Resonance theory is used to represent the different ways that the same molecule can distribute its electrons. So what that means is that even though the connectivity or how atoms are connected isn't going to change, the electrons between them can move sometimes. And that's what resonance theory is all about. So I'm going to teach you guys some rules and you guys are going to get the hang of it as I go along. All right. So the first thing to know is that atoms will never ever move. The reason is because remember that I said the connectivity of those atoms, how they're connected to each other, doesn't change. The only thing that changes is the kinds of electrons that are in between them that are keeping them linked together. The only thing that moves is the electrons. When I talk about electrons, what I'm talking about is pi bonds. Pi bonds move and I'm also talking about lone pairs. So what that means is that literally I'm not moving any atoms. All I'm moving is double bonds around or triple bonds around and I'm also moving where lone pairs are at. That has to do with the electrons that are moving throughout the molecule. Now, something about resonance structures is we're going to find out that there's something called contributing structures. Contributing structures are structures that both contribute to the actual representation of the molecule because they average together. What we're going to find out is that none of these contributing structures are actually going to look like the actual molecule. So what that means is the molecule is a blend of all the different possible resonance structures that a molecule can have. So let's go ahead and learn some rules. First of all, we're going to use curved arrows to represent electron movement. Just so you know, these rules are going to apply to the rest of organic chemistry. We're going to keep using these rules any time that we're moving electrons, which is pretty much all the time. So what a curved arrow would look like is like this. So notice that I'm using a full arrow and I'm curving it around. What that means is that 2 electrons that represents 2 electrons are moving from one place to another. What we're going to do is after we've built our resonance structures, we're going to use double-sided arrows and brackets to link related structures together. So that means that once I figure out my resonance structures, I link them together using those double-sided arrows like I have here and then brackets like I have here. Okay? Then finally well, not finally, but arrows are always going to travel from regions of high density, high electron density to low electron density. Like I said, this is a rule that applies for the rest of organic chemistry. Any time we're moving electrons, we always start from the area of the highest density and move to the area of lowest density. So what that means is that for example, a positive charge would be an area of low density because that means that you have an electron missing. Right? So what that means is you would never start an arrow from a positive charge. In fact, you would always go towards the positive because that's the area of low density. Then finally, the net charge of all the structures that we make must be the same. The reason for that is that remember that resonance structures are different ways to represent the same molecule and what that means is that all of them should have the same net charge because we're just distributing the electrons differently, but we're not adding any electrons or subtracting any electrons. What that means is they should really all have the same charge.
- 1. A Review of General Chemistry5h 9m
- Summary23m
- Intro to Organic Chemistry5m
- Atomic Structure16m
- Wave Function9m
- Molecular Orbitals17m
- Sigma and Pi Bonds9m
- Octet Rule12m
- Bonding Preferences12m
- Formal Charges6m
- Skeletal Structure14m
- Lewis Structure20m
- Condensed Structural Formula15m
- Degrees of Unsaturation15m
- Constitutional Isomers14m
- Resonance Structures46m
- Hybridization28m
- Molecular Geometry16m
- Electronegativity22m
- 2. Molecular Representations1h 14m
- 3. Acids and Bases2h 46m
- 4. Alkanes and Cycloalkanes4h 19m
- IUPAC Naming29m
- Alkyl Groups13m
- Naming Cycloalkanes10m
- Naming Bicyclic Compounds10m
- Naming Alkyl Halides7m
- Naming Alkenes3m
- Naming Alcohols8m
- Naming Amines15m
- Cis vs Trans21m
- Conformational Isomers13m
- Newman Projections14m
- Drawing Newman Projections16m
- Barrier To Rotation7m
- Ring Strain8m
- Axial vs Equatorial7m
- Cis vs Trans Conformations4m
- Equatorial Preference14m
- Chair Flip9m
- Calculating Energy Difference Between Chair Conformations17m
- A-Values17m
- Decalin7m
- 5. Chirality3h 39m
- Constitutional Isomers vs. Stereoisomers9m
- Chirality12m
- Test 1:Plane of Symmetry7m
- Test 2:Stereocenter Test17m
- R and S Configuration43m
- Enantiomers vs. Diastereomers13m
- Atropisomers9m
- Meso Compound12m
- Test 3:Disubstituted Cycloalkanes13m
- What is the Relationship Between Isomers?16m
- Fischer Projection10m
- R and S of Fischer Projections7m
- Optical Activity5m
- Enantiomeric Excess20m
- Calculations with Enantiomeric Percentages11m
- Non-Carbon Chiral Centers8m
- 6. Thermodynamics and Kinetics1h 22m
- 7. Substitution Reactions1h 48m
- 8. Elimination Reactions2h 30m
- 9. Alkenes and Alkynes2h 9m
- 10. Addition Reactions3h 18m
- Addition Reaction6m
- Markovnikov5m
- Hydrohalogenation6m
- Acid-Catalyzed Hydration17m
- Oxymercuration15m
- Hydroboration26m
- Hydrogenation6m
- Halogenation6m
- Halohydrin12m
- Carbene12m
- Epoxidation8m
- Epoxide Reactions9m
- Dihydroxylation8m
- Ozonolysis7m
- Ozonolysis Full Mechanism24m
- Oxidative Cleavage3m
- Alkyne Oxidative Cleavage6m
- Alkyne Hydrohalogenation3m
- Alkyne Halogenation2m
- Alkyne Hydration6m
- Alkyne Hydroboration2m
- 11. Radical Reactions1h 58m
- 12. Alcohols, Ethers, Epoxides and Thiols2h 42m
- Alcohol Nomenclature4m
- Naming Ethers6m
- Naming Epoxides18m
- Naming Thiols11m
- Alcohol Synthesis7m
- Leaving Group Conversions - Using HX11m
- Leaving Group Conversions - SOCl2 and PBr313m
- Leaving Group Conversions - Sulfonyl Chlorides7m
- Leaving Group Conversions Summary4m
- Williamson Ether Synthesis3m
- Making Ethers - Alkoxymercuration4m
- Making Ethers - Alcohol Condensation4m
- Making Ethers - Acid-Catalyzed Alkoxylation4m
- Making Ethers - Cumulative Practice10m
