Why are most things in this world made from carbon?

Because it an extremely versatile atom, if you notice it is in group IV on the periodic table. Group 4 atoms need 4 additional electrons to complete its outer valence shell. Carbon is capable of forming many different bonds (single, double, and triple) with itself as well as forming stable bonds with many other atoms. Carbon's hybridization capability (sp,sp2,sp3) also allows many different geometric configurations (the sp3 tetrahedral geometry for instance). This versatility allows for extreme complexity making Carbon the ideal atom as the backbone for biologically active molecules. That's my explanation in a nut-shell, I hope it helped.

At 2:54 (and other places as well), how come the carbon has a -1 formal charge? I thought the formal charge should be 0, because it typically has 4 bonds, and it currently has 2 bonds and 2 lone electrons which results in 4-4 = 0.

The carbon with the formal charge of -1 has 3 bonds and 2 unshared electrons for a total of 5 electrons. 4 - 5 = -1

Why don't we put the electrons on the Benzene ring to the next available carbon like in the phenoxide ion example? I don't' understand why with benzene, you make a pi bond right away and with phenoxide you move it to a carbon and form a carbanion.

In that case carbon would get -1 formal charge. In the previous video on resonance pattern he mentioned that the charges should be conserved while drawing resonance structures. So from neutral we cannot make carbon negative. Only the formal charge can be transferred from one atom to another, It cannot be created. I hope it helps

At 2:26, why cant the pi bond shift over to the next c-h bond and form another pi bond straight away? must it form a lone pair on the next carbon first?

Yes, it can shift directly to the next c-c bond. But in this case the blue pi bond would have to shift to (because a carbon atom can't have more than 4 bonds at a time and it already has 3 sigma bonds to the carbons next to it and the hydrogen). Then the green bond would have to shift as well and that at last *must* go onto the carbon atom. The carbon on top already has 4 bonds and can't take another one.

I know carboxylic acid is more acidic than phenol, but phenoxide ion has more resonance structures than ethanoate ion this makes phenoxide ion more stable so phenol should be more acidic than ehanoic acid. But it is not, whyy ??

Good question! The stability of an ion does not depend on the number of resonance structures that can be drawn for it, but rather on the stability of its resonance hybrid structure. Remember, the hybrid structure is the one that accurately represents the actual molecule. Of course, when comparing a molecule that has one or more resonance structures to one that has none, the former will have a more stable ion because the resulting negative charge can be delocalized. :)

How likely is it that in the video at ~5:30 the electrons would travel clockwise again and the double bond resonance would stay in one area of the molecule rather than making the full circle? Or in other words, how likely is it that this molecule would complete the full sequence of "resonance" back to the starting point without backtracking any number of times?

The electrons aren't travelling at all. We just have to draw the resonance contributors in some order that makes sense. The actual structure isn't part of the time one structure and part of the time another structure. The structure is at all times a single resonance hybrid of all the structures.

Why does oxygen form a pi bond, i mean why doesnt it just stay how it is. Why is isn't it satisfied with 3 lone pairs and 1 sigma bond?- What would happen if it stayed that way? Does resonance happen for stability?

The negative charge on the oxygen generally means it's going to do something if it can. Phenol is quite a weak acid (pKa ~10) so it's likely phenoxide would react with a hydrogen ion or water if given the chance. But besides that, the whole point of this video was to show there is partial double bond character in that C-O bond due to the resonance structures Jay drew. Resonance does imply a certain amount of stability.

This is a very stupid question but I'm just curious: Can't the resonance backfire? (At 4:10 it is clearly seen that the carbon lone pairs are next to two pi bonds, so it can go either way right?)

Sure. But going the other way would not give us a new resonance structure would it?

What is the delocalized charge over benzene?

What charge are you talking about?? There are six electrons delocalised over the ring of carbon atoms but it's not usually thought of as a charge.

