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# Stabilization of a conjugate base: resonance

How resonance affects the stability of a conjugate base.

## Want to join the conversation?

• At , why does a carbon have -1 formal charge ? It is bonded to 2 hydrogen atoms and 2 carbon atoms so it's sharing 4 electrons, and 4-4=0. I don't get it, please help. :)
• It has 5 electrons: 1 from the H, 1 from each of the C atoms on either side of it, and the two lone pair electrons.
4 - 5 = -1
• Why can't the second molecule structure (bottom one) have a resonance structure? Can the pair electron between O and C on the bottom molecule move to the O to create a formal charge on O and would that be a resonance structure for the molecule structure on the bottom?
• It can't because if it had resonance, the molecule would have one formal charge of -1 on oxygen, +1 on the first carbon and -1 on the third carbon. This resonance is unusually unstable because you have three formal charges on a molecule.. On the other hand, resonance appears when there is a pi bond Next to astigmatism bond then followed by a lone pair of electron. Hope this help
• At about , when referring to the carbon that is drawn in navy blue, why is this charge considered localized and why are there no resonance structures for it? Thanks in advance!
• You can't draw resonance structures for the structure with the carbon in blue, because the adjacent carbon atoms already have four bonds and a complete octet, so the pair of electrons are localized. That pair of electrons cannot be moved to form a double bond with the carbon in blue and one of the adjacent carbons, because any of these carbons would get five bonds, which cannot be.
• At , I have a question referring to the second molecule (at the bottom):
is it not possible that the fourth carbon donates one of his protons to the negatively charged carbon right next to him? In this way, the third carbon is not negative anymore, while the fourth carbon gets the negative charge. At this point, is this not what we call delocalization?

I am sorry for my English, but I hope one can still understand my question.
• So, I wanna check something in my understanding, do we consider an acid to be a strong one by looking at the degree of stability the conjugated base it forms has, because if it's stable then it's most likely to be formed, which means that the acid is more likely to lose its proton, is this correct, am I getting this right?
• Yes. If the conjugate base is more stable, it's less likely to accept a proton to reform the acid.
(1 vote)
• Guys a small doubt. Doesn't acetic acid show resonance even b4 it gave up its H+.
So won't acetic acid be very stable so that it doesn't donate its H+ ??
• Resonance stabilization only operates on anions because it allows the overall negative charge to be dispersed, thus stabilizing the compound. Since acetic acid doesn't have an overall negative charge, resonance doesn't act to stabilize it. That's why we usually look at the conjugate base when judging the strength of the acid.
(1 vote)
• At when drawing the conjugate base of the acid, why is the other hydrogen left off of the structure? It isn't the hydrogen being donated so shouldn't it appear in the conjugate base structure?
(1 vote)
• We usually don’t include hydrogens in Lewis structures because a majority of them do not participate in reactions. The hydrogens which do we do include. In reality there are more hydrogens than were shown on the other carbons in the ring, but again we omit them because they are unimportant to the reaction. This is done for the sake of brevity. We assume that each carbon is saturated with the maximum number of hydrogens even if they are not explicitly shown. Usually what is done to show missing hydrogens on carbons is to include formal charges.

Hope that helps.
• what is the priority of the phenomenon(resonance, induction, hybridization, solvation) in deciding the stability of the conjugate base
(1 vote)
• Resonance is typically the strongest by far, next up might be hybridization but not always. The order of the other two will vary depending on conditions.
• when the charge is delocalized shouldn't it make it unstable? why does it make it more stable
(1 vote)
• Why should it be destabilising?

You are spreading the electrons over a greater area, logically that should be stabilising right?
(1 vote)
• how to find out whether the electrons r localised or declocalised
(1 vote)
• If there is a π orbital next to a p or π orbital, the electrons will be delocalized.
(1 vote)