- Ether Cleavage8m
- Alcohol Protecting Groups3m
- t-Butyl Ether Protecting Groups5m
- Silyl Ether Protecting Groups10m
- Sharpless Epoxidation9m
- Thiol Reactions6m
- Sulfide Oxidation4m
- 13. Alcohols and Carbonyl Compounds2h 17m
- 14. Synthetic Techniques1h 26m
- 15. Analytical Techniques:IR, NMR, Mass Spect6h 50m
- Purpose of Analytical Techniques5m
- Infrared Spectroscopy16m
- Infrared Spectroscopy Table31m
- IR Spect:Drawing Spectra40m
- IR Spect:Extra Practice26m
- NMR Spectroscopy10m
- 1H NMR:Number of Signals26m
- 1H NMR:Q-Test26m
- 1H NMR:E/Z Diastereoisomerism8m
- H NMR Table21m
- 1H NMR:Spin-Splitting (N + 1) Rule17m
- 1H NMR:Spin-Splitting Simple Tree Diagrams11m
- 1H NMR:Spin-Splitting Complex Tree Diagrams8m
- 1H NMR:Spin-Splitting Patterns8m
- NMR Integration18m
- NMR Practice14m
- Carbon NMR4m
- Structure Determination without Mass Spect47m
- Mass Spectrometry12m
- Mass Spect:Fragmentation28m
- Mass Spect:Isotopes27m
- 16. Conjugated Systems6h 13m
- Conjugation Chemistry13m
- Stability of Conjugated Intermediates4m
- Allylic Halogenation12m
- Reactions at the Allylic Position39m
- Conjugated Hydrohalogenation (1,2 vs 1,4 addition)26m
- Diels-Alder Reaction9m
- Diels-Alder Forming Bridged Products11m
- Diels-Alder Retrosynthesis8m
- Molecular Orbital Theory9m
- Drawing Atomic Orbitals6m
- Drawing Molecular Orbitals17m
- HOMO LUMO4m
- Orbital Diagram:3-atoms- Allylic Ions13m
- Orbital Diagram:4-atoms- 1,3-butadiene11m
- Orbital Diagram:5-atoms- Allylic Ions10m
- Orbital Diagram:6-atoms- 1,3,5-hexatriene13m
- Orbital Diagram:Excited States4m
- Pericyclic Reaction10m
- Thermal Cycloaddition Reactions26m
- Photochemical Cycloaddition Reactions26m
- Thermal Electrocyclic Reactions14m
- Photochemical Electrocyclic Reactions10m
- Cumulative Electrocyclic Problems25m
- Sigmatropic Rearrangement17m
- Cope Rearrangement9m
- Claisen Rearrangement15m
- 17. Ultraviolet Spectroscopy51m
- 18. Aromaticity2h 31m
- 19. Reactions of Aromatics: EAS and Beyond5h 1m
- Electrophilic Aromatic Substitution9m
- Benzene Reactions11m
- EAS:Halogenation Mechanism6m
- EAS:Nitration Mechanism9m
- EAS:Friedel-Crafts Alkylation Mechanism6m
- EAS:Friedel-Crafts Acylation Mechanism5m
- EAS:Any Carbocation Mechanism7m
- Electron Withdrawing Groups22m
- EAS:Ortho vs. Para Positions4m
- Acylation of Aniline9m
- Limitations of Friedel-Crafts Alkyation19m
- Advantages of Friedel-Crafts Acylation6m
- Blocking Groups - Sulfonic Acid12m
- EAS:Synergistic and Competitive Groups13m
- Side-Chain Halogenation6m
- Side-Chain Oxidation4m
- Reactions at Benzylic Positions31m
- Birch Reduction10m
- EAS:Sequence Groups4m
- EAS:Retrosynthesis29m
- Diazo Replacement Reactions6m
- Diazo Sequence Groups5m
- Diazo Retrosynthesis13m
- Nucleophilic Aromatic Substitution28m
- Benzyne16m
- 20. Phenols55m
- 21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
- Naming Aldehydes8m
- Naming Ketones7m
- Oxidizing and Reducing Agents9m
- Oxidation of Alcohols28m
- Ozonolysis7m
- DIBAL5m
- Alkyne Hydration9m
- Nucleophilic Addition8m
- Cyanohydrin11m
- Organometallics on Ketones19m
- Overview of Nucleophilic Addition of Solvents13m
- Hydrates6m
- Hemiacetal9m
- Acetal12m
- Acetal Protecting Group16m
- Thioacetal6m
- Imine vs Enamine15m
- Addition of Amine Derivatives5m
- Wolff Kishner Reduction7m
- Baeyer-Villiger Oxidation39m
- Acid Chloride to Ketone7m
- Nitrile to Ketone9m
- Wittig Reaction18m
- Ketone and Aldehyde Synthesis Reactions14m
- 22. Carboxylic Acid Derivatives: NAS2h 51m
- Carboxylic Acid Derivatives7m
- Naming Carboxylic Acids9m
- Diacid Nomenclature6m
- Naming Esters5m
- Naming Nitriles3m
- Acid Chloride Nomenclature5m
- Naming Anhydrides7m
- Naming Amides5m
- Nucleophilic Acyl Substitution18m
- Carboxylic Acid to Acid Chloride6m
- Fischer Esterification5m
- Acid-Catalyzed Ester Hydrolysis4m
- Saponification3m
- Transesterification5m
- Lactones, Lactams and Cyclization Reactions10m
- Carboxylation5m
- Decarboxylation Mechanism14m
- Review of Nitriles46m
- 23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
- 24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
- Tautomerization9m
- Tautomers of Dicarbonyl Compounds6m
- Enolate4m
- Acid-Catalyzed Alpha-Halogentation4m
- Base-Catalyzed Alpha-Halogentation3m
- Haloform Reaction8m
- Hell-Volhard-Zelinski Reaction3m
- Overview of Alpha-Alkylations and Acylations5m
- Enolate Alkylation and Acylation12m
- Enamine Alkylation and Acylation16m
- Beta-Dicarbonyl Synthesis Pathway7m
- Acetoacetic Ester Synthesis13m
- Malonic Ester Synthesis15m
- 25. Condensation Chemistry2h 9m
- 26. Amines1h 43m
- 27. Heterocycles2h 0m
- Nomenclature of Heterocycles15m
- Acid-Base Properties of Nitrogen Heterocycles10m
- Reactions of Pyrrole, Furan, and Thiophene13m
- Directing Effects in Substituted Pyrroles, Furans, and Thiophenes16m
- Addition Reactions of Furan8m
- EAS Reactions of Pyridine17m
- SNAr Reactions of Pyridine18m
- Side-Chain Reactions of Substituted Pyridines20m
- 28. Carbohydrates5h 53m
- Monosaccharide20m
- Monosaccharides - D and L Isomerism9m
- Monosaccharides - Drawing Fischer Projections18m
- Monosaccharides - Common Structures6m
- Monosaccharides - Forming Cyclic Hemiacetals12m
- Monosaccharides - Cyclization18m
- Monosaccharides - Haworth Projections13m
- Mutarotation11m
- Epimerization9m
- Monosaccharides - Aldose-Ketose Rearrangement8m
- Monosaccharides - Alkylation10m
- Monosaccharides - Acylation7m
- Glycoside6m
- Monosaccharides - N-Glycosides18m
- Monosaccharides - Reduction (Alditols)12m
- Monosaccharides - Weak Oxidation (Aldonic Acid)7m
- Reducing Sugars23m
- Monosaccharides - Strong Oxidation (Aldaric Acid)11m
- Monosaccharides - Oxidative Cleavage27m
- Monosaccharides - Osazones10m
- Monosaccharides - Kiliani-Fischer23m
- Monosaccharides - Wohl Degradation12m
- Monosaccharides - Ruff Degradation12m
- Disaccharide30m
- Polysaccharide11m
- 29. Amino Acids3h 20m
- Proteins and Amino Acids19m
- L and D Amino Acids14m
- Polar Amino Acids14m
- Amino Acid Chart18m
- Acid-Base Properties of Amino Acids33m
- Isoelectric Point14m
- Amino Acid Synthesis: HVZ Method12m
- Synthesis of Amino Acids: Acetamidomalonic Ester Synthesis16m
- Synthesis of Amino Acids: N-Phthalimidomalonic Ester Synthesis13m
- Synthesis of Amino Acids: Strecker Synthesis13m
- Reactions of Amino Acids: Esterification7m
- Reactions of Amino Acids: Acylation3m
