Synthesis of substituted benzene rings I
Synthesis practice problems. Created by Jay.
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- At4:40, I'm confused on how to determine what step is last? Why is nitration last? Can you please go into more detail about determining the last step? Thanks!(5 votes)
- because Br is an o.p. director and (NO2) as well as (C2H3O) happen to be at the o.p. positions they can be added precisely at those positions if Br (bromination) is the first step.
When all this is true you can think about adding the (NO2) OR (C2H3O) after you have added Br to your benzene. Your last reaction has to be a nitration because an acyl can not be added when there is a moderately strong de activator a.k.a. the (NO2) with the partial positive moment. Someone correct me if I'm wrong. :)
3. nitration(5 votes)
- At6:30, Jay says that "Since this is a weakly deactivating group, you can still do this (acylation)." But from my understanding and my textbook, Friedel-Crafts Alkylation and Acylation do not occur in presence of a vinylic halide or aryl halide, which is what we have here in the video... Is my textbook wrong?(1 vote)
- I think your confusing substrates with substituents. Jay is correct and so is your textbook.
Friedel–Crafts reactions usually give poor yields when powerful electron-with- drawing groups are present on the aromatic ring or when the ring bears an -NH2, -NHR, or -NR2 group. This refers to substitutes on the starting material and in this case benzene with the Br attached is acceptable because Br is not a powerful electron withdraw group.
Now, for substrates if they contain vinylic halides and aryl halides. Jay used a Ch3C(=O)Cl for the substrate. If he would have used a benzene with a Cl attached instead, then this would have prevented the FC reaction from occurring. Reason is they don't for carbcations readily.(5 votes)
- @2:40Wouldn't adding the "moderate to strongly deactivating nitro group" not allow bromination to occur because the molecule is so deactivated?(3 votes)
- Br+ is a very good electrophile, thus bromination of yr nitrobenzene is possible. Carbocation derived from acetyl chloride is not as good an electrophile because the positive charge on C is reasonable stabilised.(1 vote)
- Wouldn't adding the Nitro group last have a better yield than adding the Br last? They're both deactivating but isn't NO2 more deactivating than Br?(1 vote)
- Yes, NO₂ is more deactivating than Br, but you can compensate for this by raising the temperature.
There are many factors that affect yield.
For example, bromination of nitrobenzene gives an 80% yield of m-bromonitrobenzene.
Nitration of bromobenzene gives a 50% yield of p-bromonitrobenzene.(3 votes)
- At1:00how do u know bromine was added last? i dont get why it is becasue Br is meta to both of the other two substituents.(2 votes)
- Why only benzylic hydrogen is reactive in ethyl benzene,but not other carbon of alkyl group.please show mechenism.(1 vote)
- what about if we want the almost same products but with a 2,2-dimethyl-1,3-cyclopentanedione instead of the Ketone that came from the acylchloride ?(1 vote)
- Reactivity of poly substituted benzene(1 vote)
- at4:40how do we know nitration is to be done last i did not get it(1 vote)
- I didn't get how he finalized the order of the steps.How he selected which has to happen first and which last.
Can anyone give me a hand?
