|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 8mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 26mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 8mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 59mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
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|EAS: Sulfonation Mechanism|
|EAS: Gatterman–Koch Reaction|
|EAS: Total Benzene Isomers|
|EAS: Polycyclic Aromatic Hydrocarbons|
|EAS: Directing Effects|
|Resonance Theory of EAS Directing Effects|
|Activated Benzene and Polysubstitutions|
|EAS: Dueling Benzenes|
|Hydrogenation of Benzene|
|EAS: Missing Reagent|
|Diazo Coupling Reactions|
|SNAr vs. Benzyne|
|Aromatic Missing Reagent|
|EAS on 5-membered Heterocycles|
In general, we refer to the products of an EAS o,p-director as a mixture – but there are some patterns we can learn.
Concept #1: Ortho, Para major products
In this video, we're going into a little bit more depth on the major products of ortho, para-directed reactions. In general, whenever we have an EAS reaction on an ortho, para-director we're just going to say that the product is a mixture of both the ortho and the para positions. If you want to play it safe, you can always just say that it's a mixture. That's the way that most textbooks teach it.
However, there are a few patterns that might be helpful for us to learn, especially later on when we start talking about specific groups called blocking groups. Let me tell you about the difference between the ortho position and the para position.
Basically, these positions are in competition with each other because they both have advantages. Let's say you have an electron donating group, that means it's pushing electrons into the ring and it's probably – it's an ortho, para director. The ortho position has a clear advantage right from the beginning because notice that it has two positions instead of one. The para position, there's only one para. But there's two ortho positions. So you might think we're going to get more ortho product because you have twice as many positions that could react. That's one way to think about it that there's two times as many positions.
But on the other hand, the para position also has an advantage, which is that the para position is usually the one that's less hindered because it's furthest away from the donating group. So it's actually easier to add to that para position even though there's less of them.
In general, when you pair these two together, when you complete them against each other, number versus steric hindrance, steric hindrance is usually going to win. Usually, there's going to be a slight advantage for the para position over the ortho positions.
If in a question, you're asking to supply one major product instead of a mixture, if it says specifically which one is the major product, here we can see a sulfonation on toluene. You can see we've got our SO3. We've got out fuming H2SO4 and look what we're going to get. We are going to get is a major product of para. You would say that your para sulfonic acid is going to be the major product and your ortho is going to be the minor. Even though there's more ortho positions, the steric hindrance, the benefit of sterics is going to make the para the major product.
Basically, that's going to be the rule that we always follow. We're going to say that para wins out over ortho if asked to supply one major product.
Now there is one noted exception. That one exception is if the final product can hydrogen bond with itself. If the final product can hydrogen bond because of the orientation of the groups, then we would say that the ortho will slightly win.
Here you guys can see the nitration of phenol. Notice that here I have my nitric acid. I have my sulfuric acid. We know that we're going to generate, what is it? It's going to be the nitronium cation or my nitronium ion. And we know that OH is a ortho, para director because it has a lone pair. So that means that it's an electron donating – I mean, it has two lone pairs. I'm just saying if it has at least one lone pair. We know that it's going to direct either next to it, which is ortho, or at the bottom, which is para.
But it turns out that in this case, ortho is slightly better. Why? Because if you make it ortho, then you can get a hydrogen bond between the phenol and the nitro group. If you make it para, it's less hindered, so it's easier to add in that way, but it can't interact with itself, so it's not going to be quite as favored. Now this ratio actually isn't as big of a difference as you might thing. It's actually only going to be a 60% to 35% ratio, meaning that it's not even winning by that much, but still, this would be the only exception where you would expect more ortho.
But for everything else, if asked to supply one major product, I mean, you can always say it's a mixture, but if you're asked to supply one, you're usually going to go with the para product because it's the one that's less sterically hindered.
All right guys. Believe me, you don't need to know these percentages. I'm just giving them as a teaching moment so that you guys can see actual examples in real life and you guys can understand how it plays out. But in general, you can just go with the rules that I'm telling you.
All right guys, so let's move on to the next video.
There is one exception to this rule – which occurs if the final product can hydrogen-bond with itself. Then, we would expect more ortho, than para product.
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