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Textbook.
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In this problem, we're asked to prepare three compounds starting from benzene, and we are told that more than one step is going to be required in each of these cases.
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So first, we are trying to prepare metaclorobenzioic acid.
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So starting from benzene, there are not a lot of ways to introduce functionality to benzene.
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So the first thing that we're going to do is going to be a friedel -crafts, alcohol.
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Collation.
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And so we will make tallyene by using chloromethane, so ch3cl.
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And then for the fidel crafts, we need aluminum chloride, or aluminum trichloride is the typical strong lowest acid there.
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And so that, as i said, is going to get us to tallyene.
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So we've added a methyl group.
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Now we can oxidize that benzolic carbon using potassium permanganate with some water here.
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And so that is going to get us to benzolic acid.
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And then now we can chlorinate the aromatic ring.
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And that carboxylic acid there is going to direct this to a metaposition.
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So we just do that with chlorine and iron trichloride.
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Okay.
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And so that is our product for part a, metaclorobenzioic acid.
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Okay.
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For part b, we are going to make parabromo benzoic acid.
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And so again, starting from benzene, we need to start by doing a free of craft's alkalation to introduce a methyl group.
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And so that's going to be exactly the same as part a.
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Okay.
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So now we have toluene.
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And this time we are going to do the bromination on the aromatic ring before we oxidize that benzellic position.
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So we'll use bromine and iron tribromine.
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And so this is going to alcalate the aromatic ring at the para position this time...