ethanol molecules, right? But there's no necessity to show them. The most important part is to indicate the formation of each product. Of course, if we want at the end of each part of the mechanism, we can state that we're also obtaining rotated ethanol as a by product. OK. So that would be it and we have our mechanism and we can conclude the video. Thank you for watching.

When the following compound undergoes solvolysis in ethanol, three products are obtained. Propose a mechanism to account for the formation of these products.

Key Concepts

Solvolysis

Nucleophilic Substitution Mechanisms

Product Formation and Rearrangement

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