- Reactions of Amino Acids: Hydrogenolysis6m
- Reactions of Amino Acids: Nihydrin Test11m
- 32. Lipids 2h 50m
- 34. Nucleic Acids1h 32m
- 35. Transition Metals5h 33m
- Electron Configuration of Elements45m
- Coordination Complexes20m
- Ligands24m
- Electron Counting10m
- The 18 and 16 Electron Rule13m
- Cross-Coupling General Reactions40m
- Heck Reaction40m
- Stille Reaction13m
- Suzuki Reaction25m
- Sonogashira Coupling Reaction17m
- Fukuyama Coupling Reaction15m
- Kumada Coupling Reaction13m
- Negishi Coupling Reaction16m
- Buchwald-Hartwig Amination Reaction19m
- Eglinton Reaction17m
Resonance Structures - Online Tutor, Practice Problems & Exam Prep
![](/channels/images/AIStars.png)
Resonance theory illustrates how electrons in a molecule can be distributed without altering atomic connectivity. Curved arrows represent electron movement, with double-sided arrows and brackets linking contributing structures. Cations adjacent to double bonds can shift with one arrow, while anions require two arrows due to octet preservation. The resonance hybrid combines these structures, indicating electron likelihood. Major contributors are typically neutral, fulfill octets, and place charges on more electronegative atoms, enhancing stability. Understanding these principles is crucial for grasping molecular behavior in organic chemistry.
Resonance is used to represent all the different ways that identical molecules can distribute electrons.
Rules for Drawing Contributing Structures
- Use curved arrows to represent electron movement
- Use double-sided arrows and brackets to link contributing structures to each other
- Arrows always travel from region of HIGH electron density to LOW electron density
The net charge of each structure must be equal
The rules you need for resonance:
Video transcript
Common Types of Resonance
Common ways to move arrows in resonance.
Video transcript
So what I want to do now is I want to talk about common forms of resonance. These are patterns that I've basically just discovered while teaching organic chemistry and I want to share these with you guys. So let's talk about basically 3 right now: movement of cations, anions, and neutral heteroatoms.
Let's talk about cations first. So cations, the way that this works is that if you have a cation next to a double bond, let's go ahead and put that next to a double bond. What that gives us the ability to do is now switch the place of those electrons. Because what I have is an area of high density on one side, which is a double bond. Remember that a double bond not only has a sigma bond but also has a pi bond. Remember that pi bonds are extra electrons that are shared between two atoms. So that's something that I can actually move. Then I have an area of low density, which is my positive charge. So if I want to move this around, what do I do? Well, it turns out that this is one of the easier examples: If you have a positive charge and a double bond next to each other, you can actually kind of swing them open like a door hinge using one arrow. So what that means is I would start from the high density, my double bond, and I would move towards the positive charge. But I wouldn't aim just towards the positive; I would make it towards that bond. So what I'm going to get now is that now I get a double bond in the place where the positive used to be, and now my positive moves over here. What that means is that now my positive is actually distributed from the left side over here on the red and then over on the blue side, it's going to the right side as well. So what I'm trying to say is that any time you have a positive charge next to a double bond, it can be represented by both of these drawings. It's not just going to stay in one place. Automatically, just by the laws of chemistry, it's going to wind up switching places at some point. The real molecule is going to look like an average of both of these or a combination of these. That's why I talked about the fact that none of them is a true representation. The best representation is by hybridizing both of these, and I'm going to talk about what that is in a little bit.
Now let's look at anions. What if I had a negative charge next to that double bond? Now I have to ask you guys, what do you think is going to be the region of the highest electron density? It turns out that the double bond has a lot, but now that we have a full negative charge, that's going to have even more electron density because a full negative charge means that it just has a lone pair just hanging out. So if we have a full negative charge, we're actually going to use two arrows. What I'm going to do is I'm going to take these electrons and push them into this bond, making a double bond, okay? But now we have an issue. If I make a double bond there, then let's look at this carbon right here. How many bonds did it already have? Well, it already had a double bond, that's 2. It already had a bond to hydrogen, that's 1. And then it already had a bond to carbon. So we had 4 bonds already. If I make another bond with that negative charge, how many bonds is that carbon going to have? It's going to have 5. That would be really, really bad. That would be basically impossible. You can't have a carbon with 5 bonds. That would be super terrible. So this is a situation where we're going to use a rule that's called make a bond, break a bond. So if I make a bond on this side, in order to preserve the octet of the middle carbon, I must break a bond. Because if I don't, then I'm going to give this carbon that I'm shading in green, I'm going to give it 5 bonds, and that just sucks. So if I make this bond, I have to break this bond. Okay. And when I break that bond, what winds up happening is that now I get a negative charge over here. So, as you can see, with a positive charge, I didn't have to actually break any bonds because I was never violating any octets. But in the movement of anions or negative charges, I do have to break a bond because I am going to violate an octet. So what's important to note here is that cations move with one arrow and then anions move with two arrows. Not so bad, right? So just remember that positive charges, they can swing like a door hinge, whereas two arrows, whereas with the negative charge, I'm going to use make a bond, break a bond because of the fact that I have to preserve that octet of the middle atom.
