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Electrophilic aromatic substitution mechanism

The general reaction and mechanism of electrophilic aromatic substitution. Created by Jay.

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  • aqualine seedling style avatar for user Venkata Sai Ram
    how do electrons move from one position to another during resonance, what is the driving force during resonance
    (6 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      Actually, in the molecule, electrons do not move during resonance. What we see is a probability cloud of electron density that is always higher at some atoms than at other atoms. We try to explain this by drawing resonance structures that put the extra electron density on the appropriate atoms.
      We draw the structures and move the electrons. The electron movements usually start with a lone pair on an atom moving between that atom and the atom to which it is σ bonded, or with a pair of π electrons moving between the atoms that are σ bonded to the atoms on either side of the π bond.
      The driving force is that this spreading out or delocalization leads to a lower energy.
      (24 votes)
  • blobby green style avatar for user dilasha23
    Can you please explain how we know which compound is more electrophilic than the other? For example, ranking propanal, 1-propanol and 1-chloropropane in increasing electrophilicity?
    (3 votes)
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    • male robot hal style avatar for user Satwik Pasani
      Electrophilicity is a result of deficiency of electrons at a particular atom in the compound. For example, BF3 is an electrophile, that is, it's octet is not complete, and hence is in search of electron donors to complete the octet. F- acts as such a donor and forms BF4-. AlCl3, Br+, NO+ are a few examples, where, owing to either their structures or to the very high electronegativity (desire for electrons) of the central atom, there is a deficiency of electrons because of which these act as electrophiles. Having said that, you can assess the electrophilicity of the compounds in your question, considering the carbon attached to the functional group as the electrophilic site. Since a double bonded carbon is very electronegative, the carbon to which it is bonded is severely deficient in electrons, as compared to -OH group which is more electronegative than -CL group. Hence, Propanal>propanol>cvhloropropane. However, if you consider the carbon adjacent to the carbon doubly bonded to oxygen in the aldeheyde to be the electrophilic site, the order might change.
      (4 votes)
  • leaf green style avatar for user svkgvk
    why we call complex as sigma complex
    (3 votes)
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  • blobby green style avatar for user Arun Acharya
    Why benzene undergo electrophilic substitution type of reaction but alkene undergo electrophilic addition reaction although both are unsaturated compound?
    (2 votes)
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  • blobby green style avatar for user Viktor Hansen
    Does this reaction categories as a Sn1 or Sn2 reaction?
    (1 vote)
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  • blobby green style avatar for user Josh Holtzman
    After drawing all of the resonance intermediates, how do you know which one to choose to proceed with?
    (1 vote)
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    • male robot hal style avatar for user mk
      hi
      tyresome has given a great answer.
      Just wanted to add an extra point to go with it.
      There is no one fixed resonance structure that could depict the entire molecule correctly on its own. The actual molecule exists as a mixture of its resonance structures.
      Hope it helps.
      (2 votes)
  • blobby green style avatar for user andyyy.fatima
    why in sigma complex carbon contains +charge
    +charge appear on it at that time when it loses its one electron.Does it lose its electron?
    I am confused over here can you please explain it for me ?
    (1 vote)
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  • starky tree style avatar for user Isaac
    why secondary carbocations are more stable than primary carbocations?
    (1 vote)
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    • leaf red style avatar for user Richard
      Carbocations have a stability pattern of: 3° >2° >1°> methyl. Where more alkyl groups bonded to a carbocation make it more stable. This is due to two effects, through the inductive effect and hyperconjugation of the unhybridized p orbital. The positively charged carbon at the center of a carbocation is electron-deficient without a completed octet and is stabilized by additional electron density.

      The inductive effect is electron density donation through sigma bonds from the alkyl groups to the positively charged carbon.

      Hyperconjugation is electron density donation from the alkyl group’s neighboring sp3 hybrid orbitals into the unhybridized p orbital of the positively charged carbon.

      Hope that helps.
      (2 votes)
  • duskpin ultimate style avatar for user Mihir Paranjape
    What is Ipso substitution? We have only been taught the desulphonation reaction, where the -SO3H group gets substituted by -H. My question is- Does this reaction happen on its own (ipso facto)? I searched online and haven't gotten a clear answer... Usually, they show the reaction in presence of water, which makes sense: a proton donor should be there... So, I'm confused. Not about the reaction but about the conditions under which it occurs.
    (1 vote)
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  • female robot amelia style avatar for user Asma Waseem
    I don't understand why do we study mechanisms in organic chemistry. Like we don't have mechanisms in inorganic chemistry (as far as i have studied).
    (1 vote)
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    • mr pants purple style avatar for user Ryan W
      There are mechanisms in inorganic chemistry, but they often aren't as involved as these.

