The directing effects of substituents on a benzene ring. Created by Jay.
Want to join the conversation?
- When we have two deactivators, we take directing effects from the one which is less deactivating, right?(12 votes)
- Right, but with two deactivators it will take some strenuous conditions to add another substituent to the ring.(10 votes)
- Why won't the sulfonic group interact with the other substituents instead of benzene?(5 votes)
- It does The reactants also react with the phenolic OH group to form p-cresyl hydrogen sulfate, but this ester rearranges under the reaction conditions to the ring -sulfonated product. The net effect is as if they had never reacted with the OH group at all.(6 votes)
- At1:50. Is it possible to get 1,3-dichloro-2-methyl-5-nitrobenzene as a side product?(4 votes)
- What would happen if you would have a strong deactivator and a strong activator in meta positions, where would the added substituent go in these conditions? Would the substituent follow the ortho para from the strong activator or would it add meta thanks to the strong deactivator?(3 votes)
- It probably depends on their relative strengths, but usually the strong activator wins.
For example. chlorination of 3-nitrophenol gives mainly 4-chloro-3-nitro phenol.(3 votes)
- Chlorine is a weak deactivator not a donor. Atleast that's what I've learnt(2 votes)
- Yes, because Cl can donate its one of the lone pair to the benzene ring in order to stabilise it but on the other hand, Cl is an electronegative element so it shows -I effect and withdraws electron through sigma bond. However, the induction of withdrawing electron density through induction is greater than donation of the lone pair. Hence, Cl is ortho- and para- deactivator. Its ortho-/para- directing only because of donation of lone pairs !(4 votes)
- If we have 2 substituents directing to 2 different positions one of them is strong deactivating and the other is a weak activating group.
According to which one the electrophile would go ?!(3 votes)
- The electrophile would go para or ortho to the weak activator, but it would not go between the two substituents.(2 votes)
- At6:14, are the products equally made or is one more stable than the other? Also does Cl being a deactivator and CH3 being an activator make any difference?(3 votes)
- Aren't Halogens ortho/para deactivators? meaning that they are Electron Withdrawing Groups, right? If yes, then don't halogens favor the meta position?
Thanks in advance.(2 votes)
- Yes, halogens are EWGs. Their inductive effect removes electrons from the ring, so they are deactivators.
But their resonance effect puts more electron density on the ortho and para positions than on the meta positions.
The deactivated ring still has more electron density at the ortho/para positions, so that's where an electrophile will attack.(3 votes)
- If I has toluene and bromo group on ortho so where will nitro will attach on this ring(2 votes)
- For the last example, why did you use the chlorine to determine where the substituents would be added? isn't chlorine considered a weak deactivator why the methyl group is considered a weak activator? therefore shouldn't the methyl group be used in this case instead of the Cl?(2 votes)
- He is actually using both.
After a lot of digging I think I have an answer for why the Cl is just as important (comments by real chemists strongly desired).
