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Ortho-para directors III

Moderate and weak activation, weak deactivation. Created by Jay.

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  • leaf green style avatar for user beste_sevil
    At , Jay says that the halogen can give its pair of electrons to form a pi bond but if that halogen is so electronegative (as Jay says previously) then why does it give them anyway? Is it because electrons are highly active and doesn't stay where they are, so they are forming the pi bond even for a slightest moment?
    (5 votes)
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  • leaf green style avatar for user khjhzw
    From to , Jay says that Chlorine and Nitrogen have approximately the same electronegativity. Is there an authoritative (or even reliable) electronegativity chart (or period table showing electronegativity) that is worth memorizing? When I googled for it, different charts have different electronegativity values for the same element.

    Also, how is the electronegativity value calculated in the first place?
    (1 vote)
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    • spunky sam blue style avatar for user Ernest Zinck
      I don’t think any electronegativity chart is worth memorizing, but you should memorize trends in electronegativity values.
      There are different charts because different people created their own scales.
      The one most commonly used is the Pauling electronegativity scale.
      He developed it using differences in bond energies and then scaling the numbers so H had a value of 2.2 and F a value of 4.0.
      There have been some minor adjustments since then.
      (5 votes)
  • leaf grey style avatar for user Harrison Wynn
    At , Jay says induction must be more important than resonance. Is induction in general a more important factor than resonance when determining the affect of a substituent on electron density in a benzene ring? Or does he mean that in that induction is stronger than resonance in this specific example?
    (1 vote)
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    • orange juice squid orange style avatar for user awemond
      It's difficult to always predict whether induction or resonance will be the more important factor. In this specific case induction is the more important effect, but in other cases resonance may be more important. Often, we look at experimental results and induce which factor must have been more important, induction or resonance, based on the results (so it's an after-the-fact explanation, rather than a prediction).
      (4 votes)
  • leafers seedling style avatar for user abhed manocha
    but resonance is more important factor than inductive effect. So here, how is it that inductive effect is given more priority?
    (2 votes)
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  • marcimus pink style avatar for user Ci Qian W
    , R-groups( in this case -CH3) are electron-donating, isn't the electron density being donated to the ring through the bond between the ring and the methyl group?
    (2 votes)
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    • aqualine tree style avatar for user Ali Hussain Al-Muslem

      When the benzene ring attacked some of electrophiles [ ether ortho or para position ] the carbon where methyl group is attached will be in it a positive charge [ due to resonance ].

      When we arrived to this situation , both methyl group and benzene will share with each other one electron [ from any hydrogen that attached to methyl group ], or to be more specific, they will share the positive charge.

      This process make alkyl groups as activator groups.

      Good Luck.
      (1 vote)
  • blobby green style avatar for user irtza_rehman
    may we not consider the halogen as a weak activator because it donates an electron to our benzene ring then why we are calling it de-activator??
    (1 vote)
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    • leaf green style avatar for user DSet
      Being particularly electronegative, halogens pull electron density away from the carbon they are attached to, which overall lowers the reactivity of the benzene ring. However, due to the minor contributing (as the halogens prefer not to give up electrons and the benzene ring prefers not to break its aromaticity) resonance structure drawn in the video, there is a semblance of additional electron density available at adjacent carbons due to the presence of the halogen, leading to ortho-para direction. Remember that resonance structures are like snapshots of potential states for a molecule, and not separate states that molecules exist in: they actually exist as a tapestry of all of the resonance structures, with some states more preferable (stable) being represented more of the time.
      (2 votes)
  • blobby green style avatar for user Sw1fty
    there was a question that said:
    consider the three isotopes of dimethylbenzene
    if it is chlorinated with one chlorine atom
    it now forms only one isotope
    what isotope is it
    (is the answer 3-chloromethylbenzene because that is the only isotope for activating groups, but methyl is a deactivating group, and if you consider that you could have para or ortho chloromethylbenzene)
    im so confused
    (1 vote)
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    • female robot grace style avatar for user tyersome
      Isotopes refer to atoms of the same element that have different masses due to differing numbers of neutrons, so I'm assuming you actually meant isomers.

      I think your question is asking something like this:
      1) You started with one of the three possible isomers of dimethylbenzene.
      2) You chlorinated this compound in such as way as to get only a single chlorine added.
      3) You then determined that this new compound had only a single isomer.
      4) What is that isomer.

      To answer this question start by drawing out the three possible isomers of dimethylbenzene.
      Then draw the result of adding a single chlorine to each possible place on each of those isomers. For one of them you will then see that all products are actually the same structure. This is the one you want!

