EAS: Retrosynthesis

Concept: Concept: Aromatic synthesis starting with benzene/benzene derivatives

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Video Transcript

All right guys, so now we're going to get some practice with proposing aromatic synthesis. So at some point in your course work 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.
So what's going on here? What are the different transformations that are taking place? One thing that's happening is that I have a ketone that at some point needs to become a benzoic acid. So that's interesting. I've also got to add a nitro group in the para position so that's p, para substitution. So there's kind of a lot going on here. I've got a few different things.
One thing that I know for sure is that to put a nitro into the para position, at some point, my ketone is going to have to become what? An ortho, para director. Right now what type of director is it? Meta. Right now this is a meta director. If I were to nitrate this ketone right now, or this acetophenone, I would actually get a nitro here. Is that right? No. So I'm going to have to use my knowledge of sequence groups to figure out how can I turn that into an ortho, para director.
Now benzoic acid. Is this an ortho, para director? No, this is also a meta director. That means that I have to turn this into a benzoic acid sometime after I've added the nitro group because if I just turn it straight into benzoic acid, then I'm going to get a meta nitro group again. So this is kind of already outlining all the things we need to do.
So now I'm going to let you guys get creative. This is the part that I can't do for you. You just have to get creative with all the reactions we've learned and see if you can figure out the right sequence of reagents to make this final product. So go ahead and try your best and then I'll answer it for you. 

Concept: Example 1: Synthesize the target molecule

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Video Transcript

Alright so this was a three step reaction so if you got three steps you're on the right track the first step would be to turn this ketone into an ortho para director and we learned that we could do that through clemmensen reduction so my first regent would be a zinc mercury amalgam and what that's going to give me is in alkyl benzene I'm going to wind up getting specifically ethyl benzene. Now this is an ortho para director so since it's an ortho para director now I can use a nitration to put a nitro group in the pair position so I'm going to use H N O 3 and H S O 4 and that's going to give me my active electrophile which is going to attack the pair position predominantly and I'm going to get a nitro group. Why did I have to do a nitration next because now I have an ortho para director. Now do I have a one step way to turn this molecule into a benzoic acid? Yeah I do guys I can use K M N O 4 I can use another sequence group which is using K M N O 4 to oxidize the ethyl. Now remember that there's quite a few regents you have to write down so there's K M N O 4 which I mean some professors will be absolute fine with that but we have to be cautious. So the real the right way to write it would be K M N O 4 in the presence of base and heat and then I guess I said 3 steps it could technically be 4 but the way you could draw is you could also write like 3 A and 3 B since it's still the same reaction right so three B would be our acid H 0 + and what that's going to do is it's going to create a benzoic acid and that's our final product it does nothing to the nitro group so our final product ends up being this if we had to name it right it would be P nitro benzoic acid.

So guys I that was a good little introduction to sequence groups notice that there's no other order of regents that would've work I had to use those regents in specifically those borders because those are the specific sequence groups that I needed. Awesome guys so now you have another question go ahead and read it for yourself try your hardest and then I'll come in and I'll solve it for you.

Concept: Example 2: Synthesize the target molecule

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Video Transcript

Alright 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 pair positions this is para substitution. Even harder than that I'm going to have to put it in an annulene in the ortho position so ortho substitution. Now I've got a few issues here first of all is ortho substitution very favored? In most cases no so 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 pair 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 annulene because I want it to be synergistic with the ethyl group to face towards the top so we should be thing are there any groups that I can add that could be a meta director and then be turned into an ortho para director later, so that's one thing another thing is do we even know how to add annulene to benzene did I ever teach you how to add N H 2 specifically to a benzene an A E S reaction? No but did I teach you a precursor that can be easily turned into annulene? Yes guys remember reduction of nitro groups we can use nitration to make annulene 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 alright you might not but let me just step in and try to help and guys this is just you have to just start getting a feel for these. So the first direction 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 so I'm going to use I'm just going to erase this I'm going to use an acid chloride right. Now this part is important guys your acid chloride needs to contain the number of carbons that you want in your end product so since I'm adding it here how many carbon 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 carbon yield I'm going to combine that with A L C L 3 and I'm going to get a molecule that looks like this.

Now see so I got an aldehyde now notice that it attached to the pair position because we said that para is pretty much predominately favored especially when you have larger groups like the A silk groups cool. So 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 so now we would take my nitric acid and myself sulphuric acid and I would now make my nitrated my nitro group along with my aldehyde at the bottom, cool. Awesome guys so now what do you want to do? Well we have two more things we have to do, we have to do a clemmensen reduction and we need to do a nitro reduction. Now these you split slightly different regents so I'm just going to draw both of them and it doesn't really matter the order because both things need to happen and we don't have any more groups to direct so if you chose to do clemmensen first and then reduce the nitro that's fine or the other way around let's just do that so lets do clemmensen zinc mercury amalgam with H C L that's going to give me a single carbon with an N O 2 and then finally I would use you know any of my reducing agents that work on nitro groups to turn it into an annulene again I've told you a few times now I prefer stannous chloride S N C L 2 and H 2 O because that's the one that's chemoselective if there was anything else here that could be reduced it would only react with the nitro group and that's going to give me my final product. So this was a four step reaction and we got it we have figured it out. So you see how sequence groups are so important awesome guys so let's go ahead and move on.

