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Course: Class 11 Chemistry (India) > Unit 12
Lesson 10: Electron displacement effects- Intro to Electronic Effects (& why should I care ?)
- The Inductive Effect
- Inductive effect and stability of intermediates-Part 1
- Inductive effect and stability of reaction intermediates- Part 2
- Inductive effect on stability of intermediates
- Stability of intermediates (all electronic effects)
- Identify aromatic, anti-aromatic and non-aromatic molecules
- Application of aromaticity
- Resonance structures
- Resonance structure patterns
- Resonance structures for benzene and the phenoxide anion
- Common mistakes when drawing resonance structures
- Identifying the correct resonating structure.
- Resonance structures and hybridization
- Stability due to Resonance
- Quantity of Resonating Structures vs Quality
- Resonance and stability of aliphatic intermediates.
- +R & -R groups in Resonance
- Relative strengths of different +R groups
- Relative strengths of different -R groups
- Resonance due to Electron Donors in a Benzyl Cation
- Effect of an electron withdrawing group in a benzyl cation
- Stability of a substituted benzyl anion
- Resonance and stability of aromatic intermediates.
- Comparing the stabilities of aromatic and aliphatic intermediates using resonance.
- Worked ex - Resonance vs Inductive Effects
- Inductive v/s Resonance.
- Hyperconjugation in a benzyl cation
- Hyperconjugation and stability of carbocations.
- Alkene stability
- Hyperconjugation - Isotopic Effect
- Heat of hydrogenation vs stability of alkenes
- Hyperconjugation and stability of alkenes.
- Worked ex : Resonance vs Hyperconjugation
- Who wins? Resonance, Hyperconjugation or Inductive? (ADVANCED)
- Carbocation stability and rearrangement introduction
- Carbocation rearrangement practice
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Resonance structure patterns
Examples showing how different types of bond configurations can be represented using resonance structures. Created by Jay.
Want to join the conversation?
- So this may seem crazy but why would I use a resonance structure if I just end up with the same formal charges on a different atom? Why not just use the original bond-line structure?(21 votes)
- It seemed pointless when I first learned about it, but the goal is to get 2+ resonance structures and then find the hybrid. The hybrid is closest to the actual molecule and is more stable because it spreads the negative/positive charge.(49 votes)
- On the H3CNO2, how does the Nitrogen have 4 bonds? Doesn't Nitrogen already have a pair on one side, and therefore only be able to have 3 bonds?(10 votes)
- The Nitrogen has a +1 formal charge on it, this means that it has one extra bond. Keep in mind that Nitrogen has 5 valence electrons.
Formal charge= valence electrons on atom – (non-bonded electrons + number of bonds).
+1= 5 - (0-4)
When it is bonded to 3 it has a 0 formal charge and 2 non-bonded electrons.
0= 5 - (2+3)(22 votes)
- I'm confused on two things.
1.) Why can't Nitrogen have five bonds? Wouldn't that give it a formal charge of 0 which is the most stable?
2.) Nitrogen has 5 valence electrons and wants three more to fill it's octet. So how can it have four bonds (given four electrons) and have a positive charge... Wouldn't four bonds give it a -1 charge?(12 votes)- 1) Nitrogen cannot form 5 bonds because it is unable to exceed it's octet. That would give it a formal charge of 0, so it wants to do that but it cannot because it doesn't have a d orbital (it doesn't have any more space for electrons)
2) Remember that in bonding, it is not given the four electrons, but it is sharing them with another atom. So in this case, it starts off with 5 valence electrons and then it forms 4 bonds, each of which decreases the formal charge by 1 so therefore you are left with a +1 charge. Formal charge = (family number) - (# unshared electrons) - 1/2(# of shared electrons) so in this case fc = (5)-(0)-1/2(8) = 1(8 votes)
- In the example started at, it seems to me that both resonance structures are indeed equal, only flipped...Am I wrong? 4:47(7 votes)
- Hey,
You are right- both resonance structures are similar. However, in resonance structures, we try to spread the electrons over different atoms (even if they are of the same element), so this is OK.
I believe you are confusing resonance structures with structural isomers, as the two examples given atare resonance structures, but are not structural isomers. 4:47(13 votes)
- When someone lists something like NO3-, how do you know where the negative charge is, or how many of the say Oxygens will have the extra valence electrons? I am working on extra questions, and from what I can see, there should only be one Oxygen with 7 valence electrons, but in the solution it shows that two of the three Oxygens have 7 valence electrons. How do we determine that, or even which element holds that negative formal charge when it's just listed like that?(8 votes)
- You're almost correct. Its important to differentiate formal charges with valence electrons, as all 4 atoms in NO3- have a full octet. In NO3-, there are two oxygens have a -1 formal charge, and the third has a double bond, with no formal charge. The N, however, has a formal charge of +1. This brings the net formal charge of the molecule to -1.