Main content

Organic chemistry

Course: Organic chemistry > Unit 2

Lesson 2: Resonance structures

Resonance structures for benzene and the phenoxide anion

Name: Resonance structures for benzene and the phenoxide anion
Uploaded: 2014-06-14T14:58:24Z
Description: Examples of how to draw resonance structures for molecules with aromatic rings.

Google Classroom

Examples of how to draw resonance structures for molecules with aromatic rings. Created by Jay.

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Leon Hook Hook
Posted 9 years ago. Direct link to Leon Hook Hook's post “Why are most things in th...”
Why are most things in this world made from carbon?
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(15 votes)
Answer
- kyo1356
  Posted 9 years ago. Direct link to kyo1356's post “Because it an extremely v...”
  Because it an extremely versatile atom, if you notice it is in group IV on the periodic table. Group 4 atoms need 4 additional electrons to complete its outer valence shell. Carbon is capable of forming many different bonds (single, double, and triple) with itself as well as forming stable bonds with many other atoms. Carbon's hybridization capability (sp,sp2,sp3) also allows many different geometric configurations (the sp3 tetrahedral geometry for instance). This versatility allows for extreme complexity making Carbon the ideal atom as the backbone for biologically active molecules. That's my explanation in a nut-shell, I hope it helped.
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  (35 votes)
Austin Chen
Posted 9 years ago. Direct link to Austin Chen's post “At 2:54 (and other places...”
At
2:54
(and other places as well), how come the carbon has a -1 formal charge? I thought the formal charge should be 0, because it typically has 4 bonds, and it currently has 2 bonds and 2 lone electrons which results in 4-4 = 0.
Button navigates to signup pageComment on Austin Chen's post “At 2:54 (and other places...”
(15 votes)
Answer
- ElectronSea
  Posted 8 years ago. Direct link to ElectronSea's post “The carbon with the forma...”
  The carbon with the formal charge of -1 has 3 bonds and 2 unshared electrons for a total of 5 electrons. 4 - 5 = -1
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  (4 votes)
Edgar.A.Barillas
Posted 9 years ago. Direct link to Edgar.A.Barillas's post “Why don't we put the elec...”
Why don't we put the electrons on the Benzene ring to the next available carbon like in the phenoxide ion example? I don't' understand why with benzene, you make a pi bond right away and with phenoxide you move it to a carbon and form a carbanion.
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(8 votes)
Answer
- siddharth.bhatia17
  Posted 8 years ago. Direct link to siddharth.bhatia17's post “In that case carbon would...”
  In that case carbon would get -1 formal charge. In the previous video on resonance pattern he mentioned that the charges should be conserved while drawing resonance structures. So from neutral we cannot make carbon negative. Only the formal charge can be transferred from one atom to another, It cannot be created.
  I hope it helps
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  (5 votes)
Muhammad Farhaan
Posted 9 years ago. Direct link to Muhammad Farhaan's post “At 2:26, why cant the pi ...”
At
2:26
, why cant the pi bond shift over to the next c-h bond and form another pi bond straight away? must it form a lone pair on the next carbon first?
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(5 votes)
Answer
- digedag
  Posted 9 years ago. Direct link to digedag's post “Yes, it can shift directl...”