Alright, then let's look at neutral heteroatoms. This is the last situation, and what this would be is that remember I said we can move double bonds and we can move lone pairs. So imagine that I have a lone pair here. Well, let's say imagine that I have my two lone pairs there for that oxygen because remember that oxygen has a bonding preference of two bonds and two lone pairs, alright, cool. Then imagine that the nitrogen has one lone pair because remember that the nitrogen has a bonding preference of three bonds and one lone pair. Remember this? This is how it's going to satisfy its octet and how it's also going to satisfy its valence. Alright. So now let me ask you guys a question. Is there any way that I can turn these lone pairs, one of these lone pairs, into a double bond and not break an octet? Okay? And it turns out, let's look at our options. Basically, the two options are this: Either I could move one of these green lone pairs down here and make a triple bond or what I could do is I could move one of these red lone pairs here and make a double bond. It turns out you guys might be thinking, Johnny, why would I only move in that direction? Why couldn't I move like this? Why wouldn't I move the electrons down and make a double bond there? Well, first of all, the reason is that double bonds and electrons are the things that usually switch places. So I would want to go in the direction that's going to go towards the double bond. I wouldn't want to go away from it. So I would not go in this direction because that's away from my double bond. But on top of that, check this out, I have a hydrogen here. Right? If I move these electrons down into this area, I would make a double bond here. I'm sorry. I actually had more than one hydrogen. I had 2. There are 2 hydrogens there because that's a CH2. If I move these electrons in here and make a double bond, I'm going to break the octet down here and there's going to be no fixing that. There's actually no bond that I can break because these are all single bonds and you can't break single bonds in resonance theory. So what that means is that I would wind up getting a double bond down here that would violate this octet and it would suck. That would not be a good resonance structure. Lone pair jumps up and makes a double bond. So let's look at the O making a triple bond. If the O made a triple bond like this, let's look at this for a second. Don't draw it. Just look at it. How many bonds would this carbon have? It would also have 5, so we would break another octet by doing that. Is there any way that we could break a bond to make that carbon feel better? No. Because it turns out that there are just single bonds on both sides, so there's nothing we could do. We would be stuck. So we're definitely not going to move this lone pair either. So now I have one last choice. The last choice is that I would move these electrons from the n up and make a double bond. If I make a double bond here, how many bonds will that center carbon have? Still 5. So it looks like I'm screwed, either way, I'm always making 5 bonds, but there's one difference with this one. If I go ahead and go up and make the double bond up towards that carbon, guess what I can do? I can break a bond. So this is a situation where I'm making a bond towards a double bond. The reason that a double bond is helpful is because double bonds actually can break, whereas single bonds you're not allowed to break. But double bonds notice that I have these electrons in this double bond that are free to move. So what I'm going to do is I'm going to make a bond and then for the sake of preserving the octet of this carbon right here, I'm going to break a bond and that would be right here. So what I would do is I would basically turn two electrons from that bond into a lone pair on the oxygen, and that's going to preserve the four bonds that I need for that carbon right there because I'm making one, but I'm also breaking one. Okay? So what that would look like after I draw the resonance structure is I would now have a double bond here. I would have no electrons in the end because I just used those electrons to make the double bond. Then instead of having two lone pairs, now I'd have the two lone pairs from before. So let's go ahead and draw those, the green ones. But now I'm going to have one more lone pair. The last lone pair comes from the bond that I broke because basically what I did was I took two electrons from that double bond and I made them into a lone pair. Now all we have to do is count formal charges and we're done. So this oxygen wants to have 6 electrons, but it turns out that it has 7. Because remember we're counting sticks and dots, so this would have a negative charge. The n wants to have 5 electrons total, but right now it just has 4 bonds. It has the double bond. It has the single bond there, and then it has the hydrogen. So that means that the nitrogen wants 5, but it only has 4, so this would have a positive charge. That is our resonance structure. So what we have effectively done is we've taken these lone pairs and we've distributed them around, and now we're showing another way that these electrons can exist in this molecule. But notice that we're never moving single bonds. Single bonds are a big no no. Don't break those. And also, we're not rearranging the way the atoms are connected.
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/c0m1w9kvSbGpSB2iZpw1_175.png)
Draw all of the contributing structures for the following molecules:
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/vUnexGNdRim7LvBF11lq_176.png)
Drawing Resonance Structures
Video transcript
So there are a few things that you should remember that I told you guys are very important about resonance structures. First of all, remember that we use curved arrows. Easy. Remember that we use brackets with little double-sided arrows to link structures. But also remember that we always start from the area of highest electron density and work our way to the areas of less density. So in that case, that has to be the nitrogen because the nitrogen has a full negative charge on it. That means that is the most negative thing. So, what I want to do here is I want to try to move those electrons. The only way that I can move them is by becoming a double bond or becoming a pi bond. I'm sorry. So if these electrons moved down here and became a pi bond, that would be great, except I have a problem. If I did that, then this carbon would have 5 electrons on it or 5 bonds. So, 5 bonds is terrible. That would break the octet rule. It would be 10 electrons by the way. I'm sorry. Not 5. Ten electrons would break the octet rule. So if I make that bond, what do I have to do? Do you guys remember? I have to break a bond. Why did I draw it at the bottom? You could have drawn it at the top too. But if you make a bond, you have to break a bond. And this is that pattern that I told you guys that oops, that was weird that anions come with 2 arrows. The reason is because any time you're making that new double bond, you're going to have to break a bond as well or a new pi bond. So what that's going to do is it's going to give me a structure that looks like this. We now have N with a triple bond, carbon, and then an oxygen. Now, nitrogen already gave up one of its lone pairs to become a triple bond, so that means it only has one lone pair left. Carbon has the same amount of electrons as before, it's just arranged a little differently. And then oxygen has one additional lone pair because the electrons from that double bond became a lone pair. So basically, the additional lone pair is this red one, and that red one came from this bond over here breaking. Does that make sense? Cool. So now we have to do formal charges. So, I would have it's funny that I put my negative there. I actually would have a negative right here on the O. And then would I have any other charges that I have to worry about? No. The carbon is fine, and the N is fine. So really, that's it. I just got my resonance structure. Are there any other things that we could do? Not really because if I make this negative, let's say that I go back and put this negative back here, once again, I'm going to have to break a bond. So what that means is that these two resonance structures are going to be basically 2 different versions of the way this molecule could look, but that's it. That's the only thing that it can do. So I'm going to put brackets around this, and that's going to be a wrap. Okay. That's going to be the end of that problem.
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/CZ9qR5BASFixzCAU32dh_177.png)
Drawing Resonance Structures
Video transcript
All right. So remember that I said that we can move electrons as long as we're not breaking octets. Okay. We can't break octets. We can't have more than 8 electrons. So, in this case, I really only have one set of electrons that has my attention. We always want to start with the most negative thing, and that would be my lone pair because my lone pair is just these free electrons. So, if I were to move these electrons and make them into a double bond, would that be okay? Would I be breaking the octets? Actually, I would be if I just left it like that. Because remember, this carbon here already has a hydrogen. So if I made that double bond, I would now have 5 bonds to that carbon; that would suck.