Now that we know all of our reactions, let's see if we can put those reactions together to synthesize some simple organic compounds. And so our goal is to make this molecule from benzene. And one approach that you can use is the concept of retrosynthesis. So you try to think backwards, and you think to yourself, what can be an immediate precursor to this molecule? Well, to do that, we have to analyze the groups that are attached to our ring. And so if I look at this bromine up here, I know this bromine is an ortho/para director, because I know it has lone pairs of electrons around it. So this bromine is an ortho/para director. But look at the nitro group. I know this is a meta director. I know it's meta because there's a plus 1 formal charge on that nitrogen. And then, over here, for this acyl group attached to our ring, I know this is also a meta director, because this carbonyl carbon right here, is partially positive, like that. And so when we try to figure out which of these groups was added last, it makes sense that the bromine was added last because this bromine right here is meta to both our nitro group and our acyl group. And so we can go ahead and draw the precursor. We go ahead and just take the bromine off. So we're left with a benzene ring. And we have an acyl group on our ring, and we also have a nitro group. So next we just have to remember how to put a bromine on a benzene ring, and of course it's a bromination reaction. So you would need some bromine and a catalyst, so something like iron bromide. So FeBr3 will work for that. All right, let's see if we can figure out the next precursor here. So what could we do to make this molecule? Well, once again, we have two groups on here. We have a nitro group, and we have an acyl group. So we could do a nitration to put the nitro group on, and we could do a Friedel-Crafts acylation to put this acyl group on our ring. So the question is which one of these comes first? And it turns out that you can't really do a Friedel-Crafts alkylation or acylation with a moderate or strongly deactivating group already on your ring. And this nitro group here is strongly deactivating, which means we can't put the nitro group on first and then add our acyl group. The acyl group must come on before the nitro group, which means in this step, we're going to put on the nitro group. So the immediate precursor to this molecule-- we just take off our nitro group, and we're left with our benzene ring and an acyl group attached to our benzene ring like that. And so we need to do a nitration, which requires, of course, concentrated nitric acid and also concentrated sulfuric acid like that. All right, now all we have to do is go from benzene to this molecule. And we know how to do that, of course. That's a Friedel-Crafts acylation reaction. And so when we think about what kind of acyl chloride we're going to use, just count the number of carbons here, so 1, and then 2. So we need a 2 carbon acyl chloride. So we're going to draw here a 2 carbon acyl chloride like that. And then we need a catalyst. Something like aluminum chloride will work for our catalyst. So our synthesis is complete. We start with a Friedel-Crafts acylation. And the acyl group is a meta director, which would direct the nitro group to the meta position. And then, finally, we have two meta directors, which we now brominate, which would direct the bromine to the final position. And we are complete. Let's do another problem here. And, actually, it's the exact same groups that we just saw in the previous problem, but this target molecule looks a little bit different. And so we're going to need to do the reactions that we did in the previous synthesis in a different order here. So once again, let's start by analyzing the groups. So once again, we know that this bromine is an ortho/para director because of the lone pairs of electrons on it. We know the nitro group is a meta director because of the plus 1 formal charge. And our acyl group is a meta director because of the partial positive charge on our carbonyl carbon, right here. So when we look at those groups, and we think about which of those reactions was done last, it makes sense that this nitration was done last. And that's because this nitro group is meta to our acyl group, because our acyl group is a meta director, and our bromine, more importantly, is an ortho/para director. And of course the nitro group is ortho to the bromine. And so it makes sense the last reaction was a nitration reaction. So, to draw the precursor to this, all we do is take off that nitro group, and we would have our benzene ring, like this. And we would have a bromine on our ring, and would already have our acyl group on our ring, like that. So our last reaction was a nitration reaction. So we need to add, once again, concentrated nitric acid and concentrated sulfuric acid for our nitration. So when we think about the precursor to this molecule-- so once again, we have an ortho/para director on our ring, and we have a meta director on our ring. And we have our groups. Our bromine and our acyl group are para to each other, which means that the ortho/para director directed the acyl group to the para position as the major product. And so we can think about doing a Friedel-Crafts acylation reaction here. So that means that we're taking off the acyl group. So we're left with bromobenzene to start with over here, like that. So we have bromobenzene, and we're doing a Friedel-Crafts acylation. And, once again, we need 2 carbons on our acyl group. So let me just point that out, 1 and 2. So we need a 2 carbon acyl chloride. So go ahead and put on a 2 carbon acyl chloride, like that. Once again, our catalyst, something like aluminum chloride, will work. And you might think to yourself that I know that the halogen, the bromine, is deactivating. So how can we do a Friedel-Crafts acylation with a deactivating group on there, even though it's an ortho/para director? Well, remember, it's only weakly deactivating. And so you can't do an alkylation or acylation with a moderate or strongly deactivating group. And so it turns out, since this is weakly deactivating, you can still do this, and you'll get the para product as your major product over here. So I'm sure you'd get a little bit of ortho as well. All right, so now all we have to do is go from benzene to bromobenzene And, of course, that's really simple. It's just a bromination reaction again. So we have our bromine, and then we have our catalyst, and then our synthesis is complete. So for this time, we start out with a bromination reaction to form bromobenzene. The bromine is an ortho/para director. This is an ortho/para director. And so it's going to put to this acyl group on our ring in the para position as our major product, here. And then, of course, we nitrate it, and we have an ortho/para director and a meta director, which means the nitro group will end up in this position. And we're done. So that's how to think about the synthesis problem, so retro synthesis, working backwards, thinking about target molecules. And we'll do two more in the next video, which are maybe a little bit harder than these two.