Show how you would synthesize the following alcohol from appro...

Question

Show how you would synthesize the following alcohol from appropriate alkene.

(a)

Accepted Answer

Hey everyone and welcome back in today's video, We are going to solve the following problem. Show how the following alcohol could be synthesized using an appropriate al king B 10 to E. O. As the problem suggests, our final alcohol to be synthesized is be tend to oh, meaning we have a four member chain and an alcohol group at position number two. So that's our final material. We want to think about possible al keys that could be formed in an elimination of that alcohol group. Using that acid catalyst dehydration, for example, it's not really important how you arrive at these two alc ins. But one of the ways to do that is to think about the acid catalyzed dehydration. So we're thinking about the elimination and we have two beta hydrogen which are unique. So this hydrogen over here and this hydrogen over here. So we first of all need to protein it is to make sure that alcohol becomes a good leaving group. Right? Water. So if we remove this hydrogen atom, let's suppose that this is a pathway in blue. We are going to form a double bond at this position, followed by the loss of the leaving group. Rights, we will have one possible alkaline at this position. We may say that one al keen that we can use as this one and the other one. If we remove this hygiene, we would form a double bond at this position, position number two. Right, so our second possible al king would be this one. We have to al keys that we can use for our synthesis. Now. Generally we can basically use the acid catalyzed hydration. Right? We can essentially say that we are creating each of these al canes with H two s water. Right? If that happens, we are essentially first of all protein. Ating each of our Elkins and then we have the nuclear attack by the water molecule. In each case we want to make sure that we satisfy the more comical rule which says, we're going to add that proton to the less substituted carbon over here. Right? And then we will add the wage group at a less more substituted. My apologies position. So we are going to form that alcohol which is indeed be 10- and that's exactly the same thing what we expect to have here. The two positions are equally substituted. So it doesn't really matter where we add that hydrogen, we will form exactly the same product. So, again, we are going to form you tend to o of any of these two Elkins. But let's think why this pathway is not really good for us. It's basically because the equilibrium of this reaction generally shifts towards Elkins and some Elkins are not really soluble. So, these reactions are not really good for us because they don't favor the formation of the product that much. In other words, we can say that this would be low yield and we want to make sure that we increase the yield of our product. We don't have any possible carbon. Karen rearrangement or anything else that would prevent us from getting a low yield of the desired product. And the solution to that is another pathway. What if we take this elk in? It doesn't really matter. We can just take the D. Elkin with the double bond position number two. And if we draw it we may essentially recall that this is this al came can undergo the oxy mercure ation Diemer curation reaction. Right in the first step of this reaction, you may recall that we are adding mercury to acetate in water and we are forming that mark uranium bridge followed by the nuclear attack by the water molecule. And what happens is that you can think of this as follows essentially we can just draw the mechanism for simplicity, the double bond will attack mercury Which contains two acetate groups and has a lone pair. So that lump air attacks back and we are forming our bridge, followed by the loss of acetate. Now we get our branch Mercury now has a formal positive charge because it still has one of the two acid groups, bonnets yet. And now this is where we are interested in the nuclear attack. Stop water will will attack the more substituted position out of the to the more substituted position is one and two, they're pretty much equally substituted. Right? So because we have two equivalently substituted positions. That doesn't really matter which one we attack. So I suppose we go with this one, we have the loss of the leaving group and we are forming our intermediate. So there will be mercury acetate and the alcohol and disposition actually, it's still protein ated wide because we're adding water molecule is still protein. It'd So we may say that we are also d protein eating it. So now the world last step is to make sure that we're adding hygiene here to get beets and two Oh and to break that bond and to replace mercury acetate with hygiene, we want to make sure that we are using sodium borrow high tribe and methanol. Essentially what happens in this step is that we are replacing the carbon mercury bond with carbon hydrogen bond and that would give us our final product. Now this reaction gives the alcohol desired and a better yield. And this is why we want to propose this reaction instead of hydration because the equilibrium shifts towards the formation of the alcohol instead of the ALC in itself. Because the yield is better for this reaction. We are going to propose this reaction and start simply, let's just summarize our reaction scheme. We said that we are starting with our Elkin right bit too in that's our Elkin and we said that we are going to perform two reactions number one. We will just add mercury to acetate in water followed by sodium borough high, tried and methanol and then we are forming B 10 20 in a high yield using this oxy Mercure Ation Diemer curation reaction. So this is our reaction scheme, and we may conclude that the correct answer choice to this multiple choice question is a. However, if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

Show how you would synthesize the following alcohol from appropriate alkene.
(a)

Key Concepts

Alkene Reactivity

Hydration Reaction

Markovnikov's Rule

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