Provide the reaction for each of the following reaction steps

Question

Diagram showing reaction steps for EAS in organic chemistry, including aromatic compounds and products.

Accepted Answer

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 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. So These are the reagents for our first step. Now, here next we're going to go from what? 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 a carboxylic acid. So here we could use KMNO4 over H plus So that'll give us that carbesilic acid. So this is side chain oxidation. Next, I want to change that Oh 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. Make it lower than room temperature. Here, making the temperature cold enough will ensure that sn2 will happen and not E2. Now, here we want to go from this alcohol to that carboxylic acid. What we need to realize here is how the heck 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, camel 4 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 really don't 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, tetrahydrofurane, over H202 and Oh minus. 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 Antonarkovnikov. So, these last 2 are a little bit 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.

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Provide the reaction for each of the following reaction steps

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