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Ortho-para directors I

Regiochemistry. Created by Jay.

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  • piceratops seed style avatar for user Salman Khan
    Why is para considered the major product while ortho the minor product?
    (5 votes)
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    • piceratops ultimate style avatar for user Bailan
      Products should always be more stable than the reactants. We determine the stability here by counting the number of resonance structures. The more the resonance structures, the more easily we can delocalise the charge within a benzene ring.
      Ortho and Para have 4 resonance structures while meta has only 3 resonance structures. This means we can delocalise charge easily in ortho and para which also means that these two are more stable comparing to meta positions.
      So now the competition is between the ortho and para.
      We know that 2 large groups will have to face steric hindrance while coming into proximity. Since the para position is opposite to the substituent of the benzene, this arrangement will be more stable and will be easily formed.
      Ortho products will have to overcome steric hindrance to form products. This is the reason why we see ortho as the minor product.
      While para ones which have only minor resistance from spatial arrangement form the product easily and the para products are the major products.
      (9 votes)
  • spunky sam blue style avatar for user Chunmun
    What are activators and de activators in the ortho para directors .
    What does it mean ?
    (4 votes)
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  • blobby green style avatar for user tomm0_
    In the meta resonant structures, why can't the oxygen form a double bond like the ortho and para resonant structures?
    (4 votes)
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    • orange juice squid orange style avatar for user awemond
      In all three resonance structures that Sal has drawn at in the meta example, the benzene carbon that is attached to the oxygen already has four bonds (one single bond to the oxygen, one single bond to another carbon in the ring, and a double bond to a second carbon in the ring). That is, that carbon already has a full octet so can't accept any more electrons to form a double bond with oxygen. (Note that it is possible to draw a resonance structure with the meta example that has a double bond between the carbon and oxygen, but this would involve three atoms with a formal charge and thus is not a good resonance structure (ie it's very unstable) so would not make a significant contribution to the resonance hybrid and thus is not considered).
      (3 votes)
  • aqualine ultimate style avatar for user Moises Jerez Schachtler
    How would you do this for seven-membered rings?
    (4 votes)
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  • aqualine ultimate style avatar for user Morgan Gardner
    At , Jay says that the lone pairs for Oxygen can form a pi bond between Oxygen and Carbon. Why can't they form pi electrons inside the Benzene ring?
    (2 votes)
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    • orange juice squid orange style avatar for user awemond
      If the electrons in the C=O pi bond (from the oxygen lone pair) were pushed into the ring, that would leave oxygen with a +2 charge and an incomplete octet. This would be very unstable so does not happen.
      (6 votes)
  • blobby green style avatar for user Felicitas Tengen
    For a Phenol molecule there are 4 different resonance structures possible, and due to the positive mesomeric effect 3 of these resonance structures have a negative charge on the ortho and para carbons but none on the meta carbon. Do these negative charges make it more likely for an electrophile to attack at the ortho and para position ?
    (3 votes)
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  • blobby green style avatar for user saramohamed.hegazy
    why ortho and para are activating group while meta is deactivating group
    (3 votes)
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  • leafers tree style avatar for user Ben Lee
    Can ortho and para substitutions occur on a same benzene ring, or do they always occur separately in different molecules?
    (3 votes)
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    • female robot grace style avatar for user tyersome
      If the added substituent is activating, then you could easily get multiple substitutions (in fact this is one problem with doing Friedel-Crafts alkylations). However, where the additional groups are added will depend on interactions among the activating groups and steric effects.
      (0 votes)
  • aqualine seed style avatar for user Bableen Kaur
    What if rather than the methoxy group there was an isopropyl group? Would the major product still be ortho? or would it be para? Would you consider isopropyl a sterically hindered group?
    (1 vote)
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    • leaf red style avatar for user Richard
      So alkyl groups (which includes isopropyl) are known as activating ortho, para-directing substituents. Activating meaning they cause substitution reactions much faster than similar substitution reactions with ordinary unsubstituted benzene. And ortho, para-directing because they direct the new substituent to the ortho or para positions relative to the substituent obviously.