So, can you guys see anything that I could do to fix that? What I could do is break a bond. So I could break this double bond and put those 2 electrons. Remember that there are 2 electrons in that double bond? In the additional π bond, we could take those 2 electrons and make them into a lone pair. So what that means is that let's just go ahead and draw this as double-sided arrows since we're going to draw a new resonance structure. What I would get is something like this where I have an NH2 here, but now I have a double bond and now I have a lone pair here. But remember that with bond line structures, usually, we don't include a lot of lone pairs. We instead want to use formal charges. So let's compute the formal charges here. By the way, that H is still there. I just didn't draw it because H's can be implied. So what would be the formal charge of this carbon right here now? Well, it wants 4 electrons and how many does it have? 5. It has 5 valence electrons, so this is going to have a negative charge. So I'm just going to erase the lone pair and I'm just going to replace it with a negative because I think that's a little easier to look at. Now let's look at the nitrogen. Does that one have a formal charge? Well, nitrogen wants 5 electrons and it has 4, so kind of like they swapped. The nitrogen has a positive charge.
All right. So there we have it. That is a resonance structure. Is there anything else that could happen? I know that some of you guys are wondering, okay, but couldn't I do something else? Couldn't I, let's say, make this negative do a double bond there? Couldn't I do that? And the answer is no, you couldn't. That would be terrible. Please don't do that. The reason is that think about it. There are already 2 hydrogens here. If I went ahead and tried to make a double bond here, first of all, that carbon would now have 5 bonds. Secondly, there's nothing else that I can break to make that work. You can never break single bonds with resonance structures. So what that means is that I would have to either break off one of the H's or I would have to cut off this carbon-carbon bond, which would suck. So that negative charge is stuck. It can't go there. You say, oh, well, what if it goes down? How would it be if I put it down here? The same exact thing. Once again, I've got 2 H's, and by making a double bond, I would be forced to break off a hydrogen or break off a carbon. It would suck. So that negative charge is stuck there. The only other thing that it can do is it can go back in the direction it came from. So if I made a double bond there, then that would be fine because then I could break this bond and make it a lone pair there. See how this works? So you basically keep going with that charge until you get stuck. Until there's nothing else you can do. So those are my resonance structures for this compound. Cool?
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/5RbgsHceTcGAxJ3rl5ZS_178.png)
Drawing Resonance Structures
Video transcript
All right. So remember that positive charges, I said they swing like a door hinge. So imagine that you're just opening up this door and you could just do that. So what I would get is, in my first resonance structure by the way, this resonance structure that I'm showing you is going to be super important for Orgo 2. But for right now, we're not going to concentrate on it too much. We're just going to do this. So now I have a double bond here and I have a positive charge here. The reason is that remember the double bond and the positive switch places when you do this resonance structure. So now what I ask myself is okay, is that positive charge stuck? Is there nothing else that it can do? Actually, no, it's not stuck because now it's next to another door hinge. So what I could do now is swing this one up like that and now I would have another resonance structure. This resonance structure is now going to have a double bond there, a double bond there, a double bond there, and a positive charge there. So now is that one stuck? Nope. It's still not stuck because it can do swing another door open, and that's going to be this one. Now some of you guys, some of you smart guys out there might be saying, Johnny, isn't that just the same thing that I did over there? Where I'm basically moving the double bond up or whatever. And it's similar, but actually with resonance structures, we want to draw every single movement that can happen, even if all of them look similar to you. So even if it looks like we're doing the same exact thing on both sides, you would still draw them because you want to indicate the motion of these electrons all over the molecule. So now what I'm going to do is draw that. I'm going to draw a double-sided arrow, and if this was actually a test, I probably wouldn't do this because it could be a little bit confusing. But I'm going to continue the resonance structure down here. So then what I would have is double bond, double bond, double bond, positive charge. And then that's it. Those are your 4 resonance structures. If you want, you could then show how you get back to the other one, and you could show that that is in resonance. That's fine. If you wanted to do that, that's fine. Now, we just have to set this off in brackets, so I'm just going to do bracket, bracket. Okay. So the resonance structures are the important part. The fact that I have double-sided arrows is important. Brackets are important. Then the way that I laid this out probably could have been better. It would have been awesome if I would have put all 4 in a horizontal row, but I couldn't fit all of them. I made my arrows too big. But don't worry about it too much. If anything, you could do something like this. If you're ever running out of space, you could just do something like this. Double-sided arrow. Double-sided arrow. That takes care of it. Your professor will know exactly what you're doing. Cool. So hopefully that helped resonance make a little bit more sense to you, and let me know if you have any questions.
Resonance Hybrids
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/rf9OMo70SFa9OGLWojyJ_179.png)
How to draw a resonance hybrid.
Video transcript
All right, guys. So we just talked about resonance structures and how one single molecule could have several different contributing structures. That's what we called each structure that has a slightly different distribution of electrons. We call that a contributing structure. Well, it turns out that what we want to talk about now is hybrids, how they blend together, and also which one would be the major structure in terms of which one would represent the way that the molecule looks the most. So, let's go ahead and begin. Basically, the resonance hybrid is going to be a mathematical combination of all the contributing structures. And what it does is it indicates where the resonating electrons within a molecule are most likely, oops, most likely to reside. So, what we do for this is we literally combine the different resonance structures into one drawing. And we combine them all into one drawing and then we try to analyze which one would be the resonance structure that would contribute the most to that hybrid.
Here's a molecule that we're going to deal with a lot in Organic Chemistry 2. It's called isocyanate. I don't really care that you guys know that much about it, but what's interesting is let's look at the contributing structures here. For one of these, I have two double bonds and then what I've done here is I've used the negative charge rule to make a bond break a bond. So, what I'm doing here is I'm taking these electrons here, making a triple bond, but then if I made that triple bond, that carbon would violate its octet. It would have five bonds. So then I'm going to break this bond and make a negative charge over there. So at the end, what I'm going to get is two different structures, one that has a negative charge on the N, one that has a negative charge on the O. What the resonance hybrid is, is it's a blend of both of these. What I would do is I would just draw the parts of the bond that are not changing. So, for example, notice that here I always have at least two bonds between the carbon and the nitrogen. In this structure, I actually have three bonds, but in this one, I have two, so I would draw those two. On the oxygen side, I always have at least one bond between the carbon and the oxygen, but in this case, I have two. Okay. Then what I would do is I would draw a partial bond from the nitrogen to the carbon and from carbon to the oxygen. What that indicates is that this bond is being created and destroyed at the same time. What it also indicates is that basically I'm in between both. Okay? Then finally, I put partial charges in all the places that have a negative charge. Why? Because the hybrid, like I said, it's not in equilibrium. This is not like we've talked about in Chemistry 2. We have a reaction that favors the right or favors the left, and it goes back and forth. No. What this is, it's a mathematical concept where I say, okay, this gets, let's say, 40% of the molecule. This is 60% and the actual molecule looks like a blend of both of them. Does that kind of make sense? So, in this case, I've drawn my hybrid. Notice that basically everything that's changing is showing on this hybrid. I'm showing that the bonds are being broken and created at the same time and I'm also showing that the negative charge is moving from one place to another. Okay?
The hybrid is the drawing of the mathematical combination of all contributing structures. Remember, you can never break single bonds!
