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Synthesis of substituted benzene rings II

Synthesis practice problems. Created by Jay.

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Video transcript

Once again, our goal is to synthesize simple organic molecules starting from benzene. And this time, our goal is to make this molecule over here on the right. Once again, we're going to use the concept of retro synthesis, so working backwards and thinking about what could be an immediate precursor to this molecule here. To do that, let's analyze the groups that are attached to our ring. Right here we have an amino group. And we know that's an ortho/para director. We know that because of the lone pair of electrons on this nitrogen right next to our ring. And then over here, we have a carboxylic acid functional group attached to our ring. And we know this carbonyl carbon here is partially positive, which makes this a meta director. Now, we haven't covered any reactions that would install an amino group meta to a carboxylic acid group. But we have talked about how to put a nitro group meta to a carboxylic acid group. So let's turn that amino group into a nitro group. So let's go ahead and draw our benzene ring here. And we'll go ahead and put our carboxylic acid down here. And instead of an amino group, let's draw a nitro group, like that. So all we need to do now is think about a way to reduce our nitro group to our amino group. And there are many ways to do it. One thing you could use would be a metal like iron and some hydrochloric acid. You could have used tin. You could have also used hydrogen and a metal Catalyst like nickel. So there are many ways to reduce a nitro group to an amino group on a benzene ring. And now I know how to put that nitro group on meta to my carboxylic acid. So we have a meta director right here. And so if I take that nitro group off, I can come up with the precursor to this molecule, which should be just benzoic acid. So I'll go ahead and draw benzoic acid here. And I have to think about the reagents necessary for a nitration reaction, which of course would be concentrated nitric acid and concentrated sulfuric acid. And so once again, because my carbonyl here on my carboxylic acid is meta directing, the nitro group ends up in the meta position. So now I have been benzoic acid. And I need to make a benzoic acid from benzene. And if you think about one of the reactions we did in the benzylic position video, we were able to create a benzoic acid molecule from an alkylbenzene. And so the immediate precursor to this molecule must be some sort of alkylbenzene, so a benzene ring with an alkyl group on it. So just to make things easy, I'll just make it a methyl group here, so a methyl group like that, so toluene as our immediate precursor to benzoic acid. To oxidize the alkyl side chain, I need to add something like sodium dichromate, so Na2Cr2O7. A source of protons, so something like sulfuric acid, and some heat. And so this is oxidation of an alkyl benzene, like that. You could have also done this reaction with permanganate and heat as well. And you could have chosen any number of carbons on your alkyl side chain. So I just chose one to make it easy. So this carbon, when this group gets oxidized, would turn into that carbon right here. Now we would need to convert benzene into toluene. And so we could do that using a Friedel-Crafts alkylation reaction. And so we need one carbon. So for our alkyl chloride, we would need a one-carbon alkyl chloride, so CH3 Cl. And our catalyst for Friedel-Crafts alkylation would be aluminum chloride, like that. So our synthesis is complete. Let's go ahead and run through the reactions very quickly. And let's make sure that everything makes sense. So we start with benzene. And we do a Fridel-Crafts alkylation to put a methyl group onto our benzene ring to form toluene. Next, we oxidized that alkyl side chain with sodium dichromate to convert it into a carboxylic acid, which is now a meta director. So when you nitrate your ring, the nitro group gets put on in a position meta to your carboxylic acid. Finally, you reduce your nitro group to turn it into an amino group, and you are done. So let's do one more synthesis problem here. And so we can see that, this time, our target molecule is over here on the right. And immediately, I see a bromine at the benzylic position, the carbon right next to our benzene ring. And so I know a reaction that will put a bromine in the benzylic position. If you remember, that's the free radical bromination reaction. And so I can go ahead and draw the precursor to that molecule, which would be a benzene ring. This bromine over here is on our ring. And we would have just an ethyl group now, like that. And so I would need some NBS, some heat. I could use carbon tetrachloride as a solvent, a peroxide to initiate the free radical mechanism. And this is a free radical bromination reaction, which puts a bromine at the benzylic position. All right, trying to figure out a precursor to this molecule. Let's analyze what sort of groups we have on our ring. So this ethyl group is an alkyl group. And remember, that's an ortho/para director. This bromine here, of course, is also an ortho/para director because of lone pairs of electrons on it. So this is an ortho/para director. We have two ortho/para directors on our ring, but those ortho/para directors are meta to each other. So we need to think about a way to turn one of ortho/para directors into a meta director. And a good way of doing that would be to turn our alkyl group into a group that has a carbonyl on it, so an acyl group. So thinking about the precursor molecule here, I draw my benzene ring. I'm going to leave my bromine here. And instead of an alkyl group, we're going to have an acyl group now, because our acyl group is a meta director. So this carbon right here is partially positive. So we need to think about how to reduce an acyl group to an alkyl group. And one way of doing that would be to do a Clemmensen reduction. So you take amalgamated zinc, so we take some zinc here, so an amalgam with mercury, and also some HCL. And so that will reduce our acyl group to and alkyl group, like that. Now that we have an acyl group on there, which is a meta director, that would allow us to install the bromine meta to that acyl group. So when I think about the precursor to this molecule, I just take off that bromine. And so I have my ring with an acyl group on that ring. And once again, since this is a meta director here, all I need to do is a bromination reaction. So once again, I would need some bromine. And I would need our catalyst, so FeBr3. And that reaction will work. Finally, I have to go from benzene to this molecule. And once again, I can do that with a Friedel-Crafts acylation reaction. I would need two carbons to do my Friedel-Crafts acylation reaction. So from my acyl chloride, I need to make sure I have two carbons. And once again, my catalyst would be something like aluminum chloride for this Friedel-Crafts acylation. And so this would be the complete synthesis. Let's once again run through each step starting with benzene. We do a Friedel-Crafts acylation to install an acyl group on our ring, like that, which is a meta director. So if we follow that with a bromination, the bromine is installed meta to our acyl group. Our acyl group is then reduced, using a zinc amalgam, and converted into an ethyl group right here. And then finally, we do a free radical bromination to put a bromine at the benzylic position. And we are done. So think retro synthesis. And then also double-check and make sure your synthesis makes sense.