      Organic chemistry leads on to synthesis (eg making molecules) at higher levels. It's helpful to us and others replicating works to know the outcomes of reactions and how the reaction proceeds.
      (2 votes)

Video transcript

Let's look at the general reaction for electrophilic aromatic substitution. So we start with the benzene ring, and we react benzene with a molecule that contains an electrophile in there. And what happens in electrophilic aromatic substitution. We're going to substitute the electrophile for a proton on our benzene ring. And so over here, we can see the electrophile is now in place of that proton. So that's where that that's where the electrophilic part comes in this. And that's where the substitution parts comes in. You're substituting an electrophile for a proton. The aromatic comes in because you are going to reform an aromatic ring in your mechanism. Electrophilic aromatic substitution requires a catalyst. And the point of a catalyst is to generate your electrophile. So down here, you can see that the catalyst is going to react to produce the positively charged electrophile. So remember, electrophile means loving electron. So if something is positively charged, it's going to love electrons. We also formed this catalyst complex over here, which is going to factor into our mechanism. So now that we formed our electrophile, let's look in more detail as to what happens in electrophilic, aromatic substitution. So we start with our benzene ring. And I'm showing one of the hydrogens on the benzene ring. It could be any of the six, since they are all equivalent. And now we formed our electrophile from our catalyst. So the pi electrons in the benzene ring can be attracted to the positively charged electrophile. Because negative charges are attracted to positive charges. And so pi electrons in your benzene ring are going to function as a nucleophile, and those electrons are going to attack the electrophile. So this is a nucleophile, electrophile attack, where those pi electrons are going to bond to that electrophile there. So those pi electrons are going to form a covalent bond with your electrophile. So let's go ahead and show that. So these pi electrons didn't do anything. The hydrogen stays there. Now, I could show the electrophile adding to either of the two carbons on the side of the double bonds. So it could be that carbon. Or it could be this carbon. Since I've drawn this hydrogen up here at the top, I'm going to go ahead and say that the electrophile adds to the top carbon there. So there's my electrophile there. Let me go ahead and highlight the electrons that are forming that covalent bond. So these pi electrons here are the ones that are functioning as a nucleophile. And those pi electrons are going to form this bond right here. Now in forming that bond, we're taking a bond away from this bottom carbon here. And so that bottom carbon is going to be left with a positive one formal charge. Therefore, we can draw a resonance structure for this cation. So let's go ahead and show a possible resonance structure here. So these pi electrons could move over to here. And let's go ahead and draw what would result if that happened. So now, we have these pi electrons up here. We have our hydrogen. We have our electrophile. And the electrons moved over to this position. Let me go ahead and highlight those in magenta. So I'm saying that these pi electrons right here moved over to here. And when those electrons moved over to there, we're taking a bond away from this carbon this time. So that is the carbon that's going to get a plus 1 formal charge like that. So we can draw another resonance structure. So let's go ahead and do that. So we could take these pi electrons and move them into here. So let's go ahead and show what that would look like. So if those pi electrons moved into there, we would now have, again, our hydrogen, our electrophile, these pi electrons, and then these pi electrons right here. So once again, let me go ahead and highlight those. This time I'll use blue. These pi electrons are going to move over to here. And once again, we're taking a bond away from a carbon. This time, it's this top carbon up here. So that's the carbon that's going to get the plus one formal charge like that. So these are all resonance structures. And remember, the actual cation would be a hybrid of these resonance structures. And we call we call that hybrid a sigma complex. So you have a positive one formal charge de-localized over three carbons in your sigma complex. So the next step in the mechanism-- I'm just going to redraw the first resonance structure that we did here. So I'm going to go and redraw that down here. So let's go ahead and show the first resonance structure. So in our first resonance structure, we had our hydrogen here, our electrophile already bonded to our ring. And we had a positive one formal charge on this carbon right here. Well remember, the catalyst had formed a complex. And I represent it like this. So something bonded to your catalyst like that. So let's just go up here and refresh our memory. So right up here, when we generated our electrophile, we also generated this catalyst complex up here. So y bonded to a catalyst, so I have y bonded to a catalyst down here. And you could think about this as functioning as a base. Or it's going to accept a proton. So I could show these electrons in here taking this proton. And if it takes that proton, that leaves these electrons behind. And those electrons are going to move in here to reform your benzene ring and take away that positive one formal charge. So let's go ahead and show that. So we now have our benzene ring back. And our electrophile is now bonded to our ring. And the proton has left. So the electrophile has completely substituted for that proton. Let's follow those electrons again. So the electrons in magenta in here, so those are the ones that are going to move in here to reform your aromatic ring. So deproteination of the sigma complex restores the aromatic ring. And so we have a stable product here. So the other product you could think about this y here is now going to be bonded to that proton. So you could have the y here bonded to that proton. And you could highlight those electrons. You could say that these electrons right here are now these electrons. And, taking those electrons away from the catalyst would of course regenerate your catalyst. And so it's free to then catalyze another reaction. And so this is the general mechanism for electrophilic aromatic substitution, which the reactions that we're going to see are pretty much going to follow this general mechanism.