      Leave a comment if you want more help :-)
      (1 vote)
  • spunky sam blue style avatar for user Artistique
    By application of the concept presented in this video, I reasoned that inductive effect prevails in +:S(CH3CH2)2 when connected to benzene ring. While the lone pair on S can participate in resonance with the ring, S comes from a different period than C does and so, the overlap between 2pC and 3pS is poor (or at least weaker than the overlap between 2pC and 2pO, 2pC and 2pN, and 2pC and 2p F. Thus, resonance effect in +:S(Et)2 isn't that effective. A reason why the inductive effect might prevail is the difference in electronegativity between C and S, but this difference is negligible (2.55 vs. 2.58). So, which effect would prevail then? What do you think?
    (1 vote)
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  • male robot johnny style avatar for user rajat497kharbanda
    in last line he said halo re op directors ?/////
    (1 vote)
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    • blobby green style avatar for user jrlim6
      He mentioned that: -
      (i) Due to the large inductive effect (due to the electronegativity of halogens), the electrons will be absorbed by the halogens (electron- withdrawing group)
      (ii) However, since halogens have lone pair electrons which can be contributed to the ring via the overlapped- orbital, it can contribute more to the resonance effect.
      (iii) Due to the strong inductive effect, when compared to the resonance effect, halogen benzene tends to be weak deactivator, they are, however, ortho para directing group due to the contributing resonance hybrids.

      Consider the following points:-
      (i) Going down the group (G 17), the electronegativity decreases.
      (ii) But, due to the different size of the orbitals, (the orbitals of halogen get bigger going down the group)
      (iii) Therefore, degree of overlapping will decrease as well.
      (iv) This, in return, prevent the resonance effect > inductive effect
      (v) So, halogens remain as weak deactivator group, although their electronegativity decreases going down the group.
      (1 vote)
  • blobby green style avatar for user marocann2
    I can't Understand why NHCOR is an activating group although the oxygen withdraws the electrons leaving a positive charge on the N atom which should make it an electron withdrawer in return
    (1 vote)
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Video transcript