Which of these structures looks the most like the hybrid?
Major Contributors
How to determine which structure is most stable.
Video transcript
Now, in terms of major contributors, that's when we determine, okay, which of these is the one that looks the most like the hybrid? Is it number 1 or is it number 2? Okay. And to figure that part out, we have to use just a few rules. Okay. So often it turns out that one of the resonance structures will be more stable. So it turns out let's say you have more than one resonance structure, one of them is the most stable. So that's going to be the one that we use and that means that it's going to contribute to the hybrid more than the others will. And major contributors will often have the following characteristics. Okay. So the first thing is that neutral structures are almost always going to be more stable than charged ones. So if I have a choice between let's say I have a resonance structure that's neutral and a resonance structure that has charges on it, I'm going to pick the neutral one to be my major contributor and to be the one that looks most like the resonance hybrid. Why? Because that's the one that's overall most stable. So even if the other one is possible, it may exist to some extent, but the one that's really going to exist in excess or not exist, but the one that's going to contribute in excess is going to be the neutral one. Does that make sense? Cool. So a good example for that would be where I showed you guys the neutral heteroatom example on the other page where there was one that had basically a neutral structure and then one that had a positive and a negative. The major contributor would be the one that was just fully neutral. The one that had positive and negative would be a minor contributor because that one already has charges. It's not as stable. So let's keep looking at this. Another rule is that if possible, every atom should fill its octet. So what that means is that we're going to look towards resonance structures that are not satisfying the octet. Let's say ones that have too few electrons. Those are usually going to be minor contributors. If I have a choice between a resonance structure that fulfills all of its octets and one that doesn't, I'm always going to pick the one that does fulfill its octets. Then finally, the electronegativity trends are going to determine the best placement of charges. Now I know it's been a really long time since you talked about electronegativity. In fact, for a lot of you guys, you haven't heard about it since general chemistry. So I just want to remind you guys that this is the electronegativity scale. What it basically says is that as you go to the right and as you go up, your electronegativity gets higher. What that means is that fluorine is the atom that is most comfortable having a negative charge or having electrons on it. So if I were to pick that the negative charge is on a fluorine or the negative charge is on a carbon, which one is going to be more stable? What do you guys think? The fluorine, right, because that's electronegative. That means that it likes to have electrons or negative charges on it. Whereas carbon is not as to the right as fluorine, so carbon is going to be a lot less comfortable having that negative charge. Does that make sense?
Often one of the resonance structures will be more stable, so it will contribute to the hybrid more than the others.
Use the octet rule and electronegativity trends to determine the best placement of charges. (i.e. Fluorine is more stable with a negative charge than oxygen).
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/1bJJBlWkRirg95VLXHDn_180.png)
How to determine which structure is most stable.
Video transcript
All right, guys. So it turns out that there were no neutral structures, so I couldn't use the neutral rule. All of these molecules fulfilled their octets, so I couldn't use the octet rule. But we had differences in electronegativity. It turns out that the O being with a negative charge is going to be more stable. So that would be my major contributor. Why? Because it turns out that O is more electronegative, I'm just going to use "for electronegative," than N. What that means is that oxygen is more comfortable having that lone pair on it than nitrogen is. So this would be less electronegative. Because of that, this is going to be the minor contributor. So when I go ahead and draw my resonance hybrid, we can draw it the same exact way, but we have to acknowledge that, let's say that I'm drawing it like this, NCO, partial bond, partial bond.
One of the ways that we could draw this is we could draw the partial negative on the O bigger. So we draw a bigger partial negative on the O and a smaller partial negative on the N. Why is that? Because remember, we just said that even though both of these could exist, the negative on the O is going to be the most stable. So that means that most of the time it's going to look more like this one—sorry, that kind of got blurry—more like this one and less like the other one. So this means that my hybrid would be a bigger share of the major contributor. Does that kind of make sense?
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/fJ8BUgFOQfKfvwd3nau3_181.png)
By applying the rules we learned to the above example, we saw that the negative charge could either rest on the nitrogen or on the oxygen. Since oxygen is more electronegative, that structure is the major contributor.
Let's practice by drawing all of the contributing structures for the following molecules. Label the major contributor if applicable and draw the resonance hybrid.
![Content](https://cdn.clutchprep.com/guide_visuals/inline_images/krsQ4sdRgmFL9TDy4r4g_182.png)
Drawing Contributing Structures
Video transcript
All right guys, so my resonance structures were as follows. I always start from the thing that's most negative and that's my negative charge. And I can actually go in 2 different directions here. I could either go in this direction or I could go in this direction. Now the reason that I know that I can go in both those directions is because my negative doesn't get stuck. Because if I make that bond, I can break a bond. So if I go towards the blue direction, I know that I would be able to break this bond in order not to violate the octet of that carbon, if I were to go in the red direction, then it could break that double bond in order not to violate the octet of this carbon. Does that make sense? So I have 2 different directions that we could go. Obviously, this notation is horrendous. You should never draw 2 different resonance structures on the same compound. So let's just go with the blue one first.
So the blue one would look like this. What I would get now is a double bond still there, but now I'd have a double bond here. The O stays the same and now I have an extra lone pair on that O or what I can just put as an O- because the negative charge has now transferred to that O. Okay? So that's one resonance structure. Is there anywhere else that that negative could go? Well, the only thing it could do is it could go back here and then what that would do is that would send these electrons back here. Now think about it. You might be thinking, well, could it go towards the O? Well, no. Think about it. If the strong bond went there, the only other option that I would have besides breaking the strong bond is to just kick off the O altogether in order to preserve the octet of that carbon, in order to make sure that it has 4 bonds still. But that's crazy. Like I said, you can't break single bonds. So my only option here is really to go back the exact way that I came. All right. So that shows you that's one set.
But I also told you that there's another possibility. What if I went in the other direction? Well, then that would lead to a structure that looks like this. So what I would have is that now I have a double bond here because remember I said that I'm going this way. And then this would break, so I would get a negative charge there. And then I would still have this double bond here, so I would have an O and an O-. Okay. So that one's a little ugly. Let me try to clean it up a little bit. There, there, there. Okay? So now I have to ask you guys, okay, is there anywhere else that that negative could go? Well, that negative could only go back where it came from and then that would just cause the first resonance structure that we had. All right. So those are our 3 major resonance structures. We basically made the negative charge go as far as it could until it got stuck and then that's it.
So now it's our job to figure out what the major contributor is going to be. So which one is the major contributor here? Which one looks like it's going to be the most stable? Remember, there were 2 rules: 1 was preserving octets, and the other one had to do with electronegativity. So in this case, do we have any octets that were breaking? No. All of them have octets. But I do have differences in electronegativity. In the first one, I had a negative charge on a carbon. In the second one, I had a negative charge on an oxygen. Which one is more electronegative? The oxygen. So what that means is that this is going to be my major contributor. Why? Because it is the one that has the negative charge on the most stable atom, the one that's most likely to be okay having a negative charge on it.
