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This is the answer to chapter 10, problem number 67, from the smith organic chemistry textbook.
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And this problem asks us to devise a synthesis for each of the products that were given in the problem from the given starting material.
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And it says more than one step is required.
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So each of these is going to be a multi -step problem.
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And so i'm just going to go ahead and get started.
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So for a, we are going to need to proceed through an alkyne intermediate.
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And so one way that we can do that is to use pbr and get to the bromide here and then eliminate the bromide with potassium terpetoxide.
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And that's going to give us the terminal alken.
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And then we can do an acid -catalyzed hydration to add an alcohol in markovnikov fashion.
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So that's going to add the alcohol to the more substituted carbon.
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And there's our product for a.
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Okay, so then for b, we're starting from the bromide.
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And we need to get to the dibromide.
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And so the easiest way to do this is to do an elimination.
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So again, we'll use potassium terpitoxide.
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We'll do this elimination.
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And then we can treat that alkyne that we made with some elemental bromine.
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And that is going to give us the dibromide.
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So there we go, di bromide.
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So that's b.
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So then for c, we're going to say.
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Start with an elimination again.
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However, in this case, we have an alcohol instead of an alkali.
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And so in order to affect an elimination, we're going to have to use concentrated sulfuric acid and some heat.
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And that's going to get us to this same intermediate that we had in b.
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And so now we need to make a term.
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Terminal methyl ether from this.
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And so what we can do is a hydroboration oxidation.
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And the purpose of the hydroboration oxidation is going to be to add a terminal alcohol here.
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So remember that's an anti -marcoffinikov addition of an alcohol.
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And so now we can treat this with a strong base like sodium hydride, followed by methyl iodide...