EAS:Retrosynthesis - Online Tutor, Practice Problems & Exam Prep

Proposing aromatic synthesis involves transforming benzene or its derivatives into more complex molecules using knowledge of substituent effects. For instance, converting acetophenone to a target molecule requires understanding the roles of ortho- and para-directors versus meta-directors. Key transformations include nitration to introduce a nitro group and subsequent conversion of a ketone to a benzoic acid. Mastery of these concepts is essential for effective organic synthesis, emphasizing the importance of sequence groups and regioselectivity in achieving desired products.

Aromatic synthesis starting with benzene/benzene derivatives

All right, guys. Now we're going to get some practice with proposing aromatic synthesis. At some point in your coursework, you're going to be asked to propose an aromatic synthesis starting only from benzene or other benzene derivatives. You're going to have to turn that benzene into something more complicated. In order to make this work, you're going to have to use your knowledge of sequence groups so that you can add the groups in the right positions. It's never going to be that easy that you don't need to use sequence groups. Let's do this first example kind of as a worked example where I'm going to give you some hints. And the second one, you'll be completely on your own.

For this first one, it says synthesize the target molecule from acetophenone and any other reagents. In this case, acetophenone is given to you as the molecule. But you should be aware of what the name acetophenone means from the naming benzene area of our clutch videos. What's going on here? What are t

Synthesize the target molecule

All right guys, there was a lot going on in this question. Notice that pretty much the only thing that stayed the same was the ethyl group. But a lot of other stuff changed. First of all, I'm going to need to find a way to put a methyl group in the para position. This is para substitution. Even harder than that, I'm going to have to put an aniline in the ortho position, so ortho substitution. Now we've got a few issues here. First of all, is ortho substitution very favored? In most cases, no. The only way that I'm going to get a high yield of the ortho position is to make sure that this methyl group is there first or something is there blocking the para position. I need to block it so that it's going to be forced into the ortho. On top of that, ideally, this should be a meta director when I add the aniline because I want it to be synergistic with the ethyl group to face towards the top. I should be thinking, are there any groups that I could add that could be a meta director and then be turned into an ortho para director later? That's one thing. Another thing is do we even know how to add aniline to a benzene? Did I ever teach you how to add NH₂ specifically to a benzene, an EAS reaction? No. But did I teach you a precursor that can be easily turned into aniline? Yes guys. Remember, reduction of nitro groups. We can use nitration to make aniline which is why it's so popular. So at some point, I need to nitrate that position. But only after I already have some kind of meta director in this position. See where this is going? You might not, but let me just step in and try to help. This is just you have to just start getting a feel for these.

The first reaction we're going to do is a Friedel Crafts acylation because we know that acylation can be turned later on into alkylation using a Clemmensen reduction. I'm going to use an acid chloride. Now this part is important guys. Your acid chloride needs to contain the number of carbons that you want in your end product. Since I'm adding it here, how many carbons should this acid chloride have? Just one because I only have one carbon here. That means that I should have an H on the other side because I only want one carbon total and that's the carbon on the carbonyl. I'm going to combine that with AlCl₃ and I'm going to get a molecule that looks like this. I got an aldehyde for right now. Notice that it attached to the para position because we said that para is pretty much predominantly favored, especially when you have larger groups like acyl groups. Cool. Now what can we do? Well, now I have a meta director here and an ortho para director here, now is the time to nitrate because they're both synergistically pointing to that top position. Now I would take my nitric acid and my sulfuric acid and I would now make my nitrated, my nitro group along with my aldehyde at the bottom.

