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    draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below.

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    Solution for Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below. CH;CH2CH2CH2 c=CHCH3 CH;CH,CH,CH, • Use wedge…

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    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below.

    Transcribed Image Text:Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below. CH3CH2CH2CH2 c=CHCH3 CH;CH2CH2CH2 • Use wedge and hash bonds ONLY for rings. Do not show stereochemistry in other cases. • If the reaction produces a racemic mixture, just draw one stereoisomer. C opy aste Previous Next

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    Chemistry 18 Nov 2019

    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below. CHCH2CH3 Use wedge and hash bonds ONLY for rings. Do not show stereochemistry in other cases. If the reaction produces a racemic mixture, just draw one stereoisomer.

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    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below:CH;CHz C=CHz CH3Use wedge and hash bonds ONLY for rings Do not sho...

    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below:CH;CHz C=CHz CH3Use wedge and hash bonds ONLY for rings Do not sho... Question

    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below:CH;CHz C=CHz CH3Use wedge and hash bonds ONLY for rings Do not show stereochemistry in other cases Ifthe reaction produces a racemic mixture, just draw one stereoisomer

    Draw a structural formula for the product formed upon hydroboration/oxidation of the alkene below: CH;CHz C=CHz CH3 Use wedge and hash bonds ONLY for rings Do not show stereochemistry in other cases Ifthe reaction produces a racemic mixture, just draw one stereoisomer

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    Answers

    Draw a structural formula for the product of each alkene addition reaction.

    So here we are, just looking at some chemical transformations where we have the final product as well as the re agents, and we are needing to work backwards in order to deduce the formula on the structure off are starting materials. So our formula being C five h 10 was listed up up over the top of the screen, as you can see there. And so for each chemical transformation 123 I've drawn out my starting material on this is the starting material that reacts with R B R. Two. Now the site of all reactivity is at that double bond where were simply adding a bro mean to each end of our double bond, and this trend is present in all three of our chemical transformations.

    So here we are looking out some chemical transformations on. We are given products as well as the materials used in terms of re agents. In order to generate those products and then given the molecular formula C five h 10 we needed to work backwards in order to determine the starting material. So I've done so below where we have our first example that gives us this 1234 length carbon chain. So where we have abused Dean where we have to me style Butte one e is our starting material. They're moving on to our next example. Again, we have are similar. A similar starting material, however, are double bond. Husband moved so hit. You can see that we can call up, be wanting the same however we can a she in the army tile is not want the third carbon three Me felt beat warning in our final product here. So we've got a cyclo hex Juan een while we have Amy Thile substitution on the first carbon and these are the starting materials that were used in order to generate the products that have been illustrated given the re agents also used within the chemical transformation

    So here we just have a chemical transformation where we're looking R C five age 10 structure, which is drawn up here on underlined in green on how exactly it might react with H c. L. So, given that we have an AL Keen, we can assume that the site of our reactivity is at that double bond. And so our product will look something like this. Well, we've broken that double bond, and instead we now have the proton attached on a chlorine atom attached. So I've drawn it out in fall so you can see where each off the components off our hydrochloric acid have added to. However, normally we would not draw out the proton. So we would simply see this where we have all to Chlor. Oh, propane. Now we have a nut, the end, and that is because we only have single bonds present. And so this is the product. As a result, off our Alka keen reaction with hydrochloric acid

    Okay, So for this problem, you're doing the dehydration reaction and in dehydration reactions. All right, this here for dehydration, basically removing water, and you're going to be forming an AL Keen, which is a carbon carbon double bond. So your water is going to come from your alcohol and this water H 20 is going to come from your alcohol, which is going to be on one carbon, and you're going to grab ah, hydrogen, one of these hydrogen ins from an adjacent carbon. So if you kind of draw it out, call it this way so you can kind of see it easier. So you have moved to call. This are for the rest of your effort cheese. And on this terminal carbon, which is the one in blue, you have a no h and you have your hydrogen. So just to kind of keep it in the back of your mind So again for a dehydration reaction. So ch three ch two ch. So now you loose this hydrogen shown here in green. You lose this alcohol, so you're going to form a new double bond between this carbon and green and the carbon in blue each que plus we'll call this each. Ohh. So this would be your out keen plus water. Okay, let's go to the next one. So again, you're losing h 20 so you're losing an alcohol or Euro age group from this carbon? And because this cyclo plantain is symmetric, it doesn't matter if you grab a hydrogen from either of these two indicated carbons as long as you're grabbing a hydrogen from the adjacent carbon. So, for example, for this problem, let's grab the one on the right. So now you would have your cyclo pen teen plus your water because now you lose this, this form your water and you're left with a double bond on this carbon. No, the last one. So now instead of your alcohol being on a ring at the end of the chain now it's in the middle of a chain. Um, for these for this problem, you can either grab this hydrogen just to make it easier to see where you can grab this hydrogen. And again, just like the cyclo plantain, it's symmetric. So no matter which hydrogen, you remove your going to form the same product. So just for the sake of argument, Let's remove the one on the left in green. So now we have ch three ch. And now this bond right here is going to be come a double bond. So ch plus for a rather ch two ch three plus now your work. So that's all you need. Just so just remember for dehydration, you were losing water and forming an LP.

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