  Yes, it can shift directly to the next c-c bond. But in this case the blue pi bond would have to shift to (because a carbon atom can't have more than 4 bonds at a time and it already has 3 sigma bonds to the carbons next to it and the hydrogen). Then the green bond would have to shift as well and that at last must go onto the carbon atom. The carbon on top already has 4 bonds and can't take another one.
  Comment on digedag's post “Yes, it can shift directl...”
  (5 votes)
Ali Arshad
Posted 8 years ago. Direct link to Ali Arshad's post “I know carboxylic acid is...”
I know carboxylic acid is more acidic than phenol, but phenoxide ion has more resonance structures than ethanoate ion this makes phenoxide ion more stable so phenol should be more acidic than ehanoic acid. But it is not, whyy ??
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(5 votes)
Answer
- Darmon
  Posted 7 years ago. Direct link to Darmon's post “Good question! The stabil...”
  Good question! The stability of an ion does not depend on the number of resonance structures that can be drawn for it, but rather on the stability of its resonance hybrid structure. Remember, the hybrid structure is the one that accurately represents the actual molecule. Of course, when comparing a molecule that has one or more resonance structures to one that has none, the former will have a more stable ion because the resulting negative charge can be delocalized. :)
  Button navigates to signup page
  (4 votes)
tlueddek
Posted 8 years ago. Direct link to tlueddek's post “Some of the resonance str...”
Some of the resonance structures look identical. Like you could just flip them over to get the other one. Is there actually 5 resonance structures? Are they all different formations? I'm guessing if they are, is there a chiral center at the carbonyl carbon? I thought it needed to be 4 different groups attached?
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(5 votes)
Answer
Alex Bridges
Posted 8 years ago. Direct link to Alex Bridges's post “How likely is it that in ...”
How likely is it that in the video at ~
5:30
the electrons would travel clockwise again and the double bond resonance would stay in one area of the molecule rather than making the full circle? Or in other words, how likely is it that this molecule would complete the full sequence of "resonance" back to the starting point without backtracking any number of times?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ernest Zinck
  Posted 8 years ago. Direct link to Ernest Zinck's post “The electrons aren't trav...”
  The electrons aren't travelling at all. We just have to draw the resonance contributors in some order that makes sense.
  The actual structure isn't part of the time one structure and part of the time another structure.
  The structure is at all times a single resonance hybrid of all the structures.
  Button navigates to signup page
  (6 votes)
Mitko Gjorgjiev
Posted 6 years ago. Direct link to Mitko Gjorgjiev's post “Why does oxygen form a pi...”
Why does oxygen form a pi bond, i mean why doesnt it just stay how it is.
Why is isn't it satisfied with 3 lone pairs and 1 sigma bond?-
What would happen if it stayed that way?
Does resonance happen for stability?
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(2 votes)
Answer
- Ryan W
  Posted 6 years ago. Direct link to Ryan W's post “The negative charge on th...”
  The negative charge on the oxygen generally means it's going to do something if it can. Phenol is quite a weak acid (pKa ~10) so it's likely phenoxide would react with a hydrogen ion or water if given the chance.
  