## Video transcript

- This proton on ethanol has a pKa value of approximately 16, while this proton on acetic acid has a pKa value of approximately five. Remember, the lower the value for your pKa, the more acidic the proton. So with the lower pKa value, acetic acid is more acid than ethanol. We can explain why by looking at the conjugate bases. If ethanol donates this proton, the electrons in this bond, the electrons in magenta are left behind on the oxygen. So let's draw in the conjugate base. This oxygen would have three lone pairs of electrons, and one of those lone pairs would be the electrons in magenta. That gives this oxygen a negative one formal charge. Let's draw the conjugate base for acetic acid. If acetic acid donates this proton, then the electrons in magenta are left behind on the oxygen, so the conjugate base would have a carbon double bonded to an oxygen here with two lone pairs of electrons. And then on the right we would have another oxygen, this one with three lone pairs, one of those lone pairs would be the electrons in magenta. So that gives this oxygen a negative one formal charge. Let's compare our two conjugate bases. Both of them have a negative charge an oxygen, so there must be some other factor to stabilize a conjugate base, and that factor is resonance. So for the conjugate base on the right, we can take those electrons in magenta, we can take these electrons here and we can move them in, and then we can kick off these electrons on to the top oxygen. So let's draw the resulting resonance structure. So we would have for our top oxygen we would now have three lone pairs of electrons around the top oxygen, giving that top oxygen a negative one formal charge. This oxygen would now have only two lone pairs of electrons around it, and the electrons in magenta move into here. And these pi electrons, let me make them blue, moved off onto the top oxygen to give the top oxygen a negative one formal charge. So the negative charge on this oxygen is not localized to this one oxygen, it's spread out, it's delocalized. So there's actually some negative charge on this oxygen too. So remember, resonance structures are not perfect. In reality, it's really a hybrid of our two resonance structures. So the negative charge is spread out or delocaliazed over two oxygens, and when you spread out a negative charge, that has a stabilizing effect for the anion. So our conjugate base is stabilized by resonance, and since our conjugate base is stabilized by resonance, that means acetic acid is more likely to donate to this proton, and that's why we see a lower pKa value. If we compare that to the conjugate base for ethanol, alright this is called the ethoxide anion, alright we can't draw a resonance structure, we can't delocalize that negative charge. That negative charge is stuck on this one oxygen here, and that means this conjugate base is not as stable, it's not stabilized by resonance. So that means ethanol is not as likely to donate its proton, and that's why we see a higher pKa value for ethanol. So when you're trying to figure out a more acidic proton, draw the conjugate base and look for resonance. Here we have an organic compound, and our goal is to determine which is the more acidic proton. Is this the more acidic proton? Or is this the more acidic proton? So let's say a base comes along and takes this top proton here. Well the electrons in this bond in magenta will be left behind on this carbon I just circled in magenta, so let's draw that conjugate base. We would have our ring, we would have this carbon double bonded to this oxygen, this oxygen has two lone pairs of electrons on it, and the electrons in magenta would remain on this carbon right here in magenta which gives that carbon a negative one formal charge. Don't forget that carbon in magenta also has another hydrogen bonded to it, but I'm not drawing it in on the conjugate base just to make things easier to see. Let's say a base comes along and takes this proton. So that would mean the electrons in dark blue would remain behind on this carbon in dark blue that I just circled, so let's draw that conjugate base. So let me draw in our ring here, and then this oxygen would have two lone pairs of electrons on it. The electrons in dark blue would end up on this carbon in dark blue which gives that carbon in dark blue a negative one formal charge. Let me draw that in. Again remember, this carbon also has a hydrogen bonded to it, but I'm not gonna draw it in on the conjugate base because it makes it easier for us to see. Which of those two conjugate bases is the most stable? Well the top conjugate base is stabilized by resonance, you could take these electrons in magenta and you could move them in to form a double bond. Let's go ahead and show that, the electrons in magenta moving here, now we kick these electrons off on to the oxygen, so we would have now... When we draw our ring we'd have a double bond here, and this oxygen would have three lone pairs of electrons around it, giving it a negative one formal charge. So the electrons in magenta moved into here, and the electrons in, let's make these light blue, electrons in light blue moved off onto the oxygen to give the oxygen a negative one formal charge. Now remember, oxygen is more electronegative than carbon, and so this oxygen here is better able to bear this negative charge, than this carbon in magenta over here. So this on the right, this resonance structure actually contributes more to the overall hybrids, but our conjugate base is stabilized by resonance. That's in contrast to this conjugate base down here. We have a negative charge on this carbon, but that negative charge is localized to that carbon in dark blue right here. We can't draw any resonance, we can't show any resonance stabilization. Which means that this conjugate base is not as stable as what we drew up here, which did have some resonance stabilization. And that means that we found our answer, the more acidic proton is the one that has the conjugate base that's resonance stabilized. So this proton that's next door to this carbon double bonded to the oxygen that's on a carbon that's next door to this carbon doubled to the oxygen, that is the acidic proton. And this is something that will come up later in the course, it's very important to understand.