Provide the reaction for each of the following reaction steps

Hey guys. Let's take a look at the following practice question. Here it says, Provide the product for each of the following reaction steps. For the first one, we're going from benzene to ethylbenzene. Now, here we're only adding a 2 carbon chain. Here if we're only adding only a 2 carbon chain, we don't have to worry about rearrangements. So, we could just simply do alkylation. Here, we have our alkyl group connected to bromine over our Lewis acid catalyst. Remember, this is going to leave so it can attach to this aluminum. So, this carbon becomes positive. It's a primary carbocation but it won't rearrange because if it rearranges it'll just go to the other side where it's still a primary carbocation. So we don't have to worry about rearrangements here. These are the reagents for our first step.Now, here next we're going to go from ethyl and here we have to add a Br group. Now, remember which carbon did that Br group add to? It added to the benzylic carbon, the carbon directly connected to benzene. There's 2 ways we could add that Br. One way is by using NBS and bromosuccinimide or we could just do Br2 and UV light. Now, next we're going from ethyl to this carboxylic acid. Remember, if your benzylic carbon has at least 1 hydrogen on it, if I use a strong oxidizer, the whole chain changes into carboxylic acid. So here we could use KMnO4 over H plus So that'll give us that carboxylic acid. So this is side-chain oxidation.Next, I want to change that Br from the Br into an OH. Now, here, if I want to ensure that the Br gets substituted with an OH, what could I do? Well, the easiest thing I could do is use NaOH. But this is important. OH- could do E2 as well because this is a secondary alkyl halide. How do I ensure SN2 will happen instead of E2? You decrease temperature. Making the temperature lower than room temperature will ensure that SN2 happens and not E2.Now, here we want to go from this alcohol to that carboxylic acid. What we need to realize here is how can we do this? Well, there's a bunch of different pathways that we could take. One potential pathway we could take here is notice here that this has 2 carbons but this here only has 1 carbon. So, one thing I could do here is this. I could do simply concentrated H2SO4 and heat. When I do that first step, what does it do? It changes this alcohol into an alkene. It gives me a double bond. Okay. So we get that structure there. Then what we could do lastly is this. You could do KMnO4 and H+. Now, KMnO4 is a strong oxidizer, but there's also another feature that it has that you learned when you dealt with alkenes. Remember, if your alkene carbon has 1 hydrogen on it, KMnO4 will cut through the double bond and that alkene carbon with 1 hydrogen gets changed into a carboxylic acid. Here, if you have 2 hydrogens on you, then this would change into CO2. But we don't really care about that. That's a gas that floats away anyways. We're mainly concerned with the major product left behind which would be that carboxylic acid.Now, here we want to change this alkyl halide chain into an alkene. So how would we do that? Remember, we want to do E2 here so just use a strong base. So we use tert-butoxide. So that would change that alkyl halide chain into an alkene by E2. Because remember the carbon with the halogen is our alpha carbon and next to it is our beta carbon. So all that would happen is we'd rip off one of these beta hydrogens. It would fall here to make a double bond and kick out the Br.Now finally, these last 2, here we have CH and here we have CH2. What did we do? Well, we still have an H there and a CH2 here. But what did we add? We added an H and we added a Br. Now, it's the opposite of what we're used to seeing. Hydrogen should've gone to the less substituted alkene carbon. Hydrogen should've gone to the double bonded carbon with more hydrogen. But here the opposite happened. That must mean we did anti-Markovnikov. How do we force anti-Markovnikov to occur? We use a peroxide. Peroxides force anti-Markovnikov addition. So we do HBr on top and peroxide on the bottom. And then, it will do the opposite of what we expect. Br will go to the double bonded carbon with more hydrogen. H will go to the one with less.Now, here the same thing happens with this alcohol. What happens is OH went to the double bonded carbon with more hydrogens for some reason instead of the hydrogen. This is also anti-Markovnikov. How do I change an alkene into an alcohol by anti-Markovnikov? We do it by hydroboration. Just remember, these are old terms that are still important now. So we do hydroboration-oxidation. So we do borane, BH3, with THF which is just our ether solvent, tetrahydrofuran, over H2O2 and OH-. This adds a hydrogen and an OH to an alkene to make an alcohol. But look, we have a peroxide. Peroxide can be written as this or as this. Because the peroxide is there, it forces anti-Markovnikov. So, these last 2 are a little tricky because they entailed things that you guys learned a long time ago, but we can still use today because we're still dealing with an alkene in a sense here with this. And we're just talking about what kind of groups can I add and how do I add them? Do I add them by Markovnikov's rule or anti-Markovnikov's rule? That's the key to this question.