These conclusions are based on data here:
So, the relative rates (compared with benzene) are:
methyl group — ortho = 43, meta = 3, para = 55
chloro group — ortho = 0.03, meta = 0, para = 0.14
If we assume that these relative rates multiply on a multiply substituted molecule, for 3-chlorotoluene you would get that the relative rates would be:
position 2: ~1 = 43 * 0.03
position 4: ~2 = 55 * 0.03
position 5: ~0 = 3 * 0
position 6: 6.0 = 43 * 0.14
This seems to conflict with what Jay says, but we are ignoring steric effects which make position 2 very unfavorable and perhaps make position 6 less favorable.(1 vote)
Now that we understand ortho para and meta directors, let's see what happens when you have multiple substituents on a benzene ring. So if I look at this first reaction, I can tell that this is a halogenation reaction. And it's going to put a chlorine onto the ring. The question is where will the chlorine go. To figure that out you have to look at your substituents and see what kind of directors they are. So we'll start right here with this methyl group, which we know is an ortho para director. So it's going to direct the chlorine to the ortho and para positions. And so this would be the ortho position. And this is also an ortho position. But by symmetry these are the same thing, if you think about the molecule here. The para position is occupied by this nitro group right here. And so we can't add anything on para in this example. If we look at the nitro group, right here, I think about what kind of a director that is. I know that it's a meta director from the reasons we talked about in previous videos. So the position that's meta to this nitro group would, of course, end up being the exact same carbons that we marked before, which again by symmetry are the same ones here. So this turns out to be a case where both of my substituents direct to the same spot. And so I can go ahead and draw my product. So I have my have benzene ring. And I have my substituents on there, my methyl group, and also my nitro group. And I could have picked either one of these two carbons, so this one or this one. It doesn't matter which one you choose again, because of symmetry, you will get an identical product here. So we're going to put a chlorine on in one of those positions, ortho to the methyl group, and meta to the nitro group. And so that's the only product that we will get. Let's do another one here. Let's look at this reaction. So I know that this is a sulfonation reaction, right. So I'm going to put an SO3H group onto my benzene ring. And the question is, where will that group go. And so once again I analyze my substituents. And first I look at the OH group, which I know is an ortho para director, right. So this is an ortho para director here. And I'd look at where that would be on my ring. Well, once again this would be the ortho position. And again by symmetry, the same ortho position over there. The para position is once again taken up by this methyl group here. When I think about the methyl group-- so let me just use a different color for that one. The methyl group is also an ortho para director. So what's ortho and para to the methyl group. Well this would be ortho, right. These two spots would be ortho, again identical because of symmetry. The para spot is taken up by this OH here. And so now we have a case where we have two ortho para directors directing to different spots. And so the way to figure out which one wins is to think about the activating strength of these two ortho para directors. So the strongest activator is going to be the directing group. And the strongest activator here is, of course, the OH group here. So this is a strong activator. And the methyl group we saw is a weak activator from some of the earlier videos here. So the spot that's going to get the SO3H is this carbon, which again by symmetry would be this one right here. So we can go ahead and draw the product of this sulfonation reaction. So let me go ahead and sketch in my benzene ring. Let me do another one here. That one wasn't very good. And we have our ring. We have our OH. We have our methyl group. And since the OH is the strongest activator, we're going to go ahead and put SO3H ortho to the OH. And that is our final product. Let's do one more example. So let's see what happens with this reaction here. So once again I can identify this as being a halogenation reaction where I add a chlorine onto my ring. And let's go ahead and analyze the substituents once again. So I have a methyl group again, which I know is an ortho para director. So I'll go ahead and mark the spots that are ortho and para to that methyl group. So this would be the ortho position. This would be the ortho position. And then this time the para position is free, so we could possibly put the chlorine there. If I look at what else is on my ring, right-- so I have another substituent here, which is a chlorine. We know that halogens are also ortho para directors because of the lone pair of electrons that are on the chlorine there. So an ortho para director for the chlorine. So what's ortho to this chlorine here. Well this spot is ortho. So is this spot and so is this spot. And so we have a couple of different possible products here. And let's go ahead and start drawing them here. So if I think about the product-- if we add on a chlorine onto this position. Let me go ahead and draw that. So we would have our benzene ring here. We would have our methyl group. We would have our chlorine. And it's possible that a chlorine would add on to that position. Let's next look at this position. So this could be another possible product here. So let me go ahead and draw that one as well. So we have our benzene ring. And we have our methyl group. And we had the chlorine that was originally on our ring. And then we're adding on a chlorine to this position. And finally, let's go ahead and think about what would happen if a chlorine tried to add to this final position right here on the right. Well, it turns out that there's just too much steric hindrance for a chlorine to add on to this position. So if you think about this methyl group being bulky and this chlorine being bulky, that makes it difficult for another chlorine to add to here. So this is not observed in large amounts. So this product, we're going to say, does not form here. And so we're left with two major products for this reaction. And that's how to think about synthesis-- directing effects in synthesis-- when you're talking about multiple substituents. First think about where the substituents will direct. And you have to think about activating strength. And then finally, think about steric hindrance for your products.