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Course: Organic chemistry > Unit 2
Lesson 2: Resonance structuresResonance structures
Introduction to resonance structures, when they are used, and how they are drawn. Created by Jay.
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- this may seem stupid.. but,in the very first example in this video..isnt the resonating structure the same as the original?its just the inverted form of it....(78 votes)
- The equivalent ressonance structures seem like the same but there are non equivalent ressonance strutures that occur when the delocalization of electrons is between qualitativity different bonds (they are different because they bond different atoms for instance a nitrogen and a carbon and two carbons)(6 votes)
- Do only multiple bonds show resonance ?(31 votes)
- Yes, resonance is a part of the pi bond system only.(51 votes)
- why delocalisation of electron stabilizes the ion(26 votes)
- Another way to think about it would be in terms of polarity of the molecule. If you have electrons that are localised on one particular atom, there would be a lot of polarity, thus the molecule would be more likely to both react and bond with other molecules. This decreases its stability.(9 votes)
- I still don't get why the acetate anion had to have 2 structures? aren't they both the same but just flipped in a different orientation? why does it have to be a hybrid? from what i understand, only one oxygen should be negative since a hydrogen nucleus left the molecule but what i'm seeing is that 2 oxygens are negative and this doesn't make sense(9 votes)
- The two oxygens are both partially negative, this is what the resonance structures tell you!
While both resonance structures are chemically identical, the negative charge is on a different oxygen in each. This is important because neither resonance structure actually exists, instead there is a hybrid. The oxygens share the negative charge with each other, stabilizing it, and reducing the charge on either atom.(26 votes)
- Where is a free place I can go to "do lots of practice?" ... Where can I get a bunch of example problems & solutions?(15 votes)
- It might be best to simply Google "organic chemistry resonance practice" and see what comes up.(3 votes)
- Are two resonance structures of a compound isomers??(5 votes)
- They are not isomers because only the electrons change positions.
Isomers differ because atoms change positions.(13 votes)
- Around8:44I don"t understand what does the stability of whats left have to do with the leaving H+?(7 votes)
- The more stable a conjugate base is the strong the acid is due to the equilibrium favoring the forward reaction a little bit more. This is relatively speaking. A non organic example are the halides, where the iodine anion is more stable than the flourine anion leading to a difference in the pKa of HF (3.2) and HI (-10). This is carried over to resonance structures, if your conjugate base has a resonance structure it's charge is delocalised and the anion is resonance stabilised, making it's corresponding acid stronger.(5 votes)
- Why at1:19does that oxygen have a -1 formal charge? It was my understanding that oxygen's atomic number was 8, and that particular oxygen has 7 electrons. Therefore, 8 - 7 = +1, not -1. Can anyone explain where I'm wrong?(4 votes)
- You're right to say that an oxygen atom has 8 electrons, but only 6 of them are valence electrons. Non-valence electrons aren't shown in Lewis structures.(7 votes)
- How do you find the conjugate acid? They were mentioned around7:55but it was not explained how he knew those were the conjugate bases.(4 votes)
- A conjugate acid/base pair are chemicals that are different by a proton or electron pair. For instance, the strong acid HCl has a conjugate base of Cl-. Remember that acids donate protons (H+) and that bases accept protons. So each conjugate pair essentially are different from each other by one proton. There's a lot of info in the acid base section too!(4 votes)
- I'm confused at the acetic acid briefing...
So a single bond naturally takes only one electron from the oxygen, but then a double bond takes two more electrons? i thought it should only take one more.(4 votes)- The single bond takes a lone pair from the bottom oxygen, so 2 electrons. The double bond gives 2 electrons to the top oxygen, forming a lone pair on the top oxygen.
Is that answering to your question ?(3 votes)
Video transcript
Voiceover: Sometimes one
dot structures is not enough to completely describe
a molecule or an ion, sometimes you need two or
more, and here's an example: This is the acetate anion,
and this dot structure does not completely
describe the acetate anion; we need to draw another
resonance structure. And so, what we're gonna
do, is take a lone pair of electrons from this
oxygen, and move that lone pair of electrons in here,
to form a double-bond between this carbon and that oxygen. And at the same time,
we're gonna take these two pi electrons here, and move
those pi electrons out, onto the top oxygen. So let's go ahead and draw a resonance, double-headed arrow here,
and when you're drawing resonance structures, you
usually put in brackets. And let's go ahead and draw the other resonance structure. So now, there would be a double-bond between this carbon and this oxygen here. This oxygen on the
bottom right used to have three lone pairs of electrons around it, now it only has two, because one of those lone pairs moved in, to form that pi bond. The oxygen on the top used
to have a double-bond, now it has only a single-bond
to it; and it used to have two lone pairs of
electrons, and now it has three lone pairs of electrons. That gives the top oxygen a
negative-one formal charge, and make sure you
understand formal charges, before you get into drawing
resonance structures, so it's extremely important
to understand that. All right, so next, let's
follow those electrons, just to make sure we
know what happened here. So, these electrons in
magenta moved in here, to form our pi bond, like
that, and the electrons over here, in blue, moved
out, onto the top oxygen, so let's say those electrons in blue are are these electrons, like that. So, we have two resonance structures for the acetate anion,
and neither of these structures completely
describes the acetate anion; we need to draw a hybrid of these two. And so, if we take a look
at, let's say the oxygen on the bottom-right here, we
can see there's a single-bond between this carbon and this oxygen. If we look at this one over here, we see there is now a double-bond between that carbon and the oxygen. So, if you think about