Also note that NO3- experiences resonance, so when drawing the structure, its important to see that these charges aren't completely stationary., and any of the three can be marked as the "double bonded" one, with the other two would bear the formal charge.(8 votes)
- In first example,how come carbon has a negative formal charge?It is bonded to two carbons and has one lone pair.Then,formal charge=4-4=0.(2 votes)
- It's bonded to three things - two carbons and a hydrogen (the hydrogen is implicit). Three bonds and a lone pair gives you -1 formal charge.(7 votes)
- At, the central carbon atom in the resonant structure has only 3 bonds. Its octet isn't complete. How is this possible? 8:25(4 votes)
- It is still a valid resonance structure, octets do not have to be complete for it to be valid. It does mean it will not contribute much to the overall structure of the molecule, but that resonance structure does show us why carbonyl carbons are reactive towards nucleophiles. Jay explains this further on in the video.(2 votes)
- Hello,
Thank you for your help. One question though, at minute 6.42 (the bond in blue) why did you move it directly to the bond and not to the carbon in green. I understood when you said that we want a distribution of the charge; it makes sense if you look at it that way. However, I just need clarification. Thanks in advance.(3 votes)- What you describe is valid, but would result in a structure with a lot of (formal) charge - positive charges on both end carbons and negative on the middle carbon. This is energetically unfavorable and so I would expect it to make a very minor contribution to the overall (hybrid) structure.(2 votes)
- At around, he says that nitrogen can't have five bonds because of its position on the periodic table. I can't quite comprehend this. Isn't nitrogen placed at the top of group 5A having five valence electrons. Having five v.electrons, I can't see how it should have any problem making five bonds. 3:08(1 vote)
- Nitrogen is too small to bond to 5 other atoms.
Elements in period 3 and on can have 5+ bonds (eg. PCl5), but not elements in period 2.
4 bonds is the most we see nitrogen form.(4 votes)
- What is the use of resonance?(1 vote)
- Copying and pasting this from above; all credit to Evan Loney.
"It seemed pointless when I first learned about it, but the goal is to get 2+ resonance structures and then find the hybrid. The hybrid is closest to the actual molecule and is more stable because it spreads the negative/positive charge."(3 votes)
Video transcript
Voiceover: Let's look
at a few of the patterns for drawing resonance
structures, and the first pattern we're gonna
look at, is a lone pair of electrons next to a pi bond. And so, here's a lone pair of electrons; I'm gonna highlight it in
magenta, that lone pair of electrons is located on this carbon, let me go ahead and put
this carbon in green, here. And I'm saying, there's a
negative-one formal charge on that carbon in green,
and so that carbon in green is also bonded to
a hydrogen, so once again, you need to be very
familiar with assigning formal charges. So we have a lone pair
of electrons next to a pi bond, because over
here, we have a double-bond between the carbon and the
oxygen, one of those bonds is a sigma bond, and one of
those bonds is a pi bond, so I'm just gonna say that
these are the pi electrons. So our goal in drawing
a resonance structure is to de-localize that
negative-one formal charge, so spread out some electron density. And so, we could take
the electrons in magenta, and move them into here,
to form a double-bond, between the carbon in green
and this carbon right here, and that'd be too many bonds
to the carbon in yellow, so the electrons in blue have to come off, onto this top oxygen here. So we go ahead, and draw in our brackets, and we put our double-headed
resonance arrow, and we draw the other resonance structure, so we have our ring, like
that, and then we have, now, a double-bond
between those two carbons, and then this top oxygen here, now has only one bond to it. The oxygen used to have two
lone pairs of electrons, now it has three,
because it just picked up a pair of electrons from that pi bond. So let's go ahead, and
follow the electrons. The electrons in magenta moved in here, to form our pi bond, like
that, and the electrons in the pi bond, in blue,
moved off, onto this oxygen, so I'm saying that they
are those electrons. That gives the top oxygen a
negative-one formal charge, and so we have our two
resonance structures for the enalate anion. We know that both resonance
structures contribute to the overall hybrid,
and if you think about which one contributes more,
for the example on the left, we have had a negative-one formal charge on the carbon in green,
so that's a carb anion; and for the resonance
structure on the right, we had a negative one
formal charge on the oxygen, so that's an oxyanion. Oxygen is more
electronegative than carbon, which means it's more likely to support a negative-one formal
charge, and so the resonance structure on the right
contributes more to the overall hybrid for an enalate anion. All right, let's do another
pattern, a lone pair of electrons next to a
positive charge, this time. So, let's look at
nitromethane, and we could look at this lone pair of electrons
here, on this oxygen, and that lone pair of electrons is next to a positive charge; this
nitrogen has a plus one formal charge on it. And, so, let's think about
drawing the resonance structure, so our goal is to de-localize
charge, to spread charge out. We could take the electrons in magenta, and move them into here,
to form a double-bond between the nitrogen and
the oxygen, but that's too many bonds to this
nitrogen; that would give us five bonds to that
nitrogen, which we know doesn't happen, because
of nitrogen's position on the periodic table. So, that means that the