      So if you did an electrophilic substitution on isopropylbenzene (or cumene), then the ortho and the para products would be considered the major products, while the meta product (which is still produced) would be considered the minor product. Now as far as the percentages go for each product, this can vary depending on the type of substitution (and even then there is some inconsistency among whether ortho- or para- are produced in larger amounts). But in general I would say that para- products are produced in greater quantity over the ortho- products becuase of the steric hindrance of the isopropyl group like you mentioned. To get an idea of the percentages, I've seen one reaction using an alkyl substituent resulting in 38% ortho, <1% meta, and 62% para products.

      Hope that helps.
      (2 votes)
  • blobby green style avatar for user Nirvana Nivi Harirajh
    what are the most common types of activators and deactivators
    (1 vote)
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Video transcript

So far we've looked at electrophilic aromatic substitution reactions involving only benzene. But what happens when you start with a substituent already on your benzene ring? So we'll look at this molecule over here on the left, which is methoxybenzene. So now we have a methoxy substituent on our benzene ring. And if we react methoxybenzene with concentrated nitric and sulphuric acids, we should recognize that as being a nitration reaction, which will install a nitro group onto your aromatic ring. The presence of that methoxy substituent is going to affect where that nitro group goes on your ring. So, for example, one of the observed products is to put the nitro group on this carbon, which is the carbon right next to the carbon that has our methoxy substituent. We say that these two groups are ortho to each other. So this would be the ortho product. So if I go back over here on this diagram I can label this carbon as being the ortho position on my benzene ring. The other product that we observe is a nitro group installed on this carbon, which is on the opposite side from the carbon containing our methoxy substituent. We call this the para product. So let me go ahead and label the para position on our benzene ring over here. So this would be the para position. Between the ortho and the para product, it turns out that the para product is the observed major product in the reaction of the nitration of methoxybenzene. And the ortho product is the minor product. And this has to do with mostly thinking about steric hindrance. And this methoxy group is having some steric hindrance, which obviously would prevent this nitro group from adding onto the ortho position easily. Obviously the para position will have much less steric hindrance. And that's the reason that you usually see for the para product being the major product for this reaction. Now this isn't always the case. Sometimes your ortho product is more than your para. But for this reaction the para product is the major product. And once again steric hindrance is one factor to think about when you're doing these kinds of reactions here. So there's of course another position on your aromatic ring. So if we installed the nitro group on this position, we would call this the meta product. And the meta product is not observed in high yield for this reaction. So we say that the methoxy group is an ortho para director. And we could also label this as being the ortho position on this side. And we could say this is the meta position because of symmetry. But the meta product is not seen in a large yield, in a high yield. And let's go ahead and look at why by drawing a bunch of resonance structures and thinking about the mechanism for electrophilic aromatic substitution. And so let's go ahead and start with an ortho attack. And so we know that when you're doing a nitration the sulfuric acid acts as a catalyst to generate the nitronium ion from nitric acid. And that functions as the electrophile in your mechanism here. So if we're going to do an ortho attack, we need to show the nitro group adding onto the ortho position. So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here. I'm showing the nitro group adding onto the ortho position. And remember there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. So that carbon gets a +1 formal charge. We can show some resonance structures. So we can show some resonance stabilization of this cation here. So I could show these pi electrons moving over to here. And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. So we get a +1 formal charge on this carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro group in the ortho position. And we have these pi electrons here. And now we show the movement of those pi electrons over to here. So let me go ahead and highlight those. These electrons in red move down to here. I took a bond away from this carbon. So that carbon is the one that gets a plus 1 formal charge now. Since the oxygen is right next to this carbon-- the oxygen has a lone pair of electrons. And so that lone pair of electrons can give us yet another resonance structure. So these electrons could move into here to draw a fourth resonance structure. So the presence of that methoxy substituent with the lone pair of electrons on that oxygen allows you to draw a fourth resonance structure. So this will give this oxygen a +1 formal charge. We have these pi electrons over here. We have our nitro group, once again, in the ortho position. And let me just go ahead and show the movement of those electrons. So these electrons-- I'll make them green-- these electrons right here are going to move in to form our pi bond like that. And we have a total of four possible resonance structures. Remember the ion is actually a hybrid of these four resonance structures, which we call our sigma complex. And so, again, the presence of this methoxy substituent with the lone pair of electrons right next to our aromatic ring gives us an extra resonance structure. This one is our extra