a hybrid of these two resonance structures, let's go ahead and draw it in here, we
can't just draw a single-bond between the carbon and that oxygen; there's some partial,
double-bond character there. So, it's a hybrid of the
two structures above, so let's go ahead and draw in a partial bond here, like that. The exact same thing for the top oxygen: Here we have a double-bond,
and then over here we have a single-bond,
so somewhere in between is going to be our hybrid. So let's go ahead and draw that in. So, we can't just draw a
single-bond in our hybrid; we have to show some partial,
double-bond character, drawing the dotted line
in there, like that. And also charge, so if
we think about charge, the negative charge is on the oxygen on the bottom-right, and then over here the negative charge is on the top oxygen. And, so that negative charge
is actually de-localized, so it's not localized to one oxygen; it's de-localized, it's
distributed evenly, over both of those oxygens, here. And so this is just one way
to represent the hybrid, here, and studies have shown that the hybrid is closer to what the
actual anion looks like. So, studies have been
done on these bond lengths here, and the bond between
this carbon and this oxygen, it turns out to be the
exact same bond length as the bond between the
carbon and this oxygen, so, it's the exact same bond length. And so, the hybrid,
again, is a better picture of what the anion actually looks like. Let's think about what
would happen if we just moved the electrons in magenta in. So if I go back to the very first thing I talked about, and you're
like, "Well, why didn't "we just stop, after moving
these electrons in magenta?" Let's go ahead and draw
what we would have, if we stopped after moving
in the electrons in magenta. So we would have this, so
the electrons in magenta moved in here, to form our double-bond, and if we don't push off
those electrons in blue, this might be our resonance
structure; the problem with this one, is, of course the fact that this carbon here
has five bonds to it: So, one, two, three,
four, five; so five bonds, so 10 electrons around it. We know that carbon can't exceed the octet of electrons, because of its position on the periodic table, so
this is not a valid structure, and so, this is one of
the patterns that we're gonna be talking about in the next video. So the pattern is, a
lone pair of electrons, so next to a pi bond, which is the example we see here for the acetate anion, and so these are the two resonance structures. The problem with the
word, "resonance," is, when you're a student,
you might think that the anion will resonate back and forth between this one and this one; that's just kind of what the name seems to imply. And that's not actually what's happening; it's just that we can't
draw, if we're just drawing one dot structure, this is not an accurate description,
and so the electrons are actually de-localized,
so it's not resonating back and forth. When you draw resonance
structures in your head, think about what that
means for the hybrid, and how the resonance
structures would contribute to the overall hybrid. So don't forget about
your brackets, and your double-headed arrows, and
also your formal charges, so you have to put those in, when you're drawing your
resonance structures. All right, let's look at an
application of the acetate anion here, and the resonance
structures that we can draw. If we look at the
acetate anion, so we just talked about the fact that one of these lone pairs here, so this is
not localized to the oxygen; it's de-localized, so we
can move those electrons in here, we push those
electrons off, onto the oxygen, we can draw a resonance structure, and so this negative-one formal
charge is not localized to this oxygen; it's de-localized. And so, because we can
spread out some of that negative charge, that
increases the stability of the anion here, so
this is relatively stable, so increased stability,
due to de-localization. So, the fact that we can draw an extra resonance structure,
means that the anion has been stabilized. And so, this is called,
"pushing electrons," so we're moving electrons
around, and it's extremely important to feel comfortable
with moving electrons around, and being able to follow them. So, the only way to get
good at this is to do a lot of practice problems,
so please do that; do lots of practice
problems in your textbook. If we compare that to the ethoxide anion, so over here, if we try
to do the same thing, if we try to take a lone pair of electrons on this oxygen, and move it into here, we can't do that, because
this carbon right here, already has four bonds;
so it's already bonded to two hydrogens, and then we
have this bond, and this bond. And so, moving those
electrons in, trying to de-localize those electrons, would give us five bonds to carbon,
and so we can't do that; we can't draw a resonance structure for the ethoxide anion. So those electrons are
localized to this oxygen, and so this oxygen has a full,
negative-one formal charge, and since we can't spread
out that negative charge, or it's going to destabilize this anion. So this is not as stable,
so decreased stability, compared to the anion on the left, because we can't draw a resonance structure. If we think about the
conjugate acids to these bases, so the conjugate
acid to the acetate anion would be, of course, acetic acid. So we go ahead, and draw
in acetic acid, like that. The conjugate acid to the ethoxide anion would, of course, be ethanol. So we go ahead, and draw in ethanol. And we think about which
one of those is more acidic. We know that acetic acid is more acidic; it's more likely to donate a proton, because the conjugate base is more stable, because, you could think about resonance, or de-localization of electrons. If you're looking at
ethanol, ethanol's not as likely to donate its proton, because the conjugate
base, the ethoxide anion is not as stable, because you can't draw any resonance structures for it. The negative charge is not
able to be de-localized; it's localized to that oxygen. So this is just one
application of thinking about resonance structures, and, again, do lots of practice. In the next video, we'll
talk about different patterns that you can
look for, and we talked about one in this video:
We took a lone pair of electrons, so right here
in green, and we noticed this lone pair of electrons
was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that
situation with ethoxide: We have a lone pair of electrons, but we don't have a pi bond next to it, And so, more in the next video on that.