So now we just have to do one more thing and that is to draw my hybrid. So my resonance hybrid is going to have all the single bonds exactly the same. So if you have a single bond, draw it the same. But then everywhere that the negative charge is moving, you have to draw a partial bond. That means that bonds are breaking and being made at the same time. So then I would have partial bond there, partial bond there, partial bond there, and partial bond there. Why? Because notice that the negative charge had double bonds moving throughout all of those atoms. And where is the negative charge at any one time? It could be in the middle or it could be on the O, or it could be on the end. And in all reality, it's going to be a mathematical combination of all three of those. So what we're going to do is we're going to put partial negatives on each of the atoms that it could be on. So that would be those 3. And just so you know, another way to notate that that is sometimes used is instead of using partial negatives, it would just be to simply use a negative charge and just draw it right in the middle. Then that would show that the negative is being distributed throughout all of those atoms. All right. Does that make sense so far?
Drawing Contributing Structures
Video transcript
All right. So we can see that this example is something called an iminium cation, which I'll explain more later. But for right now, that doesn't really mean anything in terms of resonance structures. But what's the first thing we always want to look at when we look at a resonance structure? And it's where to start the arrow from. So right now, what do I have going for me? Well, I've got a positive charge and I've got 2 double bonds. So if I had to start my arrow from somewhere, where do you think we would start from? One of the double bonds, right? Because double bonds have electrons. We just want to start from high density to low density. So I'd want to start from one of the double bonds and then go to where? Go to the positive charge because the positive charge is the thing that's missing electrons. Okay? On top of that, there's one other pattern that we talked about that might be helpful here. Remember that positive charges tend to move with how many arrows? What do you guys remember? 1. So most likely, we're going to be using 1 arrow and we're going to be moving from negative to positive. So where will we start? Well, we've got let's say double bond A and double bond B. And even though I could start from either of these, I think B is the easiest one to visualize because it's the closest to the positive charge. So how could we move the electrons from double bond B towards that positive? And well, we learned that there's 2 things that double bonds can do. 1 is that they can donate electrons directly to an atom that they're adjacent to. So what could happen is that the double bond becomes a lone pair on the end. Okay. So we'll explore that. We'll see. But we also learned that double bonds can move, swing like a door hinge to other neighboring carbons or another neighboring atoms. So we kind of want to evaluate both of these possibilities. Is it possible to move it over as it like open it like a door or would it be just add it to the nitrogen? Well, in order to figure out if you can move it like a door, you need to look at the atom that you would be attaching it to. And what we see is that for example, this carbon here, we learned how to calculate how many hydrogens it has. How many does it have? It has 3. 1-2-3. So if I were to swing this double bond over like a door hinge, would I run into any problems? Yes, guys. Because now you have a double bond on that carbon, you'd be breaking the octet, right? So actually even though I kind of I'm thinking I want to swing it open, that's not possible there, okay? But maybe you're saying, but Johnny, there's another carbon at the top. How about that one? I have a carbon here. This one has how many H's? 2, right? 1, 2. Couldn't I maybe try to swing it open up to here? Well, if I did that, check it out. I'd be breaking the octet again because once again, now this carbon has 4 bonds with the double bond here, it would have 5. So both of those motions aren't possible. Okay? So that means what can I do with my double bond? Well, what I could do is I could take the electrons and I could donate them directly to the N, making a lone pair. So what that means is that for this resonance structure, what it would look like is like this: N draw the ring just like before. But now what changed? Well, this double bond stayed exactly the same. There's still a methyl group there or just a carbon, a CH3, right? That's what we'll call it for now. But now instead of having a double bond, now I'm going to get a lone pair on this N. That lone pair came from the electrons being donated to the N. Now notice that guys, remember I always like to count hydrogens when I'm doing these resonance structures. At least at the beginning because you're still getting your feet wet. You're still trying to understand these. So we can't be too careful with the way we calculate these. Notice that this carbon here only has 1 H. Why? Because it's got 3 bonds to carbon3 bonds. So it can only have 1 H. Well now it still only has 1 H. So what kind of charge should that carbon now have? Well, going based on our rules of formal charges, we know that carbon wants 4 bonds. How many does it actually have? It has 3. This carbon that I'm looking at right here only has 3. So 4 minus my sticks and my dots, which is equal to 3, equals positive 1. That means I should have a positive charge here. What that means is that my first resonance structure, I'll just erase this H now, looks like this. I took my electrons from the double bond and made a lone pair on the end and a positive charge on the carbon. If you guys want to verify the charge of the nitrogen, you'll find that it's neutral because nitrogen with a lone pair and 3 bonds is always neutral. So that's one resonance structure. Great job, guys. So now I'm just going to move this over so we have more space. Here I'm going to grab this and move it over here. So now guys, what is the next step? Do we have any other resonance structures possible? Well, what I like to say is let's take that positive and keep moving it all the way down until it can't move anymore. So is there anything else that it could possibly move with? And what I see is that I haven't used this double bond yet. So is there a way that that double bond could perhaps react with or resonate to the positive? And what I could try to do is swing it like a door hinge and see if that's going to help me. So what if I were to swing it like a door hinge? Would I break the octet? Well, right now, remember this hydrogen I mean this carbon has 1 H. So if I draw that, what I'm going to get is this. I'm going to get N, but now methyl or CH3, my bad, CH3. And then what I have is an H here, so it has 3 bonds, But now I just added a double bond here. So did I violate the octet of that carbon? It's perfect. Now it has 4 bonds. So actually, in this case, I actually can move the double bond down. And notice it's because it's next to a carbon with a positive charge, which we said when you have that specific situation, you can swing your door open like a door hinge. So that just shows that you could do that. Now, what should be the charge on this atom here? What should be the charge on that one? Well, we can just use the same method. How many hydrogens did it originally have? 1. How many does it have now? Still 1. So what would be the charge? Well, it only has 3 bonds so it should be a positive. So this is another resonance structure. So now I've drawn 3 resonance structures. I've drawn the original, Now the positive at the bottom and the positive now resonated to the left side. Now notice, guys, that there's these two rules that are like the most important rules of resonance structures which is 1, you can't move atoms. Have I moved any atoms so far? Have I moved moved any hydrogens, any carbons, any nitrogens? No. So that's good. That means I'm probably on the right track. 2, are all the net charges of my structures the same? Net charges, meaning they all add up to the same number of charges. Yes. The original cation was plus 1. My second structure is plus 1. My third structure is plus 1.
Great job! Not the easiest of topics but we got through it!
Do you want more practice?
More setsYour Organic Chemistry tutors
- For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers. e. o...
- In the following sets of resonance forms, label the major and minor contributors and state which structures wo...
- Use resonance structures to identify the areas of high and low electron density in the following compounds: a...