electrons in this pi bond here, are gonna come off, onto the oxygen so these electrons in blue,
come off, onto this oxygen, and we draw our other resonance structure for nitromethane, so we have a CH three. We now have a double-bond between nitrogen and this oxygen; this
oxygen used to have three lone pairs of electrons,
but the electrons in magenta moved in here, to form this
bond, and so that means we have only two lone
pairs left, on this oxygen. For the oxygen on the
bottom-right, there's only one bond now, between the
nitrogen and the oxygen, because the electrons in blue moved off, onto this oxygen, and
that means this oxygen has two more lone pairs of electrons. And so, when we go ahead
and put in our resonance bracket here, you always
need to think about assigning formal charge, so
what happened to the charge? Well, this oxygen now, has a
negative-one formal charge, and this nitrogen still has
a plus-one formal charge, so we've de-localized
that negative charge; it's actually over both of those oxygens. And notice that the overall
charge for nitromethane is zero, for both resonance structures. So we have one positive charge and one negative charge on the
left, so that gives us zero; and we have one positive
charge and one negative charge on the right, so that gives us zero: So conservation of charge. All right, let's do another
example for a pattern that we might see. So, for this one, we
have a positive charge next to a pi bond, so
let's look at this carbon. So I'm saying it has a
plus-one formal charge, and if it has a plus-one
formal charge, it must have only three bonds,
and since it's already bonded to another carbon,
so it's already bonded to- Let me go ahead and label these. So the carbon in yellow there is bonded to this carbon in green,
because it has a plus-one formal charge, it must
have only two other bonds, and so those must be to hydrogen. So I draw in those hydrogens. So now, it make a little
more sense why it's a plus-one formal charge;
it be four minus three, giving us plus one. The carbon in green has
a formal charge of zero, so it already has three
bonds, so it needs one more, two hydrogen, and let's go ahead and make this carbon, over here,
in red, already has two bonds, it has a formal charge of zero, so it needs two more hydrogens. So, once again, our pattern
is a positive charge next to a pi bond, so let
me go ahead and highlight these things here, so we
have a positive charge, next to a pi bond, and so here, let's say this one is our pi bond like that. So, when you're drawing
resonance structures, again, your goal is to
de-localize that charge, and so we could spread out
that positive charge by taking the electrons in
blue, the pi electrons, and moving them into here. So let's draw the resonance structure. So, we now have, let's see, we
would now have a double-bond between the two carbons on the right. The hydrogens haven't
moved, right, so I'm gonna leave those hydrogens in
there, so there's still one hydrogen on the carbon in the middle, two hydrogens on the carbon in the right, and two hydrogens on
the carbon on the left. So the electrons in blue
moved to here, like that, so let me go ahead and
highlight those carbons. So the carbon in green, right
here, and the carbon in red. So what happened to the
plus-one formal charge? Well, you can see that it's actually moved to the carbon in the
red; the carbon in red right here, has only three
bonds, so four minus three gives us a plus-one formal charge. And, let's go ahead and finish
our resonance bracket here, so I put that in, and so
when you're doing this for cations, you're not
gonna move a positive charge, so when you're drawing
your arrows, you're showing the movement of electrons,
so the arrow that I drew over here, let me go ahead
a mark it in magenta. So this arrow in magenta
is showing the movement of those electrons in blue,
and when those electrons in blue move, that creates
a plus-one formal charge on this carbon, and so
don't try to move positive charges: Remember, you're
always pushing electrons around. Then finally, let's do one more. So, for this situation,
this is for acetone, so we have a carbon
right here, double-bonded to an oxygen, and we know that there are differences in
electronegativity between carbon and oxygen: Oxygen is
more electronegative. So what would happen if we
took those pi electrons? Let me go ahead an highlight
those; I've been using blue for pi electrons, so
these pi electrons right here, and we move those pi
electrons off, onto the more electronegative atom, like
that, so let's go ahead and draw our resonance structure. So this top oxygen would
have three lone pairs of electrons: one of those
lone pairs are the ones in blue, those pi
electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that
one negative charge and one positive charge give you an overall charge of zero,
so charge is conserved, and over here, of course,
the charge is zero. So if you're thinking
about the resonance hybrid, we know that both structures contribute to the overall hybrid,
but the one on the right isn't going to contribute
as much, so this one on the right is pretty
minor, and that's because you have a positive and a negative charge, and the goal, of course, is
to get to overall neutral. But, what's nice about drawing
this resonance structure, and thinking about this
resonance structure, is it's emphasizing the
difference in electronegativity, so, for this one, you
could just say oxygen get a partial negative,
and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance
structure is just another way of thinking about,
emphasizing the fact that when you're thinking
about the hybrid, you're thinking about a little
more electron density on that oxygen. All right, so once again,
do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the
easier the patterns become.