Draw structures for compounds A–F.

Question

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and provide compounds one through six. Here we have a multi step synthesis in which the products after each step are labeled from 1 to 6. So to understand the products for each part, we have to understand not only the starting material but also the reagents provided in each step. So four compound one which represents the first reaction. We have a primary alcohol reacting with PBR three in the presence of purity. Recall that this reaction is going to convert an alcohol into an o bromide because PBR three is in the presence of pyridine. When the alcohol is primary or secondary, the reaction mechanism is SN two. If it's tertiary, then no reaction occurs. So in this case, product one here would have the same carbon chain as the reagent. But instead of having a hydroxy group, we would have bromine instead. So because there's no chiral center, I would not showcase stereo chemistry. So now product one here, which is an al bromide is then going to react with turbo to side to produce product two here. Now, in this case, the, the ocho bromide produced in the first reaction here is primary and therefore relatively unhindered. So this coupled with the fact that chirpy oxide has a negative charge and therefore, is considered strong, would normally favor an SN two reaction. But recall that tut to side is characteristically very bulky. It is hysterically hindered. So because of this, it's more effective as a base than a nucleo file. So the second set of reagents provided is actually going to favor an E two reaction as opposed to SN two. So here, turp oxide, which is the very strong and bulky base removes an available proton attached to the carbon that's next to the carbon that contains the bromine because bromine is our leaving group here. Now, typically, sometimes there may be multiple different carbon atoms that contain protons next to the carbon that bears a leaving group. But in this case, there is only one. So because of this, only one elimination product is possible, so we would have the same carbon chain still. But this time there is no bromine present. And instead there's a double bond between the carbon that contains bromine and the carbon directly next to it. So product two is now an all keen which is then going to react with sulfuric acid and water to produce product three. Now, product three is worthy of a side note. So I'm going to go ahead and scroll down so that I can redraw product two here because I want to point out here that the set of reagents described to produce product three Rh two. So four and water recall that the reaction of an alkene with sulfuric acid and water is an example of acid catalyzed dehydration or excuse me, acid catalyzed hydration. This means that water is added across both carbons of the double bond. So if I scroll down here to my little side note, the first step of acid catalyzed hydration is when the pi bond, specifically the pi electrons interact with a proton that is in solution after the strong acid dissociates. So the pi bond creates a connection between carbon and hydrogen, which means that the less substituted carbon of the starting all keen gain the hydrogen and the more substituted position becomes a carbo cion because a carbo cion is present in solution, we must always check for rearrangement because if a carbo cion can achieve a more stable configuration, it's going to do precisely that. So if we consider the structure provided of the carbo carrion, there is a tertiary position right next to the secondary Kbo Carra, this means that a hydride shift is going to take place to move the positive charge from the secondary carbon over to the tertiary one. So finally, after the more stable configuration or more stable carbo cion has been obtained is when the Nile namely water can attack. And after a subsequent dero nation step is when we get our final product, which would be an alcohol. So here the hydroxy group would be on that tertiary carbon. And so a tertiary alcohol is going to be product three. And I wanted to give that little side note because the presence of or possibility of rearrangement makes the product of acid catalyzed hydration a little bit less straightforward. So now product three is a tertiary alcohol which will then react with sulfuric acid and water. Now product three was formed from acid catalyzed hydration. No product four is the result of acid catalyzed dehydration. Water is removed from two carbons to create an all key. Now, for the purposes of terminology, the carbon atom here, the tertiary one that contains the hydroxy group is known as the alpha position because that's where the hydroxy group, the leaving group is so adjacent to the alpha carbon are three beta positions or three beta carbons. The resulting pi bond for product four is going to be in between the alpha carbon and either of the beta carbons. It just so happens that all three beta hydrogens are equivalent. So a double bond between either of them would produce the same product. And because the carbon atom that contains a leaving group is already tertiary, there is no possibility of rearrangement. So product four would be an A K. So our last step is actually a two step sequence to produce compounds. Five and six. Here step one is ozone at a temperature of negative eight °C. And in the second step, we have dimethyl sulfide. Now, the reason why I wanted to highlight these reagents is because if you recall the combination of these sets of reagents with an All Keen is an example of ozone ays. Now, ozono assis is also referred to as oxidative cleavage. So if we consider the structure of the All Keen here for compound four oxidative cleavage means that the two carbons from the alkene are going to be separated from each other and then both carbons from the now separated alkene are going to be oxidized to produce carbos. So after we split our all keen into two distinct halves, the first half would contain a five carbon chain with a car bottle on position three. So that would be compound five and then compound six would be a two carbon chain with a car bottle on position one and there you have it. So with all this in mind, we have gone through the entire multi step synthesis and have produced all six products. So if you watch this video all the way through, thank you so very much for watching. And I hope we found this helpful.

Draw structures for compounds A–F.

Key Concepts

Structural Representation

Functional Groups

Isomerism

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