- Draw the important resonance forms to show the delocalization of charges in the following ions. In each case, ...
- b. Draw the resonance forms for ozone (bonded OOO ) c. Sulfur dioxide has one more resonance form than ozone. ...
- a. Draw the resonance forms for SO2 (bonded OSO )
- Draw the important resonance forms of the following cations and anions: [H2C=CHCNH2(=OH)]+ [H3CCHC(=O)CH=CHCN...
- Draw the important resonance forms for the following molecules and ions. NO3- NO2- HCOCH=CHCH2-
- Draw the important resonance forms for the following molecules and ions. CO32- H2C=CH—CH2 H2C=CH—CH2
- Determine whether the following pairs of structures are actually different compounds or simply resonance forms...
- Determine whether the following pairs of structures are actually different compounds or simply resonance forms...
- The following compound can become protonated on any of the three nitrogen atoms. One of these nitrogens is muc...
- Two resonance structures are shown for each molecule. Use the arrow-pushing formalism to represent the electro...
- Each pair of structures represents two valid resonance structures. Use the arrow-pushing formalism to justify ...
- Each pair of structures represents two valid resonance structures. Use the arrow-pushing formalism to justify ...
- Draw the resonance structure that would result from the indicated movement of electrons. (a)
- For each of the molecules shown, do the following: (i) Identify all pushable pairs. (ii) Identify all places w...
- (••) In the following molecules, identify all pushable electron pairs. (a)
- Two resonance structures are shown for each molecule. Use the arrow-pushing formalism to represent the electro...
- Two resonance structures are shown for each molecule. Use the arrow-pushing formalism to represent the electro...
- Two resonance structures are shown for each molecule. Use the arrow-pushing formalism to represent the electro...
- Identify the resonance structure that will be produced given the molecule shown and the electron flow indicate...
- Identify the resonance structure that will be produced given the molecule shown and the electron flow indicate...
- Draw the important resonance forms of the following free radicals. c. CH3COO⋅ d.
- Draw the important resonance forms of the following free radicals. b.
- Draw the important resonance forms of the following free radicals. a. CH2=CH—⋅CH2
- Acetonitrile (CH3C≡N) is deprotonated by very strong bases. Write resonance forms to show the stabilization ...
- Allylic halides have the structure H2C=CHC(X)H2 b. Draw the resonance structures of the allylic cations fo...
- 3-Bromocyclohexene is a secondary halide, it undergo SN1 substitution about as fast as most tertiary halides. ...
- Benzyl bromide is a primary halide. It undergo SN1 substitution about as fast as most tertiary halides. Use ...
- Which of the following has delocalized electrons? g. h. i.
- b. Draw the resonance contributors for anilinium ion.
- c. Draw the resonance contributors for the phenolate ion.
- Draw resonance contributors for the following ions: d.
- a. Predict the relative bond lengths of the three carbon–oxygen bonds in the carbonate ion. b. What is the ch...
- Draw resonance contributors for each of the following species and rank them in order of decreasing contributio...
- Draw resonance contributors for each of the following species and rank them in order of decreasing contributio...
- Draw resonance contributors for each of the following species and rank them in order of decreasing contributio...
- Draw resonance contributors for each of the following species and rank them in order of decreasing contributio...
- Draw resonance contributors for each of the following species and rank them in order of decreasing contributio...
- Which of the following has delocalized electrons? a. b. c.
- Draw all resonance structures of the following carbanion and carbocation. (a) (b)
- How many atoms share the unpaired electron in semiquinone?
- Draw the resonance contributors for the following radicals: a.
- Another way the preceding might be presented is to say that the minor resonance structure(s) reveal the reacti...
- A carboxylic acid has two oxygen atoms, each with two nonbonding pairs of electrons. (a) Draw the resonance f...
- Draw the resonance contributors of the peptide bond in the less stable configuration.
- Draw all possible resonance structures for the reactive intermediates shown. (b)
- Draw the important resonance contributors for both resonance-stabilized cations (in brackets) in the mechanism...
- Draw two important resonance structures involving the lone pair on oxygen for the molecule shown. Which carbon...
- Draw two important resonance structures involving the C―O π bond for the molecule shown. To which carbons woul...
- Use resonance forms to show delocalization of the negative charge in the Ruhemann's purple anion.
- Draw all possible resonance structures for the reactive intermediates shown.(c) <IMAGE>
- Draw the resonance contributors of the cyclooctatrienyl dianion.a. Which of the resonance contributors is the ...
- Which is longer, the carbon–oxygen single bond in a carboxylic acid or the carbon–oxygen bond in an alcohol? W...
- Draw resonance contributors for the following ions:a. <IMAGE>
- Draw the resonance contributors for the following radicals:c. <IMAGE>
- Draw the resonance contributors for the following radicals:d. <IMAGE>
- <IMAGE> (Solved Problem 20-1)(a) The solution given for Solved Problem 20-1 was missing some important r...
- All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance f...
- Which of the following has delocalized electrons?j. <IMAGE>k. <IMAGE>l. <IMAGE>
- For each of the following compounds, draw the important resonance forms. Indicate which structures are major a...
- Use resonance structures to identify the areas of high and low electron density in the following compounds:HC=...
- For each of these ions, draw the important resonance forms and predict which resonance form is likely to be th...
- For each of the following compounds, draw the important resonance forms. Indicate which structures are major a...
- Use resonance structures to identify the areas of high and low electron density in the following compounds:HC=...
- For each of these ions, draw the important resonance forms and predict which resonance form is likely to be th...
- For each of these ions, draw the important resonance forms and predict which resonance form is likely to be th...
- For each of the following compounds, draw the important resonance forms. Indicate which structures are major a...
- For each of the following compounds, draw the important resonance forms. Indicate which structures are major a...
- Use resonance structures to identify the areas of high and low electron density in the following compounds:g. ...
- Use resonance structures to identify the areas of high and low electron density in the following compounds:i. ...
- Draw the resonance contributors of the cyclooctatrienyl dianion.b. Which of the resonance contributors makes t...
- a. Draw resonance contributors for the following species, showing all the lone pairs:3. NO2-b. For each specie...
- i) Which of the following resonance structures represents the 'actual' structure of the molecule shown? (ii) W...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- Which resonance contributor in each pair makes the greater contribution to the resonance hybrid? a. b.
- Which is a correct statement? A. The delocalization energy of an ester is about 18 kcal/mol, and the delocali...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- a. Draw resonance contributors for the following species. Do not include structures that are so unstable that ...
- Which resonance contributor in each pair makes the greater contribution to the resonance hybrid? c.
- Which resonance contributor in each pair makes the greater contribution to the resonance hybrid? d.
- Draw another resonance form for each of the substituted allylic cations shown in the preceding figure, showing...
- Draw the Lewis structure of azidomethane (CH₃N₃) [Show two important resonance structures.]