resonance structure. And so we have a total of four resonance structures for an ortho attack. And the more resonance structures you can draw, the more that positive charge is delocalized. And the more stable your sigma complex is, the more likely it is to form in your mechanism for electrophilic aromatic substitution. And so because we can draw four resonance structures for an ortho attack, that is a favored carbocation. That's a stable sigma complex. And that's going to form much more easily. And, of course, the last step in your mechanism is to deprotonate your sigma complex to reform your aromatic ring. But I'm not going to show that step here. I just wanted you to see the four possible resonance structures for an ortho attack. That's different from a meta attack. So let's go ahead and look at what would happen if we added on our nitro group meta. So the meta position would be, of course, this one right here. So we would use these pi electrons, so nucleophilic attack. And it kicks these electrons off onto our oxygen. So let's go ahead and show the result of our nucleophilic attack and adding on our nitro group in the meta position. So if we're going to show the nitro group in the meta position like that-- let's highlight these electrons here-- so these pi electrons come off onto and form a bond, I should say, with that nitrogen, taking a bond away from that top carbon. So that top carbon is going to get a plus 1 formal charge. And, of course, we have our other pi electrons in our ring like that. So resonance structure-- I could draw a resonance structure for this ion here. I could take these pi electrons and move them over to here. So we'll go ahead and show another resonance structure. So I have those pi electrons moving over to there. I have these pi electrons here. I still have my nitro group in the meta position to my original methoxy substituent like that. And now my positive 1 formal charge is on this carbon right here. So to save time I'm not going to color coordinate these resonance structures. I can draw one more, right. I could show these pi electrons moving over to here. And let's go ahead and draw that resonance structure. So I have those pi electrons move over to here. I have these pi electrons. I have my methoxy substituent. I have, once again, my nitro group. And, of course, now my positive charge moves down to this carbon down here. And that's it. I can only draw three resonance structures. So, once again, the actual sigma complex is a hybrid of these three resonance structures. And since I can only draw three, only a total of three resonance structures for this situation, this sigma complex is not as stable as the one that we saw for an ortho attack. So the sigma complex for an ortho attack was more stable because we could draw four resonance structures. And here we can only draw three. Let's go ahead and show a para attack. So if I want to show my nitro group adding on to the para position. So here is the para position. So I'm going to use these pi electrons, right. So nucleophilic attack kicks these electrons off. So this would be our para attack, adding on our nitro group in the para position. So, once again, I have my methoxy group. So this time I'm going to show my nitro group in the para position. So let's go ahead and follow these pi electrons. So these pi electrons in here. Those pi electrons formed a bond with that nitrogen. And I took a bond away from this carbon. So that is where my plus 1 formal charge is going to be. I still have these pi electrons in my ring like that. So a resonance structure for this one, I could show these electrons in here moving over to there. And let's see what we would get. We would still have, once again, this substituent. Our pi electrons moved over to here. These pi electrons were here. I had my nitro group para. And now I took a bond away from this carbon. So that is the carbon that gets the plus 1 formal charge. So I can draw another resonance structure. I could show these electrons in here moving to there. So let's get a little more room here for more resonance structures. So I can show my ring and I can show these pi electrons have moved over to here. I still have my substituent. And, of course, my nitro group is still in the para position. I took a bond away from this carbon. So I get a plus 1 formal charge. And once again the presence of that lone pair of electrons on that oxygen right next to my benzene ring allows me to draw another resonance structure. So I could think about these electrons on my oxygen moving into here to form a pi bond and pushing these electrons down to here. And so I can draw a fourth resonance structure showing the oxygen now double bonded to my ring. So there is still a lone pair of electrons on that oxygen, which would give it a plus 1 formal charge. I still have pi electrons here. I showed pi electrons moving over to here. And, once again, my nitro group is still in the para position. And so I have a total of four resonance structures if I show the nitro group adding on in a para fashion as well. So the para attack also has four resonance structures. And the ortho had four resonance structures. And so that helps to explain why the methoxy substituent functions as an ortho para director. If you show an ortho or a para attack, you can draw a total of four resonance structures, which stabilizes the sigma complex more than a meta attack. And that's the reason for the regiochemistry that we see in the nitration of methoxy benzene. Now I drew these resonance structures because I started with my benzene ring. I started with my electrons in this position right here. Now if you started with a different benzene ring-- so a resonant structure of that-- then your resonance structures might look a little bit different from mine. And you'll see different versions in different textbooks. So make sure to, once again, always look at what your professor does in class or your textbook. And think about that on exams if they have you draw resonance structures for an ortho, para, or a meta attack.