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Organic chemistry
Course: Organic chemistry > Unit 2
Lesson 2: Resonance structuresCommon mistakes when drawing resonance structures
How to avoid common mistakes when drawing resonance structures.
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- In both examples, the carbon ring has one atom that is bonded to two hydrogens instead of one. How does he know which carbon atom has two hydrogens on it based only on the simple hexagon diagram? Why are there two hydrogens there in the first place?(9 votes)
- For the second problem, he says that carbon can never exceed an octet, but when you move the lone pair onto the carbon (in the correct version) isn't the carbon still exceeding an octet?(3 votes)
- No, it isn't because it is "trading" its pi/double bond for a lone pair on the single carbon. It gains a formal charge but does not exceed the octet rule.(8 votes)
- So are our H stationary in this case?(1 vote)
- When drawing resonance structures, ONLY the electrons move.
EVERY atom remains stationary.(11 votes)
- why can carbon not exceed an octet??(3 votes)
- Carbon only has access to 4 orbitals in its shell (a s orbital and three p orbitals). Each orbital takes two electrons, so it can only have 8 electrons in its shell (an octet). Too exceed an octet, an atom needs access to more orbitals or hybrid orbitals. Thus, they need d orbitals, but atoms do not have access to d orbitals until the third row of the periodic table.(6 votes)
- at, the two resonance structures ARE NOT enantiomers, right? if you flip one across the mirror plane, you get the other structure but they're just rotations of each other right? 3:42(1 vote)
- The two resonance structures are not enantiomers because they do not exist.
They are drawings that we use to explain the properties of compounds.
The actual structure is a resonance hybrid of the two structures.(6 votes)
- How do you identify wheater or not resonance you need to do a resonance structure or not? For, example would we draw resonance for H2O if yes then why so?(1 vote)
- You could draw a resonance contributor for water as HO⁻ H⁺, but it would be so high in energy that it would be an extremely minor contributor.
Resonance usually involves π electrons in conjugated double bonds, an arrangement like
A=B-C=D or A=B-C:
In this course, the atoms in the chain are usually carbon atoms, but they can be other elements as well.(6 votes)
- Inwhy didn't we (instead of moving the pie bond to the positive carbocation) move the 2 electrons in the pie bond to form a lone pair on any carbon?? 2:28(0 votes)
- Let's number the double bond carbons as C-1 and C-2 and the cationic carbon as C-3.
We could move the π electrons onto C-1, as you suggest.
That would put a lone pair and a negative charge on C-1, a positive charge on C-2, and a positive charge on C-3.
This is a legitimate resonance contributor, but it is a very minor contributor because:
1. It now has three charges instead of one, and
2. It puts positive charges on adjacent carbon atoms.
Both factors make the structure a high-energy, and therefore a VERY minor, contributor.
Remember, electrons tend to move towards a positive charge.(6 votes)
- In the 2nd example, Why isnt the first resonance structure( which sal states it to be wrong) correct? Because if we moved the pi electrons ( as the carbon is eexceeding octet) over the carbon givin it a negative charge? It would fit right.(0 votes)
- the reason for this is because the bottom hydrogen in the structure (opposite the nitrogen) would have 5 bonds due to having 2 hydrogen, a double and a single bond. Having a hydrogen leave that carbon would then make it a different compound, not a resonance structure.(6 votes)
- Hi, in the first example of a cation, is it possible to move the electrons from the pi bond to one of the C atoms involved in C=C double bond? This would mean that one of the C atom in C=C (now C-C) will have a + charge and one with a - charge. The overall charge of the molecule though is still + as the + and - would cancel out.(1 vote)
- It’s a valid resonance structure but it’s not a good one because it now has 3 formal charges instead of 1. Additionally it has a positive charge directly next to a negative charge and the most likely situation from that is for them to cancel out.
And you can move them directly to another atom, but again you are going to be creating unfavourable resonance structures, it’s really best to stick to the “good“ ones.(2 votes)
- At, are we assuming the carbon at the bottom point of the hexagon has a H bond too, or a lone pair of electrons? How can you tell? 3:41(1 vote)
- You should always assume the atoms have the maximum number of bonds unless a charge is shown.