Beginning from Benzene, synthesize the following compound.

Hey guys. Let's take a look at the following question. Here it says, beginning from Benzene, synthesize the following compound. Now, if we're trying to synthesize from benzene, it's most important first to determine what kind of directing groups we have. Here we have an alkyl group. Alkyl groups are ortho-para directors. Then, we have a halogen. Halogens are ortho-para directors. Now, what's the issue though? What you should notice here is that even though both are ortho-para directors, they are meta to each other, which doesn't make very much sense. How can we get 2 ortho-para directors to add meta to each other? That's not going to happen. What this indicates is that one of these groups was a meta group beforehand, then I added the other group which made it go to the meta position, Then, I changed that meta group into an ortho-para group. Now, halogens are halogens. You add them all. You don't manipulate them. You don't change them from a meta group to a halogen, at least not yet. So, we're going to say that the alkyl group must have been a meta director first. Then I did bromination which added it to the meta position. Then I changed that meta group into this final alkyl group. So think about it. Alkylation adds an alkyl group which is ortho-para. But acylation adds an acyl group which is a meta director. What we do first is we're going to do acylation. This halogen leaves to attach the Lewis acid catalyst. That carbonyl carbon becomes positive, so it'll attach to benzene. Now, realize here that this carbonyl is directly connected to benzene so it's a meta group. Since it's a meta group, we want to add bromine. The Br will want to add right there. So what we do next is we do bromination. So now the bromine group adds to the meta position. So great. We got our 2 groups to go meta to each other. Now, we need to change the acyl group into an alkyl group. So you just have to think what are the 2 different ways that we know of to change a carbonyl into a CH2 group. The two ways are we could do Clemmensen reduction which would be this. Or, we could do Wolff-Kishner. Wolff-Kishner here is through basic means. Both of them would change that carbonyl into a CH2 group. And just like that, we've isolated our product. So remember, first look to see what kind of directing groups we have. Are they ortho-para directors or meta directors? Then look to see what their spacings are. Are they meta to each other? Are they para to each other? Are they ortho to each other? This is key to knowing what to do. Since they both were ortho-para but they were meta to each other, that meant that one of them must have been a meta director first. Then, I added the other one and then I came back and changed that meta group into an ortho-para group like we did here.

Beginning from Benzene, synthesize the following compound. 

1-Phenylethanol

Hey guys, let's take a look at the following practice question. Here it says, beginning from benzene, synthesize the following compound. Here we need to synthesize 1-phenylethanol. Now, here, ethanol is a 2 carbon chain. Ethanol means that we have an OH group. Phenyl would mean that on carbon number 1, we attached a benzene. Now, where the OH is is carbon number 1. This is carbon number 2. One phenylethanol would look like this. Starting from benzene, we need to create this structure here. Here what we need to realize is what is directly attached to benzene. Directly attached to benzene is a carbon with an oxygen. So think about it. We do know of a reaction that can add a carbon directly to benzene connected to an oxygen. We say that that reaction is acylation. All we're going to do here is add an acyl group first to benzene. The acyl group is going to be 2 carbons long because this chain is 2 carbons long. Cl leaves, so we're going to attach to the aluminum. There we have a carbon directly connected to oxygen that's connected to benzene. This group here is a ketone group. Now, we need to reduce this ketone to a secondary alcohol because that's what this is. So we're going to say here, which reducing agents do we know that can reduce a ketone into a secondary alcohol? Well, the reducing agents that we know are NaBH4 which is sodium borohydride. This one's a weak reducing agent. It only reduces aldehydes and ketones. Then we have the stronger LiAlH4 which shorthand is LAH which stands for lithium aluminum hydride. This is a stronger reducing agent. It works on any carbonyl, any C double bond O group, on carboxylic acids, on esters, aldehydes, ketones, amides, all of them. So here, either one would work and that will reduce my ketone into a secondary alcohol. So although this is not exactly what we're used to seeing, just understand the fundamentals. We have a carbon connected to benzene. That carbon is connected directly to an oxygen. What's similar to that? Similar to that would be an acylation. Once you get that group there, change that group into an alcohol. That's key to this question, looking at similarities.

