Ch. 18 - Reactions of Aromatics: EAS and BeyondWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Sections
Electrophilic Aromatic Substitution
Benzene Reactions
EAS: Halogenation Mechanism
EAS: Nitration Mechanism
EAS: Friedel-Crafts Alkylation Mechanism
EAS: Friedel-Crafts Acylation Mechanism
EAS: Any Carbocation Mechanism
Electron Withdrawing Groups
EAS: Ortho vs. Para Positions
Acylation of Aniline
Limitations of Friedel-Crafts Alkyation
Advantages of Friedel-Crafts Acylation
Blocking Groups - Sulfonic Acid
EAS: Synergistic and Competitive Groups
Side-Chain Halogenation
Side-Chain Oxidation
Birch Reduction
EAS: Sequence Groups
EAS: Retrosynthesis
Diazo Replacement Reactions
Diazo Sequence Groups
Diazo Retrosynthesis
Nucleophilic Aromatic Substitution
Benzyne
Additional Practice
EAS: Sulfonation Mechanism
EAS: Gatterman–Koch Reaction
EAS: Total Benzene Isomers
EAS: Polycyclic Aromatic Hydrocarbons
EAS: Directing Effects
Resonance Theory of EAS Directing Effects
EAS: Badass Activity Chart
Activated Benzene and Polysubstitutions
Clemmensen Reduction
EAS: Dueling Benzenes
Hydrogenation of Benzene
EAS: Missing Reagent
EAS: Synthesis
Diazonization of Aniline
Diazo Coupling Reactions
SNAr vs. Benzyne
Aromatic Missing Reagent
Aromatic Synthesis
Aromatic Retrosynthesis
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

Transcript

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.