Concept: Practice 1: Synthetic Road Map

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Hey guys, take a look at the following practice question. So here it says provide the product for each of the following reaction steps. So for the first we were going from benzene to ethylbenzene. Now here we're only adding a two carbon chain and here we're only adding only a 2 carbon chain, we don't have to worry about rearrangements so we can just simply do alkylation. So here we have our alkyl group connected to bromine over our Lewis acid catalyst. Remember this is going to leave so I can attach his 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 would that BR group add to? It added to the benzylic carbon, the carbon directly connected to benzene and there's two ways we could add that BR. One way is by using NBS, N-Bromosuccinimide, or we could just do B R 2 and UV light. Now next we're going from ethyl to this carboxylic acid. Remember if your benzylic carbon has at least one hydrogen on it, if I use a strong oxidiser the whole chain changes into a carboxylic acid so here we could use K M N O 4 over H plus. So that would give us that carboxylic acid. So this is side chain oxidation. Next I want to change that B R into an O H. Now here if I want to ensure that the B R gets substituted with an O H, what could I do? Well the easiest thing I could do is use N A O H but this is important. O H minus could do E 2 as well because this is a secondary alkyl halide. So how do I ensure S N 2 will happen instead of E 2? You decrease the temperature, make it lower than room temperature.

So here making the temperature cold enough will ensure that S N 2 will happen and not E 2. 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 two carbons but this here only has one carbon. So one thing I could do here is this I could do simply concentrated H 2 S O 4 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. So we get that structure there. Then what we could do lastly is this. You could do K M N O 4 and H plus. Now K M N O 4 is a strong oxidiser but there's also another feature that it has that you learned when you dealt with alkenes. Remember if you're alkene carbon has one hydrogen on it, K M N O 4 will cut through the double bond and that carbon, that alkene carbon with one hydrogen gets changed into a carboxylic acid. Here if you have 2 hydrogens on you, then this would change into C O 2 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 will we do that? Remember we want to do E 2 here so just use a strong base. So we use tert-butoxide. So that would change that alkyl halide chain into an alkene by E 2 because remember the carbon with the halogen is our alpha carbon and next to it is our beta carbon so what would happen is we'd rip off one of these beta hydrogens so it'd fall here to make a double bond and kick out the B R. Now finally these last two here we have C H and here we have C H 2.

What do we do? Well we still have an H there and a C H2 here but what did we add? We added an H and we added a B R. Now it's the opposite of what we're used to seeing hydrogen should have gone to the less substituted alkene carbon. Hydrogen should have gone to the double bond in carbon with more hydrogen but here the opposite happened so that must mean we did anti-Markovnikov. So how do we force anti-Markovnikov to occur? We use a peroxide. Peroxide force anti-Markovnikov addition. So we do H B R on top and peroxide on the bottom and then it would do the opposite of what we expect. B R will go to the double bond of carbon with more hydrogen, H will go to the one with less. Now here the same thing happens with this alcohol. What happens is O H went to the double bond of carbon with more hydrogens for some reason instead of the hydrogen so this is also anti-Markovnikov. So how do I change an alkene into an alcohol by anti-Markovnikov? We do it by hydroboration. Remember these are all terms that are still important now so we do hydroboration-oxidation. So we do borane, B H 3 with T H F which is just our ether solvent, tetrahydrofuran over H 2 O 2 and O H minus. This adds a hydrogen and an O H to an alkene to make an alcohol but look we have a peroxide. Peroxide can be written as this or is this because the peroxide is there with forces anti-Markovnikov. So these last two are a little bit tricky because they entail 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. We're just talking about what kind of groups can I add and how do I add them? Do I add the by Markovnikov's rule, or anti-Markovnikov's rule? That's the key to this question.

Concept: Practice 2: Aromatic Synthesis

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Hey guys, let's take a look at the following question. So here it says beginning from benzene synthesise the following compound. Now if we're trying to synthesise 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 two 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. So what we do first is we're going to do acylation. So this halogen leaves to attach to 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 it wants to add 1,3. So the B R I 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've got our two 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 two different ways that we know of to change a carbonyl into a C H 2 group? The two ways are we could do Clemmensen reduction which would be this or we could do Wolff-Kishner and Wolff-Kishner here is through basic means. Both of them would change that carbonyl into a C H 2 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 but 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.