Beginning from Benzene, synthesize the following compound.

Let's take a look at the following practice question. Here it says, beginning from benzene, synthesize the following compound. Here, we have benzene. All we have to think about is what is directly connected to benzene. Directly connected to benzene is just one carbon. So think about it. How can I add one carbon directly to benzene? Here, what we could do is CH3 connected to a halogen over the Lewis acid catalyst. What we're doing here is we're doing alkylation. The Br will leave, and that carbocation will form. It won't rearrange because it's just a methyl carbocation that attaches directly to benzene. Then what happens? Here we need to replace one of the hydrogens with an OH. Now, we don't know how to replace one of the hydrogens with an OH directly, but we do know how to replace one of the benzylic hydrogens with Br. If we use NBS or Br2HV, that will replace one of the benzylic hydrogens with a bromine. So we get this structure here. Now, here, we have an OH,

Here’s what students ask on this topic:

What is the role of ortho-para directors and meta directors in EAS reactions?

In Electrophilic Aromatic Substitution (EAS) reactions, substituents on the benzene ring influence the position where new groups are added. Ortho-para directors, such as alkyl groups and hydroxyl groups, activate the ring and direct incoming electrophiles to the ortho and para positions relative to themselves. Meta directors, like nitro groups and carbonyl groups, deactivate the ring and direct electrophiles to the meta position. Understanding these directing effects is crucial for planning the sequence of reactions in aromatic synthesis to achieve the desired substitution pattern.

How do you convert a ketone to a benzoic acid in aromatic synthesis?

To convert a ketone to a benzoic acid in aromatic synthesis, you typically use an oxidation reaction. A common reagent for this transformation is potassium permanganate (KMnO₄) under acidic or basic conditions. The reaction proceeds as follows:

$R_{\mathrm{CO}}+{\mathrm{KMnO}}_4\to R_{\mathrm{COOH}}$

This reaction converts the ketone group (R-CO) to a carboxylic acid group (R-COOH), resulting in benzoic acid.

What are sequence groups in aromatic synthesis?

Sequence groups in aromatic synthesis refer to the order in which substituents are introduced to the benzene ring to achieve the desired product. The sequence is crucial because the first substituent affects the reactivity and position of subsequent substituents. For example, if you need a para-substituted product, you might first introduce an ortho-para director before adding the second group. Understanding the directing effects and reactivity of each substituent helps in planning the correct sequence of reactions to synthesize complex aromatic compounds.

How do you introduce a nitro group into a benzene ring?

To introduce a nitro group into a benzene ring, you typically use a nitration reaction. This involves treating benzene with a mixture of concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄). The reaction proceeds as follows:

$C{6}_H+{\mathrm{HNO}}_3+H_2\to C_6{\mathrm{NO}}_2+H_2$

This reaction introduces a nitro group (NO₂) into the benzene ring, typically at the meta position if the ring is already substituted with a meta-directing group.

What is the importance of regioselectivity in aromatic synthesis?

Regioselectivity in aromatic synthesis is crucial because it determines the position at which new substituents are added to the benzene ring. This is important for achieving the desired structure and properties of the final product. Regioselectivity is influenced by the nature of existing substituents, which can be ortho-para directors or meta directors. Proper understanding and control of regioselectivity allow chemists to design efficient synthetic routes and avoid unwanted by-products, making the synthesis process more predictable and efficient.