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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
Ch. 26 - Transition Metals
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

When two or more substituents are already on benzene, there are multiple new factors we must take into account. These include Steric Effects, Synergistic Groups, and Competitive Groups.

Concept #1: EAS with Polysubstituted Benzene


By now we're really good at adding EAS reactions to unsubstituted benzene by itself and we're even pretty good at figuring out where to put the second group once you have the first one on there when you react with a monosubstituded benzene but what if you have two or more groups already on that benzene, then what happens? Guys so now there's going to be multiple new factors you have to take into account because now you have two or more groups so this is called poly substituted benzene. So when you have those two or more substituents, there's multiple new factors that you have to think about of where to add that third one or that fourth or that fifth one. The first one is steric effects. Now that you have two or more groups you may have some sites on that benzene that are completely unreactive due to the fact that they're in between two sterically hindered groups. So for example the group in between two tert-butyl's like here, that position, or the position between two rings like maybe a ring that's on the chain and a phenol group, P H stands for phenol. That wouldn't be very reactive so these are sites that just would not react they would literally be the very last site to react in an EAS reaction because all the other sites are going to be more favoured than that one so that's something you have to think about when you are using EAS poly-substituted you have to avoid those spots that are between sterically hindered groups but guys we also have directing effects because it turns out that the groups, now that you have more than one you have to consider what the directing effects of all of them not just one of them. So the first type of effect that you can see is what's called synergistic groups, synergistic directing of groups. That's when multiple directing groups, uh oh not what I meant to do, that's when multiple directing groups all point to the same position so here you notice that I'm adding a B R 2 or F E B R 3, so this is going to be an EAS bromination right. However we've got an issue we've got two groups already on this benzene so we have to analyse what are the directing effects of both of those. Well nitro we know is a strong electron withdrawing group and O H C, what is that? Is that an alcohol? No guys you should know this from ORGO 1, that is a aldehyde. We know that the C H O condensed formula stands for an aldehyde so it's actually an aldehyde so that's also a strong withdrawing, not strong, it's a moderate withdrawing group as well so then you would expect these to both be meta directors so the nitrile group I'm just going to draw its effects with a circle shape. So the nitrile group is going to direct towards this meta position and this meta position.

See how those are both meta to the carbon that it's on and then the aldehyde is going to direct towards this meta position and this meta position because those are the ones that are meta to the carbon that it is on. Notice that two of these positions aren't going to work in both situations because they both already have occupants. This one can't work and this one can't work right there's only one that can work and that's this site right here because notice that both of them are directing meta to the very bottom carbon. When you have two groups that agree with each other, two or more, we call this synergistic. We call that synergistic because they work in synergy, that's a word to mean that they're working together, they're cooperating. So they're synergistic and that means that we're going to get a high yield of whatever we're trying to yield, whatever we're trying to react because everything is pointing in the same direction so I would expect to get a very high yield of my bromine, I'll draw the aldehyde out so you guys can see what it looks like like that, C H O and you're going to wind up getting a high yield of the bromine adding to the bottom because it's meta to both of them. So that's synergy and that's one of the more important facts that you can get when you have multiple directing groups but guys we also have another situation which would be competitive groups. How about if you have multiple directing groups that actually disagree on where to substitute? Then what do you get? Well then you're going to get a mixture of products, you're going to get lower yield, mixed products because they're not agreeing with each other they're not synergistic. So let's look at this example. We've got once again a bromination and notice that I've got my same two groups, my nitrile and my aldehyde but now they're in different positions let's see what happens. Well what we see is that the nitrile group, I'm going to make it blue, is going to direct towards this position and this position, the two circles it's going to direct it two bottom carbons those are meta but the aldehyde since it's at the bottom is going to direct towards this position and this position. So notice that they're disagreeing the nitrile actually wanted to go here, the aldehyde wanted to go here so what do we do? Well sorry I'm trying to pick green. Well the short answer is you're going to get both so let's go ahead and draw both. We're going to wind up getting a group or a molecule where some of the bromine added meta to the nitrile and we're also going to get that mixed with a benzene product that has the bromine meta to the aldehyde.

So we're going to get a mixture, see how one of them wins and one loses or you know they're basically competing with each other. One's going to be happy, one's going to be sad. In this case my aldehyde is sad, in this case my nitrile's sad. So the answer is which one's going to be the major product? And actually you can tell which ones major guys. The major product is going to be determined by the strongest activator. The strongest activator will determine the major product. Will we still get a mixture? Yes but you can determine that one's going to be greater and one's going to be smaller. One's major and one's minor. So my question to you is based on that definition, which one do you think is going to form in higher yield? Am I going to get more of the blue compound making the nitrile happy or more of the green compound making the aldehyde happy? You can talk to the screen, no one's judging you. You've probably been doing it this whole time anyway. So maybe you're right. The answer is actually going to be that the green is major guys and the blue is minor. So you have to be thinking but Johnny you said the greatest activator I thought that N O 2 was stronger wait, you remembered it wrong. N O 2 is the strongest deactivator, it's the weakest activator. C H O our aldehyde, it's not an activator but it's a little bit more active nitrile. Nitrile's at the very bottom of my badass activity chart and then we learn that carbonyls are moderately deactivating so that means that in terms of absolute activity the carbonyl is actually higher than the nitrile group so the carbonyl is going to win and that would happen with every single group on that chart if you go all the way up to annulene, all it has to do with is the relative activity of both groups. So it's not about which one is the strongest it's about which one is the highest up on that chart, which one is the most active compared to the other. So that's why my green is going to win that's why my aldehyde is going to win because it's higher on my activity chart and my nitrile is going to lose because it's the very lowest one. Alright guys, so I hope that made sense. You guys know about synergistic effects and competitive effects. So let's move on to the next video.

Practice: Which is NOT a possible product of the reaction?

Practice: Which is NOT a possible product of the reaction?