Concept: Practice 3: Aromatic Synthesis

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Hey guys, let's take a look at the following practice question. So here it says beginning from benzene synthesise the following compound. So here we need to synthesise one phenol ethanol. Now here ethanol is a two carbon chain. Ethanol means that we have an O H group. Phenol would mean that on carbon number one we have attached a benzene. Now where the O H is is carbon number one. So this is carbon number two.

So one phenol ethanol will look like this. So starting from benzene we need to create this structure here. Now 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. So all we're going to do here is add an acyl group first to benzene. The acyl group is going to be two carbons long because this chain is two carbons long. C L leaves, so it can attach to the aluminium. So there we have a carbon directly connected to oxygen that's connected to benzene and 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 N A B H 4 which is sodium borohydride, this one's a week reducing agent. It only reduces aldehydes and ketones and then we have the stronger L I A L H 4, which shorthand is L A H 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. You have a carbon connected to benzene, that carbon is connected directly to an oxygen. What's similar to that? Similar to that would be acylation. Once you get that group there, change that group into an alcohol. That's the key to this question. Looking at similarities.

Concept: Practice 4: Aromatic Synthesis

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Video Transcript

Hey guys, let's take a look at the following practice question. So here it says beginning from benzene synthesise the following compound. So 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? Well here what we could do is C H 3 connected to a halogen over the Lewis acid catalyst. So we're doing here is we're doing alkylation. The B R 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 O H. Now we don't know how to replace one of the hydrogens with an O H directly but we do know how to replace one of the benzylic hydrogens with a halogen, a B R. If we use N B S or B R 2 over H V, that will replace one of the benzylic hydrogens with a bromine so we get this structure here.

Now here we have an O H and it's a primary alkyl halide and remember primary alkyl halides love to do S N 2. So we have O H minus since it's primary it'll do S N 2 where this shoots out, hits this and kicks out the B R and in that way we'll get our answer. So this is one possible way of doing it. So this is option one. Here if you think about it there's a second way we could do this. If we can do alkylation then we can also do acylation. Here we have this structure here so a chlorine would leave and attach to the aluminum. So we just added an aldehyde to benzene and here we can reduce an aldehyde into that alcohol. How? By using lithium aluminum hydride or by using sodium borohydride. Either one will reduce that aldehyde into a primary alcohol. So here option one and option two both give us the same exact answer. Since option two requires less steps we'd say that option two would be more effective, more efficient. So option two would be the better choice but option one would still be a correct answer in itself too. So the thing about synthesis is that it's not always just one way to get to a possible answer as long as you can reason your way through it and prove why this way works, then that's a viable option.

Now for those of you who have to do fill in, you have basically a blank canvas to figure out a way to get to the answer. For those of you doing multiple choice exams, your answers are more rigid so you just have to check to see are the steps they're are giving you do they make sense do they follow any type of rule any type of limitation? That way you can determine what your final answer will be.

EAS: Retrosynthesis Additional Practice Problems

Propose a synthesis of compound B starting from benzene and acetyl chloride (CH3COCl ) as the only sources of carbon. Several steps may be required. Assume that all reagents needed for the synthesis are available. Indicate clearly the reagents used and the products obtained after each step. Mechanisms are not required.

 

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For the reaction below, draw the structure of the appropriate compound in the box. Indicate stereochemistry where it is pertinent. For Friedel Crafts reactions, assume that the major product is the least sterically hindered one.

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Using any necessary reagents, show how you would accomplish the following transformations.

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Perform a retrosynthetic analysis on the molecules below (work backwards) from the given commercially available starting materials.

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Provide the reagents to accomplish the transformation below. More than one step might be required for the transformation. 

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Which is the best method for carrying out the following reaction?

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Provide stepwise synthesis for the following transformation:

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Propose the sythetic plan complete with reagents and proper order of reactions. You must show the product of each step in your scheme, however no detailed mechanism is required. 

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Circle the stating material for the following synthetic sequence.

 

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Which sequence of reactions would give (some of) the correct product? 

1) A and B

2) A and C

3) B and C

4) B only

5) A, B, and C 

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Which sets of reagents would best be used to perform the following reaction?

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Provide a scheme for the synthesis of the molecule below, using the indicated starting material and reagents of your choice.

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Propose a multi-step synthetic scheme complete with reagents and proper order of reactions (no mechanisms required) for the following molecule starting from the indicated compounds. You may use any reagents neccessary. Show the product of each step in your scheme.

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The following synthetic pathway does NOT sufficiently produce the indicated product. Provide an acceptable synthetic pathway. 

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Predict the major product for each of the following reactions paying attention to the region- and stereochemistry. If there is no reaction, write just “No Reaction.”

 

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Using any necessary reagents, show how you would accomplish the following transformations.

 

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Which set of reagents will best accomplish this transformation?

 

 

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