In the previous video, we saw how induction and resonance affect ortho-para directors that are strong activators. This video is really just a continuation of the previous video so make sure you've seen that one first. In that video, we saw that when you have an atom with a lone pair of electrons that's right next to the carbon on your benzene, ring that lone pair of electrons can participate in resonance and it can be conjugated into the pi system of your ring. That increases the electron density of your ring, which activates the ring towards electrophilic aromatic substitution since the ring can be a better nucleophile, and the positively-charged sigma complex is better stabilized by the increased electron density. And so that's where the activated portion comes in for this molecule so that lone pair of electrons can activate the ring towards electrophilic aromatic substitution. However, it's only a moderate activator and not a strong one like we saw on the previous video. And the explanation for that observation comes from the fact that this lone pair of electrons can participate in resonance outside of the ring. So if I go ahead and draw a resonance structure here. I have my benzene ring. I now have my nitrogen double bonded to this carbon. That carbon is bonded to an R group. The nitrogen is bonded to a hydrogen, and now my oxygen has three lone pairs of electrons around it giving it a negative 1 formal charge. The nitrogen now is a plus 1 formal charge. So the electrons in magenta moved in here to form a pi bond, and because those electrons in magenta are tied up in resonance outside of the ring, they can't donate as much electron density to the ring, and so that's why it's only a moderate activator instead of a strong one. Let's look at an example of a weak activator so something like toluene here, so a methyl group on a benzene ring. All of the carbons on our benzene ring are SP2 hybridized so all of those carbons have free P orbitals. Let me go ahead and sketch a free P orbital on all of my carbons. So I have my six pi electron system here in my benzene ring, and so those electrons are delocalized in my pi system like that. I also have a methyl group, right? So let's go ahead and sketch in the methyl group here. So I have a carbon bonded to three hydrogens, and it turns out one of those carbon hydrogen sigma bonds can actually interact with the pi system of our ring. So we could think about this sigma bond right here interacting with the pi system, and therefore, increasing or donating some electron density to that ring. And since that sigma bond can interact weakly and donate some electron density to that ring, we can think about the methyl group as being an activator activating the ring towards electrophilic aromatic substitution. So you could call this interaction sigma conjugation, you could call it hyperconjugation, whatever you want to call it, but there is a little bit of electron donation to the pi system, therefore, making it an activator. However, this sigma conjugation is nowhere near as strong as the full conjugation that we saw in the previous video with the oxygen and nitrogen. And so since sigma conjugation is not as strong of an effect, that's why this is only a weak activator. Let's look at one more example of an ortho-para director, and this one is the exception. This is an ortho-para director that is a deactivator here. So a halogen on a benzene ring will still direct substituents ortho and para to it because of the resonance structures that you can draw. However, it turns out to be a deactivator most the time. The reactions are slower than that of benzene on its own. Let's first examine the inductive effect that the halogen has on our benzene ring. So let me go ahead and write induction here. If we think about the sigma bond between the carbon on the ring and the halogen-- so this is our generic halogen here with an X-- halogens a are relatively electronegative, more electronegative than carbon so they're going to withdraw some electron density from that ring. Whenever you withdraw electron density from a benzene ring, you deactivate it towards electrophilic aromatic substitution. So the inductive effect would lead you to think that a halogen is a deactivator. However, the halogen does have lone pairs of electrons on it so you could also think about the resonance effect. So let me just go ahead and write the resonance effect here. And if I took one of these lone pairs of electrons and moved them in here to form a pi bond between that carbon and that halogen, right? That would push these electrons off onto this carbon so we could go ahead and draw our resonance structure here. So we would have our halogen now double bonded to our carbon. The halogen has two lone pairs of electrons, a plus 1 formal charge, and we now have a lone pair of electrons out on this carbon with a negative 1 formal charge like that. So let me just go ahead and highlight one of these lone pairs of electrons can participate in resonance, and that would increase the electron density of your ring. And again, I'm not going to draw the other resonance structures, but increasing the electron density of your ring implies activation towards electrophilic aromatic substitution. So we have these two competing factors here. Induction says that the halogen is electron withdrawing and therefore a deactivator. Resonance suggests that the halogen might be electron donating, and therefore an activator. Experimental results show that these halogens are actually deactivators here. So induction must be a more important effect than resonance. Let's see if we can think about why using different halogens. So let's start with fluorine here. So if your halogen is fluorine. Fluorine is, of course, extremely electronegative, and so you could think about what the inductive effect if the halogen is fluorine, it's withdrawing a lot of electron density from that benzene ring, deactivating the ring towards electrophilic aromatic substitution and so the inductive effect is very strong when fluorine is on your benzene ring. When you think about resonance, so the fluorine forming a pi bond with a carbon on your ring. So let me show this carbon right here forming a pi bond with a halogen. I'm saying the halogen is fluorine this time so I can go ahead and show carbon bonded to fluorine. So we have P orbitals here, which are overlapping, and so we can draw in the P orbitals on carbon. Carbon and fluorine are in the same period on the periodic table. Therefore, you can think about their P orbitals being pretty much the same size, and that allows for good overlap. And so the fluorine can donate some electrons and increase the electron density of your ring. And so because there is some good overlap of your orbitals there, fluorobenzene turns out to be the most reactive of your halogens because of that overlap, but because fluorine is so electronegative you could think about the inductive effect winning most the time, and still giving you a deactivator. However, there are some exceptions for fluorobenzene, but, in general, we consider the halogens to be deactivates. Let's look at another halogen here. So for that example, induction wins because of the large electronegativity difference between fluorine and carbon. However, if the electronegativity difference decreases, such as when your halogen is something like chlorine, then you have to come up with a little bit different explanation because an atom like nitrogen has pretty much the same electronegativity as chlorine, and so the inductive effect would be the same for nitrogen as it is for chlorine, approximately. And so there must be a slightly different explanation as to why chlorine is a deactivator. And if you think about the resonance effect where your chlorine forming a pi bond with your carbon. So let's go ahead and show chlorine bonded to carbon here. Carbon being in the second period of the periodic table has a P orbital of a certain size, chlorine of course is in a different period, it's in a third period, so it has a larger P orbital. And the size mismatch of those P orbitals means that you don't get as good of an overlap as you would with orbitals that are the same size, and so therefore, the chlorine can't donate as much electron density via resonance. So resonance decreases in importance because of the poor overlap of those orbitals. And so that's a way of thinking about the fact that the resonance effect is decreased, and so therefore, you have more of an inductive effect and the chlorine is a deactivator. I was talking about comparing chlorine to nitrogen, and the two, of course, have very similar electronegativities, but the nitrogen-- aniline in the previous video was observed to react much, much faster. And again, we can think about the size of those P orbitals. Nitrogen being in the same period as carbon, and so you get better overlap of those orbitals, the nitrogen is better able to donate electron density to that ring, making the nitrogen an activator here. So you could think about, again, large differences in electronegativity favoring induction. You could think about size mismatch of orbitals, but overall, you need to think about halogens as being weak deactivators, but still ortho-para directors because of the resonance structures that you can draw.