  But besides that, the whole point of this video was to show there is partial double bond character in that C-O bond due to the resonance structures Jay drew. Resonance does imply a certain amount of stability.
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  (2 votes)
Swarna
Posted 7 years ago. Direct link to Swarna's post “This is a very stupid que...”
This is a very stupid question but I'm just curious: Can't the resonance backfire? (At
4:10
it is clearly seen that the carbon lone pairs are next to two pi bonds, so it can go either way right?)
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(2 votes)
Answer
- Ryan W
  Posted 7 years ago. Direct link to Ryan W's post “Sure. But going the othe...”
  Sure. But going the other way would not give us a new resonance structure would it?
  Comment on Ryan W's post “Sure. But going the othe...”
  (2 votes)
Farhan Ali
Posted 6 years ago. Direct link to Farhan Ali's post “What is the delocalized c...”
What is the delocalized charge over benzene?
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(2 votes)
Answer
- Ryan W
  Posted 6 years ago. Direct link to Ryan W's post “What charge are you talki...”
  What charge are you talking about??
  
  There are six electrons delocalised over the ring of carbon atoms but it's not usually thought of as a charge.
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  (2 votes)

Video transcript

Voiceover: Here is the dot structure for benzene C6H6 and we can draw a resonance structure for this. We could take these electrons right here. Move them over to here. That would mean too many bonds to this carbon. We have to take these electrons and push them to here, which would mean too many bonds to this carbon and so finally, we take these pi electrons and move them over to here. We draw our resonance brackets and go ahead and draw our other resonance structure for benzene. The electrons move over to here, to here and then finally, to here. And so, let's follow those electrons. Let's make the ones on the top left here red. These electrons in red. I'm showing them move to over here and let's make these electrons over here green. The electrons in green move down to here and then finally, we'll use blue. These electrons in blue. I showed them moving over to here and remember that the actual benzene molecule is a hybrid of these two resonance structures. If you're just drawing on a sheet of paper, you could use one or the other but remember that it's actually the hybrid because our dot structures are just not perfect ways to represent molecules or ions. Those are our resonance structure for benzene. Next, let's look at the phenoxide anion. Here is the phenoxide anion down here and I'm gonna try to color code the electrons. Let me go ahead and make these electrons in here, red and let's make these right here, green and then let's make these blue. Just like we did with the benzene ring up above and I could start off for the resonance structures for the phenoxide anion by doing the other resonance structure just like we did for benzene like that but I'm gonna save that for the end and so, let's think about what we would do first. We know one of our patterns is a lone pair next to a pi bond and that's what we have here. If you think about a lone pair of electrons on this oxygen. I'll make it magenta. That lone pair is next to the pi bond. The one in red and so, we can go ahead and draw a resonance structure and we take these electrons in magenta and move then into here. That would mean too many bonds to this carbon. We take the electrons in red and we push them off onto this carbon. Let's go ahead and draw our resonance structure. We have our ring here and we have now a double bond between the oxygen and the carbon. Only two lone pairs of electrons on this oxygen now. The electrons in magenta move in here to form a pi bond and the electrons in red move off onto this carbon right here. That's gonna give that carbon a -1 formal charge. Let's go ahead and draw a -1 formal charge here. The electrons in blue have not moved and the electrons in green, I haven't showed moving yet either. We put those in there like that. All right, next, we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond. The lone pair of electrons are the electrons in red right here. Next to a pi bond, the electrons in blue. Let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here. That would mean too many bonds to this carbon. If you take these electrons in blue and push them off onto this carbon. Let's draw that resonance structure. Once again, we have our carbon double bonded to an oxygen up here. We said that these electrons were the ones in magenta and the electrons in red moving here to form a pi bond. The electrons in blue move off onto this carbon and that gives this carbon a -1 formal charge. This carbon has a -1 formal charge. This one right here the one that has the blue electrons on it. We still have our electrons in green over here and we have the exact same pattern. We have a lone pair next to a pi bond. The lone pair are the ones in blue and this time, the pi bond are the electrons in green here. We can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon. To take the electrons in green and push them off onto that carbon and let's draw that resonance structure. Once again, we have our ring. We draw our ring in here. We have this double bond up here. Put in lone pairs of electrons on the oxygen and these electrons were the ones in magenta and we go around the ring. We had our electrons in red right here. The electrons in blue move into here and finally, the electrons in green move off onto this carbon. This carbon right here in green. Therefore, that carbon gets a -1 formal charge now. Thinking about it again, we once again have a lone pair of electrons next to a pi bond. We have the electrons in green. Those electrons could move into here and then that would mean too many bonds to this carbon. We can take these electrons in magenta and push them off onto the oxygen. Let's go ahead and draw our last resonance structure here. We have now a single bond to this top oxygen and three lone pairs of electrons giving that top oxygen a -1 formal charge. The electrons in magenta, let's say that those electrons are these electrons right here. Going around our ring, we had electrons in red. We had electrons in blue right here and then finally, the electrons in green move into here. We have five total resonance structures for the phenoxide anion. I can go ahead and put brackets around all five of these and since we're talking about resonance structures in the benzene ring, we can think about going back and forth between these two as a final thought here. We could take these electrons in red, push them into here, which would take these electrons In green over to here, which would take those electrons in blue over to here and then that would give us the one that we started with as well. It's also important to think about the hybrids. The hybrid has the negative charge delocalized. The negative charge is delocalized over. We could see in this resonance structure and this one, the negative charge is on the oxygen and this one, it's on this carbon. And this one is on this carbon and this one is on this carbon. The negative charge is delocalized over oxygen and three carbons when you're thinking about the resonance hybrids.