That carbon at the bottom would have implied bonds to 2 hydrogen atoms.(2 votes)
Video transcript
- [Voiceover] I see a lot of mistakes when students draw resonance structures, and so I wanted to make a video on some of the more common
mistakes that I've seen. So let's say we wanted to draw a resonance structure
for this carbocation. Some students would take these electrons and move them down to here and say, all right, so on the right, now, I would have this, and this is my resonance structure. Let me highlight those
electrons in blue here, so these electrons here move down to here. But this is incorrect, so let me write "no" here. So the resonance structure on the right, this is an incorrect resonance structure, why is this resonance
structure not possible? Well, let's draw in the hydrogens on the carbons, and it
will be much more obvious. So this carbon right here
has one hydrogen on it, same with this carbon, and this carbon right here
has two hydrogens on it, and the carbon with a
plus one formal charge must have one hydrogen. So let's put in those hydrogens for the resonance structure on the right, and it should be obvious
why this resonance structure is incorrect. Let's focus in on this carbon right here, the one I marked in red. How many bonds are there to that carbon? Well here's one bond, two, three, four, and five, that's five bonds to a carbon, that does not happen,
you can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed
an octet of electrons. Because of carbon's position on the periodic table,
in the second period, there's four orbitals, and each orbital can hold a maximum of two electrons, which gives us four times
two, which is eight. So carbon can never exceed an octet. There's another reason why this is wrong. If we go to this top carbon here, there's only three bonds
around that carbon, so that carbon would have
a plus one formal charge. So we added another formal charge, and we have carbon with five bonds, so this is incorrect,
this is not a correct resonance structure. So what is the proper resonance structure to draw? Well, let's show that down here. You take your electrons, and you move them in the direction of the positive charge, of the positive one formal charge, and so let's show that. The electrons in, let
me make them blue again, the electrons in blue move over to here, like that. And that moves the positive formal charge over to this carbon. If we draw in our hydrogens, it'll be clear why this is correct. So we put in a hydrogen here, we put in a hydrogen here, and we put in a hydrogen here. So let me draw in those three hydrogens on the right. Okay, now it's very obvious, let me point this out in red. It's obvious that this carbon here in red has a plus one formal charge, it has three bonds around it. So one, two, and three. And this carbon, this carbon over here on the right that had the
plus one formal charge, now its formal charge is zero, because there are four bonds around it. So one, two, three and four, so now the formal charge is zero. So this is the correct
resonance structure. Now, it looks a little bit confusing when I have those
hydrogens drawn in there, which is why we leave them off. Let me go ahead and draw it again on the right just for clarity. That's why we leave off those hydrogens when we\re drawing our
resonance structures, because they get in the way, and once you understand what's going on it's not necessary to
draw in those hydrogens. Let's do another example, and again, I'll start with the wrong way to do it, and then we'll talk about the correct way. So, a student might say, all right, I have a negative one formal charge on this nitrogen, so I
could take this lone pair of electrons and move into here, which would push these electrons over to here, so let me go ahead and draw what some students might think is a correct resonance structure. So let me put in my
lone pair of electrons, and let's follow some electrons along. So electrons in light
blue on this nitrogen move into here, and electrons in, let's say dark blue, move down to here. And then finally,
electrons in magenta remain behind on the nitrogen. So on the right, why is this not a correct resonance structure? So again, this is the wrong way to do it. Well, think about your hydrogens. So we'll start with
this carbon right here, this carbon has one hydrogen, this carbon has one hydrogen, and this carbon down here has two. So if we put in those hydrogens over here on the right,
hopefully it's obvious why this is incorrect. Let's look at this
carbon down at the bottom of the ring, so this carbon right here I just marked in red. How many bonds do we have? Well, here's one, two, three, four, and five, so there are five bonds to that carbon, and we know carbon can never have five bonds. Carbon can never exceed
an octet of electrons. So immediately we know that this is not a correct
resonance structure. All right, let's talk about
the right way to do it. So you take these electrons, and you move them into here, and then these (mumbles) electrons have to go somewhere, and they move out onto this carbon, so now, let's draw the correct other
resonance structure here. So we'll put in our double bond, we'll put our electrons on this carbon, that gives this carbon a
negative one formal charge, and then we have some electrons, a lone pair of electrons
left on the nitrogen. I'll use the same colors as before, so these electrons
right here in light blue move in to form our double bond. The electrons in dark blue move off onto this carbon, so the electrons in dark blue are on this carbon that I just marked in dark blue which gives that carbon a
negative one formal charge. And the electrons in magenta remain behind on the nitrogen. All right, the reason why the carbon -- now I'll go ahead and mark
it in a different color. This carbon I just marked in magenta has a negative one formal charge, is because, remember, there's one hydrogen on that carbon, so let me draw in that hydrogen over here. Let me see if I can squeeze
it in over here like that. This carbon right here has three bonds to it and a lone pair of electrons, which gives that carbon in magenta a negative one formal charge. Now this nitrogen over here would have a formal charge equal to zero. And so this on the right, would be the correct resonance structure, and again, drawing in hydrogens is a waste of time, it gets in the way. Let me go ahead and draw the
resonance structure again, I'll take out that hydrogen so it'll look cleaner, and it also takes less time when you're not drawing
in all of your hydrogens. So you just put a lone pair of electrons and write a negative one formal charge, and you have to know that there's still a hydrogen on this carbon
that I marked in magenta. So if you're having trouble
drawing resonance structures, usually the problem is not thinking about your hydrogens, forgetting about putting in your hydrogens. And once you put those in, it's a lot easier to see
if your dot structure, if your resonance structure, I should say, is correct, so be careful about that. And resonance structures
are just practice. The more you draw, the
better you're going to get and if you make a mistake, it's not a big deal. You learn from your mistake, and you keep on practicing.